Lectures in Optics, Vol. 2, Geometrical Optics 2019001142, 9781510619456, 1510619453, 9781510619463, 1510619461

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Geometrical Optics LECTURES IN OPTICS

Volume 2

George Asimellis

By the Author

Lectures in Optics, Vol. 1, Introduction to Optics Lectures in Optics, Vol. 2, Geometrical Optics Lectures in Optics, Vol. 3, Wave Optics Lectures in Optics, Vol. 4, Visual Optics Lectures in Optics, Vol. 5, Ocular Imaging

Geometrical Optics LECTURES IN OPTICS

Volume 2

George Asimellis

SPIE PRESS Bellingham, Washington USA

Library of Congress Cataloging-in-Publication Data Names: Asimellis, George, 1966- author. Title: Geometrical optics / George Asimellis. Description: Bellingham, Washington, USA : SPIE Press, [2020] | Series: Lectures in optics ; vol. 2 | Includes index. Identifiers: LCCN 2019001142| ISBN 9781510619456 (softcover) | ISBN 1510619453 (softcover) | ISBN 9781510619463 (pdf) | ISBN 1510619461 (pdf) Subjects: LCSH: Geometrical optics. | Refraction. | Reflection (Optics) Classification: LCC QC381 .A85 2019 | DDC 535/.32--dc23 LC record available at https://lccn.loc.gov/2019001142 Published by SPIE P.O. Box 10 Bellingham, Washington 98227-0010 USA Phone: +1 360.676.3290 Fax: +1 360.647.1445 Email: [email protected] Web: http://spie.org Copyright © 2020 Society of Photo-Optical Instrumentation Engineers (SPIE) All rights reserved. No part of this publication may be reproduced or distributed in any form or by any means without written permission of the publisher. The content of this book reflects the work and thought of the author. Every effort has been made to publish reliable and accurate information herein, but the publisher is not responsible for the validity of the information or for any outcomes resulting from reliance thereon. Printed in the United States of America. First Printing. For updates to this book, visit http://spie.org and type “PM290” in the search field.

COVER IMAGE: ‘UMBRELLAS’ – MODERN ART CREATION BY GEORGE ZOGGOPOULOS, IN THESSALONIKI, GREECE, IMAGED VIA A BALL LENS. IMAGE CREATION: EFSTRATIOS I. KAPETANAS.

FACEBOOK.COM/PHOTOSTRATOSKAPETANAS/

“Μηδείς ἁγεωμέτρητος εἱσίτω” Πλάτων.

300 BC Greek mathematician Euclid (Εὐκλείδης), the Father of Geometry, authored Optica (Οπτική). Euclid asserted that light travels in straight lines and proposed mathematical formulae for reflection and refraction. Shown above is one of the oldest preserved papyrus sheets with his writings.

GEOMETRICAL OPTICS

TABLE OF CONTENTS Table of Contents .............................................................................................................................................................................. i Foreword ............................................................................................................................................................................................ vii Preface ................................................................................................................................................................................................. ix About this Series .................................................................................................................................................................. ix About this Book ..................................................................................................................................................................... x Acknowledgments ............................................................................................................................................................. xiii

1

REFRACTION AT A SPHERICAL INTERFACE .................................................................................... 1-1 1.1 Reflection and Refraction ................................................................................................................................................1-1 1.1.1

The Laws of Reflection ...................................................................................................................................1-1

1.1.2

The Refractive Index ........................................................................................................................................1-2

1.1.3

The Laws of Refraction ...................................................................................................................................1-4

1.1.4

Critical Angle and Total Internal Reflection ..........................................................................................1-6

1.2 The Single Spherical Refracting Interface .................................................................................................................1-9 1.2.1

Curvature and Radius of Curvature ..........................................................................................................1-9

1.2.2

The Convex and Concave SSRI ................................................................................................................ 1-11

1.2.3

Refraction by an SSRI .................................................................................................................................. 1-11

1.2.4

Optical Power and Focal Length ............................................................................................................. 1-13

1.2.5

The Flat SSRI .................................................................................................................................................... 1-19

1.2.6

The Nodal Point in an SSRI ....................................................................................................................... 1-20

1.2.7

The Listing Eye Model as an SSRI ........................................................................................................... 1-21

1.3 Advanced Practice Examples ...................................................................................................................................... 1-22 1.4 Refraction Quiz ................................................................................................................................................................. 1-25 1.5 Refraction Summary ....................................................................................................................................................... 1-34

2

LENS REFRACTION AND POWER ............................................................................................... 2-37 2.1 What is a Lens? ................................................................................................................................................................. 2-37 2.2 Principles of Lens Operation ....................................................................................................................................... 2-40 2.2.1

Refraction by Two Surfaces ....................................................................................................................... 2-40

2.2.2

Two Prisms ....................................................................................................................................................... 2-40

2.2.3

Principle of Least Time ................................................................................................................................ 2-42

2.2.4

Wavefront Transformation ........................................................................................................................ 2-44

2.2.5

The Gravitational Lens ................................................................................................................................. 2-45

2.3 The Thin Lens ..................................................................................................................................................................... 2-48

i

LECTURES IN OPTICS

2.3.1

Radii of Curvature and Material .............................................................................................................. 2-48

2.3.2

Primary and Secondary Focal Points ..................................................................................................... 2-49

2.3.3

Focal Planes and Optical Axis................................................................................................................... 2-51

2.3.4

Lens Shape ....................................................................................................................................................... 2-53

2.4 Lens Optical Power ......................................................................................................................................................... 2-56 2.4.1

Lens-Maker’s Formula ................................................................................................................................. 2-56

2.4.2

Dependence on Orientation, Media, and Geometry ...................................................................... 2-58

2.5 Advanced Practice Examples ...................................................................................................................................... 2-63 2.6 Lens Power Quiz ............................................................................................................................................................... 2-65 2.7 Lens Power Summary ..................................................................................................................................................... 2-68

3

IMAGING DEFINITIONS ............................................................................................................ 3-71 3.1 Object and Image ............................................................................................................................................................ 3-71 3.1.1

Real and Virtual Object; Real and Virtual Image ............................................................................. 3-73

3.1.2

Object and Image in a Plane Mirror ...................................................................................................... 3-75

3.2 Sign Conventions ............................................................................................................................................................. 3-77 3.2.1

Object / Image Height and Angle Sign Conventions..................................................................... 3-78

3.3 Magnification .................................................................................................................................................................... 3-79 3.3.1

Angular Magnification ................................................................................................................................ 3-80

3.4 Vergence ............................................................................................................................................................................. 3-82 3.4.1

Wavefront Vergence .................................................................................................................................... 3-82

3.4.2

Vergence and Propagation ....................................................................................................................... 3-84

3.4.3

Vergence and Optical Interfaces............................................................................................................. 3-87

3.5 Vergence in Imaging ...................................................................................................................................................... 3-89 3.5.1

Vergence of a Real and a Virtual Object ............................................................................................. 3-89

3.5.2

Vergence of a Real and a Virtual Image .............................................................................................. 3-90

3.5.3

Upstream and Downstream Vergence in Lens Imaging ............................................................... 3-91

3.5.4

Vergence and a Flat Refracting Interface ............................................................................................ 3-94

3.5.5

Vergence and the SSRI Power ................................................................................................................. 3-97

3.6 Advanced Vergence Examples ................................................................................................................................... 3-98 3.7 Vergence and Imaging Concepts Quiz .................................................................................................................3-103 3.8 Vergence and Imaging Concepts Summary .......................................................................................................3-106

4

IMAGING WITH LENSES ......................................................................................................... 4-109 4.1 Lens Imaging Relationship .........................................................................................................................................4-109

ii

4.1.1

Imaging: Left to Right or Right to Left? ............................................................................................4-113

4.1.2

Image Magnification ..................................................................................................................................4-114

4.1.3

Image Reversal .............................................................................................................................................4-116

GEOMETRICAL OPTICS

4.1.4

Newton’s Imaging Relationship ............................................................................................................4-117

4.2 Imaging by a Plus-Powered Lens ............................................................................................................................4-118 4.2.1

Summary .........................................................................................................................................................4-122

4.3 Ray Diagrams...................................................................................................................................................................4-124 4.3.1

Ray Diagrams for a Plus-Powered Lens .............................................................................................4-124

4.3.2

Some Considerations About Construction Rays ............................................................................4-125

4.3.3

Ray-Tracing Examples with Plus Lens Imaging ...............................................................................4-128

4.4 Imaging by a Minus-Powered Lens........................................................................................................................4-130 4.4.1

Ray Diagrams for a Minus-Powered Lens .........................................................................................4-131

4.5 Imaging by an SSRI .......................................................................................................................................................4-134 4.6 Notes on Imaging ..........................................................................................................................................................4-137 4.6.1

The Optical Invariant ..................................................................................................................................4-137

4.6.2

The Wild Ray .................................................................................................................................................4-140

4.6.3

Optical Infinity ..............................................................................................................................................4-141

4.7 Virtual Object Imaging ................................................................................................................................................4-143 4.7.1

Virtual Object Imaging with a Plus Lens ............................................................................................4-144

4.7.2

Virtual Object Imaging with a Minus Lens........................................................................................4-147

4.8 Advanced Lens-Imaging Examples ........................................................................................................................4-153 4.9 Lens Imaging Quiz .........................................................................................................................................................4-156 4.10 Lens Imaging Summary...............................................................................................................................................4-161

5

IMAGING WITH MIRRORS ...................................................................................................... 5-165 5.1 Plane Mirror Principles ................................................................................................................................................5-165 5.1.1

The Cartesian Convention in Mirrors ..................................................................................................5-167

5.1.2

Multiple Plane Mirror Surfaces ..............................................................................................................5-168

5.2 Spherical Mirrors ............................................................................................................................................................5-170 5.2.1

Geometry of a Spherical Mirror ............................................................................................................5-170

5.2.2

Focal Points and Focal Lengths in a Spherical Mirror ..................................................................5-174

5.2.3

Nodal Point in a Spherical Mirror .........................................................................................................5-175

5.2.4

Optical Power in a Spherical Mirror.....................................................................................................5-176

5.2.5

Vergence and Spherical Mirror Power ...............................................................................................5-178

5.3 Imaging with a Spherical Mirror ..............................................................................................................................5-180 5.3.1

Spherical Mirror Imaging Relationship ..............................................................................................5-180

5.3.2

Convex Mirror Imaging Examples ........................................................................................................5-182

5.3.3

Ray Diagrams for Convex Mirrors ........................................................................................................5-183

5.3.4

Concave Mirror Imaging Ray Diagrams and Examples ...............................................................5-188

5.3.5

The Virtual Object in Mirror Imaging ..................................................................................................5-196

iii

LECTURES IN OPTICS

5.4 Advanced Mirror Imaging Examples .....................................................................................................................5-202 5.5 Mirror Imaging Quiz .....................................................................................................................................................5-205 5.6 Mirror Imaging Summary ...........................................................................................................................................5-210

6

THICK LENSES AND LENS SYSTEMS ......................................................................................... 6-215 6.1 The Thick Lens .................................................................................................................................................................6-215 6.1.1

Equivalent Optical Power and Focal Length ....................................................................................6-215

6.1.2

Specialty Lenses ...........................................................................................................................................6-221

6.1.3

The Cornea Equivalent Power ................................................................................................................6-221

6.2 Cardinal Points: Concept and Applications.........................................................................................................6-223 6.2.1

Principal Points and Principal Planes ..................................................................................................6-223

6.2.2

Nodal Points ..................................................................................................................................................6-229

6.2.3

Cardinal Points in an SSRI and a Mirror .............................................................................................6-231

6.3 Vertex Powers in a Thick Lens ..................................................................................................................................6-233 6.3.1

The Concept of Front and Back Vertex Power and Focal Length ...........................................6-233

6.3.2

The Measure of Vertex Powers ..............................................................................................................6-234

6.4 Imaging with a Thick Lens ..........................................................................................................................................6-240 6.4.1

Ray Propagation in a Thick Lens ...........................................................................................................6-240

6.4.2

Ray Diagrams in a Thick Lens .................................................................................................................6-242

6.4.3

The Wild Ray in a Thick Lens ..................................................................................................................6-244

6.4.4

Thick Lens Imaging Relationship ..........................................................................................................6-245

6.5 Lens Systems....................................................................................................................................................................6-249 6.5.1

Optical Power in a Lens System ............................................................................................................6-249

6.5.2

Cardinal Points in a Thin Lens System ................................................................................................6-252

6.5.3

The Afocal System.......................................................................................................................................6-254

6.5.4

Cardinal Points in a Thick Lens System ..............................................................................................6-257

6.5.5

Intermediate Image Technique in Two or More Lenses .............................................................6-258

6.5.6

The Thick Lens as a Two-Element System ........................................................................................6-263

6.6 Advanced Practice Examples ....................................................................................................................................6-265 6.7 Thick Lens and Cardinal Points Quiz .....................................................................................................................6-278 6.8 Thick Lens and Cardinal Points Summary ...........................................................................................................6-289

7

FINITE TRANSVERSE OPTICS................................................................................................... 7-295 7.1 Aperture Stop and Pupils ...........................................................................................................................................7-296

iv

7.1.1

The Aperture Stop.......................................................................................................................................7-296

7.1.2

Significance of the Aperture Stop ........................................................................................................7-304

7.1.3

Entrance and Exit Pupil .............................................................................................................................7-306

7.1.4

Numerical Aperture and F-number .....................................................................................................7-315

GEOMETRICAL OPTICS

7.2 Principal / Chief and Marginal Rays ........................................................................................................................7-317 7.2.1

The Principal / Chief Ray ..........................................................................................................................7-317

7.2.2

The Marginal Rays.......................................................................................................................................7-318

7.3 Fields, Stops, and Related Effects............................................................................................................................7-325 7.3.1

Field of View ..................................................................................................................................................7-325

7.3.2

The Field Stop ...............................................................................................................................................7-327

7.3.3

Entrance and Exit Windows / Ports ......................................................................................................7-331

7.3.4

Size of the Field of View ...........................................................................................................................7-334

7.3.5

Fields of Half and Full Illumination ......................................................................................................7-342

7.3.6

Vignetting and Glare..................................................................................................................................7-344

7.4 Depth of Field and Depth of Focus ........................................................................................................................7-347 7.5 Brightness, Contrast, and Resolution ....................................................................................................................7-353 7.6 Geometrical Image Blur ..............................................................................................................................................7-358 7.7 Advanced Practice Examples ....................................................................................................................................7-363 7.8 Transverse Optics Quiz ................................................................................................................................................7-377 7.9 Transverse Optics Summary ......................................................................................................................................7-390

8

OPTICAL ABERRATIONS ........................................................................................................ 8-395 8.1 Imaging to an Idealized Point ..................................................................................................................................8-395 8.1.1

The Origin of Optical Aberrations ........................................................................................................8-397

8.1.2

The Paraxial Approximation ....................................................................................................................8-398

8.1.3

Classification of Optical Aberrations ...................................................................................................8-400

8.2 Chromatic Aberration ..................................................................................................................................................8-402 8.2.1

Management of Chromatic Aberration .............................................................................................8-405

8.3 Monochromatic Aberrations .....................................................................................................................................8-407 8.3.1

Spherical Aberration ..................................................................................................................................8-407

8.3.2

Coma ................................................................................................................................................................8-415

8.3.3

Oblique / Radial Astigmatism .................................................................................................................8-420

8.3.4

Field Curvature and Distortion ..............................................................................................................8-423

8.4 Aberrations Quiz ............................................................................................................................................................8-428 8.5 Aberrations Summary ..................................................................................................................................................8-431

APPENDIX ...................................................................................................................................... 433 Conventions and Notation ...................................................................................................................................................... 433 Conventions ....................................................................................................................................................................... 433 Object-Space versus Image-Space Notation ....................................................................................................... 433 The Cartesian Sign Convention ................................................................................................................................. 434 Frequently used Notation ............................................................................................................................................ 435 v

LECTURES IN OPTICS

Useful Notes ...................................................................................................................................................................... 436 Geometrical Optics Formulation ........................................................................................................................................... 437 Refraction ............................................................................................................................................................................ 437 Vergence, Optical Power and Focal Lengths, and Imaging ........................................................................... 439 Imaging Relationships ................................................................................................................................................... 443 Optics of the Human Eye ............................................................................................................................................. 445 Answers To Quiz Questions .................................................................................................................................................... 447 Index ................................................................................................................................................................................................. 449

vi

GEOMETRICAL OPTICS

FOREWORD At the very beginning of my optics classes, I promise my students that I will probably turn a number of them into optics geeks. At the very least, I hope that I am able to make them more aware that optics is everywhere. By sending them out into the world with their cameras and the goal of finding and capturing optical phenomena, they do indeed start to develop an optics awareness and perhaps a bit of optics geekiness. It is, after all, the core of what we do as Optometrists. When Dr. Asimellis asked me to provide feedback on his Geometric Optics, I took it on thinking that it wouldn't hurt to see another approach to the subject. I’d already amassed a sizable collection of optics texts spanning at least 50 years of approaches, always looking for different perspectives. Once I started reviewing the lectures, I realized that George’s take on the subject was more of a journey through the concepts with a narrative that draws you in, rather than a series of derivations and formulas. It turns out I had found a kindred spirit. The interactions, discussions and, yes, even sometimes disagreements on the topic were enjoyable for us and I believe translated into a valuable resource for the educator and the student of optics. The material is presented in a manner that provides a way to visualize the concepts, while still following well-established notation and formulation. The real value of this book is its didactic approach: emphasis is given to understanding the effects; the mathematical formulation then follows naturally and is easier to comprehend. Step by step, students are guided from the simple effects of refraction and reflection to the refractive effects of lenses and prisms, then on to the more complex concepts of thick lenses and lens systems, and the optics of lateral restriction. Educators will be pleased to see the complete coverage of the material prescribed in most optics curriculae in optometric education.

Corina van de Pol, OD, PhD, FAAO Assistant Professor, Southern California College of Optometry, Marshall B. Ketchum University Fullerton, California June, 2020

vii

GEOMETRICAL OPTICS

PREFACE “Geometrical Optics is either very simple or else it is very complicated.” Richard Feynman, The Feynman Lectures on Physics, Volume I, 27-1

About this Series Optics is fundamentally simple. At first glance, optics can be, indeed, formidably complicated. Sign conventions are difficult to memorize, the reciprocal of (meter-converted) distances involved in imaging equations are hard to rationalize, and focal distances are impossible to add. These are just a few of the hurdles encountered in the ostensibly easier part of optics, that of geometrical optics. Throw into the mix the wave nature of light, the complicated integrals involved in the description of light propagation through a small aperture, or some aspects of interference and polarization, and you have the perfect recipe for confusion. This perspective is permeated by the fact that optics instruction is fragmented, most often as part of a Physics 102 course or sometimes as part of a classic electromagnetism curriculum. The presentation of optics as a whole is rare. As a graduate student, I enrolled in two courses, one in Fourier Optics and another in Teaching Methodology. The recommended books were Introduction to Fourier Optics by Joseph W. Goodman and The Feynman Lectures on Physics by Richard Feynman. Albeit uncorrelated, these two courses changed my view of optics forever. I appreciated how certain phenomena can be explained in a straightforward manner, for example, through a simple Fourier transform, or by the connection between quantum physics, phase diagrams, and interference. Geometrical optics can be vastly simplified if we adhere to the Cartesian convention and the vergence method. Breaking away from the traditional approach, this formulation provides a much simpler and unified tool to address imaging in geometrical optics, and, to a substantial extent, in visual optics. The philosophy that optics is simple permeates this book series. Once the reader appreciates this essential simplicity, it is a lot easier to build the foundation of fundamental knowledge, from the basics all the way to the more esoteric topics. I feel that without an understanding of this basic simplicity on which to develop, the structure of accumulated knowledge is unsteady at best, and at worst, will crumble under its own weight. I hope that this second volume of the Lectures in Optics series will be appreciated by those seeking a bottom-up textbook, fitting the needs for any college-level optics or optometry optics course.

George Asimellis, PhD Pikeville, Kentucky June 2020 ix

LECTURES IN OPTICS

About this Book Geometrical optics is perhaps among the most challenging courses in many programs, including optometric education. The material comes, unfairly in my opinion, with a reputation of being hard and challenging. I side with the ‘challenging,’ topping it with ‘rewarding.’ Optics is the foundation of how the eye works, how we image the eye for diagnosis, and, progressively, how we use many laser-based therapeutic and cosmetic applications. Optics helps to develop critical thinking skills that are necessary in a successful diagnostic career such as that of an Optometrist or an Optical Engineer. This book builds on the previous volume 1, Introduction to Optics, mainly on the topics of refraction and its applications to prism deviation and simple optical instruments. The present volume 2, Geometrical Optics, completely develops the instructional requirements pertaining to the foundation of the topic, including: refraction at a spherical surface (SSRI), lens refraction, and imaging by lenses, SSRIs, and mirrors; thick lenses and optics of stops and pupils; and optical aberrations. The material is presented at a level applicable to medical students with a limited optical science background and covers the rubric presented by the National Board of Examiners in Optometry. Emphasis is placed on conceptual development, with ample examples ranging from very simple to advanced practice exercises. Using the two books, Introduction to Optics (ItO) and Geometrical Optics (GO), the following brief curriculum structure can be used as a general guideline in order to deliver an introductory and foundational 4-credit optics course (50 + 2 lecture hours). Unit 1 (4 hours): Light, Rays, Wavefronts, Vergence, Reflection and Refraction ItO, Chapter 1 (§ 1.3 Propagation of Light, § 1.4 Index of Refraction, § 1.5 Light-Matter Interactions) ItO, Chapter 2 (§ 2.1 Angle Measurement) ItO, Chapter 3 (§ 3.1 Reflection, § 3.2 Refraction, § 3.4 Refraction Applications, § 3.5.1 Refractive Atmospheric Phenomena)

Unit 2 (4 hours): Prisms and Color Dispersion ItO, Chapter 3 (§ 3.3 Prisms, § 3.5.2 Prismatic Atmospheric Phenomena)

Unit 3 (4 hours): The Single Refracting Spherical Interface GO, Chapter 1: Refraction in a Spherical Interface

Unit 4 (4 hours): Lenses and Lens Power GO, Chapter 2: Lens Refraction and Power GO, Chapter 6: Thick Lenses and Lens Systems (§ 6.1 The Thick Lens)

Unit 5 (4 hours): Imaging Definitions and Vergence GO, Chapter 3: Imaging Definitions and Vergence

Unit 6 (8 hours): Lens Imaging GO, Chapter 4: Imaging with Lenses x

GEOMETRICAL OPTICS

Unit 7 (4 hours): Mirror Imaging GO, Chapter 5: Imaging with Mirrors

Unit 8 (6 hours): Imaging with Thick Lens and Lens Systems GO, Chapter 6: Thick Lenses and Lens Systems (§ 6.2 Cardinal Points: Concept and Applications, § 6.3 Vertex Powers in Thick Lens, § 6.4 Imaging with a Thick Lens, § 6.5 Lens Systems)

Unit 9 (10 hours): Pupils, Stops, and Related Effects GO, Chapter 7: Finite Transverse Optics (§ 7.1 Aperture Stop and Pupils, § 7.2 Principal / Chief and Marginal Rays, § 7.3 Fields, Stops, and Related Effects) ItO, Chapter 5 (§ 5.1.2 Microscope Principle of Operation, § 5.2.2 Telescope Principle of Operation)

Unit 10 (4 hours): Simple Optical Instruments ItO, Chapter 4 (§ 4.1 Camera Obscura, § 4.2 The Human Eye, § 4.3 The Magnifying Lens)

Unit 11 (8 hours): Image Quality and Optical Aberrations GO, Chapter 7: Finite Transverse Optics (§ 7.4 Depth of Field and Depth of Focus, § 7.5 Brightness, Contrast, and Resolution, § 7.46 Geometrical Image Blur) GO, Chapter 8: Optical Aberrations

xi

LECTURES IN OPTICS

Acknowledgments I would like to thank the following colleagues for their helpful suggestions and recommendations: Natalie Hutchings, PhD, MCOptom Associate Professor & Associate Director of Academics and Student Affairs, Optometry and Vision Science, University of Waterloo, Canada Pantazis Mouroulis, PhD, Fellow, OSA, SPIE Senior Research Scientist, Jet Propulsion Laboratory, California Institute of Technology, Pasadena, California Corina van de Pol, OD, PhD, FAAO Assistant Professor, Southern California College of Optometry, Marshall B. Ketchum University, Fullerton, California

I would also like to thank the following members of the Frank M. Allara Library, University of Pikeville for their contribution in providing grammatical and English-language corrections: Karen Evans, Director of Library Services Edna Fugate, University Archivist and Reference Librarian Jerusha Shipstead, Reference and Instruction Librarian Katherine Williams, Instructional Design Librarian & Coordinator of Instruction

Image Credits Most of the artwork in this book was created by the author. Any figure that was not created by the author is attributed to the source provided in the caption.

xiii

GEORGE ASIMELLIS

1

LECTURES IN OPTICS, VOL 2

REFRACTION AT A SPHERICAL INTERFACE

1.1 REFLECTION AND REFRACTION 1.1.1 The Laws of Reflection Reflection is perhaps the most notable phenomenon associated with light propagation, despite the fact that it is not limited to waves since this phenomenon is encountered in both waves and particles. The laws of reflection can be explained equally well by either the wave or the corpuscular theory of light. In the illustration of a corpuscular reflection by the tennis ball in Figure 1-1, we notice that the ball is reflected because it encounters something ‘else’ along its course—in this case, the grassy lawn—and then bounces off.

Figure 1-1: Reflection of a tennis ball off of grass. 1-1

GEOMETRICAL OPTICS

This bounce is described by two angles and employs two planes. We use subscript i for incidence and subscript r for reflection, and define the following: • The reflecting surface is a real surface, considered flat even if it is not, at least for a small surface area near the point of reflection. • The plane of incidence is defined by the incident beam and the normal to the point of incidence. This is a virtual plane. These two planes are perpendicular to each other. The two angles involved in reflection are: • the angle of incidence ϑi, defined by the path of the incident ray just before the incidence and the normal to the point of incidence, and • the angle of reflection ϑr, defined by the path of the reflected ray just after the incidence and (again) the normal to the point of incidence. The rules governing reflection are: ❶

The ray path remains on the same plane: the plane of incidence.

❷

The angle of incidence ϑi equals the angle of reflection ϑr.

Law of Reflection:

i

=

angle of incidence

r

(1.1)

angle of reflection

Figure 1-2: Laws of reflection: (1) The ray path for both the incident and the reflected beam are on the plane of incidence and (2) the angle of incidence equals the angle of reflection.

1.1.2 The Refractive Index Light travels in vacuum at the speed of light, denoted by c, which applies to all electromagnetic waves. The value of c, considered to be constant, is Speed of Light in Vacuum:

1-2

c = 2.99×108 m/s

(1.2)

REFRACTION IN A SPHERICAL INTERFACE

Light travels in transparent media, such as air, certain plastics, water, and glass. The key difference is how fast light travels in each of them. The speed of light in these media, even though they are transparent, is less than c. The property of a medium that relates the speed of light within it is the refractive index n or index of refraction. The refractive index value is obtained by dividing the magnitude of the speed of light in vacuum by that in the medium u: refractive index n =

speed of light in vacuum c speed of light in the medium u

(1.3)

By rule, the refractive index is a dimensionless quantity, i.e., a plain number, because it is a ratio of speed to speed, and its value is larger than unity. The larger the refractive index, the slower the light propagates in that medium. In vacuum, the refractive index has its lowest possible value, 1.0; in water n ≈ 1.33, and in diamond n ≈ 2.4. Because light changes speed when it travels in different media, either frequency ν or wavelength λ must change, since their product, the propagation speed u = ν · λ, changes. The frequency of the wave is a characteristic of the source and does not change as the wave propagates in different media. Thus, the wavelength is greater in a medium with a smaller refractive index (where the light travels faster):

c

n1 u 1 u 2 v  2  2 = = = = c n2 u 1 v  1  1

(1.4)

u2

The notion of optical density essentially expresses the value of the refractive index of a medium. A medium that has a greater refractive index in comparison to another is called optically more dense. For example, water (n = 1.33) is optically more dense than air (n ≈ 1.0) but is optically less dense than glass (n ≈ 1.5). We call a medium optically less dense if, again, in comparison to another medium, it has a lower refractive index; for example, air is optically less dense than glass. Along the path of propagation of any ray, the product of the length of the path length traveled LAX and the index of refraction of that medium n is the optical path length (OPL, Lo) or optical thickness. The optical path length is a measure of the propagation ‘difficulty’ between two points along a given curve. Along a path traveled in a uniform and isotropic (same refractive index, regardless of direction or location) medium, Optical Path Length:

LoAX = n · LAX

(1.5)

1-3

GEOMETRICAL OPTICS

1.1.3 The Laws of Refraction When a beam of light (generally, any wave) encounters a refracting surface, part of the beam returns to the initial medium; this is the reflection that we just discussed. Another part of the beam propagates in the new medium. Refraction describes the phenomena associated with the transition of light to the second medium. Therefore, refraction occurs when light propagates from one medium to another medium whose refractive index differs from that of the first. A refracting surface is an interface that separates two media that have different values of refractive index, such as air from water, plastic, glass, etc. The defining effect in refraction is the change in propagation speed. The bending of the optical rays is not the defining effect but is perhaps the most distinct observable effect in refraction. Rectilinear propagation comes to an abrupt change exactly at the refracting surface. This is why a pencil in a water tank appears broken. When shining a thin beam of light (such as from a laser pointer) onto a surface such as air–water or air–glass, the beam obviously changes direction, bending directly into a refracting interface.

Figure 1-3: A pencil inside a water container appears to be broken. The multiple line segment images are due to refraction.

Figure 1-4 illustrates a refracting ray from medium 1 (air, refractive index n1) to medium 2 (glass, refractive index n2). The dividing surface is a refracting surface. Another plane that is perpendicular to the dividing surface is the plane of incidence, which is defined by the incident beam and the normal to the surface at the point of incidence. The two angles involved in refraction are: • the angle of incidence ϑi, defined, as in reflection, by the path of the incident ray just before the incidence and the normal to the point of incidence, and • the angle of refraction ϑt, defined by the path of the refracted ray just after the incidence and (again) the normal to the point of incidence. The rules governing refraction are:

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REFRACTION IN A SPHERICAL INTERFACE

❶

The ray path remains on the plane of incidence.

❷

The sine of the angle of incidence ϑi (also denoted as ϑ1) times the refractive index n1 equals

the sine of the angle of refraction ϑt (also denoted as ϑ2) times the refractive index n2. The above rules are summarized by the law of refraction, or Snell’s law, named after the Dutch astronomer Willebrørd Snellius (1580–1626): Law of Refraction:

n1  sin ( i ) same side (1) of refractive interface

=

n2  sin ( t )

(1.6)

same side (2) of refractive interface

Figure 1-4: Laws of refraction: (1) The ray path for both the incident and the refracted beam are on the plane of incidence and (2) the sine of the angle of incidence multiplied by the refractive index at the first medium equals the sine of the angle of reflection multiplied by the refractive index at the first medium. Example ☞: A ray refracts from air (n1 = 1.0) into water (n2 = 1.333) with an angle of incidence ϑi = 45°. What is the angle of refraction in water? We apply Snell’s law to find the angle of refraction: 1.0 · sin(45°) = 1.33 · sin(ϑt) ⇒ 0.1 · 0.707 = 1.33 · sin(ϑt) ⇒

sin(ϑt) = 0.53. This is the sine of the angle. The angle can be calculated by the inverse sine: ϑt = sin– 1(0.53) = 32°. The ray propagates in water with a refraction angle of 32° with respect to the normal, compared to the incidence angle of 45° in air; the ray is now approaching the normal. This applies to any refraction from an optically rare medium, which causes the ray to move faster, to an optically more dense medium.

The angle formed between the refracted beam and the incident beam is called the deviation angle ϑE. The deviation angle is therefore the difference between the angle of refraction and the angle of incidence: ϑE = ϑt − ϑi. In the example above, ϑE = 32° − 45° = −13° (the negative sign simply means that the deviation angle is clockwise; see § 3.2.1). 1-5

GEOMETRICAL OPTICS

In the case of normal incidence, which means that the angle of incidence ϑi = 0°, the angle of refraction is also ϑr = 0°, which means that the deviation, in this case, is simply zero: ϑE = 0°. There is no apparent change in the ray propagation, despite the fact that the ray is refracted (which physically means that the ray enters a medium in which the propagation speed of light is different from the medium the ray exited).

Figure 1-5: Refraction from an optically less dense medium (air) to an optically more dense medium (glass) for decreasing angles of incidence. The smaller the angle of incidence, the smaller the angle of refraction; the refracted ray approaches the normal. For normal incidence (right side), the angle of refraction is also 0°, so there is no ray deviation.

Laws of Refraction

• The incident ray, the refracted ray, and the normal to the point of incidence are coplanar, as they all lie on the plane of incidence. • The product of the sine of the angle of incidence ϑi and the refractive index in the initial medium equals the product of the sine of the angle of refraction ϑt and the refractive index in the second medium.

1.1.4 Critical Angle and Total Internal Reflection In most examples presented in most texts, refraction occurs when light travels from a medium with a lesser refractive index (optically less dense) n1, such as air, into a medium with a greater refractive index (optically more dense), n2 > n1, such as glass or water; this is called external refraction, although the term is rarely used. The examples illustrated in Figure 1-5 correspond to external refraction. Note that the angle of refraction becomes smaller than the angle of incidence and the ray bends closer to the normal at the point of incidence. However, it is almost as likely in nature that refraction occurs when light travels from a medium with a greater refractive index (such as glass or water) into a medium with a lesser refractive index; this is called internal refraction (n1 > n2). Note that when refraction is internal, the angle of refraction becomes larger than the angle of incidence, and the ray bends away from the normal at the point of incidence—just the opposite of what occurs when the refraction is external. 1-6

REFRACTION IN A SPHERICAL INTERFACE

Example ☞: Calculate the angle of refraction of a ray that propagates: (a) from air (n1 = 1.0) to glass (n2 = 1.5) with angle of incidence ϑi = 40° and (b) from glass (n1 = 1.5) to air (n2 = 1.0) with angle of incidence ϑi = 40° (see Figure 1-6). (a) External refraction from air to glass: Here n1 = 1.0 and n2 = 1.5. So 1.0 · sin(40°) = 1.5 · sin(ϑt) ⇒

sin(ϑt) = sin(40°)/1.5 = 0.428 ⇒ angle of refraction is ϑt = sin–1(0.428) = 25.37°. (b) Internal refraction from glass to air: Here n1 = 1.5 and n2 = 1.0. So 1.5 · sin(40°) = 1.0 · sin(ϑt) ⇒

sin(ϑt) = sin(40°) · 1.5 = 0.965 ⇒ angle of refraction is ϑt = sin–1(0.965) = 74.62°.

Figure 1-6: (left) Refraction from an optically less dense medium to an optically more dense medium; the ray attains a smaller angle in the second medium. (right) Refraction from an optically more dense medium to an optically less dense medium; the ray attains a larger angle in the second medium.

Although the angle of refraction ϑt increases, we know that it may not exceed 90°. There is an angle of incidence, called the critical angle ϑCR, for which the angle of refraction ϑt equals 90°, which means that n1· sin(ϑCR) = n2 · sin(90°) = n2 · 1.0 = n2. We use this relationship to produce the equation for the critical angle: Critical Angle:

sin CR =

n  n2 n  sin ( 90 ) = 2  CR = sin −1  2  n1 n1  n1 

(1.7)

If the angle of incidence equals the critical angle (ϑi = ϑcr), the sine of the angle of refraction equals 1.0 and the angle of refraction is 90° [Figure 1-7 (center)]. The transmitted ray is parallel (tangent) to the interface, in a grazing emergence along the dividing interface. If the angle of incidence becomes larger than the critical angle (ϑi > ϑCR), the sine of the angle of refraction becomes larger than 1.0, and auntie Trigonometry rises in protest! She says that this is unacceptable, so there can be no refraction angle [Figure 1-7 (right)]. The light wave is then completely reflected in a condition called total internal reflection (TIR). TIR occurs only for internal refraction and involves 0% refraction and 100% reflection of the incident light beam. Gonioscopy is an application of TIR in the examination of the eye, as light from the anterior chamber angle undergoes TIR at the air–tear-film interface. The gonioscopy lens is a special

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lens with a refractive index similar to that of the cornea that prevents TIR and allows visualization of the irido-corneal angle when placed in contact with the eye.

Figure 1-7: Internal refraction from a water–air interface. (left) When the angle of incidence is less than the critical angle (ϑi < ϑCR), there is refraction in the optically less dense medium. (center) When the angle of incidence equals the critical angle (ϑi = ϑCR), there is grazing refraction. (right) When the angle of incidence is greater than the critical angle (ϑi > ϑCR), there is no refraction but only total internal reflection.

Figure 1-8: Angle of refraction as a function of angle of incidence for external refraction (dashed lines) and internal refraction (solid lines) for three pairs of refractive indices. Example ☞: What is the critical angle of incidence for light refracting internally from (a) water to air, (b) low-index glass to air, and (c) high-index glass to air? nair = 1.0, nwater = 1.33, nlow-index = 1.5, nhigh-index = 1.7. (a) From water to air, n2 = 1.0 and n1 = 1.33; then ϑCR = sin–1(1.0/1.33) = 48.59°. (b) From low-index glass to air, n2 = 1.0 and n1 = 1.5; then ϑCR = sin–1(1.0/1.5) = 41.81°. (c) From high-index glass to air, n2 = 1.0 and n1 = 1.7; then ϑCR = sin–1(1.0/1.7) = 36.83°.

Note

: In the case of external refraction, as the angle of incidence approaches 90° (ray is ‘grazing’ on

an interface), the angle of refraction approaches the value of the critical angle for internal refraction! For example, from air (n1 = 1.0) into water (n2 = 1.33), if ϑi = 90°, then 1.0 · sin(90°) = 1.33 · sin(ϑt) ⇒ ϑt = sin–1(1.0/1.33) = 48.59°, which is the critical angle of incidence for internal refraction. 1-8

REFRACTION IN A SPHERICAL INTERFACE

1.2 THE SINGLE SPHERICAL REFRACTING INTERFACE A single spherical refracting interface (SSRI) is a single surface (S) that has a spherical shape (S) and operates on refraction (R) at the interface (I) that separates two media of different refractive indices.

1.2.1 Curvature and Radius of Curvature An SSRI may be (a small) part of a perfect spherical surface. The case of a curved surface is fundamental in optics and involves the concepts of curvature and radius of curvature. The center of the sphere is the center of curvature, and the geometric radius of this sphere is the radius of curvature r. The reciprocal of the radius of curvature is called curvature C. A spherical surface is rotationally symmetrical with respect to a line called the optical axis. We usually represent the optical axis horizontally, crossing the center of the sphere. The point of intersection of the optical axis and the surface is the vertex point V. The optical axis is therefore a virtual line connecting the vertex and the center of curvature. It is also called the primary optical axis. Every other line passing through the surface and the center of curvature that is not the optical axis (i.e., is not crossing at the vertex) is called a secondary optical axis. The radius of curvature r of an SSRI is the geometrical radius of the spherical surface. An important aspect in optics is that the radius (as well as other distances) is directional. Thus, the radius has a specific direction and therefore a well-defined algebraic sign. The radius of curvature is drawn from the vertex V at the interface toward the center of curvature.

Figure 1-9: Algebraic sign definitions for the radii of curvature.

The conventions for the algebraic sign of the radius of curvature are simple and follow the Cartesian vector convention (analytically defined in § 3.2.), which stipulates that positive is the direction of light propagation; unless otherwise stated, light travels from left to right. Thus, a positive vector (r > 0) is a vector drawn from the vertex of the interface with its tip to the 1-9

GEOMETRICAL OPTICS

right. This is the case of a radius of curvature in a spherical interface that is oriented like an opening parenthesis. (A negative radius of curvature (r < 0) is a vector drawn with its tip to the left, as in an interface that is oriented like a closing parenthesis.) The curvature C, which is a measure of how ‘sharply’ the surface bends, is inversely proportional to the radius of curvature r. Curvature is small when the radius of curvature is large, and conversely, a sharply curved surface has a small radius of curvature. A plane surface has an infinite radius of curvature and zero curvature.

Figure 1-10: Radius of curvature and curvature: a reciprocal relationship.

Radius of Curvature r • Unit: meter (m) • The longer the radius, the smaller the curvature • Has same algebraic sign as curvature

Curvature C • Unit: diopter (D = m–1) • The larger the curvature, the shorter the radius • Has same algebraic sign as radius of curvature

In the world of Optometry and Vision Science, the unit for curvature is the diopter (D or dpt), which is the reciprocal of the meter (1 D = 1 m–1). The fundamental unit of length used to express the radius of curvature is the meter (m). Indeed, while length quantities such as the radius of curvature may be reported in other units, such as centimeters or millimeters (or even inches), the length quantity has to be converted to meters when engaging in relationships that require its reciprocal, such as curvature and vergence, which relate to object and image locations. Other disciplines and branches of optics, such as optical engineering, do not restrict the unit for curvature to the diopter. In these disciplines, the radii of curvature is often expressed in other units such as millimetes; then curvature is expressed simply in inverse millimeters (mm–1). 1-10

REFRACTION IN A SPHERICAL INTERFACE

1.2.2 The Convex and Concave SSRI The shape of the spherical surface comprising an SSRI can be convex [Figure 1-11 (left)] or concave [Figure 1-11 (right)]. By definition, a convex interface wraps around a medium of higher refractive index (center of curvature situated in the higher-index medium, e.g., glass), while a concave interface wraps around a medium of lower refractive index (center of curvature situated in the lower-index medium, e.g., air).

Figure 1-11: (left) A convex SSRI. (right) A concave SSRI.

1.2.3

Refraction by an SSRI

As in the case for any refracting surface, a ray incident on an SSRI is refracted. To properly draw the refracted ray from a spherical surface, we apply the law of refraction. We assume the simple case of a convex SSRI that separates air from glass and a ray that propagates parallel to the optical axis, striking the SSRI from the air side (Figure 1-12). The first step is to identify the center of curvature, which is the center of the hypothetical spherical surface that defines the SSRI. We then identify the normal to the surface at the point of incidence; this line draws along the spoke that connects the point of incidence to the center of curvature.

Figure 1-12: Identification of the center of curvature and the normal to the surface at the point of incidence.

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GEOMETRICAL OPTICS

The normal to the surface helps identify the angle of incidence, which is formed between the normal and the incident ray. The angle of refraction is then calculated by application of the law of refraction [Eq. (1.6)]. Obviously, it does not equal the angle of incidence. In the example shown here, in which the SSRI wraps around glass with n = 1.5, the angle of refraction ϑt is less than the angle of incidence ϑi. For example, if ϑi = 35°, then ϑt = is 22.5°. Therefore, in this convex SSRI, the ray bends toward the optical axis [Figure 1-13 (right)].

Figure 1-13: Identification of the angles of incidence and refraction, and the refracted ray in a convex SSRI for a ray that propagates from air to glass.

Now consider the opposite orientation: a convex SSRI separating glass (first medium) from air (second medium). The incident ray emerges from within the glass section of this interface; just assume that there is a source within this medium, but it is quite far away [Figure 1-14)]. The procedure is similar: We still draw the normal to the surface, identify the angle of incidence, and complete the angle of refraction. The difference here is that the angle of refraction is now larger. For instance, if the angle of incidence is ϑi = 35°, the angle of refraction can be ϑt ≈ 60°.

Figure 1-14: Refraction by a convex SSRI when the ray is incident from the glass side.

The result in either SSRI orientation is that, when the incident ray is parallel to the optical axis, the refracted ray bends toward the optical axis.

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REFRACTION IN A SPHERICAL INTERFACE

1.2.4

Optical Power and Focal Length

1.2.4.1 The Two Focal Points in an SSRI Now consider a collimated pencil of rays (plane wavefront) traveling along the optical axis (Figure 1-15). This is a group of multiple rays that are all parallel to each other and to the optical axis. The collimated bundle of rays can originate from an object far away, at optical infinity (§ 4.6.3), thereby forming a plane wavefront incident on the interface. Refraction by the SSRI causes the rays to meet at (converge to) a unique point, the focus or focal point F΄. The distance from the vertex to the focal point F΄ is called focal length f΄ (or focal distance). The algebraic sign of the focal length is positive when the vector defining the focal length f ΄ is to the right of the surface, as in the SSRI shown in Figure 1-15.

Figure 1-15: Focal length in a convex SSRI: A collimated bundle of rays converges to the focal point F΄.

Another interesting case is when the pencil of rays incident on the surface appears to originate from an axial focal point F, which is positioned such that, after refraction, all rays are collimated, propagating parallel to the optical axis. The emerging, collimated rays thus form a plane wavefront leaving the interface, as in the SSRI shown in Figure 1-16.

Figure 1-16: Rays that originate from the axial focal point F leave the SSRI collimated.

The SSRI therefore has two different focal points, both of which are situated on the optical axis at either side of the SSRI. Point F is called the primary focal point [Figure 1-17 (right)], and point F΄ is the secondary focal point [Figure 1-17 (left)].

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Figure 1-17: The two focal points in a convex SSRI. (left) A collimated ray bundle converges to the secondary focal point F΄. (right) All incident rays originate from a single point—the primary focal point F. After refraction by the SSRI, the ray bundle is collimated parallel to the optical axis.

The primary focal point F is the unique object-space point serving as the origin of the diverging rays that become collimated (forming a plane wave) in image space following refraction by the interface. The secondary focal point F΄ is the unique image-space point to which rays, initially collimated in object space (deriving from a plane wave), come to focus following refraction by the interface.

1.2.4.2 Optical Power of an SSRI The most important aspect of an SSRI is its dioptric, optical, or refractive power F, which expresses its ability to bend rays. The optical power is positive if the SSRI has a converging property (convex), or negative if the SSRI has a diverging property (concave). The optical power for a spherical interface with a radius of curvature r, in which the medium after the interface has refractive index n΄ and the medium before the interface has refractive index n, is −

n΄

SSRI Optical Power:

F

=

refractive index after the interface

optical power, expressed in D

n refractive index before the interface

r

(1.8)

radius of curvature, expressed in m

Note

: Some variations in the terminology: In some textbooks, point F΄ is denoted as the principal

focal point and point F as secondary focal point. Also, optical power is often denoted by the letter P, while some texts use the letter D.

The unit of power is the diopter (D), the reciprocal of the meter (m–1). The following quantities share the same unit: curvature (§ 1.2.1), optical power of an SSRI and a lens (§ 2.4), optical power of a mirror (§ 5.2.2), and vergence (§ 3.4.1). This is by no means a coincidence! In these relationships, the length quantities (f, r) are expressed in meters (m) and optical power in diopters (D). Often enough, the SSRI is made of glass (a medium of index n΄ ), and the space in front of it is simply air (refractive index 1). In this case, the optical power of the SSRI is SSRI Optical Power, when refracting from air to n΄:

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F =

n΄ − 1.0 r

(1.9)

REFRACTION IN A SPHERICAL INTERFACE

Example ☞: In the following SSRIs, consider air- (n = 1) to-glass (n΄ = 1.5) interfaces. Given the radius of curvature, calculate the optical power.

Both of these SSRIs are convex because they wrap around glass. The radius of curvature is positive because the arrow to the center of curvature points to the right. First SSRI radius of curvature: +12.5 cm. Optical Power: F = (n΄ – n)/r = (1.5 – 1.0)/( +0.125 m) = +4.0 D. Second SSR, radius of curvature: +25 cm. Optical Power: F = (n΄– n)/r = (1.5 – 1.0)/( +0.25 m) = +2.0 D.

Example ☞: Consider air- (n = 1) to-glass (n΄ = 1.5) spherical refracting interfaces. Given the radius of curvature, calculate the optical power.

These SSRIs are concave because they wrap around air. In this configuration, the radius of curvature is negative (arrowhead to the left) in both SSRIs. First concave SSRI radius of curvature: –12.5 cm. Optical Power: F = (n΄ – n)/r = (1.5 – 1.0)/( –0.125 m) = –4.0 D. Second concave SSRI radius of curvature: –25 cm. Optical Power: F = (n΄– n)/r = (1.5 – 1.0)/( –0.25 m) = –2.0 D.

Typically, we assume that the material after the surface has a higher refractive index; i.e., is optically more dense. But we also encounter cases in which the second medium is optically less dense. Regardless of the orientation, a convex interface always has a positive optical power, while a concave interface has a negative optical power.

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Example ☞: In the following cases, consider glass- (n = 1.5) to-air (n΄ = 1.0) interfaces. Given the radius of curvature, calculate the optical power.

Convex SSRI radius of curvature: –12.5 cm. Optical Power: F = (n΄– n)/r = (1.0 – 1.5)/( –0.125 m) = +4.0 D. Concave SSRI radius of curvature: +12.5 cm. Optical Power: F = (n΄– n)/r = (1.0 – 1.5)/( +0.125 m) = –4.0 D.

Example ☞: What would be the refractive index of glass such that an air–glass convex SSRI with +40 cm radius of curvature has power F = +2.0 D? We note that in the relationship F = (n΄–n)/r, there is just one unknown, the refractive index of glass n΄. We substitute F = +2.0 D, n = 1.0, and r = +0.4 m (suggesting that the direction of the radius of curvature is to the right of the interface). Then, 2.0 D = (n΄–1) / 0.4 m ⇒ (n΄–1) = 2.0 · 0.4 = 0.8 ⇒ n΄ = 1.8.

1.2.4.3 Relationships Between Optical Power and Focal Length(s) In an SSRI, the two media separated by the SSRI are, by rule, of different refractive indices. Therefore, the two focal lengths across the two sides of an SSRI, the secondary and the primary focal lengths, always have different magnitudes. The longer focal length is along the side of the larger refractive index. The ratio of the refractive index to the focal length, however, equals the optical power. The more general relationship between the optical power and the focal length is Optical Power and Focal Length:

Optical Power =

Refractive Index Focal Length

(1.10)

Figure 1-18: (left) Smaller optical power and longer focal length. (right) Larger optical power and shorter focal length.

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The optical power of an SSRI is its defining property and has one unique value. The two sides of an SSRI have different focal lengths because of the different refractive indices across the interface. Therefore, the relationship in Eq. (1.10) can be explicitly stated as follows:

Optical Power F and Focal Length(s) f :

F =

n΄

n

image space refractive index

object space refractive index

f΄

= −

secondary focal length

f

(1.11)

primary focal length

Their reciprocal forms are also used to produce the focal length(s) f from the optical power F:

f΄

=

n΄

n

image space refractive index

object space refractive index

secondary focal length

F

and

f primary focal length

= −

F

(1.12)

Because the optical power is positive in a convex SSRI and negative in a concave SSRI, the secondary focal length f΄ is positive in a convex SSRI and negative in a concave SSRI, as the formula f΄ = n΄/F suggests. The directional vector f΄ is drawn from the vertex to the secondary focal point F΄. Therefore, its arrow points to the right in a convex SSRI and to the left in a concave SSRI. Τhe primary focal length f is drawn, again, from the vertex to the primary focal point F, but this time the arrowhead points to the left in a convex SSRI and to the right in a concave SSRI. Since, apparently, this vector is pointing against the direction of light propagation, it should be negative, and the formulas are f = –n/F and F = –n/f; note the – sign. Therefore, when using the primary focal length, we must pay attention to the algebraic signs, as they are opposite to those of the secondary focal length. We can also use the simpler version of the formula, f = n/F, if we are mindful to indicate that the primary focal point is located on the opposite side of the optical element (an SSRI or a lens, as we will see in Chapter 2) with respect to the secondary focal point. Finally, the location of the primary or the secondary focal point is not necessarily indicative of its function. In convex SSRIs (and plus lenses), the order of appearance along the optical axis is first the primary, then the secondary focal point. In concave SSRIs (and minus lenses) the order of appearance is first the secondary, then the primary focal point. Notes

: In the relationships between the secondary focal length and the optical power, the numerator

should be n΄—the refractive index of the image space to the right of the interface—regardless of the location of the secondary focal point, i.e., whether it is to the right or left of the interface. In other words, it is not the location but the function of the focal point that matters. Likewise, in the relationships

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between the primary focal point and the optical power, the numerator should be n, the refractive index of the object space to the left of the interface. The (primary) focal length f does not normally function as a ‘focal length.’ It does so only when light travels from right to left. If, for example, a collimated pencil of rays intersects the lens from the right side, then it will be brought to focus at the ‘primary’ focal point (which then functions as the secondary focal point for that configuration). In other words, we only use the primary focal point as an object point to produce an image at optical infinity (the light beam exiting the SSRI is collimated).

Focal Length f΄ • Unit: meter (m) • The longer the focal length, the lower the optical power • Has the same algebraic sign as the optical power

Optical Power F • Unit: diopters (D = m–1) • The higher the optical power, the shorter the focal length • Has the same algebraic sign as the focal length

Optical Power Optical power is an expression of how close or far the refracting surface can focus a collimated beam. It is proportional to the curvature and inversely proportional to the radius of curvature. A large curvature is associated with a large optical power, while a nearly plane surface (small curvature) has a small optical power. In an SSRI, the optical power is proportional to the difference between the refractive index of the material (n΄ or nm) and that of the surrounding medium (n or no).

So far we have dealt primarily with convex SSRIs, which have a positive optical power and therefore a positive secondary focal length. A concave SSRI has a negative optical power and therefore a negative secondary focal length. In such a case, a collimated beam incident on the SSRI diverges. Because we draw the rays following refraction by the concave SSRI, the emerging rays appear to originate from the secondary focal point F΄ (Figure 1-19), which is situated to the left of the refracting interface. As will be discussed in § 3.1.1, this secondary focal point is a virtual image—the optical conjugate of optical infinity—formed by the SSRI. 1-18

REFRACTION IN A SPHERICAL INTERFACE

Figure 1-19: A collimated beam that is refracted from a concave surface appears to be diverging, as if it originates from the focal point F ΄, which is situated to the left of the SSRI. Example ☞: Consider an air- (n = 1) to-glass (n΄ = 1.5) spherical refracting interface with a radius of curvature r = –0.2 m. Calculate the location of the secondary focal point.

The optical power of the SSRI is F = –2.5 D, so the secondary focal length is f΄ = n΄/F = 1.5/–2.5 D = –0.6 m. The focal point F΄ is located 60 cm to the left of the surface, as is suggested by the – algebraic sign.

1.2.5 The Flat SSRI Hello, flat Earthers! The earth is flat, right? I tend to agree with you; it looks like it. The radius of the earth is the approximate distance from its center to its surface, about 6371 km. The term radius is a property of a true sphere, and even though the earth is not a proper sphere, this value is generally accepted in geophysics as the radius of the earth and is often denoted by R⊕. Thus, the surface of an earthly lake or ocean can be considered a convex SSRI with a radius of curvature of 6371 km. Example ☞: What would be the power and the focal lengths of an SSRI formed by an earthy lake? Assume the following media and refractive indices: air (n = 1.0) and water (n΄ = 1.33). Optical Power of the ‘Earth SSRI’:

F = (n΄ – n)/r = (1.33 – 1.0)/( +6,371,000 m) = +0.0000000518 D ≈ +0.052×10–6 D. Focal Length, Earth Side:

f΄ = n΄/ F = 1.33/ +0.0000000518 D ≈ 25,677 km. Focal Length, Atmosphere Side:

f = –n/ F = –1.00/ +0.0000000518 D ≈ 19,306 km.

1-19

GEOMETRICAL OPTICS

This SSRI has a very small (nearly zero) optical power and nearly infinite focal length(s) and radius of curvature. The larger the radius, the lesser the power. A flat SSRI (technically noted as an SFRI) is an interface that has an infinite radius of curvature. Conversely, its optical power is zero.

Figure 1-20: A single flat refracting interface (SFRI) is an SSRI with an infinite radius of curvature.

1.2.6 The Nodal Point in an SSRI Consider a ray that is incident on a spherical interface such that it is perpendicular to it. Geometrically, this means that the ray is directed toward the center of curvature of the SSRI. The angle of incidence is zero because the ray is normal to the surface, so if forms an angle of zero degrees (ϑi = 0°) with the normal. As a result, the refracted ray also forms a zero degree angle (ϑt = 0°), which means that the ray is undeviated in its path inside the SSRI. In Figure 1-21, we note that this occurs for any ray directed toward the center of curvature of the interface. Therefore, we conclude that any ray directed toward the center of curvature continues to propagate inside the SSRI undeviated. For this reason, the center of curvature in an SSRI is its nodal point. In general, a nodal point N is a point for which a ray directed toward it continues to propagate undeviated. We will also discuss nodal points in thick lenses and lens systems (§ 6.2.2).

Figure 1-21: A ray incident on an SSRI with a zero degree (0°) angle of incidence (normal incidence) continues to propagate undeviated to the nodal point, which is the center of curvature. 1-20

REFRACTION IN A SPHERICAL INTERFACE

Nodal point in an SSRI

• Is situated at the center of curvature. • A ray directed toward it continues to propagate undeviated.

1.2.7 The Listing Eye Model as an SSRI The Listing eye model is a very simple eye model1 consisting of just one refractive surface (radius of curvature r = 5.6 mm) separating air (n = 1.0) from a liquid (n΄ = 4/3 = 1.333). This eye model is, in essence, a convex SSRI by virtue of its spherical anterior surface.

Figure 1-22: The simplified Listing eye model is a convex SSRI with a 5.6 mm radius of curvature.

The refractive index to the right of this surface (inside the eye) is n΄ = 1.333 and to the left of the surface (in air) is n = 1.0. Therefore, the optical power is F = (n΄– n)/r = (1.33 – 1.0) / (+0.0056 m) = +59.5 D ≈ +60.0 D. We now calculate the focal lengths: Focal Length Inside the Eye (n΄ = 1.33):

f΄ = n΄/ F = 1.33/60 D = +22.2 mm (meaning 22.2 mm to the right of the surface). Focal Length in Air (n = 1.0):

f = –n/ F = –1.00/60 D = –16.6 mm (meaning 16.6 mm to the left of the surface). The nodal point N is at the center of curvature of this surface (as is the case in any SSRI), which in this model is situated 5.6 mm to the right of the vertex point V.

1

Visual Optics § 1.5.2 Schematic Eye Models & Introduction to Optics § 4.2.2 Schematic Eyes. 1-21

GEOMETRICAL OPTICS

1.3 ADVANCED PRACTICE EXAMPLES Example ☞: Which of the following SSRIs has the greatest optical power? Which has the longest secondary focal length?

The SSRI with the greatest optical power is the one separating the medium with n΄ = 1.8 from the medium with n = 1.0, with +0.4 m radius of curvature (top row, center). The corresponding optical power is F = (1.8 – 1.0) / 0.4 m = +2.0 D. All of the other SSRIs have optical power values of +1.0 or +0.5 D. The longest focal length (secondary, meaning to the right of the interface) corresponds to the SSRI separating the medium with n΄ = 1.8 from the medium with n = 1.4, with +0.8 m radius of curvature (bottom row, right). In this SSRI, the optical power is F = (1.8 – 1.4) / 0.8 m = +0.5 D, and the focal length is

f΄ = 1.8 / F = 1.8 / 0.5 D = +3.6 m.

Example ☞: An SSRI has the following focal lengths: primary focal length f = –0.2 m and secondary focal length f΄= 0.3 m. The medium to the left of the SSRI is air, n = 1.0, and the medium to the right is glass,

n΄ =1.5. What is the radius of curvature for this interface? Optical power using the primary focal length: F = –n/f = – 1/(–0.2 m) = +5.0 D. Optical power using the secondary focal length: F = n΄/f΄ = 1.5/(+0.3 m) = +5.0 D. To calculate the unknown radius of curvature r, we use the SSRI–curvature relationship: Power F = (n΄ – n)/r ⇒ Radius of curvature r = (n΄ – n)/F = (1.5 – 1.0) / 5.0 D = 0.5 / 5.0 D = +0.1 m.

1-22

REFRACTION IN A SPHERICAL INTERFACE

Example ☞: Calculate the optical power and focal lengths of a convex SSRI that separates air from glass (n΄ = 1.5). Consider a radius of curvature r of absolute value |R|. This SSRI is convex because it wraps around a medium of higher refractive index (glass). We examine two cases of possible orientation. ✔ Light is incident from the air side. Therefore, the radius of curvature is positive: r = +|R|. To calculate the optical power, we apply Eq. (1.8): F = Secondary focal length: F =

n΄ f΄

 f΄ =

n΄ − n 1.5 − 1.0 0.5 = = . r |R| |R|

n΄ 1.5 = 0.5 F

= 3 | R |.

|R|

The secondary focal point is situated to the right of the interface, at distance +3|R|. Primary focal length:

F = −

n n 1.0  f = − = − = − 2| R | . 0.5 f F R

The primary focal point is situated to the left of the interface, at distance –2|R|. The power of this convex SSRI is positive because (1) the radius of curvature r is positive and (2) the medium to the right is glass (refractive index n΄ = 1.5) and to the left is air (refractive index n = 1.0); therefore, the difference (n΄ − n) is positive (= 1.5 – 1.0 = +0.5).

Figure 1-23: Positive secondary focal length in an air– glass convex SSRI with a positive radius of curvature. ✔ We now ‘flip’ the SSRI. Light is incident from within the glass side. There is glass to the left (n = 1.5) and air to the right (n΄ = 1.0): The radius of curvature has a negative sign: r = −|R|. Optical Power:

F =

n΄ − n 1.0 − 1.5 0.5 = = . r −R |R|

Figure 1-24: Positive secondary focal length in a glass– air convex SSRI with a negative radius of curvature.

1-23

GEOMETRICAL OPTICS

We observe that the change in orientation of the SSRI does not affect its optical power. The optical power maintains its magnitude and sign. The second case is simply a flipped version of the first. It is still a convex SSRI wrapping around a medium of higher refractive index. Secondary Focal Length:

F =

n΄ f΄

 f΄ =

n΄ 1.0 = 0.5 F

= 2 | R |.

|R|

The secondary focal point is situated to the right of the interface, at distance +2|R|. Primary Focal Length:

F =

n −f

 f =−

n 1.5 = − 0.5 F

= − 3 | R |.

|R|

The primary focal point is situated to the left of the interface, at distance –3|R|.

Example ☞: What is the optical power of a convex interface with a radius of curvature r = 7.8 mm separating air (n = 1.0) from a medium with refractive index n΄ = 1.376? Optical Power: F = (n΄ – n)/r = (1.376 – 1.0)/0.0078 m = + 48 D. This is the anterior surface of the cornea.

Example ☞: A fictional, extraterrestrial (ET) species living on an exoplanet has eyes filled with a fictional aqueous, whose index is n΄= 1.66. The exoplanet has the same atmospheric air as Earth, and the eye of the ET species has the same optical power as the human eye. The radius of curvature of the simplified ET Listing eye model has to be... The key is the difference in refractive indices, which is the refractive index of the second medium (aqueous) minus the refractive index of the first medium (air). In the human eye, n΄ – n = 1.33 – 1.0 = 0.33, while in the ET eye, n΄ – n = 1.66 – 1.0 = 0.66. Then, power F = the difference in refractive indices / radius of curvature. Since the power maintains magnitude but the difference in refractive indices (numerator) doubles, the radius of curvature (denominator) also has to double. Thus, the ET eye radius of curvature = 2× the human eye radius of curvature = 11.2 mm.

Food for thought

: Would that ET eye have the same focal length(s) as the human eye? If not, which

focal lengths would have the same value and which ones would differ? 1-24

REFRACTION IN A SPHERICAL INTERFACE

1.4 REFRACTION QUIZ These quiz questions include evaluation of material presented in this chapter, as well as in Introduction to Optics § 2.1.1 Angle Units, § 3.3 Prisms, and § 3.4 Refraction Applications.

Angle Measurement 1)

Which of the following angles is the greatest (radian = rad, degree = °, arcmin = ΄ )? a) b) c) d)

2)

π/4 radians or 45° π/2 radians or 90° π radians or 180° 2π radians or 360°

What is the angle ϑ subtended by a full-face quarter-dollar coin (2.5 cm in diameter) held at a distance of 100 m? a) b) c) d)

6)

90° 180° 270° 360°

An arc of one-quarter of a circle subtends an angle measuring _____ radians or ____°. a) b) c) d)

5)

1 rad 0.1 rad 0.01 rad 1° 0.1° 0.01°

8)

ϑ = 0.025 m/100 m = 0.25×10–3 rad ≈ 0.014° ϑ = 0.025 m/100 m = 0.25×10–3 ° ≈ 0.000044 rad ϑ = 100 m/0.025 m = 4000 rad ≈ 229,000° ϑ = 100 m/0.025 m = 4000° ≈ 70 rad

When the angle ϑ is small, which of the following is an acceptable approximation? a) b) c)

opposite side / adjacent side adjacent side / opposite side adjacent side / hypotenuse hypotenuse / adjacent side hypotenuse / opposite side

Assume that you just calculated an angle in radians and need to quickly convert to degrees but have no access to a calculator. Your best chance to get the measure of the angle in degrees ϑ [°] is to ... a) b) c) d)

9)

sinϑ = ϑ [expressed in radians]

When an angle ϑ prescribed in a right angle triangle is small, what ratio best describes the measure of the angle, expressed in radians? a) b) c) d) e)

What is the equivalent in degrees (°) to π radians? a) b) c) d)

4)

7)

The measure of an angle subtending a 1 cm arc, inscribed in a circle with 1 m radius, is … a) b) c) d) e) f)

3)

300΄ 3° 45° 1 rad

d)

multiply ϑ [rad] by 10 multiply ϑ [rad] by 60 divide ϑ [rad] by 10 divide ϑ [rad] by 60

An angle ϑ prescribed in a circle of radius r = 0.5 cm is subtending an arc of length s = 0.2 cm. What is the angle measure? a) b) c) d) e)

ϑ = 0.1° = 0.0017 rad ϑ = 10° = 0.17 rad ϑ = 0.1 rad = 5.72° ϑ = 0.4 rad = 22.91° ϑ = 2.5 rad = 143.24°

10) The accepted resolution limit of the ‘standard’ vision for the human eye is 1΄ = º /60 ≈ 0.29 mrad. At a viewing distance of 6 m, what would be the line stroke width of a letter for that visual acuity? a) b) c) d)

1.746 cm 0.01746 mm 1.746 mm 17.46 mm

sinϑ = ϑ [expressed in degrees] cosϑ = ϑ [expressed in degrees] cosϑ = ϑ [expressed in radians] 1-25

GEOMETRICAL OPTICS

Light Propagation 11) Light travels in a transparent uniform medium at a speed of 2.0×108 m/s. What is the refractive index n of that medium? a) b) c) d) e)

0.666 1.5 2.0 3.0 6.0

12) Light traveling from air (nair = 1.0) enters glass (nglass = 1.5). If the wavelength in air is λair = 0.630 μm (micrometers), the wavelength λglass in glass is ... a) b) c) d)

λglass = 0.630 μm λglass = 0.945 μm λglass = 0.420 μm λglass = 0.360 μm

13) Light traveling from medium 1 (refractive index n1) enters medium 2 (refractive index n2). If the wavelength in medium 1 is λ1 = 0.5 μm and the wavelength in medium 2 is λ2 = 0.666 μm, then which of the following is possible (two correct answers)? a) b)

Medium 1 is glass (n1 = 1.5), and medium 2 is oil (n2 = 2.0). Medium 1 is oil (n1 = 2.0), and medium 2 is glass (n2 = 1.5).

c) d)

Medium 1 is water (n1 = 1.33), and medium 2 is air (n2 = 1.0). Medium 1 is air (n1 = 1.0), and medium 2 is water (n2 = 1.33).

14) What is the speed of light in water (nwater = 1.33)? a) b) c) d)

2.25 ×108 m/s 2.50 ×108 m/s 3.00 ×108 m/s 6.00 ×108 m/s

15) What is the optical path length traveled along a ray that transcends a medium with refractive index n = 1.5 over a length of 2.0 cm? a) b) c) d)

1.33 cm 1.5 cm 2.0 cm 3.0 cm

16) In which case is the optical path length the greatest? a) b) c) d)

light traveling 3.5 mm into a medium with refractive index 1.5 light traveling 4.0 mm into a medium with refractive index 1.4 light traveling 4.5 mm into a medium with refractive index 1.33 light traveling 5.0 mm into a medium with refractive index 1.15

Refraction at a Surface – Snell’s Law 17) In reflection and refraction, the angle of incidence is defined by the … a) b) c) d) e) f)

incident ray and the normal to the surface incident ray and the parallel to the surface normal to the surface and the refracted ray refracted ray and the parallel to the surface reflected ray and the incident ray reflected ray and the parallel to the surface

18) Light refracts from water (n1 = 1.33) into air (n2 = 1.0) with an angle of incidence of 30°. What follows? a) b) c)

1-26

refraction into air with an angle of refraction of about 20° refraction into air with an angle of refraction of about 30° refraction into air with an angle of refraction of about 40°

d)

no refraction but only total internal reflection with an angle of reflection 30° 19) Light refracts from air (n1 = 1.0) into polycarbonate plastic (n2 = 1.4). The angle of incidence is ϑi = 10°. What is the approximate angle of refraction ϑt into this plastic material? a) b) c) d)

7° 9° 10° 14°

20) A ray perpendicularly hits an interface separating air (n1 = 1.0) from water (n2 = 1.33). The angle of refraction is ... a) b) c) d)

90° 45° 30° 0°

REFRACTION IN A SPHERICAL INTERFACE

21) Total internal reflection (TIR) occurs for ... a) b) c) d) e) f)

any angle of incidence when light propagates from air to glass some angles of incidence when light propagates from air to glass normal incidence when light propagates from air to glass any angle of incidence when light propagates from glass to air some angles of incidence when light propagates from glass to air normal incidence when light propagates from glass to air

22) When the angle of incidence equals the critical angle, then … a)

b)

c)

d)

light from an optically more dense medium emerges into an optically less dense medium normal to the surface light from an optically less dense medium emerges into an optically more dense medium normal to the surface light from an optically more dense medium emerges into an optically less dense medium parallel to the surface light from an optically less dense medium emerges into an optically more dense medium parallel to the surface

23) The critical angle for light refracting from air (n1 = 1.0) into glass (n2 = 1.5) is … a) b) c) d)

There is no critical angle from air into glass. 90° 41.8° 0.66 rad

24) Light refracts from water (n1 = 1.33) into air (n2 = 1.0). The refracted ray emerges parallel to the surface. The angle of incidence is … a) b) c) d)

0° 41.4° 48.6° 90°

25) Total internal reflection along an interface of air to glass occurs when light is incident … a) b) c)

from air into glass with an angle of incidence greater than the critical angle from air into glass with an angle of incidence smaller than the critical angle from glass into air with an angle of incidence smaller than the critical angle

d)

from glass into air with an angle of incidence greater than the critical angle

26) When light travels from glass into air, incidence is at the critical angle such that light barely starts to produce total internal reflection. With no change in the incident light path, we drop some water over the glass surface. What is the likely outcome regarding the light path at the glass–water interface (nglass = 1.5; nwater = 1.333; nair = 1.0)? a)

b)

c)

d)

Light refracts from glass into water with an angle of refraction of ϑt = 48.6°; it then emerges parallel to the water–air interface. Light refracts from glass into water with an angle of refraction of ϑt = 41.8°; it then emerges normal to the water–air interface. Light refracts from glass into water with an angle of refraction of ϑt = 90°; it then emerges normal to the water–air interface. Light refracts from glass into water with an angle of refraction of ϑt = 0°; it then refracts at the water–air interface with an angle of refraction ϑt = 48.6°.

27) Which of the following interfaces has the greatest critical angle of incidence (nglass = 1.5, nwater = 1.33, nair = 1.0.)? a) b) c) d) e) f)

air to glass glass to air glass to water water to glass water to air air to water

28) The angle of incidence for a light ray traveling from water into glass is 0°. Select the correct statement regarding the refracted ray. a) b) c) d)

There is no refraction, only reflection. There is refraction; the angle of refraction is 0°. There is no refraction because light does not change its path when propagating in glass. There is refraction; the angle of refraction is 90°.

29) A ray of light propagates from medium 1 (refractive index n1) into medium 2 (refractive index n2). If the angle of incidence is ϑi = 40° and the angle of refraction is ϑt = 25.37°, then … a) b) c) d) e) f)

n1 = 1.0 and n2 = 1.33 n1 = 1.33 and n2 = 1.0 n1 = 1.33 and n2 = 1.5 n1 = 1.5 and n2 = 1.33 n1 = 1.5 and n2 = 1.0 n1 = 1.0 and n2 = 1.5 1-27

GEOMETRICAL OPTICS

30) A ray is refracted from medium 1 into medium 2, and we gradually bring the incident ray closer to the normal to the surface. The refracted ray gradually … a)

comes closer to the normal if medium 2 is optically more dense than medium 1 (n2 > n1)

b) c)

d)

comes closer to the normal if medium 2 is optically less dense than medium 1 (n2 < n1) comes closer to the normal regardless of medium 2 being optically less dense than medium 1 moves away from the normal regardless of medium 2 being optically less dense than medium 1

Refraction at Two Surfaces – Prism and Parallel Interfaces 31) The prism deviation angle equals the angle … a) b) c) d)

between the emerging ray and the incident on the first surface ray of reflection at the second prism surface of refraction at the second prism surface between the emerging ray and the second prism surface

32) A ray is incident from air on a right-angle glass prism as shown:

Which of the following most likely represents the ray path through this prism?

a) A

b) B

c) C

d) D

e) E

33) A ray is incident from air on an equilateral glass prism as shown. If the critical angle for this air– glass material interface is 40°, then the angle of incidence at the second interface ϑi2 equals …

a) b) c) d) e)

10° 20° 30° 40° 50°

34) A glass prism with apical angle A = 30° is surrounded by air. A ray is incident at the first

1-28

interface as shown. The angle of refraction ϑt2 emerging from the second interface is ….

a) b) c) d) e)

greater than 60° equal to 60° greater than 30° equal to 30° smaller than 30°

35) For a thin glass (n = 1.5) prism, the deviation angle is approximately (A = Apical angle) … a) b) c) d)

0.5 · A 1·A 1.5 · A 2·A

36) In a thin glass (n =1.5) prism, the deviation angle ϑE is approximately 10°. The apical angle A of that prism is ... a) b) c) d) e)

5° 10° 15° 20° 40°

37) A prism of prism power P[Δ] = 2 Δ (Δ = prism diopter) at a distance of 0.5 m causes a deviation equal to … a) b) c) d) e)

0.2 cm 0.4 cm 1 cm 2 cm 4 cm

38) A P[Δ] = 6 Δ prism deviates a ray of light by what amount at 3 m?

REFRACTION IN A SPHERICAL INTERFACE

a) b) c) d) e)

1.5 cm 3 cm 6 cm 12 cm 18 cm

39) What is the prism power P[Δ] when a prism deviates a ray by 2 cm over a distance of 50 cm? a) b) c) d) e)

0.25 Δ 0.50 Δ 1.0 Δ 2.0 Δ 4.0 Δ

40) At what distance does a P[Δ] = 10 Δ prism deviate a ray by 6 cm? a) b) c) d)

0.006 m 0.06 m 0.6 m 6m

41) The minimum angle of deviation in a prism is 63°. Based on the information provided in the graph, the apical angle in that prism is …

a)

b)

c)

d)

The minimum angle of deviation is about 52°, which occurs for an angle of incidence of about 46°. The minimum angle of deviation is about 46°, which occurs for an angle of incidence of about 52°. The maximum angle of deviation is about 52°, which occurs for an angle of incidence of about 46°. The maximum angle of deviation is about 46°, which occurs for an angle of incidence of about 52°.

43) The prism deviation angle (select two correct answers) … a)

b)

c) d)

is the angle between the ray emerging from the second surface and the ray incident on the first surface is dependent on the prism geometry (apical angle), the (initial) angle of incidence, and prism refractive index equals the angle of refraction at the second surface of the prism equals the angle of reflection at the second surface of the prism

44) In a glass prism surrounded by air, the ray deviation is always … a) b) c) d)

a) b) c) d)

75° 70° 60° 50°

42) In the diagram below, the prism deviation angle (dependent variable) is plotted against the angle of incidence (independent variable). Select the correct statement in relation to this diagram.

toward the prism base toward the prism apex parallel to the first refracting prism surface parallel to the second refracting prism surface

45) A ray enters a slab of glass with an angle of incidence (with respect to the normal) of ϑi1 = 30°. The slab has parallel surfaces. The ray, upon exiting the slab on the other side, will form an angle with respect to the normal ϑt2 = ... a) b) c) d)

20° 30° 45° This cannot be determined, as we need the refractive index of the glass slab.

46) A fish appears to be 20 cm deep in water (nwater = 1.33). What is the actual depth of the fish? a) b) c) d)

15.0 cm 20.0 cm 26.66 cm 30.0 cm

47) A fossilized insect is trapped halfway within an amber cube, whose sides measure 2 cm (namber = 1-29

GEOMETRICAL OPTICS

1.54). What is the depth at which the insect appears to be trapped? a) b) c) d)

0.65 cm 1.0 cm 1.54 cm 2.0 cm

48) A clear glass slab with parallel surfaces is surrounded by air. Name the pairs of equal angles (select two correct answers).

a) b) c) d)

ϑi1 and ϑt1 ϑt1 and ϑi2 ϑi2 and ϑt2 ϑt2 and ϑi1

49) We now hold the glass slab from Q 48 over water, creating a glass–water interface. The angle of refraction in water ϑt2 (nglass = 1.5, nwater = 1.33, nair = 1.0) is …

a) b) c) d) e)

equal to the angle of incidence ϑi1 greater than the angle of incidence ϑi1 less than the angle of incidence ϑi1 greater than the angle of incidence ϑi2 less than the angle of incidence ϑi2

50) Two clear glass slabs with parallel surfaces are surrounded by air. Slab 1 is made of low-index glass (n1glass = 1.4), while slab 2 is made of highindex glass (n2glass = 1.8). In both cases, the angle of incidence is ϑi1 = 25°. The emergent angle of refraction ϑt2 has the following characteristics: a) b) c) d) e)

in slab 1, ϑt2 = 25°; in slab 2, ϑt2 > 25° in slab 1, ϑt2 < 25°; in slab 2, ϑt2 > 25° in slab 1, ϑt2 < 25°; in slab 2, ϑt2 = 25° in slab 1, ϑt2 = 25°; in slab 2, ϑt2 = 25° in slab 1, ϑt2 < 25°; in slab 2, ϑt2 = 25°

a) b) c) d) e)

converge to the secondary focal point diverge from the primary focal point remain collimated converge to the primary focal point diverge from the secondary focal point

Single Spherical Refracting Interface 51) Light traveling in air (n = 1.0) is incident on a glass (n΄ = 1.5) interface that has a radius of curvature of +0.5 m. What is the power F of this air–glass interface? a) b) c) d)

–1.0 D –2.0 D +2.0 D +1.0 D

52) Light traveling in glass (n = 1.5) is incident on an air (n΄ = 1.0) interface that has a radius of curvature of –0.25 m. What is the optical power F of this glass–air interface? a) b) c) d)

+1.0 D +2.0 D –2.0 D –1.0 D

53) A bundle of parallel light rays (a collimated beam) is incident on a convex refracting interface. Upon leaving the interface, the rays …

1-30

54) The power of an SSRI is given by the formula F = (1.5 – 1.0) / (–0.20 m) = –2.5 D. Which of the following is the SSRI that corresponds to this expression?

a) A

b) B

c) C

d) D

REFRACTION IN A SPHERICAL INTERFACE

55) A convex SSRI separating air (first medium, n = 1.0) from glass (second medium, n΄ = 1.6) has optical power +6.0 D. Its radius of curvature is ... a) b) c) d) e)

+60 cm +15 cm +10 cm –16 cm –40 cm

56) An SSRI has an optical power of +4.0 D. It separates air (first medium, n = 1.0) from glass (second medium, n΄ = 1.5). Its focal lengths are … a) b) c) d)

focal length in air: f = –0.375 m; focal length in glass: f΄ = +0.375 m focal length in air: f = –0.375 m; focal length in glass: f΄ = +0.25 m focal length in air: f = –0.25 m; focal length in glass: f΄ = +0.25 m focal length in air: f = –0.25 m; focal length in glass: f΄ = +0.375 m

57) A fictional ET species living on an exoplanet has eyes filled with a fictional aqueous whose refractive index is n΄ = 1.66. That exoplanet has the same atmospheric air as Earth (n = 1.0), but the eye of the ET species has twice the optical power compared to the human eye. Using the simplified ET Listing eye model (the refractive index of the aqueous of the human eye is n΄ = 1.33), the radius of curvature of the ET eye is … a) the same as that of the human eye, 5.6 mm b) twice that of the human eye, 11.2 mm c) half that of the human eye, 2.8 mm d) four times that of the human eye, 22.4 mm 58) A space mission finally lands on Planet X to discover that the X-lings have an eye with optical power +30.0 D, instead of +60.0 D of the human eye. Everything else is the same: The X-Planet air has refractive index n = 1.0, and the X-ling eye has aqueous with refractive index n΄ = 4/3. What would the radius of curvature of the anterior SSRI in the Listing model be for the X-ling eye? a) the same as that of the human eye, 5.6 mm b) twice that of the human eye, 11.2 mm c) half that of the human eye, 2.8 mm d) four times that of the human eye, 22.4 mm

a) b) c) d) e)

–1.5 0.66 1.0 1.5 3.3

60) An SSRI has the following focal lengths: objectspace primary focal length f = –0.2 m and imagespace secondary focal length f΄ = 0.3 m. The object space is air with, n = 1.0, and the image space is glass with n΄ = 1.5. What is the refractive power F of this interface? a) b) c) d) e) f)

–5.0 D –3.3 D –2.0 D +2.0 D +3.3 D +5.0 D

61) An SSRI has the following focal lengths: objectspace primary focal length f = –0.2 m and imagespace secondary focal length f΄ = 0.3 m. The object space is air with, n = 1.0, and the image space is glass with, n' = 1.5. What is the radius of curvature r of this interface? a) b) c) d) e) f)

–0.5 m –0.2 m –0.1 m +0.1 m +0.2 m +0.5 m

62) What is the refractive index of glass for which an air–glass convex SSRI with a +40 cm radius of curvature would have power F = +2.0 D? a) b) c) d) e)

1.2 1.4 1.6 1.8 2.0

63) Which of the following SSRIs has the greatest optical power? (The material refractive index and radius of curvature are as shown. Light travels from left to right in all cases.)

59) An SSRI has the following focal lengths: objectspace primary focal length f = –0.2 m and imagespace secondary focal length f΄ = 0.3 m. The object space is air with n = 1.0. What is the image-space refractive index n΄?

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GEOMETRICAL OPTICS

d)

The ray does not intersect the optical axis because it propagates parallel to it.

67) The object shown below is a semi-circular piece of glass, surrounded by air. Consider the spherical part of the surface as an SSRI. Select the correct description of this surface.

a) b) c) d) e) f)

A B C D E F

64) Which of the following SSRIs has the longest secondary focal length? (The material refractive index and radius of curvature are as shown in Q 63. Light travels from left to right.) a) b) c) d) e) f)

d)

b) c) d)

It is a concave SSRI because it wraps around a medium with a higher index of refraction. It is a convex SSRI because it wraps around a medium with a higher index of refraction. It is a concave SSRI because it wraps around a medium with a lower index of refraction. It is a convex SSRI because it wraps around a medium with a lower index of refraction.

68) The single flat refracting interface shown below has …

A B C D E F

65) A laser ray is directed at a huge glass ball that has a spherical shape and a diameter of 10 m. If we want the laser ray to continue propagating undeviated into the glass, it should be directed at … a) b) c)

a)

the focal point of the glass ball the nodal point of the glass ball a point far away considered to be optical infinity the vertex point of the glass ball

a) b) c) d)

zero curvature and zero radius of curvature infinite curvature and zero radius of curvature zero curvature and infinite radius of curvature infinite curvature and infinite radius of curvature

69) The interface shown below has a radius of curvature that is …

66) Light refracts from air (n = 1.0) into glass (n΄ = 1.5) via a single spherical refracting interface. If the angle of incidence is 0°, where is the ray going to intersect the optical axis (suggestion: draw a sketch!)? a) b) c)

1-32

The ray intersects the optical axis at the primary focal point. The ray intersects the optical axis at the secondary focal point. The ray intersects the optical axis at the center of curvature.

a) b)

positive, regardless of the values of the refractive index surrounding the surface negative, regardless of the values of the refractive index surrounding the surface

REFRACTION IN A SPHERICAL INTERFACE

c)

d)

positive, if the value of the refractive index to its right is greater than the value of the refractive index to its left negative, if the value of the refractive index to its right is lesser than the value of the refractive index to its left

a) b) c) d)

70) Compare the two interfaces shown below and select the correct statement.

focal length in air: f = –0.6 m; focal length in glass: f΄ = +0.6 m focal length in air: f = –0.6 m; focal length in glass: f΄ = +0.4 m focal length in air: f = –0.4 m; focal length in glass: f΄ = +0.4 m focal length in air: f = –0.4 m; focal length in glass: f΄ = +0.6 m

73) Where is the nodal point situated in this SSRI?

a) b) c) d)

Interface A has a larger radius of curvature and a larger curvature than B. Interface A has a larger radius of curvature and a smaller curvature than B. Interface A has a smaller radius of curvature and a larger curvature than B. Interface A has a smaller radius of curvature and a smaller curvature than B.

71) In a certain SSRI, the optical power is +4.0 D. From this, it can be inferred that (select two correct answers) ... a) b) c) d)

This is a convex SSRI. This is a concave SSRI. Its focal length (secondary) has a negative magnitude. Its focal length (secondary) has a positive magnitude.

72) An SSRI has an optical power of +2.5 D. It separates air (n = 1.0) from glass (n΄ = 1.5). What are the focal length values?

a) b) c) d)

25 cm to the left of the SSRI 50 cm to the left of the SSRI 75 cm to the right of the SSRI 25 cm to the right of the SSRI

74) A convex SSRI has power +2.0 D. It separates air (n = 1.0) from glass (n΄ = 1.5). The secondary focal length is ... a) b) c) d)

+0.25 m +0.50 m +0.75 m +1.00 m

75) A convex SSRI has power +4.0 D. It separates air (n = 1.0) from glass (n΄ = 1.5). The primary focal length is ... a) b) c) d)

–0.25 m –0.50 m –0.75 m –1.00 m

1-33

GEOMETRICAL OPTICS

1.5 REFRACTION SUMMARY Refraction Effects The law of refraction, or Snell’s law, describes the relationship between the angle of incidence and the angle of refraction when light passes through a boundary between two refracting media. If, within the first medium with refractive index n1, the angle formed between the incident ray of light and the normal to the incidence at the interface is ϑi, and, within the second medium with refractive index n2, the angle formed between the refracted (transmitted) ray of light and the normal to the incidence at the interface is ϑt, then Law of Refraction:

n1  sin ( i )

n2  sin ( t )

=

same side (1) of refractive interface

same side (2) of refractive interface

The critical angle ϑCR is a unique angle of incidence from an optically more dense medium (such as water with n1) into an optically less dense medium (such as air with n2), for which the angle of refraction ϑt equals 90°:

n  CR = sin −1  2   n1 

Critical Angle:

For any angle of incidence greater than the critical angle, there is total internal reflection (TIR). SSRI Optical Power and Focal Length An SSRI may be (a small) part of a perfect spherical surface. The center of the sphere is the center of curvature, and the geometric radius of this sphere is the radius of curvature r. The reciprocal of the radius of curvature is called the curvature C. The optical power F of a spherical interface with a radius of curvature r in which the medium after the interface has a refractive index n΄ and the medium before the interface has a refractive index n is −

n΄

SSRI Optical Power:

F

=

refractive index after the interface

n refractive index before the interface

r

optical power, expressed in D

radius of curvature, expressed in m

If the SSRI is a piece of glass in air and light travels from air into glass (refractive index n΄), then SSRI Optical Power, from air to n΄: 1-34

F =

n΄ − 1.0 r

REFRACTION IN A SPHERICAL INTERFACE

Refraction by the SSRI causes the rays to meet at (converge to) a unique point, the focal point F΄. The distance from the vertex to the focal point F΄ is called the focal length f΄ . The SSRI has two different focal points. The primary focal point F is the unique objectspace point serving as the origin of the diverging rays that become collimated (forming a plane wave) in image space, following refraction by the interface. The secondary focal point F΄ is the unique image-space point to which rays, initially collimated in object space (deriving from a plane wave), come to focus, following refraction by the interface.

Figure 1-25: Power and focal lengths in a convex air–glass SSRI of radius of curvature r. Note that, in this case, r is positive. Here, n΄ = nlens and n = nair = 1.0.

The two focal lengths of the SSRI have different magnitudes. Along each side of the interface, the ratio of the refractive index to the focal length (absolute value) is the same—equal to the optical power of the SSRI.

n΄ Optical Power F and Focal Length(s) f :

F =

n

image-space refractive index

f΄

= −

object-space refractive index

secondary focal length

f primary focal length

These relationships can also be written as

f΄ secondary focal length

=

n΄

n

image-space refractive index

object-space refractive index

F

and

f primary focal length

= −

F

Figure 1-26: Power and focal lengths in a concave air–glass SSRI of radius of curvature r. Note that, in this case, r is negative. Here, n΄ = nlens and n = nair = 1.0. 1-35

GEOMETRICAL OPTICS

Ray-Tracing Rules in an SSRI 1. The ray parallel to the optical axis refracts to the secondary focal point. 2. The ray originating from (or crossing through) the primary focal point refracts to become parallel to the optical axis. 3. The ray targeting the center of curvature refracts without any ray deviation (it crosses the center of curvature, which is the nodal point). The topic of ray-tracing rules in an SSRI is further discussed in § 4.5.

Figure 1-27: Summary of ray tracing in a convex SSRI.

Figure 1-28: Summary of ray tracing in a concave SSRI.

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GEORGE ASIMELLIS

LECTURES IN OPTICS, VOL 2

2 LENS REFRACTION AND POWER

2.1 WHAT IS A LENS? A lens (from the word lentil) is an optical element that, through refraction, can converge or diverge a ray bundle. It is usually a piece of glass or plastic, and its shape resembles a lentil or even a glass ball.

Figure 2-1: Optical diagram showing light being refracted by a spherical glass container full of water [image from Roger Bacon’s book de Multiplicatione Specierum (On the Multiplication of Species)].

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GEOMETRICAL OPTICS

You might have noticed that a droplet of water behaves like a lens. The droplet can converge a collimated ray bundle to a point and form an image (Figure 2-2).

Figure 2-2: A droplet of water behaves like a simple converging lens.

In this lens, the optical medium is water (n lens = 1.33), while the surrounding medium is air (next = 1.0). Its defining surfaces are spherical and have radii of curvature that are half of the droplet diameter (assuming a circular droplet), which is a few millimeters. In 1850, at an Assyrian palace in the city of Nimrud, which lies on the banks of the Tigris River in modern day Iraq (then known as Mesopotamia), archaeologist Austen Henry Layard unearthed the Nimrud lens (also called the Layard lens), a 3000-year-old crystal quartz lens that is considered to be the oldest known lens. This discovery suggests that glass lenses were in use in early civilizations, most likely as rudimentary magnifying glasses (with a focal length of about 12 cm) for focusing sunlight.2 The Nimrud lens is a lens that is thicker at the center than at the edges. As we will discuss, this is a converging lens; in other words, it has a positive optical power and is termed a positive or plus lens.

Figure 2-3: (left) The Nimrud lens on display at the British Museum in London, UK. It has a diameter of 1.25 cm and a center thickness of 2.5 mm. (right) Photograph of André Kuipers, an astronaut at the International Space Station, who, in June 2012, created an air bubble inside a water droplet, thereby forming two lenses, one positive and one negative, that created two images, one inverted and one erect. (Left image by Geni from Wikimedia Commons under license CC BY-SA 4.0; right image © ESA/NASA.) 2

Gasson G. The oldest lens in the world: a critical study of the Layard lens. The Ophthalmic Optician. 1972; 1267-72.

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LENS REFRACTION AND POWER

The definition of a lens has not changed since the ancient times of the Nimrud lens. A lens is still defined by two refracting surfaces that surround a clear optical medium. While the surface may have any shape, we prefer to deal with rotationally symmetrical surface shapes, the simplest of which is a spherical surface.

Figure 2-4: (left and center) High-quality lens surface grinding [images from Karamouzis Optical (Greece) used with permission]. (right) Preparation of the lens for the Dark Energy Spectroscopic Instrument (DESI) (image courtesy of DESI used with permission).

Not all lenses are positive, nor do they all have the same shape. A positive lens converges a ray bundle, while a negative, or diverging, lens diverges a ray bundle. The property of being positive (the lens is called positive, converging, or a plus lens) or being negative (the lens is called negative, diverging, or a minus lens) depends on two parameters: the lens shape and the ratio of the lens refractive index to the index of the surrounding medium. Two questions address the sign property of a lens: (1) Is the lens thicker or thinner at the center? (2) Is the lens material optically more dense or less dense in relation to the surrounding medium?

Figure 2-5: (left) Converging (plus) and (right) diverging (minus) lenses.

For example, the Nimrud lens (and any lens labeled ‘convex’ in a trial set) is physically thicker at the center, and its material (glass) is optically more dense than the surrounding medium, which is air. A droplet of water suspended in air is thicker at the center and its medium is optically more dense than the surrounding air. On the other hand, a bubble of air inside a piece of glass is thicker at the center, but its medium is optically less dense than the surrounding medium, which is glass. 2-39

GEOMETRICAL OPTICS

2.2 PRINCIPLES OF LENS OPERATION There are different models to describe the working principle of a lens.

2.2.1

Refraction by Two Surfaces

A lens is most commonly described as two refracting interfaces sharing the same optical axis. For example, the first interface is the air–glass interface, and the second is the glass–air interface. These two consecutive refractions result in each ray bending at two points.

Figure 2-6: Lens principle of operation: An incident ray is subject to two refractions.

Consider a ray that is initially parallel to the optical axis. In a converging lens, the ray bends toward the optical axis, while in a diverging lens, it deviates (bends away) from it.

Figure 2-7: In a converging lens (left), the ray, after two refractions, converges toward the optical axis, while in a diverging lens (right), the ray, after two refractions, deviates away from the optical axis.

2.2.2

Two Prisms

The lens action in a converging lens resembles that of a pair of prisms whose bases are on the optical axis, while the lens action in a diverging lens resembles that of a pair of prisms whose bases are on the periphery, with the apex at the optical axis. Prisms are discussed in Introduction to Optics, § 3.3.

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LENS REFRACTION AND POWER

Figure 2-8: The two refractions in a lens resemble the two refractions in a prism.

Figure 2-9: A lens as two prisms. In positive lenses (left), the prism base is on the optical axis; in negative lenses (right), the prism base is on the periphery, the apex being on the optical axis.

Lenses that are ‘nicely’ curved may be thought as a series of stacked small prisms, each having a slightly different apical angle. The lens function is none other than a controlled and gradual deviation of the incident ray bundle from zero at the center to a specific maximum toward the periphery. In other words, a lens is a graded prism with a parabolic progression. A certain type of lens can be formed into a compact Fresnel lens in a manner similar to the formation of a Fresnel prism.3 The Fresnel lens reduces the amount of material required compared to a conventional lens by dividing the lens into a set of concentric annular sections, each resembling a corresponding prism, with steeper prisms on the edges and a flat or slightly convex center.

Figure 2-10: Conventional lens (left) versus a Fresnel lens (right).

This design allows a significant reduction in thickness (and therefore in mass and volume) at the expense of a reduced imaging quality of the lens; diffraction effects from the sharp edges contribute to this. A Fresnel lens can be much thinner than a conventional lens of comparable optical power, in some cases taking the form of a nearly flat, thin sheet. To

3

Introduction to Optics § 3.3.7 Fresnel and Risley Prisms. 2-41

GEOMETRICAL OPTICS

optimize the imaging quality, the grooves and ridges are combined as a tilted pair (blazed) to create a linearly varying phase change across the groove and ridge pair width, producing a sawtooth pattern of lens grooves (Figure 2-11).

Figure 2-11: Formation of a blazed Fresnel zone lens from a full-thickness planoconvex lens.

2.2.3 Principle of Least Time Another lens function description employs the principle of least optical path (or least time).4 Lenses (and mirrors) are useful in bringing the rays from an incident bundle to a point, the focus. To reach that specific point, each ray must follow a different path. The principle requires that in order for this to occur, all paths must have the exact same travel time. This is not at all easy. The above paragraph presents the problem of stigmatic imaging. Path AF΄ (Figure 2-12) corresponds to a longer length and therefore a longer period of travel time in relation to the straight path CF΄. Since we cannot ‘accelerate’ path AF΄, we have to ‘slow down’ path CF΄. This can be achieved by inserting along the straight path a piece glass of appropriate thickness for delaying light, to compensate for the extra time it takes light to travel through the longer path AF΄. Then the time needed to travel the direct, straight CF΄ line becomes equal to the time needed to travel the bent ΑF΄ line. For ray BF΄, the path is not as long as AF΄, so we need to insert less (a thinner piece of) glass.

Figure 2-12: The optical path lengths along any rays that converge to point F΄ are equal.

By virtue of the Pythagorean theorem, we find that the additional optical path length away from the center by amount y increases by y2. There is, in other words, a geometric path 4

Introduction to Optics § 3.2.1 Refraction: an Application of the Principle of Least Time.

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LENS REFRACTION AND POWER

reduction for rays close to the center of the lens. To compensate for this, the optical path has to increase, accordingly, in a parabolic fashion, toward the center. In general, a parabolic surface is formed by rotating a parabola (a surface defined by y = α × x2, α > 0) 360° around its origin (x = 0, y = 0) (more on parabolas in § 8.1.2). The parabola function applies to many natural phenomena, such as orbits in oblique projection and architectural structures (arches), and it approximates the shape of a spherical surface, especially in the region around the optical axis.

Figure 2-13: The trajectory in an oblique projection is described by a parabolic function.

In the simplest case, the material is homogeneous; i.e., it has the same refractive index thoughout its volume. To achieve a longer optical path at the center, the glass thickness at the center increases in a parabolic fashion. This shape enables all rays to converge to the focal point

F΄ and is the ideal converging lens. The same applies to a reflecting surface: A parabolic shape (parabolic shapes are also discussed in § 8.3.1) can bring all of the reflected rays from a single object point to a single image point. Is the parabolic shape of a lens the only shape that provides proper compensation to achieve stigmatic imaging? Or, can we achieve this using a glass plate with parallel surfaces?

Figure 2-14: A graded-index lens has a constant thickness, but the refractive index increases in a paraboloidal fashion toward the center. Therefore, the optical path through the center increases in relation to the periphery.

The objective is to compensate for the longer optical path (peripherally) by increasing the optical path centrally. This can be achieved if the refractive index increases toward the 2-43

GEOMETRICAL OPTICS

center in a parabolic fashion. Such an element is called a graded refractive index (GRIN) lens. GRIN lenses have many applications, including in optical fibers, and even in the crystalline lens of the human eye.5

2.2.4

Wavefront Transformation

The ray bending from a lens corresponds to a change in the wavefront curvature or, equivalently, corresponds to a change in the vergence (more on this concept in § 3.4.1). For example, an ideally plane wavefront (zero vergence), after passing through a lens, is altered and becomes spherical, obtaining a vergence equal to the optical power of the lens (Figure 2-15).

Figure 2-15: A lens alters the vergence of the incident wavefront.

After a converging lens, a flat wavefront converges to a point, with a radius of curvature equal to the focal length f, which is positive. After a negative lens, the wavefront diverges with a negative radius of curvature that equals the focal length f of the negative lens. In every case, we can state that the optical power of the lens is added to the incident vergence to produce the vergence of the ray bundle emerging from the lens.

Lens Wavefront transformation

Application of least time

5

Visual Optics § 2.2.2 Refractive Indices in the Eye.

2-44

Two refractions

Two prisms

LENS REFRACTION AND POWER

2.2.5

The Gravitational Lens

Einstein’s theory of general relativity introduced the idea that space and time can be curved (warped) by gravity.6, 7 In warped space, a straight line is no longer the shortest path. Over such a surface, the nearest thing to a straight line is what is known as a ‘great circle.’ The shortest path between two points on a sphere, known as an orthodrome, is a great circle segment. A perfect example of a great circle is the surface of the earth. The routing of an aircraft flying, for example, between Athens, Greece, and Washington, D.C. is part of a great circle.

Figure 2-16: The shortest distance between two points in warped space is a segment of a great circle.

Massive objects such as clusters of galaxies bend the space in their vicinity. Light passing close to a massive object should be noticeably bent. The amount of bending increases as the mass increases, distorting, and in some cases magnifying, the objects behind them. This effect is called gravitational lensing (or gravitational bending).

Figure 2-17: Apparent position of a star due to gravitational bending by the Sun.

During a solar eclipse, we can observe that the stars along the same line of sight as the Sun are shifted outward. This is because the light path from the star is bent. Light from the star appears to originate from a direction that differs from where the star really is. (Newton's law of

6

Einstein A. Lens-like action of a star by the deviation of light in the gravitational field. Science. 1936; 84(2188): 506-7.

7

Klein M, Kox A, Renn J, Schulmann R. The collected papers of Albert Einstein. Princeton: Princeton University Press; 1996. 2-45

GEOMETRICAL OPTICS

gravity and Einstein's theory of special relativity also predict light deflection, but only partially. The time warp proposed by the theory of general relativity predicts the correct deflection.)

Figure 2-18: Gravitational lensing. Gravitational bending induced by the galaxy curves the light away from the quasar.

In this space–time reality, light still follows the principle of least optical path, and a massive celestial object functions as a lens. It is possible to see two or more identical images of a background quasar. In some cases, light’s straight propagation path from a background celestial body is bent (warped) by another massive celestial body, such as a foreground galaxy, thereby forming an image that has the shape of a ring.

Figure 2-19: Composite Hubble Space Telescope image focused on the galaxy cluster MACS J1149.2 + 2223, more than 5 billion light years from Earth. The cluster is so massive that its gravitational field bends light from an even more distant supernova into four points (indicated by arrows in the inset), forming an Einstein cross. The image combines data from three months of observations taken in visible light with the Advanced Camera for Surveys and in near-infrared light by the Wide Field Camera 3. (Image by S. Rodney / JHU / NASA / ESA / FrontierSN.8)

Since the amount of warping depends on the mass of the foreground galaxy, this effect is employed to estimate the total mass of the foreground galaxy. Gravitational lenses are used Kelly PL, Rodney SA, Treu T, Foley RJ, Brammer G, Schmidt KB, Zitrin A, Sonnenfeld A, Strolger L-G, Graur O, Filippenko AV, Jha SW, Riess AG, Bradac M, Weiner BJ, Scolnic DS, Malkan MA, von der Linden A, Trenti M, Hjorth J, Gavazzi R, Fontana A, Merten JC, McCully C, Jones T, Postman M, Dressler A, Patel B, Cenko SB, Graham ML, Tucker BE. Multiple images of a highly magnified supernova formed by an early-type cluster galaxy lens. Science. 2015; 347(6226): 1123-6. 8

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LENS REFRACTION AND POWER

for studying young galaxies and to gain a better understanding of the total amount of matter in the universe. Just recently, Hubble Space Telescope imaging9 led to the discovery of the largest sample of the faintest, smallest, and earliest-known galaxies in the Universe.

Figure 2-20: Schematic explaining the lens effect on galaxy images (image credit: NASA/ESA).

Figure 2-21: The gravity of a luminous red galaxy (LRG 3-757) has gravitationally distorted the light from a much more distant blue galaxy. (Image taken with the Hubble Space Telescope's Wide Field Camera 3 from Wikimedia Commons under public domain.)

9

Introduction to Optics § 5.2.6 Extraterrestrial Telescopes. 2-47

GEOMETRICAL OPTICS

2.3 THE THIN LENS A thin lens is a lens with a physical thickness that is negligible compared to the radii of curvature of the lens surfaces. The physical thickness is the distance between the vertex points of the two surfaces. A lens whose thickness is not negligible is called thick. In reality, every lens is a thick lens (this concept is presented in detail in § 6.1).

Figure 2-22: Even what we call a thin lens might not be ‘thin’ by definition, as it has a certain thickness. We ignore this fact due to the thin lens approximation. (left) A glass lens. (right) A lens made of ice (right image by Joseph A. Shaw, Montana State University, used with permission).

2.3.1

Radii of Curvature and Material

A lens has two refractive surfaces and therefore two radii of curvature. The surface order is identified with respect to the direction of light propagation (typically from left to right). Thus, the first surface is the one to the left, and, accordingly, r1 is the radius of curvature of the first (left) surface, and r2 is the radius of curvature of the second (right) surface. The radius of curvature r is defined exactly as it is defined in all refracting surfaces: It is the geometrical radius of the spherical surfaces that form the lens. The center of curvature is the center of the spherical surfaces. The radius of curvature is drawn from the vertex V of the surface toward the center of curvature.

Figure 2-23: Examples of the algebraic signs for various radii of curvature in a lens.

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LENS REFRACTION AND POWER

It is customary to assume that a lens is made of glass, and we generally take this for granted. But in many applications today, glass has been replaced by plastic and other polycarbonate materials. A lens can also be made of any transparent material, even water in air [Figure 2-3 (right)].

Figure 2-24: Traditionally, a lens is made of glass (left). However, a lens can be made of any transparent material, even water (center and right). The three lenses shown here are positive, as indicated by the inverted images that they form.

2.3.2 Primary and Secondary Focal Points Just as in single refracting interfaces, a collimated ray bundle propagating along the optical axis incident on a positive lens is refracted, converging to a point on the optical axis. This point is the secondary focal point F΄, situated to the right of the positive lens (or a convex SSRI). If the rays do not converge, but instead diverge, as in the case in a minus, or negative, lens (or a concave SSRI), the focal point is the point (before the lens) from which the rays appear to originate. This lens has a negative optical power; the focal point F΄ is situated before (to the left of) the lens; the focal length f΄ is drawn as a vector to the left of the minus lens and has a negative algebraic sign.

Figure 2-25: Secondary focal points in a positive lens (left) and in a negative lens (right).

A lens has two foci, the focal point F΄ (situated after a positive lens or before a negative lens), called the secondary, or image-space focal point, and the focal point F (situated before a positive lens or after a negative lens), called the primary, or object-space focal point. 2-49

GEOMETRICAL OPTICS

Figure 2-26: Primary focal points in a positive lens (left) and in a negative lens (right).

For a positive (converging) lens: The primary focal point is the (real) object point that produces a collimated, parallel-to-the optical-axis (plane wave) ray bundle leaving the lens. The secondary focal point is the (real) image point that is produced when a collimated, parallel-to-the-optical-axis (plane wave) ray bundle enters the lens.

When incident light is diverging:

When incident light is collimated:

• Exiting light is collimated • Object point is the Primary focal point

• Exiting light is converging • Image point is the Secondary focal point

For a negative (diverging) lens: The primary focal point is the (virtual) object point to which a ray bundle appears to converge, prior to being refracted by the lens as a collimated beam (parallel plane wave). The secondary focal point is the (virtual) image point from which an originally collimated pencil of rays, when refracted by the lens, appears to originate as a diverging beam leaving the lens.

2-50

When incident light is converging:

When incident light is collimated:

• Exiting light is collimated

• Exiting light is diverging

• Virtual object point = Primary focal point

• Virtual image point = Secondary focal point

LENS REFRACTION AND POWER

2.3.3

Focal Planes and Optical Axis

The focal planes are surfaces that are perpendicular to the optical axis at the focal points. Because a lens has two focal points, there are two focal planes. The focal length is drawn as a vector whose arrowhead points to the image-space, secondary focal point. Thus, the focal length is a positive vector for a positive lens and a negative vector for a negative lens.

Figure 2-27: Focal planes in a positive lens. If the lens is surrounded by the same medium, the focal points and the corresponding focal planes are equidistant from the lens.

The example illustrated in Figure 2-27 implies that the two focal points are equidistant from the lens. This is the case if the lens is surrounded by the same medium, for instance, air on both sides. In this simple case, the focal length is the reciprocal of the optical power of the lens (as discussed in § 2.4). If, however, the lens is not surrounded by the same medium, then, while the optical power may have a single value, the two focal lengths have different magnitudes.

F =

Lens Optical Power and Focal Length:

n  of the medium in which f is calculated f  along the side having medium n

(2.1)

Example ☞: The Nimrud lens has a focal distance of f΄=+12 cm in air. What is its optical power? We use the simple relationship F = 1/f΄, converting 12 cm to 0.12 m. So F = 1/0.12 m = +8.33 D.

Example ☞: The human eye has optical power ≈ +60 D. The medium on the left side is air (n = 1.0), and the medium to the right, which is in the eye (mostly the aqueous/vitreous), has a refractive index of 1.336. What are the respective focal lengths?

F = 60 D =

Optical power of the eye: Focal lengths:

Note

fair = −

1.0

F

= −

1.0 +60 D

nair fair

=

nin the eye fin the eye

 − 0.017 m and

fin eye =

1.336

F

=

1.336 +60 D

 + 0.0226 m

: Is the focal length f or f΄? Are they equal? Do they share the same sign? Which to use?

2-51

GEOMETRICAL OPTICS

There are two focal points in a lens, and, of course, in an SSRI, as well as in a lens system and (surprise!) in a mirror. These are the object-space focal point F (primary) and the image-space focal point F΄ (secondary). Hence, there are two focal lengths in a lens as well. The image-space secondary focal length

f΄ is measured from the lens to the secondary focal point F΄. Conversely, the object-space primary focal length f is measured from the lens to the primary focal point F. In principle, these two focal lengths are not equal in magnitude. They are equal only if the medium surrounding the lens has the same refractive index value (e.g., air on both sides). In a plus lens, the secondary focal length is positive, drawn to the right of the lens, while the primary focal length is negative. Likewise, in a minus lens, the secondary focal length is negative, drawn to the left of the lens, while the primary focal length is positive. The focal length that ‘matters’ in imaging is the secondary focal length. This is where the lens converges to (in a plus lens) or deviates from (in a minus lens) if the rays incident on the lens are collimated. When we state simply ‘the’ focal length of a lens without specifying primary or secondary, we are referring to the secondary focal length. We proceed with the following simplifications: Unless otherwise stated, we assume that the lens is surrounded by the same medium, so the two focal lengths are equal in magnitude. We often omit the prime due to the fact that the focal length to be considered in imaging relationships is the image-space secondary focal length.

2.3.3.1 Optical Axis: Principal and Secondary Every line, other than the principal optical axis, that passes through the center of the lens is called a secondary axis. There are infinite secondary axes. A ray bundle incident on the lens and collimated along a secondary axis converges to a focal point on this secondary axis. The surface defined by all of these focal points—not only by a ray bundle collimated along the principal optical axis that converges to ‘the’ focal point, but from all beams collimated along all secondary axes—is, in reality, the focal ‘plane.’

Figure 2-28: Focusing a collimated beam along a secondary axis and the focal ‘plane.’ 2-52

LENS REFRACTION AND POWER

This surface is, in approximation, flat (plane) only close to the principal optical axis. For this reason, it is termed the focal plane. The surface is often curved. This effect is related to the field curvature aberration (presented in § 8.3.4). Consider the human eye: The focal plane of the eye is the retina,10 which is a curved surface, not a flat surface. The naturally curved shape of the retina matches the naturally curved focal plane.

Figure 2-29: The sensor mosaic on the focal plane of the Kepler Space telescope consists of 42 CCDs, each 50 mm × 25 mm, formed by 2200 × 1024 pixels. The focal plane is a curved, rather than a flat, surface. (Image credit: Kepler Science/NASA.)

2.3.4

Lens Shape

2.3.4.1 The Spherical Lens A spherical lens has surfaces, each of which is part of a sphere; thus, a lens surface has a specific radius of curvature. The first surface to be intersected by rays is given the subscript 1 (radius of curvature r1), and the second surface is given the subscript 2 (radius of curvature r2). For either surface, there is a single value for the radius of curvature over any part or along any orientation of that surface.

Figure 2-30: Two spherical surfaces defining a spherical lens.

Depending on the shape of the surface, a lens can be biconvex, planoconvex, biconcave, planoconcave, or a combination of convex and concave surfaces, called meniscus (Figure 2-31) from the diminutive Greek word for moon μηνίσκος, μήν- -νός. The flat surfaces in 10

Visual Optics § 4.1 Retinal Structure, Geometry, and Optics. 2-53

GEOMETRICAL OPTICS

planoconvex and planoconcave lenses correspond to r = ∞ (no sign). The radii of curvature of the convex and concave surfaces carry an algebraic sign. To identify the sign, we consider the extended spherical surface that forms the lens and identify the center of curvature of each surface. If the vector of the radius points toward the right, the sign is is positive, and if it points to the left, the sign is negative.

Figure 2-31: Cross-sections of different lens shapes and their corresponding radii of curvature. Notes

: If we flip a biconvex (biconcave) lens, the first surface still has a positive (negative) radius of

curvature and the second surface has a negative (positive) radius of curvature. However, if we flip a planoconvex (planoconcave) lens so that the plano surface is second, the first surface will have a positive (negative) radius of curvature. Likewise, if we flip a meniscus lens (according to the presentation in Figure 2-31), all surfaces will have a negative radius of curvature. If the lens is made of material that is optically more dense than the surrounding medium, a lens that is thicker at the center (such as a biconvex or a planoconvex lens) is positive, while a lens that is thinner at the center (such as a biconcave or a planoconcave lens) is negative.

Figure 2-32: (left) Biconvex lens that is part of the Inamori-Magellan Areal Camera and Spectrograph (IMACS) on the Magellan telescope Baade at Las Campanas Observatory (LCO), Chile. (right) Production of ophthalmic lenses. [Left image from LCO used with permission; right image from Karamouzis Optical (Greece) used with permission.] 2-54

LENS REFRACTION AND POWER

2.3.4.2 The Cylinder Lens It has been assumed so far that a given lens surface has a specific radius of curvature; hence, the notation of r1 for the first surface and r2 for the second. In a cylinder lens, the radii differ along the orientation of the (same) surface. Just as a spherical lens has surfaces that are part of a spherical surface, a cylindrical lens has a surface that is part of a cylindrical surface. For this reason, the refractive power exists only along a specific orientation in space, causing the incident light to be focused in only a single dimension.

Figure 2-33: A cylinder lens has two distinct radii of curvature, depending on the lens orientation.

In contrast to a spherical lens, when a collimated beam is focused by a cylinder lens, there is no focal point. Instead, there is a succession of points that line up along a line, parallel to the cylinder axis. This is the stigmatic or focal line of the lens. A minus cylinder [Figure 2-34 (left)] diverges the beam, which appears to be originating from a virtual focal line situated before the lens. A plus cylinder converges a collimated beam on this focal line, which is situated in front of the lens [Figure 2-34 (right)].

Figure 2-34: Focusing in (left) a minus cylinder lens and (right) a plus cylinder lens. The cylinder axis is parallel to the meridian with zero optical power. The meridian with the largest optical power lies perpendicular to the meridian with zero optical power, so the former is parallel to the refractive meridian.

2-55

GEOMETRICAL OPTICS

2.4 LENS OPTICAL POWER 2.4.1 Lens-Maker’s Formula A lens is a combination of two refracting surfaces with radii of curvature r1 and r2 that separate a medium with refractive index n lens from a surrounding medium with index next.

Figure 2-35: Positive (left) and negative (right) radius of curvature in a lens.

To calculate the lens optical power, each surface that defines the lens may be considered as an independent surface. Then, we just add the individual optical powers F1 and F2 of each refracting surface, using Eq. (1.8), which expresses the SSRI optical power. The SSRI optical power is proportional to the difference of the refractive indices and is inversely proportional to the radius of curvature. Consider a lens made of glass with refractive index n lens surrounded by a medium with next: F1 =

F = F1 + F2

nlens − next r1

 F =

and

F2 =

next − nlens r2

nlens − next n −n + ext lens = r1 r2

(2.2) 1 1 −   r1 r2 

( nlens − next )  

(2.3)

If the lens is surrounded by air, then, simply, next = 1.0. In addition, the optical power (reported in diopters, D) is simply the reciprocal of the lens focal length f (expressed in meters, m), which is the distance at which the lens focuses a collimated beam to a single point. Then, if the lens refractive index is denoted by n, the lens power F in air is expressed as Lens Power in Air:

F =

1 1 = ( nlens − 1.0 )   −  f  r1 r2 

1

This relationship is known as the lens-maker’s formula, which is an approximate relationship, based on the following assumptions: 2-56

(2.4)

LENS REFRACTION AND POWER

• The lens is thin; i.e., we disregard its physical thickness. • We only consider rays close to the optical axis (this is the paraxial approximation). • The lens is surrounded by air.

Figure 2-36: Combining the refracting surfaces of Figure 2-35 to produce a lens.

• F is the sum of the optical powers of each single refractive surface.

Lens Optical Power F

Note

• In air, F = 1/f. • If the lens is surrounded by a medium with n, then F = n/f.

: When the lens is surrounded by air, the (secondary) focal length is simply denoted as f.

Example ☞: A plus lens, when surrounded by air, has optical power F = +10.0 D. Its focal length is f = 1/F = 1/10 D = +0.1 m = 10 cm.

Example ☞: A minus lens surrounded by air has focal length f = −40 cm = −0.4 m. Its optical power is F = 1/f = −2.5 D.

Example ☞: A symmetric biconvex lens, whose material refractive index is n = 1.5, is surrounded by air and has radii of curvature r = 10 cm = 0.1 m. What is the power of this lens? The key term here is ‘symmetric biconvex,’ where ‘symmetric,’ means that both interfaces have the same magnitude, and ‘biconvex’ means that both interfaces are convex. The signs for the radii of curvature are positive for r1 (+0.1 m) and negative for r2 (−0.1 m). We calculate the lens optical power using the lensmakers’ formula:

F =

1

  1 − 1  = 0.5   2  = + 10.0 D  ( )   = (1.5 − 1.0 )    +0.1 m −0.1 m   0.1 m   r1 r2 

( n − 1)  

−

1

2-57

GEOMETRICAL OPTICS

Alternatively, we can calculate the lens optical power as the sum of the individual optical powers of the two SSRIs that form it (F = F1 + F2):

F1 = (n – 1.0)/r1 = (1.5 – 1.0)/(+0.1 m) = +5.0 D and F2 = (1.0 – n)/r2 = (1.0 – 1.5)/(–0.1 m) = +5.0 D Therefore, the optical power of the lens is F = F1 + F2 = +5.0 D + 5.0 D = +10.0 D. We note that both surfaces have an equal and positive optical power because this is a symmetric biconvex lens.

2.4.2

Dependence on Orientation, Media, and Geometry

Both the optical power and its reciprocal, the lens focal length, depend on specific geometrical parameters: the refractive indices of the lens material and of the surrounding medium, and the radii of curvature of the lens surfaces. Here we explore the dependencies of the lens power and focal length on certain parameters.

2.4.2.1

Dependence on Orientation

The lens focal length does not depend on orientation. In the example of a planoconvex lens, we assume a convex interface that bulges out of the surface with a higher refractive index, with a radius of curvature r and a flat, plano surface with an infinite (∞) radius of curvature.

Figure 2-37: Sign conventions for calculating the optical powers and focal lengths in two planoconvex lens orientations.

The question is, how exactly do we orient this lens? Which is the first surface, and which is the second surface? There are two options. ✔ First option: Light strikes the flat interface first, so r1 = ∞; then the second interface is the convex one, and according to the sign convention, r2 = –|R|:

F1 =

2-58

n − 1.0 n −1 = =0 r1 

and

F2 =

1− n

r2

=

1− n

−R

=

n −1  R

F =

n −1 R

LENS REFRACTION AND POWER

✔ Second option: Light strikes the convex interface first, so r1 = +|R|; then the second interface is the flat one, which means that r2 = ∞:

F1 =

n − 1.0 n −1 = r1 R

and

F2 =

1− n

r2

=

1− n



=0  F =

n −1 R

These results indicate that the lens optical power and therefore the corresponding focal lengths are not dependent on the lens orientation. This is a general conclusion for all thin lenses, not just the planoconvex type, as long as they are surrounded by the same medium (specifically, a medium with the same refractive index on both sides of the lens). Note

: The plano surface of a planoconvex (or a planoconcave) lens has zero optical power. Therefore,

the power of a planoconvex / planoconcave lens equals the power of the convex / concave interface alone. Note

: The lens-maker’s formula in the above cases (orientations) produces the same optical power:

1

 1  = ( n − 1)   +   − R   R

Case 1: F = ( n − 1)  

2.4.2.2

−

1 

 1 1  1 −  = ( n − 1)   +    R  R   

Case 2: F = ( n − 1)  

Dependence on Surrounding Medium

The medium surrounding the lens surfaces might not be the same. In this case, in the second equation of Eq. (2.2), instead of applying next, we apply the refractive index of the second medium, n΄ext. The cornea of the human eye is such a meniscus-shaped lens with a refractive index ≈ 1.376. Internally (on the posterior surface), there is the aqueous fluid with index ≈ 1.336, while the anterior surface is in contact with air.11 The focal length of a lens may change magnitude or even flip algebraic signs if the lens is surrounded by a different medium. Consider the example (Figure 2-38) of a glass biconcave lens with n lens = 1.4. The radii of curvature are 16 cm = 0.16 m.

Figure 2-38: Optical power and focal length of a lens are dependent on the refractive indices of both the lens material and the surrounding medium. 11

Visual Optics § 2.1.3 Corneal Optical Power. 2-59

GEOMETRICAL OPTICS

When the lens is surrounded by air (next = 1.0), it has −5.00 D optical power and −20 cm focal length. When it is submerged in a liquid medium with refractive index next = 1.75, its refractive distance (and optical power) change considerably. ✔ First case: Surrounding medium is air. For a radius of curvature of 0.16 m, n lens = 1.4 and next = 1.0. We can calculate the optical power of this lens if we treat the lens as two SSRIs. The optical power of the lens is the sum of the two individual optical powers:

F1=

nlens − next n −n 1.4 − 1.0 1.0 − 1.4 = = − 2.5 D and F 2 = ext lens = = − 2.5 D  F = F1 + F2 = −5.0 D r1 −0.16 m r2 +0.16 m

The optical power is F = −5.0 D, and the focal length is f = next/F = 1.0/(−5.0 D) = −0.2 m = −20 cm. Similarly, we can calculate the lens optical power using the lens-maker’s formula:

1 1 1  −2  1 F = ( nlens − next )   −  = (1.4 − 1.0 )   − = 0.4  = − 5.0 D  0.16 m  −0.16 m +0.16 m   r1 r2  ✔ Second case: Surrounding medium is liquid with next = 1.75:

nlens − next n −n 1.4 − 1.75 1.75 − 1.4 = = + 2.1875 D and F 2 = ext lens = = + 2.1875 D r1 −0.16 m r2 +0.16 m  F = F1 + F2 = +4.375 D F1=

Here, the lens-maker’s formula takes the following form:

1 1 1  −2  1 F = ( nlens − next )   −  = (1.4 − 1.75)   − = − 0.35  = + 4.375 D  0.16 m  −0.16 m +0.16 m   r1 r2  The lens in the second case has exactly the same shape and material as the lens in the first case. However, because it is surrounded by different media (it’s submerged in an environment that is optically more dense than its own material), it has a different optical power not only in magnitude but also in sign: F = +4.375 D from −5.0 D, and, consequently, a different focal length f = next/F = 1.75/(+4.375 D) = +0.4 m = +40 cm. This effect applies in intraocular lenses (IOLs). These artificial lenses are implanted in the eye to replace a cataractous crystalline lens that has been removed. IOLs are typically biconvex lenses made of an acrylic material. When implanted in the eye, an IOL is surrounded by the aqueous. Example ☞: A negative (minus) lens made of glass (nglass = 1.5) has optical power in air −10 D when surrounded by air. This lens is immersed in water (nwater = 1.333). Its optical power is…

2-60

LENS REFRACTION AND POWER

The lens shape does not change, so we concentrate on the difference between nglass and nmedium. When the lens is surrounded by air: 1.5 − 1.0 = 0.5, and the lens power is F = −10 D. When the lens is surrounded by water, the difference in refractive indices is 1.5 − 1.33 = 0.166. The lens power F is now proportional to 0.166, not 0.5, so F = − 3.33 D. It is still a negative (minus) lens but is weaker.

Example ☞: A positive (plus) IOL lens made of a plastic material of refractive index (nglass = 1.4) has optical power +32 D when surrounded by air. This lens is immersed in the eye (naqueous = 1.333). Its optical power is… The lens shape is the same, so we concentrate on the difference between nglass and nmedium. When the lens is surrounded by air: 1.4 − 1.0 = 0.4, and the lens power is F = +32 D. This implies that the ‘shape’ factor of the lens power has a numerical value of 80 D. When the lens is surrounded by the watery aqueous, the difference in refractive indices is 1.4 − 1.33 = 0.066. The lens power is F = 0.066 · 80 D = +5.333 D. It is still a positive (plus) lens but is significantly weaker.

2.4.2.3

Dependence on Geometry

Another way to change the lens focal length is to modify the radii of curvature. Consider a biconvex lens submerged in a medium with next = 4/3 on either side. The material of the lens has n lens = 1.416. The first radius of curvature is r1 = +10 mm, and the second is r2 = –6 mm. In other words, this lens accommodates changes in the radii of curvature to r1 = +5 mm and

r2 = –5 mm. We calculate the optical power for both cases next. ❶ First case: State of Relaxation:

nlens − next n −n 1.416 − 1.33 1.33 − 1.416 = = + 8.267 D and F 2 = ext lens = =+13.77 D r1 +0.010 m r2 −0.006 m  F = F1 + F2 = +22.04 D F1=

We can also calculate the lens optical power using the lens-maker’s formula: 1 1 1 1   F = ( nlens − next )   −  = (1.416 − 1.33)   −  = 0.08266  266.66 D = + 22.044 D r r + 0.010 m − 0.00 6 m    1 2

Figure 2-39: Dependence of the lens optical power on the radii of curvature. 2-61

GEOMETRICAL OPTICS

❷ Second case: State of Accommodation:

nlens − next n −n 1.416 − 1.33 1.33 − 1.416 = = + 16.53 D and F2 = ext lens = =+16.53 D r1 +0.005 m r2 −0.005 m  F = F1 + F2 = +33.066 D F1 =

We can also calculate the lens optical power using the lens-maker’s formula: 1 1 1 1   F = ( nlens − next )   −  = (1.416 − 1.33)   −  = 0.08266  400 D = + 33.066 D r r + 0.005 m − 0 . 005 m    1 2

We realize that this specific lens ‘accommodates’ its power by about 11 D (from 8.3+13.8 = 22.1 D to 16.5+16.5 = 33 D) with a simple shape change. This is exactly the principle of

Refractive power and focal length of a lens depend on…

human eye crystalline lens accommodation.12

✔ The geometrical shape of the lens, i.e., the 'curvature' of the lens ✔ The material refractive index in relation to the surrounding medium ✘ Not dependent on lens orientation.

A practical and easy way to measure the focal length of a converging lens is to focus light from a ceiling fixture (even better, the sun) onto the floor. The distance from the ceiling to the lens is usually large, so the incident beam is collimated (from optical infinity). Then, the distance from the lens to the floor is a good approximation of the lens focal length.

Figure 2-40: A practical way to measure a positive lens’ focal length (or to burn your optics notes!).

12

Visual Optics § 7.1 The Nature of Accommodation.

2-62

LENS REFRACTION AND POWER

2.5 ADVANCED PRACTICE EXAMPLES Lens Power Relationships

Example ☞: A symmetric biconvex lens, whose material refractive index is n = 1.5, is surrounded by air and has optical power F = +20.0 D. What are the radii of curvature of this lens? The key term here is ‘symmetric biconvex.’ This indicates that the radii of curvature have equal magnitudes and opposite signs; in other words, r1 = +R and r2 = −R. We now use the lens-maker’s formula [Eq. (2.4)]:

1 1 1 1   2  1  R = 1 = + 0.05 m F = ( n − 1)   −   20 D = (1.5 − 1)   −  = ( 0.5 )    = +20 D  R −R  R R  r1 r2  Therefore, the radii of curvature are r1 = +0.05 m and r2 = −0.05 m.

Example ☞: If the lens-maker’s formula has the following values: F =

(1.7 − 1.0 )  

 ,  +0.05 m +0.10 m  1

−

1

what is the lens material and what are the lens surfaces? What is the lens power? This is a lens of a glass material with n = 1.7 surrounded by air (n = 1.0). The two surfaces have different curvatures. Both surfaces point to the right, since both carry a positive algebraic sign. The first surface, with a radius of curvature of +0.05 m, is convex because it wraps around glass. The second surface, with a radius of curvature of +0.10 m, is concave because it wraps around air. This is a meniscus lens with a positive optical power F = +7.0 D.

Example ☞: If the lens-maker’s formula has the following values: F =

(1.7 − 1.0 )  

1

 +0.10 m

−

1

 ,

+0.10 m 

what is the lens material and what are the lens surfaces? What is the lens power? As in the previous example, this is a lens of a glass (n = 1.7) surrounded by air (n = 1.0). The two surfaces have identical curvatures both in radius magnitude and orientation (sign). The first surface, with a radius of curvature of +0.10 m, is convex because it wraps around glass. The second surface, with a radius of curvature of +0.10 m, is concave because it wraps around air. This describes a meniscus lens with zero optical power F = +0.0 D. It is an example of a Höegh meniscus, which, in reality, has optical power if it is treated as a thick lens (see § 6.6).

2-63

GEOMETRICAL OPTICS

Example ☞: A lens, whose material refractive index is n1 = 1.5, is surrounded by air and has an optical power of −5.0 D. A second lens with exactly the same shape is made of a high-index material (n2 = 1.9). What is the optical power of the second lens? The key words here are ‘the same shape.’ The second lens has radii of curvature that are identical to those of the first lens; therefore, the quantity

1 1  −   r1 r2 

is the same in both lenses, regardless of the actual type of

lens (biconcave, planoconcave, meniscus). We know that F(lens 1) = −5.0 D, and we seek the power of the second lens F(lens 2). We write the lens-maker’s formula for both lenses:

1

Lens 1: F(lens 1) = ( n1 − 1)  

 r1 1

Lens 2: F(lens 2) = ( n2 − 1)  

 r1

−

−

1

 = r2 

1

(1.5 − 1)  

 r1

−

1

 = r2 

1

( 0.5 )  

 r1

−

1

1 1  = − 5 D   −  = − 10.0 D r2   r1 r2 

1

1 1 1 1  = (1.9 − 1)   −  = ( 0.9 )   −  = ( 0.9 )  ( −10 D ) = − 9.0 D r2   r1 r2   r1 r2 

The power of the second lens is F(lens 2) = −9.0 D.

Example ☞: Consider a symmetric biconvex lens with an optical power of +10.0 D. If we construct another lens of the same glass but each new surface has half the radius of curvature, what is the optical power of that second lens? The key words here are ‘each new surface has half the radius of curvature.’

  1 1  1  1 − Lens 2:   = r −r 2  r΄1 r΄2   1  2 2 second lens

1 1 Lens 1:  −   r1 r2  first lens

  =  2 − 2  = 2 1 − 1        r1 r2   r1 r2    first lens

Therefore, ‘half the radius of curvature’ means ‘double the curvature.’ If the second lens is made of the same material, it is expected that it will have twice the optical power. Indeed, the two lenses are of the same material; therefore, the parameter (n −1) is equal in both lenses. We know that F(lens 1) = +10.0 D, and we seek the power of the second lens F(lens 2):

 1

F(lens 2 ) = ( n − 1)   lens 2

 r΄1

−

1 

1 1 1 1  = ( n − 1)  2   −  = 2  ( n − 1)   −  = 2  F(lens 1) = + 20.0 D r΄ 2  r r2   r1 r2  lens 1 lens 1  1

lens 2

2-64

lens 1

lens 1

LENS REFRACTION AND POWER

2.6 LENS POWER QUIZ 1)

A converging (plus) lens can be approximated by two prisms that are …

e)

a) b) c)

f)

d)

2)

 1 − 1    +0.1 m +0.2 m 

a) b) c) d) e) f) 7)

the primary focal point the secondary focal point the lens center optical infinity

8)

 1 − 1    +0.05 m +0.10 m 

F = (1.7 − 1.0 )   a) b) c) d)

the lens material has refractive index 1.7 the lens is surrounded by a material with refractive index 1.7 the first surface is concave; the second surface is convex the first surface is convex; the second surface is convex

A symmetric biconcave lens has −4.0 D power. The surface powers of the two SSRIs that constitute the lens are …

c) d) 9)

first surface −2.0 D; second surface +2.0 D first surface −2.0 D; second surface −2.0 D first surface +2.0 D; second surface +2.0 D first surface +2.0 D; second surface −2.0 D

A minus lens made of glass (nglass = 1.5) has −10.0 D power when surrounded by air. This lens is immersed in water (nwater = 1.333). Its power is … a) b)

f΄ = +3.3 cm f΄ = +4.4 cm f΄ = +33 cm f΄ = +44 cm

Select two correct statements in relation to the following lens-maker’s formula (hint: draw a sketch before attempting to answer):

positive (plus) meniscus lens negative (minus) meniscus lens positive (plus) biconcave lens positive (plus) biconvex lens negative (minus) biconcave lens negative (minus) biconvex lens

a) b) c) d)

A lens of power +30.0 D is surrounded by water (next = 1.333). Its focal length is … a) b) c) d)

5)

first surface positive; second surface positive first surface positive; second surface negative first surface negative; second surface negative first surface negative; second surface positive

What is the lens shape if the lens has the following lens-maker’s formula values?

F = (1.6 − 1.0 )  

Which point serves as the object point that produces a collimated beam leaving a lens? a) b) c) d)

4)

6)

The surface radii of curvature in this lens are ...

a) b) c) d) 3)

joined with bases at the center of the lens joined with apexes at the center of the lens stacked so that one base is at the center and the other is at the edge of the lens side-by-side so that their apexes cross the center of the lens

the first surface is concave; the second surface is concave the first surface is convex; the second surface is concave

same as when surrounded by air, −10.0 D less than when surrounded by air, but still negative (for example, −3.33 D) more than when surrounded air, but still negative (for example, −15.0 D) positive (for example, +3.33 D)

A glass (n = 1.5) lens has +10.0 D power when surrounded by air. What is the power of a sameshaped lens if we use a higher-index glass (n = 1.7)? a) b) c) d)

+8.0 D +10.0 D +12.0 D +14.0 D

10) A glass lens has +10.0 D power. What is the optical power of a second lens made of the same glass but with surfaces that are shaped with twice

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GEOMETRICAL OPTICS

the radius of curvature (compared to the surfaces of the first lens)? a) b) c) d)

+5.0 D +10.0 D +20.0 D +40.0 D

11) A lens has +10.0 D power. What is the power of a second lens made of the same material (same index of refraction) but whose shape is different, with both surfaces having half the radius of curvature (in comparison to the surfaces of the first lens)? a) b) c) d)

+5.0 D +10.0 D +20.0 D +40.0 D

12) Consider a planoconvex lens with +5.0 D power. What is the power of the convex surface of this lens? a) b) c) d) e) f)

0.0 D –2.5 D –5.0 D +2.5 D +5.0 D +10.0 D

13) I am situated at the secondary focal point of a plus lens that is illuminated with collimated rays parallel to the optical axis. I am … a) b) c) d)

sending collimated rays to the lens sending diverging rays to the lens receiving converging rays from the lens receiving diverging rays from the lens

14) A thin glass lens consists of two surfaces with identical radii of curvature, both positive = +0.10 m. The lens optical power is (draw a sketch!) ... a) b) c) d)

0.0 D +5.0 D +10.0 D +20.0 D

15) What is the power of a lens made of glass (n = 1.6), whose front surface radius of curvature is +6.00 cm and back surface radius of curvature is –4.00 cm, if it is surrounded by air? a) b) c) d) e) 2-66

+5.0 D +10.0 D +15.0 D +20.0 D +25.0 D

16) What type of lens is a glass (n = 1.6) with front surface radius of curvature +6.00 cm and back surface radius of curvature –4.00 cm? a) b) c) d) e) f)

biconcave symmetric biconcave biconvex symmetric biconvex plus meniscus minus meniscus

17) What is the power of a glass lens (n = 1.6), whose front surface radius of curvature is +6.00 cm and back surface radius of curvature is –6.00 cm, if it is surrounded by water (n = 1.33)? a) b) c) d)

+4.45 D +6.675 D +8.90 D +11.125 D

18) What is the power of a lens made of glass (n = 1.6), whose front surface radius of curvature is +4.00 cm and back surface radius of curvature is –6.00 cm, if it is surrounded by air? (Hint: this is the same lens as in Q 15, but flipped.) a) b) c) d) e)

–25.0 D –15.0 D 0.0 D +15.0 D +25.0 D

19) A plus meniscus lens can have what types of front (F1) and back (F2) surface powers (select two)? a) b) c) d)

F1 positive; F2 negative F1 negative; F2 positive F1 positive; F2 positive F1 negative; F2 negative

20) A lens has the following surface powers: front surface +8.0 D, back surface –5.0 D. Describe the form of this lens. a) b) c) d)

a plus biconcave lens; power +13.0 D a plus biconcave lens; power +3.0 D a plus meniscus lens; power +13.0 D a plus meniscus lens; power +3.0 D

21) The following two (otherwise identical) planoconvex glass lenses, whose orientations are reversed, are surrounded by air. Which two of the following answers are true?

LENS REFRACTION AND POWER

26) The equation that best describes the power of this glass (n = 1.5) lens surrounded by air (n = 1.0) is …

a) b) c) d)

A: flat surface zero power; convex plus power A: flat surface zero power; convex minus power B: convex surface plus power; flat zero power B: convex surface minus power; flat zero power

22) The optical power of a thin lens is (SSRI = single spherical refracting interface): a) b) c) d)

the sum of the powers of each individual SSRI that make up the lens larger than the sum of the powers of each individual SSRI that make up the lens the difference of the powers of each individual SSRI that make up the lens smaller than the difference of the powers of each individual SSRI that make up the lens

23) The power of the spherical surface in a –10.0 D planoconcave lens is … a) b) c) d) e) f)

+10.0 D 0.0 D –2.5 D –5.0 D –10.0 D –20.0 D

24) The radii of curvature for the lenses below are (two correct answers) …

a) b) c) d) e) f)

A: first surface positive; second surface positive A: first surface positive; second surface negative A: first surface negative; second surface positive B: first surface positive; second surface negative B: first surface negative; second surface negative B: first surface negative; second surface positive

25) A minus lens surrounded by air has −40 cm focal length. Its power is … a) b) c) d) e)

−4.0 D −2.5 D −2.0 D −0.4 D −0.25 D

 1 − 1    +0.05 m +0.20 m  1 1  − (1.5 − 1.0 )     −0.05 m +0.20 m  1 1  − (1.5 − 1.0 )     −0.05 m −0.20 m  1 1  − (1.0 − 1.5 )     +0.05 m −0.20 m  1 1  − (1.0 − 1.5 )     −0.05 m +0.20 m  1 1  − (1.0 − 1.5 )     −0.05 m −0.20 m 

a)

F = (1.5 − 1.0 )  

b)

F =

c)

F =

d)

F =

e)

F =

f)

F =

27) A lens with __________ has the longest focal length. a) b) c) d)

power F = +20.0 D focal length f = +20 cm power F = +5.0 D focal length f = +2.0 m

28) A glass (nglass =1.5) lens has +5.0 D power when in air. We dip this lens in a liquid whose refractive index is also 1.5. The lens power is now … a) b) c) d) e)

+10.0 D +5.0 D 0.0 D –5.0 D –10.0 D

29) What is the form of a lens with +5.0 D front (F1) and –10.0 D back (F2) surface powers? a) b) c) d) e)

plus meniscus minus meniscus biconvex biconcave planoconcave

30) What is the form of a lens with –8.0 D front (F1) and –4.0 D back (F2) surface powers? a) b) c) d)

minus meniscus minus meniscus biconvex biconcave 2-67

GEOMETRICAL OPTICS

2.7 LENS POWER SUMMARY Focal Points A lens has two focal points. The secondary, or image-space focal point F΄ is situated after a positive lens or before a negative lens. The primary, or object-space focal point F is situated before a positive lens or after a negative lens. For a positive (converging) lens: The secondary focal point is the (real) image point if a collimated, parallel-to-the-optical-axis (plane wave) ray bundle enters the lens. The primary focal point is the (real) object point that produces a collimated, parallel-to-theoptical-axis (plane wave) ray bundle leaving the lens.

Figure 2-41: Secondary (left) and primary (right) focal points in a plus lens.

For a negative (diverging) lens: The secondary focal point is the (virtual) image point from which an originally collimated pencil of rays, when refracted by the lens, appears to originate as a diverging beam leaving the lens. The primary focal point is the (virtual) object point to which a ray bundle appears to converge, prior to being refracted by the lens as a collimated beam (parallel plane wave).

Figure 2-42: Secondary (left) and primary (right) focal points in a minus lens.

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LENS REFRACTION AND POWER

Lens Power The lens-maker’s formula is an approximate relationship for the optical power of a thin lens. If the lens is placed in air, the optical power takes the form Lens Power in Air:

1 1 F = ( nlens − 1.0 )   −   r1 r2 

where nlens is the refractive index of the lens, r1 is the radius of curvature of the first (left) surface, and r2 is the radius of curvature of the second (right) surface. The lens-maker’s formula can be derived by adding the two independent surface powers, using the SSRI power formula [Eq. (2.2)] for each surface. The lens power F is dependent on the material it is made of (refractive index nlens), the surrounding medium it is placed in (external refractive index, which here, is just 1.0, but in general can be different), and the geometry of the lens, which is expressed by its two radii of curvature. The lens power is not dependent on the orientation of the lens; i.e., if the lens if flipped, its power is the same (within the limits of the lens being thin and the rays being close to the optical axis). Lens Power and Focal Length:

F = 1.0/f

where f is the (secondary) focal length, which has the same measure as, but a different orientation to, the primary focal length, their only difference being the orientation toward the primary and secondary focal points. If the lens is submerged in a medium with a refractive index other than air (next), the lens power and focal length are not simple reciprocals. Now, instead of the value of 1.0, we use the refractive index of the medium: Lens Power and Focal Length (medium other than air):

F = next/f

Radii of Curvature The radius of curvature r is the geometrical radius of the spherical surfaces that form the lens from the vertex V of the surface toward the center of curvature and follows the notation that stipulates that positive is the direction of light propagation regardless of the order of the surface. Thus, for light traveling from left to right, a radius of curvature that points to the right is positive and a radius of curvature that points to the left is negative.

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GEOMETRICAL OPTICS

Figure 2-43: Τhe algebraic sign for the radii of curvature in a lens is dependent on the orientation of the vector that connects the vertex to the center of curvature of that surface.

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GEORGE ASIMELLIS

LECTURES IN OPTICS, VOL 2

3 IMAGING DEFINITIONS

3.1 OBJECT AND IMAGE Imaging, the most important aspect of geometrical optics, involves the concepts of the object and the image. An object can be any entity that either emits light (in which case it is also called a source) or reflects light. Practically, anything can be an object. From every point of the object, we may assume an infinite number of rays propagating in all directions. The simplest object is a point object, an object that occupies no volume and has no surface area. This is an idealistic concept, of course. A realistic object is an extended object that may be considered as consisting of an infinite number of adjoining point objects. Light from optically ‘processed’ rays forms the image. An image, in other words, is the distribution of light following the interaction between the object rays and the optical system. It is considered to be an optical counterpart of an object—how an object appears through the optical system. An optical system may be a reflecting or refracting surface (discussed in § 1.2), a lens (§ 2.1), a prism, or any combination of these. Every point in the image results from rays that originated from a point (or points) of the object and interacted with the optical system. For each object point there exists, ideally, a unique image point. These two points are called optical conjugates. 3-71

GEOMETRICAL OPTICS

The process of image formation from an object through an optical system is called optical imaging. The set of all points or rays associated with the light incident on the optical system is object space. The set of all points or rays associated with light leaving the optical system is image space. Quantities in object space such as vergence, distance, and size, as well as points such as focal, nodal, and principal points (§ 6.2) are nonprimed; the corresponding quantities and points in image space use the same letters, but primed.

Figure 3-1: Object-space and image-space definitions. Note

: The object and image spaces are not a priori separable physical spaces but are rather associated

with the space of light incident to (object space) or exiting (image space) the optical element. They may be different (e.g., in Figure 2-27, object space is to the left of the lens, while image space is to the right to the lens), or they may be the same physical space (e.g., in Figure 3-6, object space and image space are both to the left of the mirror).

Point Object • An ideal point in object space that emits or reflects an infinite number of rays in all directions. • We only need two rays to locate the image.

Extended Object • A real object that may be considered as an infinite number of point objects. • Its image is identified by examining each point separately. It suffices to image just a few points.

Table 3-1: Notation for object space and image space.

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Object space

Image space

Focal length

f (primary)

f ΄ (secondary)

Vergence

L

L΄

Location

x

x΄

Height

h

h΄

Focal points

F (primary)

F΄ (secondary)

IMAGING DEFINITIONS

3.1.1 Real and Virtual Object; Real and Virtual Image The concept of object is directly tied to rays ‘leaving a point.’ Naturally, these rays (and, by association, the wavefronts) are diverging. An object is placed in front of (before) the optical element, and the rays reach the optical element (a positive lens, a negative lens, a mirror, or a single refracting interface) in a divergent configuration. This is a real object.

Figure 3-2: A real object can be placed in front of a positive lens (left) or in front of a negative lens (right).

A real object is the ‘common sense’ physical object, which is how ‘object’ has been described so far. We can conceptually extend the notion of an object to associate it with any light formation incident on the optical system. In optics, therefore, the object is associated with light incident on the optical system. However, light formation is not necessarily diverging. It is possible that the wavefront incident on the optical system is converging, although this does not occur naturally.13,14 Assume, for a moment, that we remove the optical element (lens or a mirror). Light would converge to a point after (to the right of) that element—the location of the object, which, in essence, exists. The object is not physically formed due to the presence of a lens (or a mirror) in the way. This light formation incident on the optical element is a virtual object.

Figure 3-3: A virtual object can be formed after a positive lens (left) or after a negative lens (right).

Virtual objects are formed in multi-lens, or, in general, multi-element, imaging systems. A converging lens or a concave mirror will form a converging beam, forming a real image, which in turn can be incident on another optical element, forming a virtual image. 13

14

If the incident wavefront is flat, the incident vergence is zero. This object is located at optical infinity (see § 4.6.3). 3-73

GEOMETRICAL OPTICS

Real objects can exist naturally, while virtual objects can only be formed by an imaging system.

A real object is associated with diverging rays: It is located at the point of origin of the rays entering the optical element (such as a lens).

If the rays reaching the system do not diverge, but instead they converge, we have a virtual object.

A virtual object is associated with converging rays: It is formed at the point where the extrapolations of the converging rays (if there was no lens) intersect.

Following optical processing (refraction or reflection), the rays leave the system to form the image. These rays may ‘meet’ at a point in image space. Then we have a real image. This image can be formed on a screen, a piece of paper, or a film.

Figure 3-4: Real image after a positive lens (left) and after a negative lens (right). Such images can be formed on a screen.

Often enough, the image-forming rays do not intersect, but instead they diverge. The extrapolation of the image-forming rays can intersect at a point from which they appear to originate. In this case, we have a virtual image. While it cannot be formed on a screen, a virtual image can be seen by another optical system, such as the eye, looking back through the imaging lens.

Figure 3-5: Virtual image before a positive lens (left) and before a negative lens (right). The image is perceived by observing the extrapolations of the refracted rays.

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IMAGING DEFINITIONS

Real images can be formed on a surface, while virtual images can only be viewed by an imaging system or our eyes.

A real image is associated with converging rays: It is located at the point of intersection of the rays leaving the optical element (such as a lens).

If the rays leaving the system do not converge, but instead they diverge, we have a virtual image.

A virtual image is associated with diverging rays: It is formed at the point where the extrapolations of the diverging rays intersect.

3.1.2 Object and Image in a Plane Mirror The simplest imaging case involves a flat (plane) mirror (see also § 5.1). The laws of reflection (§ 1.1.1) are quite simple: The angle of incidence ϑi (defined by the incident ray path and the normal to the point of incidence) and the angle of reflection ϑr [defined by the reflected ray path and (again) the normal to the point of incidence] are equal. Consider two (divergent) rays from a given point on the object (for instance, the top/head of the boy in Figure 3-6). These rays reflect off of the plane mirror. For each ray, we apply the law of reflection, which states that the angle of reflection = the angle of incidence.

Figure 3-6: Real object and virtual image in plane mirror imaging. The image is perceived by observing the extrapolations of the reflected rays.

After reflection, the rays are diverging as well. The extrapolations of these rays meet (intersect) at a point behind the mirror; the configuration of the reflected rays is identical to the configuration we would have if we simply drew the reflected rays as straight line segments originating from this behind-the-mirror point. In the equal triangles ΑΒΓ and ΓBΔ (Figure 3-7), the ratio of opposite sides (ΑΒ = x and BΔ = x΄ ) equals the ratio of their common adjacent side

3-75

GEOMETRICAL OPTICS

(BΓ), so x = x΄: The origin of the reflected rays is at a distance equal to the distance from the object to the mirror.

Figure 3-7: Point source image as a result of reflection. The object is a point source from which an infinite number of rays originate.

What we perceive is two rays originating from a point behind the mirror. This is the image because it is associated with rays leaving the mirror (following reflection). The image is located in a virtual space behind the mirror, in which there are no rays and no light at all! We may trace any number of rays from this object point and follow them after reflection. The reflected rays still appear to originate from the same ‘image’ point. Therefore, it suffices to consider only two rays to find the image of any object point. We can repeat this for any other object point and find its corresponding image point, which will always be behind the flat mirror.

Figure 3-8: Image of an extended object. An object point corresponds to a unique image point.

To locate the reflection image of a point object: • We need two rays from the same object point. We then apply the law of reflection and trace the reflected rays. • A real image is formed at the intersection of the reflected rays. • A virtual image is formed at the intersection of the extrapolations of the reflected rays.

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IMAGING DEFINITIONS

3.2 SIGN CONVENTIONS By applying the laws of reflection (Figure 3-8), it is easy to find that an object placed 50 cm to the left of (in front of, or before) the mirror forms an image 50 cm to the right of (behind, or after) the mirror. Thus, it seems that the mirror itself can be a ‘before / after’ reference point, and distances can be expressed with respect to an origin located there. When measuring distances, we need to pay attention to the locations of the object and its image, which my often be opposite to each other (relative to the origin); they are actually vectorial distances. This is why we use signs (plus or minus) for distances such as focal length, image and object location from origin, and radii of curvature. The rules to be followed constitute the sign conventions. In the Cartesian sign convention, also known as the ‘English system,’ the positive algebraic sign is determined by the direction of light. Unless otherwise stated, light travels from left to right. Thus, a directional distance to the right is positive and to the left is negative.

Figure 3-9: Cartesian sign convention. Light travels from left to right.

The origin, or point zero (0, 0), is the intersection of the optical axis (often, the horizontal axis x) and the optical element (a lens, a refracting interface, or a mirror). In a lens, point zero is the lens center, and in an SSRI or a mirror, point zero is the vertex. The object and image locations, x and x΄, respectively, are drawn parallel to the optical axis and are positive when the arrow points to the right of (is after) the origin, and negative if the arrow points to the left of (is before) the origin. If an object is placed in front (to the left) of a reflecting surface (such as a mirror), a refractive surface (such as an SSRI), or a lens in object space (as is often the case), the object location x has a negative sign, and the arrow is drawn from point zero to the left. For example,

x = −10 cm means that the object is placed 10 cm in front (to the left) of the origin. Put it in terms of light, light needs another 10 cm to reach the origin. The image location x΄ is positive if the image is formed to the right of the origin. For example, x΄ = +10 cm means that the image is formed 10 cm after (to the right of) a lens or a refracting surface. In terms of light, light has already traveled 10 cm after leaving the surface. 3-77

GEOMETRICAL OPTICS

Note

: In reflection, however, light changes direction. The directional distance to an image point

formed to the left of a mirror is along the (new) direction of light after reflection, and it is therefore positive. If the image is situated to the right of a mirror (as in Figure 3-6 and Figure 3-8), its directional distance is negative. Example ☞: Locate the focal points in a convex glass SSRI (material refractive index 1.5), preceded by air, if the radius of curvature is +0.10 m. Optical power is calculated using Eq. (1.8): F = (n΄ – n)/r = (1.5 – 1.0)/(+0.1 m) = +5.0 D. Secondary focal length: f΄ = –n΄ / F = 1.5/+5.0 D = +0.3 m. It is at +0.3 m, to the right of the SSRI. Primary focal length: f = –n/ F = –1.0/+5.0 D = –0.2 m. The primary focal point is situated at a negative distance of –0.2 m, which means that it is 20 cm to the left of the SSRI vertex. While we would avoid stating that a focal length in this convex SSRI is negative, it is fairly common to state that the primary focal point is at a negative distance.

Figure 3-10: In an SSRI, directional distances pointing to the left of the vertex V are negative, while directional distances pointing to the right of the vertex are positive.

3.2.1 Object / Image Height and Angle Sign Conventions Object and image heights h and h΄, respectively, which are always drawn normal to the optical axis, are therefore situated along the –y axis, and they, too, carry an algebraic sign. An upward height is positive, while a downward height is negative. Angles also carry an algebraic sign: A positive sign is assigned to an angle that is measured counter-clockwise with regard to the optical axis; an angle that is measured clockwise has a negative sign.

Figure 3-11: Some examples of sign conventions for height and angle.

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IMAGING DEFINITIONS

3.3 MAGNIFICATION The image is a scale model of the object, differing in size and orientation. These properties are described by the lateral, translinear, linear, or transverse magnification m, which is defined as the ratio of the linear size of image height h΄ to the corresponding linear size of object height h:15 Linear, Lateral, or Transverse Magnification:

m 

image height h΄ = object height h

(3.1)

Simply called magnification, m indicates how large or small the image is compared to the object, when comparing a linear dimension that is perpendicular (therefore, transverse, translinear, or lateral) to the optical axis. If the magnification is greater than 1.0, the image has a larger linear size, while if the magnification is less than 1.0, the image is smaller, or minified. Magnification is a unitless ratio (a plain number) because we divide linear sizes; therefore, it does not have units. However, it does have an algebraic sign. In an inverted image (h > 0 and h΄ < 0, or h < 0 and h΄ > 0), the magnification is negative (such as in a real image from a converging lens or a concave reflecting surface). If the image is erect, the magnification is positive (h > 0 and h΄ > 0, or h < 0 and h΄ < 0). • If the image is a scaled-up version of the object, the magnification is greater (in absolute value) than 1.0. • If the image is a scaled-down version of the object, the magnification (minification), has an absolute value less than 1.0. • If the image is erect, the magnification is positive. If the image is inverted, the magnification is negative.

Magnification and Minification

Figure 3-12: "Mirrored Blenny" photograph © Adriano Morettin 2016 for National Geographic (used with permission). A horned blenny (Parablennius tentacularis) reflects off of a flat filter on top of the lens, which thus acts as a mirror. Magnification is m = +1.0; i.e., the image has the same linear dimensions as the object and is erect. Obviously, it does not have to be the full ‘height’; the ratio can involve any dimensions that correspond to the same aspect, as long as this aspect is perpendicular to the optical axis. 15

3-79

GEOMETRICAL OPTICS

Figure 3-13: If the image is erect, the magnification is positive, while if the image is inverted, the magnification is negative. The magnification is greater than 1.0 if the image has larger dimensions than the object, equal to 1.0 if the dimensions are equal to the object dimensions, and less than 1.0 if the dimensions are smaller than those of the object.

The axial or longitudinal magnification μ or ML describes the linear image size along the optical axis. The object is now three-dimensional and has a physical dimension along the optical (assumed to be the horizontal) axis. The image, consequently, also has a physical dimension along the horizontal axis. Consider a tree with an object base l, imaged to an image base l΄. The axial magnification is defined using quantities along the optical axis, for example, image base l΄ to object base l [Figure 3-15 (center row)]. Note that in considering axial magnification we are not just restricted to a base; the ratio can involve any dimensions that correspond to the same aspect, as long as this aspect is parallel to the optical axis, which is usually drawn as a horizontal line. Axial or Longitudinal Magnification:

 or ML =

l΄ (image base) l (object base)

(3.2)

3.3.1 Angular Magnification We can achieve significant magnification through a set of binoculars. The problem is that it is quite challenging to measure linear magnification (either transverse or longitudinal) because the object may be so far away, at so-called optical infinity (see § 4.6.3). In such cases, we employ angular magnification Mϑ. Angular magnification is defined as the ratio of two angles: the perceptual angular subtense ϑi under which the image is viewed (e.g., through an optical instrument) to the objective angular subtense ϑo, in other words, as seen with the naked eye.

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IMAGING DEFINITIONS

Angular Magnification:

M =

perceptual angular subtense object angular subtense

=

i o

(3.3)

Figure 3-14: Viewing an object through an optical instrument.

In many optical instruments such as binoculars, the angular magnification is noted by ×; an instrument rated as 20× views an image that appears 20× larger than it would appear to the naked eye.

Figure 3-15: Definitions for transverse magnification (top row), longitudinal magnification (center row), and angular magnification (bottom row).

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GEOMETRICAL OPTICS

3.4 VERGENCE 3.4.1

Wavefront Vergence

Divergent rays spread apart, just like the light rays from a candle. If, on the other hand, rays travel toward a single point, they are called convergent. Likewise, a wavefront may converge or diverge, or remain parallel to itself. The physical entity that expresses the degree of convergence is vergence L. A flat wavefront [the collimated ray bundle in Figure 3-16 (left)] has zero vergence; a converging wavefront has a positive vergence, and a diverging wavefront has a negative vergence.

Figure 3-16: (left) Flat wavefront with zero vergence. (center) Converging wavefront with positive vergence. (right) Diverging wavefront with negative vergence.

Vergence is reported in diopters (D), the reciprocal of a meter (m–1). In air, vergence is the reciprocal of the distance to convergence. This is the distance the wavefront needs to converge (focus) and is denoted by either l or x (Figure 3-16), expressed in meters. Vergence and Distance to Convergence (air): Note

L [D] =

1

x

=

1 distance to convergence [m]

(3.4)

: Even if certain length expressions are not reported/available in meters, before engaging in any

calculations that involve vergence or optical power, distances should be converted to meters.

Vergence carries an algebraic sign that is compatible with the Cartesian sign convention. Within a ray bundle, all rays have the same vergence at a given point of reference. If we draw the wavefronts instead, we use the wavefront curvature, the reciprocal of the radius of the wavefront (more generally, the refractive index of the medium divided by the radius of the wavefront). In a converging wavefront, the radius is measured from the wavefront to the point

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of convergence. In a diverging wavefront, the radius is measured from the wavefront to the apparent origin (considered to be a point source). In [Figure 3-16 (center)], the rays converge to point 2. The vergence, measured from point 1, is the reciprocal of the distance x: L = 1/x. For example, if x = +0.5 m, then L = +2.0 D, which is positive, indicating convergence. However, the vergence of the diverging rays [Figure 3-16 (right)] is negative because there is no distance to focus, but there is a negative distance to point 3 from which the wavefront diverges. For example, if x = −0.5 m, then L = −2.0 D. The negative sign means that the tip of the vector, which is to the left, points against the direction of light propagation. If the medium has a refractive index n, the generalized relationship is Reduced Vergence:

L [D] =

n refractive index = x distance to convergence [m]

(3.5)

Figure 3-17: Vergence in a medium with refractive index n.

Vergence describes light as it is converging or diverging. The vergence L of a wavefront or a ray bundle: • Is an expression of the wavefront curvature. • In converging rays/wavefront, the vergence is positive and is the reciprocal of the axial distance x to the point to which the wavefront converges (in air; otherwise, L = n/x). • In diverging rays/wavefront, the vergence is negative and is the reciprocal of the axial distance x to the point from which the wavefront diverges (in air; otherwise, L = n/x). • In collimated rays / flat wavefront, the vergence is zero.

The unit of vergence is the diopter (D), which is the reciprocal of the meter (1 D = m −1).

The algebraic sign of the vergence (and wavefront curvature) depends only on geometry with respect to the direction of light propagation. Vergence is always positive for converging rays, always negative for diverging rays, and always zero for collimated rays / flat wavefront. 3-83

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Figure 3-18: Vergence for right-to-left propagation. (left) The parallel, collimated bundle has zero vergence, (center) the diverging ray bundle has negative vergence, and (right) the converging ray bundle has positive vergence.

Distance to Focus • Unit: meter (m) • The longer the distance, the less the vergence. • Has the same algebraic sign as the vergence.

Vergence • Unit: diopter (D = m–1) • The larger the vergence, the shorter the distance to focus. • Has the same algebraic sign as the distance to focus.

3.4.2 Vergence and Propagation Waves emitted by any point object or by an extended source are, by nature, diverging. Therefore, the vergence of a diverging ray bundle from virtually any source or object is negative. Closer to the source, the vergence is more negative (in absolute value) because the distance x from the source is shorter.

Figure 3-19: The wavefront radius of curvature gradually increases away from the source, and the vergence, accordingly, becomes smaller (less negative).

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IMAGING DEFINITIONS

Example ☞: What is the vergence at point A, situated 0.1 m to the right of a point source, and at point B, situated 0.5 m to the right of the source? Assume propagation in air (Figure 3-20). At point A, which is 0.1 m after (to the right of) the source, the wavefront radius of curvature is −0.1 m. Alternatively, we can call this the ‘distance from the source.’ The sign is negative because the directional distance to this point is pointing to the left. The vergence at point A has magnitude 1/(−0.1 m) = −10.0 D. At point B, which is 0.5 m after (to the right of) the source, the wavefront radius of curvature is −0.5 m. The vergence at point B is also negative with a magnitude of 1/(−0.5 m) = −2.0 D.

Figure 3-20: Vergence at different points away from the source.

As the wave propagates away from the source, the radius of curvature increases and the wavefront curvature flattens. The vergence becomes smaller, and at an infinite distance, it becomes zero. Therefore, we must keep in mind that the value of the vergence changes as the wavefront propagates. Calculation of the vergence at different points involves the notions of downstream and upstream vergence. The vergence calculated at a point that has ‘traveled’ with the stream, along the direction of light propagation, is called downstream. The vergence calculated at a point that has to ‘travel’ against the stream, in a direction against that of light propagation, is called upstream. Example ☞: We examine a ray bundle at location A. It is converging to point B, which is 20 cm to the right, along the direction of light propagation (Figure 3-21). What is the vergence at point A? What is the vergence at point C, which is 70 cm to the right of point A? Assume propagation in air.

Figure 3-21: As the beam converges and then diverges, the vergence flips algebraic signs.

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At point A, the ray bundle needs +0.2 m to converge. The sign is positive because the directional distance to the point of convergence is to the right, along the direction of light propagation. At point A, the vergence has a magnitude of 1 / (+0.2 m) = +5.0 D. At point C, the ray bundle is diverging. It appears that the origin, the source, or the point of divergence is 0.5 m before it, against light propagation. Therefore, the distance to this point is −0.5 m. At point C, the vergence has a magnitude of 1 / (−0.5 m) = −2.0 D.

A formula that provides downstream vergence when the vergence is known at a certain point is Downstream Vergence (in air):

L΄ =

L 1− d  L

(3.6)

where L΄ is the (unknown) vergence at a point d [m] downstream of the initial (known) vergence

L. The distance d is positive, since downstream means along the direction of light propagation. The more generalized form that accounts for propagation in a medium with refractive index n is Downstream Vergence (medium with n):

L΄ =

L d 1−  L n

(3.7)

Example ☞: A wavefront exits an aperture of an optical black box with a vergence of L = −5.0 D. What is the downstream vergence at a screen 30 cm to the right? (Propagation is in air.) The known vergence is L = −5.0 D, and we seek vergence L΄ at d = +0.3 m downstream (note the + sign!).

L΄ =

Note

L 1− d  L

=

−5.0 D 1 − ( +0.3 m )  ( −5.0 D )

=

−5.0 D 1+1.5

=

−5.0 D 2.5

= − 2.0 D.

: The same relationship can be used for the computation of upstream vergence. The formula is

identical; the only difference is that distance d is negative, since it is against the direction of light propagation.

Example ☞: Calculate the upstream vergence L΄ at 30 cm to the left of a point A that has a vergence L = −2.0 D. Assume that between the two points the medium is air. The vergence for an upstream distance for d = −0.3 m (note the − sign!) is

L΄ =

L 1− d  L

=

−2.0 D 1 − ( −0.3 m )  ( −2.0 D )

=

−2.0 D 1 − 0.6

=

−2.0 D 0.4

= − 5.0 D.

Note that this is exactly the same problem as presented previously in the form of downstream vergence.

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3.4.3 Vergence and Optical Interfaces Vergence also changes as the wave interacts with optical surfaces and elements. This is the most useful property of vergence and is exactly why this notion is fundamental in optics. As we will see in detail later on, the vergence is added algebraically to the optical power of any optical surface or element (SSRIs § 1.2.3, lenses § 4.1, and mirrors § 5.3.1). In plane mirrors, where the optical power is zero, the vergence does not change. For example, if the incident beam is collimated, the reflected beam is collimated as well (Figure 3-22).

Figure 3-22: Conservation of vergence after reflection off a plane mirror.

In spherical mirrors, however, a collimated ray bundle, after reflection by a convex mirror, is altered and becomes spherical. The reflected wavefronts are concentric about the focal point of the mirror. Therefore, the vergence of the ray bundle becomes equal to the optical power of the mirror, as we shall see in detail in § 5.3.1. Consider a collimated beam (flat wavefront): After reflection off a convex mirror, the wavefront is diverging (negative vergence), while after reflection off a concave mirror, the wavefront is converging (positive vergence).

Figure 3-23: The optical power of the mirror is added to the incident vergence. In a convex mirror (left), the ray bundle acquires a negative vergence, while in a concave mirror (right), the ray bundle acquires a positive vergence.

In lenses, the only difference is that is there is no change in the direction of light propagation. The optical power of the lens is added to the incident vergence. The vergence leaving the lens is the sum of the incident vergence and the optical power of the lens. 3-87

GEOMETRICAL OPTICS

Figure 3-24: The optical power of the lens is added to the vergence of the incident beam. The result is the vergence of the refracted beam. (left) Positive lens. (right) Negative lens.

A collimated beam incident on a positive lens acquires a positive vergence [Figure 3-24 (left)], while a collimated beam incident on a negative lens acquires a negative vergence [Figure 3-24 (right)]. In general, for a positive (plus) lens, the vergence leaving the lens toward image space increases, while for a negative (minus) lens, the vergence leaving the lens toward image space decreases. The fact that the vergence of an incident wavefront changes can be explained very easily by a simple observation: The bending of rays by a reflecting or refracting surface corresponds to a change in the curvature of the incident wavefront. Therefore, the vergence changes as it interacts with optical elements. As we will see in many examples, the most likely form of a wavefront in nature has a diverging shape. Any reflecting object or emitting source will produce diverging wavefronts, which have a negative vergence. Collimated pencils of rays with zero vergence are often an approximation of a diverging wavefront originating from a point located so far away that it is said to be at optical infinity. Converging wavefronts do not exist alone in nature. They are usually produced from a (concave) mirror or a (converging) lens.

Vergence: Is an expression of the wavefront curvature (reciprocal of the radius of curvature). It changes as the beam propagates or interacts with optical elements.

Note

: As a converging ray bundle nears the focal point, the vergence changes (increases), and after

the focal point, the vergence flips signs as the beam becomes diverging. At the point of convergence, the vergence is ill-defined, as we have to perform division by zero; therefore, numerically, the vergence is infinite.

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IMAGING DEFINITIONS

Figure 3-25: Vergence changes that may occur when a propagating beam encounters a lens.

3.5 VERGENCE IN IMAGING 3.5.1 Vergence of a Real and a Virtual Object In a real object, as presented in § 3.1.1, the wavefront is diverging; the center of curvature is the point of convergence at a distance to the left, against the direction of light propagation. Thus, a real object has a negative vergence.

Figure 3-26: A real object placed before (to the left of) a converging lens (left) and a diverging lens (right).

The value of the object vergence (expressed in meters) is simply the reciprocal of the directional distance from the optical element (e.g., the lens or mirror) to the object location. This simple reciprocal relationship may differ as follows: If the medium of propagation is other than air, the vergence is the ratio of the refractive index of the medium to the directional distance from the lens to the object. In a virtual object, light converges to a point after (to the right of) the lens (Figure 3-27). This is the location of this object, which, in essence, exists: The light formation owing to these rays is an object because it is incident on (is entering) the optical system. This wavefront is converging; the center of curvature is the point of convergence, at a directional distance to the right of the lens, along the direction of light propagation. Therefore, a virtual object has a positive vergence. 3-89

GEOMETRICAL OPTICS

Figure 3-27: A virtual object formed after (to the right of) a plus lens (left) and after a minus lens (right).

3.5.2 Vergence of a Real and a Virtual Image An image (defined in § 3.1) is a formation of light produced by an imaging system such as a lens. It relates to rays that exit (leave) the lens; these rays can be either converging or diverging. If converging, upon leaving the lens, the rays converge to a point located after (to the right of) the lens, along the direction of light propagation, to form a real image. This image can be formed on a screen.

Figure 3-28: A real image formed after (to the right of) a converging lens (left) and after a diverging lens (right).

Therefore, the rays that form a real image have a positive vergence. The image vergence (in air) is the reciprocal of the directional distance from the optical element to the image. Likewise, the image location (expressed in meters) is the reciprocal of the image vergence. A virtual image, in essence, exists, as it is associated with light exiting the system. The difference with respect to a real image is that the image-forming rays are diverging; the extrapolations of the image-forming rays appear to originate from a point situated to the left of the lens (Figure 3-29). In other words, the wavefront leaving the optical system has as its radius of curvature a directional distance that is pointing against the direction of light propagation. Therefore, the diverging rays forming a virtual image have a negative vergence.

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IMAGING DEFINITIONS

Figure 3-29: A virtual image formed before a converging lens (left) and before a diverging lens (right).

It should be emphasized that an object or an image is characterized as real or virtual solely by its corresponding vergence. A virtual image may be erect or inverted, and magnified or minified. The same is true for a real image, as well. Factors that influence image properties are the optical power (positive or negative) of the imaging element, the propagation medium (value of refractive index), the object type (real or virtual), and the object position (from lens / mirror) in relation to the focal length. These cases are presented in § 4.2, § 4.4, and § 5.3.

Figure 3-30: Real versus virtual object / image. Negative location and vergence: A real object placed to the left of the lens and a virtual image formed to the left of the lens. Positive location and vergence: A virtual object (to be) formed to the right of the lens and a real image formed to the right of the lens.

3.5.3 Upstream and Downstream Vergence in Lens Imaging Because the lens center is the origin of the coordinate system, the vergence, if computed at a point before the lens, is upstream because the new location point is upstream, against the direction of light propagation. Likewise, downstream vergence is computed after the lens. It is downstream because the new location point is along the direction of light propagation. Example ☞: A real object with vergence L = −6.0 D (Figure 3-31) is imaged by a plus lens to form a real image with vergence L΄ = +4.0 D. Assume light propagation in air. What are the object and image locations? What are the image traits? 3-91

GEOMETRICAL OPTICS

The object location is simply the reciprocal (propagation in air) of the object vergence: x = 1 / (−6.0 D) = −0.166 m. This is the distance to the object point from the lens; think of it as the source of the wave reaching the lens. The image location is the reciprocal (again, propagation in air) of the image vergence: x΄ = 1 / (+4.0 D) = +0.25 m. This is the distance to the image point from the lens; think of it as the destination of the wave leaving the lens. The image is real (positive vergence), inverted (negative magnification), and magnified (magnification larger than 1.0); m = L/L΄ = (−6.0 D) / (+4.0 D) = −1.5.

Figure 3-31: Object and image vergence involving a real object, a real image, and a positive lens. Example ☞: In the above case, what is the vergence 5 cm before and 5 cm after the lens? Before the lens, we consider the rays reaching (entering) the lens, diverging from the (real) object. At 5 cm before the lens, the distance from the object point source is xd=−0.05 = −11.6 cm (was −16.6 cm exactly at the lens; Figure 3-32). The vergence is the reciprocal of xd=−0.05 : Ld=−0.05 = 1/xd=−0.05 = 1/(−0.116 m) = −8.57 D. This is upstream vergence because it is computed at a point, which is, with respect to the origin (the lens center), situated against the “stream,” or the direction of light propagation (d = −5 cm = −0.05 m). Alternatively, using Eq. (3.6):

Ld=−0.05 =

Ld=0 1 − d  Ld=0

=

−6.0 D 1 − ( −0.05 m)  ( −6.0 D )

=

−6.0 D 1 − 0.3

=

−6.0 D 0.7

= − 8.57 D.

Figure 3-32: Computation of upstream and downstream vergence for a real object and a real image. 3-92

IMAGING DEFINITIONS

After the lens, we consider the rays leaving the lens and converging to the (real) image point. At 5 cm after the lens, the distance to convergence is x΄d=+0.05 = +20 cm (was +25 cm exactly after the lens). The vergence is the reciprocal of x΄d=+0.05 : L΄d=+0.05 = 1/x΄d=+0.05 = 1 / (+0.20 m) = +5.0 D. This is downstream vergence because it is computed at a point, which is, with respect to the origin (the lens center), located along the direction of light propagation (d = +5 cm = +0.05 m). Alternatively, using Eq. (3.6):

L΄d=+0.05 =

L΄d=0 1 − d  L΄d=0

=

+4.0 D 1 − (+0.05 m)  (+4.0 D )

=

+4.0 D 1 − 0.2

=

+4.0 D 0.8

= +5.0 D.

Upstream or downstream vergence: Is the reciprocal of the distance from the upstream or downstream location (not the lens location any longer) to the object or image point.

Figure 3-33: Upstream vergence for a diverging (left) and a converging (right) wavefront. Upstream vergence is compared to the vergence that is incident on the optical element.

Figure 3-34: Downstream vergence for a converging (left) and a diverging (right) wavefront. Downstream vergence is compared to the vergence leaving the optical element.

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GEOMETRICAL OPTICS

Upstream vergence:

Downstream vergence:

• This is the vergence that is computed before the lens (an optical element, or a point of reference, in general).

• This is the vergence that is computed after the lens (an optical element, or a point of reference, in general).

• It is upstream because the new location is against the direction of light from the lens.

• It is downstream because the new location is along the direction of light from the lens.

• Compared to the vergence at the lens: • It is less positive for a converging wave. • It is more negative for a diverging wave.

• Compared to the vergence at the lens: • It is more positive for a converging wave. • It is less negative for a diverging wave.

Figure 3-35: Upstream vergence versus downstream vergence.

3.5.4 Vergence and a Flat Refracting Interface At the limit at which a surface is plane (flat), its curvature tends to zero, the radius of curvature tends to infinity (r → ∞), and the optical power is zero. There is no vergence change because of refraction from such a surface. In Figure 3-36, the vergence entering (incident) and that leaving (refracted) the flat interface are negative because both distances x and x΄ are negative: Vergence:

incident L = n/x

refracted L΄ = n΄/x΄

(3.8)

Figure 3-36: Diverging beam incident on a plane (flat) refracting surface. Despite the fact that there is a change in apparent point of ‘origin’ of the beam, the vergence is constant before and after the surface.

In this example, the vergence may appear to change because the ray bundle bends (slightly converges). This is not the case, however. Here, n΄ > n and x΄ > x (in absolute value), so the vergence before refraction (L = n/x) and after refraction (L΄ = n΄/x΄) remains unchanged. The simplest case is the incidence of a ray bundle with zero vergence (Figure 3-37). The vergence before the surface (incident) and after the surface (refracted) remains zero.

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IMAGING DEFINITIONS

Figure 3-37: Collimated beam incident on a plane surface. The beam before the interface and after the interface maintains zero vergence.

This is true even if the collimated beam is incident on the plane surface angled with respect to the surface normal. The vergence after refraction is zero, but there is a tilt (Figure 3-38).

Figure 3-38: Collimated beam incident on a plane surface angled with respect to the surface normal. After refraction, the beam maintains zero vergence, but there is a deviation. On a flat interface: Exiting (image-forming) vergence = Entering (object-forming) vergence.

The SSRI ‘imaging’ case can help explain the apparent depth.16 This is the effect of perceiving the image at a location other than the object location when viewing it though a flat refracting interface. For example, when viewing a fish in water, we have a water–air flat SSRI, better yet an SFRI (flat-Earthers, please join me in celebration!). The fish appears to be bigger and closer to the waterline. We can investigate this effect using arguments based on vergence. Consider an SFRI with n = 2.0 and n΄ = 1.0 (Figure 3-39). We place an object at –2.0 m, which is to the left of the interface. The object (incident) vergence is L = n/x = 2.0/(– 2.0 m) = –1.0 D. Immediately after refraction, the image (refracted) vergence is L΄ = L because the incident vergence is added to the flat interface power, which is simply zero. The vergence of the

16

Introduction to Optics § 3.4.2 Apparent Depth. 3-95

GEOMETRICAL OPTICS

rays exiting the interface then equals the vergence entering the interface: L΄ = –1.0 D ⇒ n΄/x΄ = 1.0/( x΄) = –1.0 D. This can be true if the image is located at x΄ = –1.0 m, which is to the left of the interface at half the object location.

Figure 3-39: Refraction by an SFRI: There is no change in vergence despite the apparent change in ray propagation. Food for thought

: What is the magnification in this case? Is it m = 0.5 or m = 1.0? But wait—our

experience with a fish in a tank is that the fish appears larger! Therefore, magnification must be > 1.0… The image of a fish in an tank is not only closer than it appears by about ¾, but also subtends a ≈ 4/3 larger angle at our eyes (the numbers ¾ result if we consider n = 4/3 instead of n = 0.5). Thus, the fish appears bigger by a factor of ≈ 4/3 (that’s a very good approximation of the refractive index of water). However, the notion of magnification can be quite hard to implement in such cases. Previously (§ 3.3), we introduced the concept of magnification and distinguished between linear and angular types. Further, in lens imaging, we show that magnification can be calculated by the ratio of object vergence to image vergence; see Eq. (4.6): m = L / L΄. The problem is that the vergence L΄ leaving the glass tank wall (the interface between water and air) exactly equals the vergence L incident on the wall. The flat surface has zero power to add or subtract to the incident beam. This means that the linear magnification would be just m = L / L΄ = 1.0. So, again, why does the fish appear bigger? The answer lies in the angular magnification. There is a 1:1 linear size relationship between the object and the image. However, the same-size fish image is formed at about ¾ of the distance. Thus, the fish image subtends a 4/3 greater angular size. The angular magnification directly relates to the retinal image size, which is 4/3. This, in turn, leads to linear size overestimation: We perceive the fish in the water as larger.

Because the vergence leaving the surface following refraction by the SFRI (L΄ = n΄/x΄) remains unchanged (equal to the incident vergence), we can write a general expression for the apparent depth, which is the image location x΄:

L = L΄ 

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n n΄ n΄ =  x΄ = x x x΄ n

(3.9)

IMAGING DEFINITIONS

Figure 3-40: Apparent depth effects can be solved by implementing flat SFRI considerations.

This result is identical to the one obtained in the apparent depth considerations,16 which was derived using simple refraction relationships (Snell’s law).

3.5.5 Vergence and the SSRI Power One of the seminal relationships in geometrical optics is the refractive power of the SSRI [Eq. (1.8)], which is restated here: Refractive Power F of an SSRI:

F =

n΄ − n r

(3.10)

where n΄ and n are the refractive indices after (to the right, image space) the interface and before (to the left, object space) the interface, respectively, and r is the radius of curvature of the interface, expressed in meters. It follows that the power is expressed in diopters. This relationship was simply introduced with no justification. It is surprising (or perhaps not surprising?) that this relationship can actually be derived using simple vergence arguments. The argument is that the vergence L΄ leaving the interface (just after it) is altered by the power F of that interface, in comparison to the vergence L entering the interface (just before it): Vergence entering the surface + Power of the surface = Vergence leaving the surface:

L + F = L΄ ⇒

F = L΄ – L

The steps in following discussion simply express the vergence entering (L) and leaving (L΄) the surface. We consider the configuration of a beam directed at the center of curvature of the interface because all rays in this beam intersect the interface at right angles, so both the angle of incidence ϑi and the angle of refraction ϑt equal 0°. The rays therefore converge to a point that is the center of curvature, also serving as the nodal point for this spherical interface.

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Figure 3-41: A beam incident on an SSRI aiming at the center of curvature is subject to normal incidence, so its path is not altered. The center of curvature of an SSRI is therefore the nodal point of that interface.

Right before the interface, the ‘distance to convergence’ r is the radius of curvature of the SSRI. Immediately after the interface, the distance is the same, again, equal to r. Before the interface, the refractive index value is n, and to the right of (after) the interface, that value is n΄. Thus, the vergence values for the rays entering the SSRI and for those leaving the SSRI are vergence entering L = n/r

and vergence leaving L΄ = n΄/r

Therefore, F = L΄ – L ⇒ F = n΄/r – n/r = (n΄ – n)/ r. This is the exact expression for the SSRI optical power given in Eq. (1.8).

3.6 ADVANCED VERGENCE EXAMPLES Example ☞: If point A has a vergence of LA = +5.0 D, what is the vergence (a) at point D, which is 10 cm from point A, (b) at point E, which is 45 cm from point A, and (c) at point C, which is 70 cm from point A?

Figure 3-42: The magnitude of the vergence is greater closer to the point of convergence. (a) At point D, the ray bundle needs +0.1 m to converge. The sign is positive because the directional distance to the point of convergence is still pointing to the right. Therefore, the vergence at point D has magnitude LD = 1 / (+0.1 m) = +10.0 D. It is more positive than the vergence at point A (+5.0 D). 3-98

IMAGING DEFINITIONS

Numerically, we can solve this as follows: LD = LA /[1 – dAD·LA] = (+5.0 D) /[1 – (+0.1 m)·(+5.0 D)] = +10.0 D. (b) At point E, the ray bundle is diverging. It appears that the point of origin is located 0.25 m before E; this is written as −0.25 m. Therefore, the vergence at point E has magnitude LE = 1 / (−0.25 m) = −4.0 D. Numerically, we can solve this as follows: LE = LA /[1 – dAE·LA] = (+5.0 D) /[1 – (+0.45 m)·(+5.0 D)] = −4.0 D. (c) At point C, the ray bundle is also diverging. Point C, 0.7 m to the right of A, is located 0.50 m from the point of origin, which is to the left of point C. Therefore, its vergence is given as LC = 1 / (–0.50 m) = −2.0 D. Numerically, we can solve this as follows: LC = LA /[1 – dAC·LA] = (+5.0 D) /[1 – (+0.70 m)·(+5.0 D)] = −2.0 D.

Note

: Point D is downstream from point A, while point E is downstream from point A but upstream

from point C. The concepts of downstream and upstream are relative; we must always state in relation to what point.

Example ☞: A wavefront exits an aperture of an optical black box having a vergence of L = −5.0 D. What is the downstream vergence at a screen placed 30 cm to the right of the black box? Assume propagation in air. We deal with this problem via the ‘black box’ approach: We examine the properties of the wavefront and the distance from its source, if diverging, or the distance to the point of convergence, if converging. In this case, the wavefront exiting the black box is diverging; it originates from a point within the black box. The wavefront reaching the screen belongs to the same wave, which propagates downstream +30 cm = +0.3 m. Therefore, this wave also originates from exactly the same point as the one exiting the black box. The difference is the radius of curvature, or the distance from the origin. This distance is now larger by the amount of displacement between the aperture of the black box and the screen.

Figure 3-43: Computation of downstream vergence from a known value at a given point. At the black box aperture, the wavefront radius of curvature is the reciprocal of the vergence: x = 1/ L = 1 / (−5.0 D) = −0.2 m. At the screen to the right, the radius of curvature is increased by 0.3 m, thus 3-99

GEOMETRICAL OPTICS

becoming x΄ = −0.5 m. In other words, the directional distance from the screen to the source is now half a meter, pointing to the left. The vergence at the screen is L΄ = 1/ x΄ = 1/ (−0.5 m) = −2.0 D.

Example ☞: A virtual object (vergence L =+6.0 D) is imaged in air by a negative lens to form a virtual image with a vergence L΄ = −2.0 D. What are the object and image locations? What are the image traits? The object location is the reciprocal of the object vergence: x = 1/(+6.0 D) = +0.166 m. This is the wavefront radius of curvature incident at the lens. The image location is the reciprocal of the image vergence: x΄ = 1/(−2.0 D) = −0.50 m. This is the wavefront radius of curvature leaving the lens. The image is virtual (negative vergence), inverted (negative magnification), and magnified (magnification larger than 1.0); m = L/L΄ = (+6.0 D) / (−2.0 D) = −3. Note

: The virtual image here is inverted and corresponds to a magnification greater than 1.0. These

aspects are presented in detail in § 4.7.

Figure 3-44: Object and image vergence involving a virtual object, a virtual image, and a negative lens. Example ☞: In the above case, what is the vergence 10 cm before and 10 cm after the lens? Before the lens (or, in general, the optical system), we consider the incident wavefront, which is directed at the (virtual) object point. At 10 cm before the lens, the distance from the object point source is xd=−0.10 = = +26.6 cm (Figure 3-45) (that distance was +16.6 cm exactly at the lens). The vergence is the reciprocal of

xd=−0.10 : Ld=−0.10 = 1/xd=−0.10 = 1 / (+0.266 m) = +3.75 D. It is upstream because it is calculated at a point that has ‘traveled’ against the direction of light propagation (d = –0.10 m). Alternatively, we can use Eq. (3.6):

Ld=−0.10 =

3-100

Ld=0 1 − d  Ld=0

=

+6.0 D 1 − ( −0.10 m)  ( +6.0 D )

=

+6.0 D 1+0.6

=

+6.0 D 1.6

= +3.75 D.

IMAGING DEFINITIONS

Figure 3-45: Computation of upstream and downstream vergence for a virtual object and a virtual image. After the lens, we consider the wavefront leaving the lens. This is a diverging wavefront whose origin is at the (virtual) image point. At 10 cm after the lens, the distance to convergence is x΄d=+0.10 = −60 cm (the distance was −50 cm at the lens). Therefore, the vergence at this point is the reciprocal of x΄d=+0.10 :

L΄d=+0.10 = 1/ (−0.60 m) = −1.66 D. It is downstream vergence because it is calculated at a point that has ‘traveled’ along the direction of light propagation (d = +0.10 m). Alternatively, we can use Eq. (3.6):

L΄d=+0.10 =

L΄d=0 1 − d  L΄d=0

=

−2.0 D 1 − (+0.10 m)  ( −2.0 D )

=

−2.0 D 1+0.2

=

−2.0 D 1.2

=

− 1.66 D.

Example ☞: A collimated ray bundle is incident on a +4.0 D positive lens. Assume propagation in air. (a)

What is the vergence immediately before and immediately after the lens?

(b)

What is the downstream vergence of the wavefront 0.05 m after (to the right of) the lens?

(c)

What is the vergence at a point 0.35 m after (to the right of) the lens?

(a)

The collimated bundle has a vergence of 0 D anywhere, up to the point at which the rays reach the

lens: L before lens = 0.0 D. The lens power is added to this value; therefore, immediately leaving the lens, the vergence becomes L΄ d=0 = +4.0 D. This means that the beam is converging (focusing) to a point x΄d=0 = 1/L΄d=0 = 1/+4.0 D = +0.25 m to the right of the lens. This is equivalent to saying that the wavefront leaves the lens with a radius of curvature of +0.25 m. (b)

At a point located 0.05 m after (to the right of) the lens, the new distance to the point of

convergence is x΄d=+0.05 = +0.20 m (Figure 3-46). The vergence at this point is L΄d=+0.05 = 1/ x΄d=+0.05 = 1/ (+0.20 m) = +5.0 D. It is positive because the wavefront is still converging. Alternatively, we can compute the vergence L΄d=+0.05 by using the downstream vergence formula [Eq. (3.6)]. The known vergence is L΄d=0 = +4.0 D, and we seek the downstream vergence L΄d=+0.05 at d =+0.05 m, which is

3-101

GEOMETRICAL OPTICS

L΄d=+0.05 =

L΄d=0 1 − d  L΄d=0

=

+4.0 D 1 − ( +0.05 m)  ( +4.0 D )

=

+4.0 D 1 − 0.2

=

+4.0 D 0.8

= +5.0 D.

Figure 3-46: Vergence changes through the convergence point of the wavefront. (c)

At a location 0.35 m after (to the right of) the lens, the new distance to the point of convergence is

x΄d=+0.35 = −0.1 m. This is the distance separating the wave ‘front’ from the point at which the wavefront converges, or comes to ‘focus.’ In reality, this is now the source of a diverging wavefront, and the distance is negative because this point is to the left of the ‘front.’ The vergence at this point is the reciprocal of this distance: L΄d=+0.35 = 1/ x΄d=+0.35 = 1/ (−0.1 m) = −10.0 D. Again, we compute the vergence at 0.35 m after the lens by using the downstream vergence Eq. (3.6). The known vergence is L΄d=0 = +4.0 D, and we seek the downstream vergence L΄d=+0.35 at d =+0.35 m, which is

L΄d=+0.35 =

Note

L΄d=0 1 − d  L΄d=0

=

+4.0 D 1 − ( +0.35 m)  ( +4.0 D )

=

+4.0 D 1 − 1.4

=

+4.0 D −0.4

= − 10.0 D.

: These examples show that there are two ways to compute downstream / upstream vergence.

The first way is to consider the new distance to convergence from the new location (which may have a different direction as well). The reciprocal of this directional distance is the new vergence. This is the ‘black box’ approach. The second way is to use the downstream (or upstream) vergence formula. In this method, we need to determine the original vergence and the displacement d [paying attention to the sign, which is positive if the new location is to the right (downstream) and negative if the new location is to the left (upstream)]. The formula then simply provides the new vergence. Obviously, if properly implemented, both ways should produce the same result. Don’t forget to convert all length units to meters!

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IMAGING DEFINITIONS

3.7 VERGENCE AND IMAGING CONCEPTS QUIZ 1)

a) b) c) d)

2)

5)

object is real; image is real object is real; image is virtual object is virtual; image is real object is virtual; image is virtual

The vergence incident on a lens is −1.00 D. The vergence leaving the lens is +4.00 D. Is the object real or virtual? Is the image real or virtual? a) b) c) d)

6)

+3.33 D +5.00 D +7.50 D –7.50 D –5.00 D –3.33 D

The vergence incident on a lens is +2.00 D. The vergence leaving this lens is −3.00 D. Is the object real or virtual? Is the image real or virtual? a) b) c) d)

object is real; image is real object is real; image is virtual object is virtual; image is real object is virtual; image is virtual

Rays enter an optical system with vergence – 5.00 D. These rays correspond to … a)

7)

8)

a real image formed 0.25 m to the left a virtual image formed 0.25 m to the right a real image formed 0.25 m to the right a virtual image formed 0.25 m to the left

Rays enter an optical system with vergence – 4.00 D. They propagate in water (n = 1.33). These rays correspond to an object placed … a) b) c) d) e)

9)

a virtual object placed 20 cm to the left a real object placed 20 cm to the right a virtual object placed 20 cm to the right

Rays leave an optical system with vergence – 4.00 D. These rays correspond to … a) b) c) d)

a real image formed +0.5 before the lens a virtual object formed +0.50 after the lens a real image formed +0.50 after the lens a virtual image formed +0.50 after the lens

Rays enter a single flat refracting surface with a vergence of L =+5.00 D. This surface separates air (n = 1.0) from glass (n΄ = 1.5). The vergence of the rays upon leaving that surface (L΄) is … a) b) c) d) e) f)

4)

a real object placed 25 cm in front of (before) the lens system a virtual object placed 25 cm in front of (before) the lens system a real object formed 25 cm to the right of (after) the lens system a virtual object formed 25 cm to the right of (after) the lens system

Rays, upon leaving a lens, have vergence +2.00 D. These rays correspond to … a) b) c) d)

3)

b) c) d)

Rays with vergence +4.00 D enter a lens system. These rays correspond to …

33.3 cm in front of (before) the system 25.0 cm in front of (before) the system 18.75 cm in front of (before) the system 12.5 cm in front of (before) the system 12.5 cm to the right of (after) the system

Rays leave a meniscus lens with vergence –2.00 D, propagating then into aqueous (n = 1.33). These rays correspond to an image formed … a) b) c) d)

50 cm in front of (before) the system 66.6 cm in front of (before) the system 50 cm to the right of (after) the system 66.6 cm to the right of (after) the system

10) Light propagating in air (n = 1.0) has vergence −4.00 D at point A. What is the vergence at point B, which is 75 cm downstream? a) b) c) d) e) f)

+4.00 D +2.00 D +1.00 D –1.00 D –4.00 D –10.00 D

11) Light propagating in air (n = 1.0) has vergence −4.00 D at point A. What is the vergence at point B, which is 15 cm upstream? a) b) c) d) e) f)

–10.00 D –4.00 D –1.00 D +1.00 D +4.00 D +10.00 D

a real object placed 20 cm to the left 3-103

GEOMETRICAL OPTICS

12) Light propagating in air (n = 1.0) has vergence +20.00 D at point A. What is the vergence at point B, which is 15 cm downstream? a) b) c) d) e) f)

–20.00 D –10.00 D –4.00 D +4.00 D +10.00 D +20.00 D

13) Light propagating in air (n = 1.0) has vergence −10.00 D at point A. What is the vergence at point B, which is 15 cm downstream? a) b) c) d) e) f)

–20.00 D –10.00 D –4.00 D +4.00 D +10.00 D +20.00 D

14) If light traveling in air (n = 1.0) has vergence +4.00 D at point A, what is the vergence at point B, which is 5 cm downstream? a) b) c) d) e) f)

–6.67 D –5.00 D –4.00 D +4.00 D +5.00 D +20.00 D

15) Light propagating in water (n = 1.33) has a vergence of −1.00 D at point A. What is the vergence at point B, which is 1.0 m upstream? a) b) c) d) e) f)

–10.00 D –4.00 D –3.00 D +0.42 D +0.57 D infinity

16) Light traveling in water (n = 1.33) has a vergence of +5.00 D at point A. What is the vergence at point B, which is 20 cm downstream? a) b) c) d)

+2.50 D +2.85 D +20.00 D infinity

17) Light propagating in air (n = 1.0) has a vergence of −10.00 D at point A. What is the vergence at point B, which is 15 cm upstream?

3-104

a) b) c) d) e) f)

–20.00 D –10.00 D –4.00 D +4.00 D +10.00 D +20.00 D

18) Regarding the ray configuration at locations A, B, and C below, state the correct vergence values at B and C if the vergence at A = +4.0 D (arrows indicate equal spacing; propagation is in air, n = 1.0).

a) b) c) d) e) f) g)

B: +4.0 D, C: +4.0 D B: +2.0 D, C: –2.0 D B: +8.0 D, C: +12.0 D B: +8.0 D, C: +6.0 D B: +8.0 D, C: –8.0 D B: +2.0 D, C: –4.0 D B: +8.0 D, C: –2.66 D

19) Given a +5.00 D vergence in air (n = 1.0), where is its point source of origin / point of convergence? a) b) c) d) e) f)

−40 cm −30 cm −20 cm +20 cm +30 cm +40 cm

20) If a wavefront propagating in water (n = 1.33) has a −4.00 D vergence, where is its point source of origin / point of convergence? a) b) c) d) e) f)

−18.8 cm −25 cm −33.25 cm +18.8 cm +25 cm +33.25 cm

21) An object is located 125 cm in front (to the left) of a lens. What is the vergence of the light rays incident upon the lens? (Propagation is in air.) a) b) c) d) e)

+8.0 D +0.8 D –1.25 D –0.8 D +8.0 D

IMAGING DEFINITIONS

22) An image is formed 20 cm to the right of (after) a lens. What is the vergence exiting the lens (L΄)? (Propagation is in air.) a) b) c) d) e)

+5.0 D +2.0 D –2.0 D –5.0 D –50.0 D

23) An object is placed 80 cm in front (to the left) of a single flat refracting interface (SFRI) that separates medium 1 (refractive index n = 2.0) from medium 2 (refractive index n΄ = 1.0). The vergence reaching the interface L and the vergence leaving the interface L΄ are … a) b) c) d) e) f)

27) Which of the following illustrations correctly indicates the type of object AND the type of image?

L = –2.5 D and L΄ = –2.5 D L = –2.5 D and L΄ = +2.5 D L = –5.0 D and L΄ = –2.5 D L = –5.0 D and L΄ = –5.0 D L = –1.25 D and L΄ = –1.25 D L = –1.25 D and L΄ = –2.5 D

a)

A

b) B

c) C

d) D

28) Which two illustrations show the formation of a real image from a virtual object?

24) An object point source is placed inside medium 1 of refractive index of n = 1.5. Specifically, the object is placed 30 cm in front (to the left) of an SFRI that separates medium 1 from medium 2 (refractive index n΄ = 1.0). The image location x΄ and linear magnification m in this case are … a) b) c) d) e) f)

x΄ = –20 cm and m = –1 x΄ = –20 cm and m = +1 x΄ = –20 cm and m = +2 x΄ = –30 cm and m = –1 x΄ = –30 cm and m = +1 x΄ = –30 cm and m = +2

25) A drawing of SpongeBob SquarePants is imaged via a lens system such that that the image has a 2 cm long arm. If the magnification is m = +4, what is the object’s arm length and orientation? a) b) c) d)

8 cm, inverted 8 cm, upright (erect) 0.5 cm, inverted 0.5 cm, upright (erect)

26) A 3 mm tall drawing of Pippi Longstocking is placed in front of a mirror such that a virtual image of 3× (m = +3.0) is produced. What is the image’s height and orientation? a) b) c) d)

1 mm, upright (erect) 3 mm, inverted 0.9 cm, upright (erect) 9 cm, upright (erect)

a)

A

b) B

c) C

d) D

29) A flashlight is dropped into sea water (n = 1.33) and descends to a depth of 66 cm. What is the vergence of the light emitted by this underwater flashlight (L = ?) upon reaching the flat water–air interface (from the water side)? a) b) c) d) e) f)

–2.0 D –1.5 D –1.125 D +1.125 D +1.5 D +2.0 D

30) Referring to Q 29, what is the value of the light vergence that was (originally) emitted by the underwater flashlight upon leaving the flat water– air interface (L΄= ?)? a) b) c) d) e) f)

–2.0 D –1.5 D –1.125 D +1.125 D +1.5 D +2.0 D

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GEOMETRICAL OPTICS

3.8 VERGENCE AND IMAGING CONCEPTS SUMMARY Cartesian Sign Convention In the Cartesian sign convention, a positive direction is determined by the direction of light propagation. When light travels from left to right, a directional distance to the right is positive, and a directional distance to the left is negative.

Figure 3-47: Cartesian sign convention. When light travels from left to right, the positive direction is to the right. This applies to any directional distance, including radii of curvature, focal lengths, and object and image locations.

Object and Image Space; Real and Virtual Object and Image The object is associated with rays entering the optical system. The set of all points or rays associated with the light incident on the optical system is object space. In object space, points such as the primary focal point, power determinants such as the object vergence, and dimensions such as the object distance are nonprimed. An object can be real or virtual. A real object [Figure 3-48 (left)] is associated with a diverging set of rays and is positioned to the left of the optical element. Its location, measured from the optical element, is negative, as is the object vergence. A virtual object [Figure 3-48 (right)] is associated with a converging set of rays and is positioned to the right of the optical element. Its location, measured from the optical element, is positive, as is the object vergence.

Figure 3-48: (left) Real object and (right) virtual object. 3-106

IMAGING DEFINITIONS

The image is associated with rays leaving the optical system. The set of all points or rays associated with light leaving the optical system is image space. In image space, points such as the secondary focal point, power determinants such as the image vergence, and dimensions such as the focal length and image distance are primed. An image can be real or virtual. A real image [Figure 3-49 (left)] is associated with a converging set of rays and is formed to the right of the optical element. Its location, measured from the optical element, is positive, as is the image vergence. A virtual image [Figure 3-49 (right)] is associated with a diverging set of rays and is formed to the left of the optical element. Its location, measured from the optical element, is negative, as is the image vergence.

Figure 3-49: (left) Real image and (right) virtual image.

Vergence In a wavefront or a set of rays, the physical entity that expresses the degree of their convergence is the vergence L. A converging wavefront has a positive vergence, and a diverging wavefront has a negative vergence. A flat wavefront (a collimated ray bundle) has zero vergence. Vergence is measured according to the following: •

In a diverging wavefront, the vergence is the reciprocal of the directional distance x from the reference point to the point of origin. This distance is negative, as is the vergence: L = 1/x.

•

In a converging wavefront, the vergence is the reciprocal of the directional distance x from the reference point to the point of convergence. This distance is positive, as is the vergence:

L = 1/x.

Figure 3-50: Diverging wavefront vergence (left) and converging wavefront vergence (right). 3-107

GEOMETRICAL OPTICS

In the above relationships, the length quantities have to be expressed in meters so that vergence is expressed in diopters. If the medium is not air but instead is a different medium with a refractive index n, the generalized relationship is Reduced Vergence:

L [D] =

n x

=

refractive index distance to convergence [m]

Downstream and Upstream Vergence Downstream vergence is the value of the vergence along the propagation of light. When the vergence L is known at a certain point, then at a point that is located d [m] downstream of the initial (known) vergence, the (unknown) vergence L΄ is Downstream Vergence (in air):

L 1− d  L

L΄ =

Distance d is positive, since downstream means along the direction of light propagation. The same formula can be used for upstream vergence; the only difference is that the distance d is negative. A more-generalized expression that includes the fact that the space between is filled with a medium with refractive index n is Downstream Vergence (medium with n):

L΄ =

L d 1−  L n

Magnification The ratio of the linear size of image height h΄ to the corresponding linear size of object height h is the lateral, translinear, linear, or transverse magnification m: Linear Magnification:

m

image height object height

=

h΄ h

Positive magnification: The image is erect in relationship to the object. •

Negative magnification: The image is inverted in relationship to the object.

•

Magnification greater than 1.0: The image is a scaled-up version of the object.

•

Magnification less than 1.0: The image is a scaled-down version of the object.

3-108

GEORGE ASIMELLIS

LECTURES IN OPTICS, VOL 2

4 IMAGING WITH LENSES

4.1 LENS IMAGING RELATIONSHIP The simplest application of a lens is imaging, the formation of an image from the object. We are interested in the lens imaging relationship, which links object and image locations in relation to the lens optical power. Consider a collimated pencil of rays along the optical axis incident on a positive lens. The vergence before the lens is zero; just after the lens vergence equals the lens optical power. Specifically, •

collimated beam vergence before (incident on) the lens: L = 0 D

•

lens optical power: F = 10 D. Their sum is therefore L + F = +10 D

•

vergence after (leaving) the lens: L΄ = +10 D.

Figure 4-1: A ray bundle from optical infinity is imaged to the secondary focal point F΄. Optical infinity and the secondary focal point F΄ are optical conjugates.

4-109

GEOMETRICAL OPTICS

The above discussion indicates that the vergence leaving the lens is the sum of the object vergence and the lens optical power. This a general principle that is also applicable to negative lenses (see § 4.4) and mirrors (§ 5.3.1). It is valid for any object location (before or after the lens), image location (before or after the lens), and lens power (positive or negative sign). The imaging relationship for an object at x (object vergence L = 1/x), an image at x΄ (image vergence L΄ = 1/x΄ ), and a lens with focal length f΄ (optical power F = 1/f΄) is Vergence and Optical Power:

L

+

object vergence

Object and Image Locations:

1 x object location

F

=

optical power

1 f΄

+

focal length

L΄

(4.1)

image vergence

=

1 x΄

(4.2)

image location

Equation (4.2) is known as the Gaussian thin-lens-imaging formula,17 named after the German mathematician Carl Friedrich Gauss. This formula can be derived directly from Snell’s law and is valid under certain assumptions, primarily the paraxial approximation, which practically dictates using small angles so that the rays stay close to the optical axis (see § 8.1.2). The image location and size (denoted with a prime) depend on the object location x with respect to the lens and on the optical power F. Points o and i in Figure 4-2 are optical conjugates because each can be an image of the other.

Figure 4-2: Imaging for an object at location x before the lens. Point o is imaged to point i. Equivalently, we state that points o and i are optical conjugates.

Figure 4-3: Principle of optical power and object vergence summation in lens imaging.

This formula, like all formulae in this book, follows the Cartesian sign convention. While this convention is widely used, not all optics textbooks adopt it. Therefore, there may be differences in the form of the imaging relationships across textbooks. However, results will be correct as long as a consistent set of formulas and sign conventions is followed. 17

4-110

IMAGING WITH LENSES

Object / Image location x & x΄ • Unit: meter (m) • If to the left of the lens, negative • If to the right of the lens, positive

Object / Image vergence L & L΄ • Unit: diopter (D = m–1) • The reciprocal of the corresponding object / image locations: L = 1/x and L΄ = 1/x΄ • Same algebraic sign as the corresponding locations

Expressed with vergence and optical power:

Lens Imaging Relationship

Entering, object vergence L + Lens optical power F = Exiting, image vergence L΄ Vergence and power are expressed in diopters (D) Expression with object / image locations and focal length: 1/x + 1/f΄ = 1/x΄ Valid for: any object / image location (before or after the lens), any lens power magnitude, Numerical values follow the Cartesian any lens power sign (positive or negative lens). convention: Light propagating to the right is positive (+) and to the left is negative (–). Location (distance) units are meters (m)

Lens Imaging Relationship Perhaps the most consequential relationship in geometrical optics.

Imaging is predicated on the object vergence and lens power, which simply add up to produce the image vergence.

Incident, object beam vergence (L = 1/x) + Lens power (F = 1/f΄) = Exiting, image beam vergence (L΄ = 1/x΄).

4-111

GEOMETRICAL OPTICS

Unless otherwise stated, we assume that the medium surrounding the lens is air with refractive index = 1.0. If not, then for the vergence and power we use L = n / x, L΄ = n΄ / x΄, and

F = n΄ / f ΄, respectively, where n is associated with object space and n΄ with image space. Vergence and Optical Power:

+

L object vergence

n Object and Image Locations:

object space

x object location

Note

F

=

optical power

L΄

n΄ +

image space

f΄ focal length

(4.3)

image vergence

n΄ =

image space

x΄

(4.4)

image location

: None of the above “n” values denote the lens refractive index; they are the refractive index of

the space surrounding the lens. Their values may be identical (lens surrounded by the same medium) and often equal 1.0, which is the simple case of the lens being surrounded by air. In principle, if the lens is surrounded by different optical media, the two focal lengths (to the primary and secondary focal points) have different magnitudes. It is emphasized, once again, that the focal length associated with imaging is the secondary focal length.

Example ☞: An object vergence L = −6 D is incident on a lens with power F = +10 D. Find the image location x΄ after the lens if the medium surrounding the lens is air. We apply the imaging relationship given in Eq. (4.1): L΄ = L + F = −6 D +10 D = +4 D. The image-forming beam immediately after the lens has a positive vergence; it is therefore converging and forms a real image. The image location is x΄ = 1 / L΄ = 1 / (+4 D) = +0.25 m.

In the following example, we assume that the lens is surrounded by air.

Figure 4-4: The lens optical power is added to the object vergence. Point o is imaged to point i. Points o and i are optical conjugates. 4-112

IMAGING WITH LENSES

✔ Vergence incident on the lens L = −6 D is equivalent to stating that the object (point o, Figure 4-4) is at x = 1/ L = 1/(−6 D) = −0.166 m = −16.6 cm before (note the − sign) the lens. ✔ Lens Power F = 10 D is related to the focal length via f΄ = 1/(+10 D) = +0.1 m = +10 cm. ✔ Vergence leaving the lens L΄ = +4 D is converted to image location x΄ = 1/ L΄ = 1/(+4 D) = +0.25 m = +25 cm. Thus, the image (point i ) is located 25 cm after (note the + sign) the lens. Note

: Unless stated otherwise, both sides of the lens are surrounded by air. Then, simply, F = 1/f΄. It

is also true that f = −f΄; this means that the distance from the primary focal point to the lens equals the distance from the secondary focal point to the lens, and that the primary focal point is on the opposite side with respect to the secondary focal point. We may simplify the terminology by using f = f΄ and F = 1/f, where the adjective ‘secondary’ and the prime (suggesting the secondary focal point)_are dropped for simplicity [as they are in the lens-maker’s formula, Eq. (2.4)].

4.1.1 Imaging: Left to Right or Right to Left? The two optical conjugate points o and i have the property that one is the image of the other. An object placed at o forms an image at i, and also, an object placed at i forms an image at o. In other words, the result does not depend on the direction of light propagation. While the majority of cases in optics assumes light propagating from left to right, it is not impossible for light to propagate from right to left. This involves, for example, a light source placed to the right of a lens or a mirror that reverses the direction of propagation. The optics principles do not change and are not dependent on the direction of light propagation. This is guaranteed by the principle of reversibility.18 Thus, the principle of adding the lens optical power to the object vergence is valid in reverse propagation, with some sign changes. Now, the positive direction is to the left, and the negative direction is to the right of the origin, which is the lens center. This is consistent with the direction of light propagation, which is, in this case, to the left. This way, in our example, the object point i vergence is negative (−4 D), and the image point o vergence is positive (+6 D). In Figure 4-5, light propagates from right to left, originating from point i (this is now the object) and reaching the lens with a vergence of −4 D. This beam meets the lens (lens power +10 D). We apply optical power summation. The image beam leaves the lens with a vergence of +6 D (= −4 D + 10 D) and is converging to point o, which is the new image.

18

Introduction to Optics § 3.2.3 From Optically More Dense to Optically Less Dense Media. 4-113

GEOMETRICAL OPTICS

Figure 4-5: The symmetry in optics laws is applicable to light propagation from left to right. The object becomes the image, and the image becomes the object! Attention to the proper right-to-left positive direction is needed. Comment •

: The same rules apply to imaging, regardless of left-to-right or right-to-left propagation.

For light propagating to the right, the positive direction is to the right. Thus, all directed distances are positive if they point to the right.

•

For light propagating to the left, the positive direction is to the left. Thus, all directed distances are positive if they point to the left.

4.1.2

Image Magnification

Figure 4-6 presents a simple formation of a real image from a real object by a plus lens surrounded by the same medium (such as air). We notice the triangle similarity between the optical axis and the object (h) and image (h΄ ) linear heights. Using the definition of linear magnification (§ 3.3) and the similar triangles ΑΒΓ and ΓΔΕ, in which the ratio of opposite sides (ΑΒ = h and EΔ = h΄ ) equals the ratio of the adjacent sides (ΑΓ = x and ΓΔ = x΄ ), we have Linear Magnification (using location):

m

h΄ x΄ = h x

(4.5)

This relationship implies that, while linear magnification is still defined as a ratio of image linear height to object linear height, in the case of lens imaging, and as long as the medium is the same on both sides of the lens, the magnification can instead be calculated by the ratio of the image location x΄ to the object location x if these positions are known. Note

: If the image is inverted as in Figure 4-6, its height is negative in relation to the object height, so

the magnification calculated as a height ratio is negative. The same sign results if we use locations: The object location is negative and the image location is positive, so the magnification is negative. 4-114

IMAGING WITH LENSES

Figure 4-6: Similar triangles formed between the image and object heights and locations. Example ☞: In the example shown in Figure 4-6, x΄ = +20 cm and x = −10 cm. What is the magnification? The magnification is m = x΄/ x = (20 cm)/(−10 cm) = −2.0. It is negative, which means that the image is inverted and its (absolute) value is more than 1.0, meaning that the image is larger than the object.

Example ☞: An object that is 40 mm tall forms an inverted image that is 10 mm tall. What is the magnification? The key word is ‘inverted,’ which implies that the algebraic signs of the object and image heights are opposite. Thus, if the object size is h = +40 mm, the image size is h΄ = −10 mm. We use the magnification Eq. (4.5): m ≡ h΄/ h = (−10 mm) / (+40 mm) = −0.25. The negative magnification sign indicates that the image is inverted, and the value of 0.25 indicates that the image is minified. Hint

: If the object height h is known, the image size is h΄ = m · h.

Example ☞: If the object has height h = 25 mm and the magnification is m = +2.0, we use Eq. (4.5) in the form m ≡ h΄ / h to find the image height h΄ : h΄ = m · h ⇒ h΄ = (+2.0) · (+25 mm) = +50 mm. Example ☞: An object 15 mm tall is located 30 cm before (in front of) a positive lens. It forms an image 60 cm after (to the right of) the lens. Find the image size and magnification. The object location is x = −0.3 m since it is in front of the lens. The image location is x΄ = +0.6 m since it is formed to the right of the lens. From the lens magnification Eq. (4.5): m = x΄/ x = (+0.6 m) / (−0.3 m) = −2.0. The negative sign indicates that the image is inverted. The magnification value of 2.0 indicates that the image is 2× larger than the object. Thus, the image height is (−2.0) · (+15 mm) = −30 mm tall. Example ☞: An object 25 mm tall is placed 5 cm in front of a lens. If m = +2.0, find the image location and size. The value of m = +2.0 suggests that the image is 2× magnified (twice as tall) in relation to the object. Note the + sign, indicating that the image is erect. The object location is x = −0.05 m since it is in front of the lens. We use Eq. (4.5) m = x΄ / x to find the image location x΄ : (+2.0) = (x΄) / (−0.05 m) ⇒ x΄ = (+2.0) · (−0.05 m) = −0.10 m = −10 cm. This means that the image is formed 10 cm before (in front of) the lens. 4-115

GEOMETRICAL OPTICS

Alternatively, magnification can be calculated using vergence. This relationship is more generally applicable than Eq. (4.5), which uses image/object location because it does not require that the media be the same at both sides of the lens. In general, m = L΄/ L, which equals the ratio of the object vergence to the image vergence: Linear Magnification (using vergence):

m 

h΄ n/ x L = = h n΄ / x΄ L΄

(4.6)

Example ☞: An object vergence of –10 D reaches an imaging lens. The image-forming beam leaves the lens with +40 D of vergence. Find the magnification and image location. The object is real (negative vergence), while the image is also real (positive vergence). We now use Eq. (4.6) to calculate the magnification: m = L / L΄ = (−10 D) / (+40 D) =−0.25. The image is minified and inverted. It is located at x΄ = 1/L΄ = 1 / (+40 D) = +0.025 m, which is 2.5 cm to the right of the lens.

Note

: When do we use m = x΄/x and when do we use m = L/L΄?

Most often, we use m = x΄/x. It is easier to identify the ratio of a length dimension associated with an image (x΄) to the length of the object (x). However, this formula is only applicable when both object and image space have the same refractive index. This is the case of a lens that is surrounded by the same medium, and is always the case for mirrors, whose object and image space are physically the same. This does not apply to SSRIs, which, by rule, have different refractive indices in object and image space. In that case, we can only use m = L/L΄.

4.1.3 Image Reversal The image in Figure 4-6 is inverted. This is the case of real images formed of real objects by a positive lens. Here, the image size is negative (facing down on the –y axis) since it ‘extends’ to negative –y values, while the object extends to positive –y values. Contrary to mirror imaging (§ 5.3.4), the inverted real image formed by a lens corresponds to image rotation by 180° about the optical axis (in a three-dimensional system, this is the –z axis). The real image formed by a lens is flipped top to bottom (inversion) and flipped left to right. This is image reversion.

Figure 4-7: Image reversal when a positive lens produces a real image corresponds to image inversion and reversion with no parity change, while in mirrors there is a parity change. 4-116

IMAGING WITH LENSES

4.1.4 Newton’s Imaging Relationship An alternative expression of the Gaussian imaging relationship given in Eq. (4.2) was derived by Sir Isaac Newton. This the Newtonian imaging formula:

− ( x + f )  ( x΄ − f΄ ) = − xo  xi = f 2

Newtonian Formula:

(4.7)

where xo = x + f is the object distance from the object-space focal point, and xi = x΄ – f΄ is the image distance from the image-space focal point (we assume the same medium surrounding the lens, so f = f΄). This formula also complies with the Cartesian sign convention. Thus, for an object to the left (in front) of a positive lens, quantities x and xo are negative, while quantity f is positive. Therefore, in absolute values, x is less than xo. Example ☞: An object is placed 100 mm before a positive lens of focal length 50 mm. Find the image location, size, and magnification.

✔ Application of the Gaussian imaging formula, Eq. (4.2). First, we convert millimeters to meters: x = −100 mm = −0.10 m (the sign is − because the object is before the lens), and f΄ = +50 mm = +0.05 m (the sign is + because the lens is positive): 1 1 1 + = x f΄ x΄



1 −0.10 m

+

1 +0.05 m

=

1 x΄



1 +0.10 m

=

1  x΄ = + 0.10 m x΄

✔ Application of the Newtonian imaging formula, Eq. (4.7). No conversion to meters is necessary at this stage (as long as all lengths are expressed in the same units): −(−100 + 50) · (x΄ − 50) = 502 ⇒ (x΄ − 50) = 50 ⇒ x΄ = +100 mm = +0.10 m Both relationships produce the same result: The image is formed to the right of the lens, at location x΄ = +100 mm = +0.10 m.

Figure 4-8: Calculation of the image location for an object placed before (in front of) a positive lens. In this example, the object location (x = −2f΄) corresponds to an equidistant image location (x΄ = +2f΄), and the magnification is m = −1.0.

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4.2 IMAGING BY A PLUS-POWERED LENS Imaging of a real object with a plus lens (positive optical power F and focal length f΄) has three distinct outcomes, depending on the location of the object in relation to the focal point. In all three cases, the object is real and is placed to the left of (before) the lens. Because in directed distances such as the object and image locations the origin is the center of the lens, the object location x carries a negative sign since its directional distance is pointing from the lens to the object, against the direction of light propagation. In these examples, unless otherwise stated, the lens is surrounded by the same medium, usually air. The three cases, based on the location of the object in relation to the focal point, are described next. ❶ The object is at a longer distance compared to the focal length and is placed to the left of

the focal point F such that ∞ > |x| > 2|f |. The image is real, inverted, and smaller than the object (minified). It is formed after the secondary focal point F΄, at a distance for which f ΄> x΄> 2f΄. Example ☞: An object is placed 100 cm before (in front of) a lens with focal length +20 cm. Find the image location and magnification.

We apply the imaging relationship given in Eq. (4.1): L + F = L΄. First, we convert all distances to meters. The object distance is therefore x = −1.0 m, the negative sign indicating that the image is to the left of the lens. The focal length is f΄ = +0.2 m, the plus sign indicating that the lens is positive. The image vergence L΄ and image location x΄ are

L΄ =

1

+

1

−1.0 m +0.2 m

= − 1.0 D + 5.0 D = + 4.0 D and

x΄ =

1

L΄

=

1

+4.0 D

= + 0.25 m = + 25 cm

Because the image location is positive (x΄ = +25 cm, plus sign), the image is formed after (to the right of) the lens. Because the image vergence is positive (L΄ = +4.0 D, plus sign), the image is real; as we will see in the ray-tracing section (§ 4.3), this image is produced by converging rays.

Figure 4-9: Formation of a real image by a positive lens. The magnification is negative (m = −0.25). 4-118

IMAGING WITH LENSES

We now use Eq. (4.5) to calculate the magnification: m = x΄/x = (+25 cm) / (−100 cm) = −0.25. If the object height is h = +20 mm, the image height is h΄ = m · h = (−0.25) · (+20 mm) = −5 mm. This indicates that the image is minified and inverted. This example approximates the conditions for the real image formed by the optical system of the human eye on the retina.

Example ☞: An object is placed 40 cm before (in front of) a lens of focal length +20 cm. Find the image location and magnification.

The object is placed in front of the lens, at twice the focal length (Figure 4-10). In this case, the object location is x = −0.40 m and the focal length is f΄ = +0.20 m. Therefore, the image is formed 40 cm after the lens: Image vergence: L΄ = L + F = Image location:

x΄ =

1

L΄

=

1

+

1

−0.4 m +0.2 m 1

+2.5 D

= − 2.5 D + 5.0 D = + 2.5 D

= + 0.40 m = + 40 cm

The magnification is m = (−0.40 m)/(+0.40 m) = −1.0. If the object height is h = +20 mm, the image height is h΄ = m · h = (−1.0) · (+20 mm) = −20 mm.

Figure 4-10: Formation of a real image by a positive lens. The magnification is negative (m = −1.0).

Note

: The example just presented is a special case, denoted by ❶a, for which x = 2f = −2 f ΄. In other

words, the object is placed at twice the focal length of the lens. It can be easily proven that, in this case, the image is always formed at a distance x΄ = 2 f΄ and the magnification is always m = −1.0.

L΄ = L + F =

1

−2f΄

+

1

f΄

=

1 2f΄

 x΄ =

1

L΄

= 2f΄

and

m =

x΄

x

=

2f΄ −2f΄

= − 1.0

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GEOMETRICAL OPTICS

❷ The object is at an intermediate distance, situated such that 2 |f |> |x| > | f |, and is still to the

left of the primary focal point F. In this case, the image is real, inverted, and larger than the object. It is formed after the secondary focal point F΄, where 2f ΄ < x΄ < ∞, and gradually increases in size as the object nears the lens. Example ☞: An object is placed 30 cm before (in front of) a lens of focal length +20 cm. Find the image location and magnification.

To find the image location x΄, we apply Eq. (4.1) with x = −0.30 m (the sign is − because the object is placed in front of the lens) and f΄ = +0.20 m.

Figure 4-11: Formation of a real image by a positive lens. The magnification is negative (m = –2.0).

Image vergence: L΄ = L + F = Image location:

x΄ =

1

+

1

−0.3 m +0.2 m

= − 3.33 D + 5.0 D = + 1.66 D

1 1 = = + 0.60 m = + 60 cm L΄ +1.66 D

Therefore, the image is formed to the right of the lens (+) at 0.60 m. The magnification is

m = (+0.60 m)/(−0.30 m) = −2.0. If the object height is h = +20 mm, the image height is h΄= m · h = (−2.0) · (+20 mm) = −40 mm. This indicates that the image is magnified and inverted. In the examples provided so far, the image is real because the image-forming rays converge. In addition, the image is inverted. The magnification is, in other words, negative. The common theme in both cases is that the object is placed to the left of the primary focal point of the lens such that |x| > |f |.

❸ The object is closer than the focal length, being situated between the focal point and the

lens such that |x| < | f |. In this case, the image is virtual and magnified, and is formed at |x΄| > f΄ in front of the lens.

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Example ☞: An object is placed 10 cm before (in front of) a lens of focal length +20 cm. Find the image location and magnification.

Figure 4-12: Formation of a virtual image by a positive lens. The magnification is positive (m = +2.0).

To find the image location x΄, we apply Eq. (4.1) with x = −0.10 m (the sign is − because the object is before the lens) and f΄ = +0.20 m. The image vergence L΄ and image location x΄ are

L΄ = L + F =

x΄ =

1

L΄

=

1

+

1

−0.1 m +0.2 m 1

−5.0 D

and

= − 10.0 D + 5.0 D = − 5.0 D

= − 0.20 m = − 20 cm

The image is formed at x΄ = −20 cm before the lens (note the − sign). The image is virtual because the image vergence is negative and is therefore diverging. In this example, the magnification is m = (−20 cm) / (−10 cm) = +2.0. Note the + sign, indicating that the image is erect. The value of 2.0 indicates that the image is 2× magnified. Therefore, in this case, the image is magnified and erect. Note that we discussed the cases of |x| > | f | and |x| < |f | but not the case of |x| = | f |, where the object is placed exactly at the primary focal point F. In this case, there is no image (specifically, the image is formed at optical infinity). This is presented in § 4.6.3 and illustrated in Figure 4-34. In this case, the lateral magnification tends to infinity (m → ∞) as the object approaches the primary focal point ( |x | = | f |). This section presents the imaging properties of a positive lens when the object is placed to its left (a diverging beam and a real object). The ray bundle incident on the lens is diverging, and the object vergence is negative. When the object is to the right of the lens (a converging beam and a positive vergence), we have a virtual object. This is presented in § 4.7.

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4.2.1

Summary

We can classify the typical cases of image formation from a positive lens. The object is placed before the lens (x is negative), and we use the absolute value |x|. ❶ For an object located such that ∞ > |x| > 2 | f |, the real image is formed at a location such that

f΄ > x΄ > 2 f΄ and is inverted, with its magnification less than 1.0. ❶a For |x| = 2 | f |, the image is formed at the location

x΄ = 2 f΄ with its magnification |m| = 1.0.

❷ As the object nears the lens, located such that 2 | f | > |x| > | f |, the real image is formed at a

location such that 2 f΄ < x΄ < ∞ and gradually increases in size. ✐ There is no image (specifically, the image is formed at optical infinity) if the object is placed exactly at the primary focal point: |x| = | f |. ❸ For smaller |x|, when the object is between the focal point and the lens such that |x| < |f |, the

image is magnified and virtual, and is formed in front of the lens (x΄ becomes negative).

Figure 4-13: Parametric curves showing image locations for the various (above listed) object locations in front of a positive lens.

In Figure 4-13, the horizontal axis is the object location x and the vertical axis is the image location x΄. The plot on the right shows the same data as on the left but in inverse (vergence) space (1/x and 1/x΄ instead of x and x΄, respectively). The horizontal axis refers to the reciprocal of the object location (1/x), which is the object vergence L, and the vertical axis is the reciprocal of the image location (1/x΄), which is the image vergence L΄. Both curves illustrate the imaging relationship 1/x + 1/f΄ = 1/x΄.

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Table 4-1: Image size, location, and magnification for various object locations for a positive lens, according to the imaging relationship given in Eq. (4.1). Figure 4-13 illustrates these data parametrically. Imaging of a real object by a positive lens (f΄ > 0) object

image

image location

magnification

∞ > |x| > 2|f |

real / inverted

f΄< x΄ < 2f΄

|m| < 1.0

|x| = 2|f |

real / inverted

x΄ = 2f΄

|m| = 1.0

2|f | > |x| >|f |

real / inverted

2f΄ < x΄ < ∞

|m| > 1.0

| x| = | f |

does not exist*

–

–

|x| 1.0

* The image is formed at optical infinity (see § 4.6.3).

Positive lens for an object distance longer than the focal length:

Real, inverted image (m < 0), formed to the right of (after) the lens.

Positive lens for an object distance shorter than the focal length:

Virtual, erect image, formed to the left of (before) the lens.

Magnification: • smaller than 1.0 (|m| < 1.0) for ∞ >|x |> 2|f |, • larger than 1.0 (|m| > 1.0) for 2|f | > |x |> |f |,

Magnification is always m >1.0.

• equal to 1.0 (|m| = 1.0) for |x |= 2|f |.

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GEOMETRICAL OPTICS

4.3 RAY DIAGRAMS In addition to the numerical solution described by Eq. (4.1), we can also find image location and size by applying ray diagrams. These are a set of three simple rules that enable the schematic solution of image location from every off-axis (i.e., slightly above the axis) object point to confirm the numerical solution. The rules employ three rays, often called construction rays. These three rays originate from (or pass through) the same (off-axis) object point. Rule ❶: A ray propagates parallel to the optical axis. At the lens, the ray is refracted, and its path after the lens crosses the secondary focal point F΄ after the lens. Rule ❷: A ray crosses the primary focal point F (situated before a positive lens). At the lens, the ray is refracted and becomes parallel to the optical axis. Rule ❸: A ray that crosses the center of the lens (the intersection of the lens with the optical axis) maintains its inclination (Figure 4-8). In other words, there is no path change at the lens. This rule is subject to the condition that the medium is the same on both sides of the thin lens. All three construction rays, if converging, (must) meet at a single point. This is the location of the real image. If they do not cross (in the case of converging rays, of course), we check for a possible drawing error. In the case of a virtual image, however, the rays actually diverge. Then we apply reverse ray extrapolation. The ray extrapolations cross at a unique point, which is the location of the virtual image.

4.3.1

Ray Diagrams for a Plus-Powered Lens

The three ray diagram rules specific to a plus-powered (positive) lens are as follows: A ray parallel to the optical axis (parallel ray) converges after the lens and crosses the optical axis at the secondary focal point F΄. A ray crossing the primary focal point F (focal ray) before the lens becomes parallel to the optical axis. A ray directed to the center of the lens (nodal ray) does not change direction; after the lens, it continues along the same path with the same inclination (undeviated).

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Figure 4-14: Ray diagrams for a converging (plus-powered) lens.

4.3.2 Some Considerations About Construction Rays ✽ Are these rays real? Yes, the construction rays in a thin lens are real, actual rays (this does not apply to a thick lens, as discussed in § 6.4.2). ‘Real’ implies that the path along their entire propagation length follows the rules of refraction and may be physically reconstructed. However, it is not necessary that construction rays actually exist (yet another existential question about rays… they do compete with Santa Claus on this…). In other words, it is not necessary that they correspond to an actual path throughout the optical system. This is because there might be an obstacle or a small aperture in front of the lens, or the actual size of the lens might be very small (a small lens diameter). What do we do in this case? We simply ignore the finite size of the lens and draw the rays as if the lens had an arbitrary large diameter. For the purpose of locating the image, we have this poetic license.

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GEOMETRICAL OPTICS

Figure 4-15: (left) Construction rays ❶ and ❷ cannot propagate through the lens due to the small lens diameter. However, we can very well use them (i.e., assume that they exist) for the purpose of locating the image! Moreover, we need only two of the three construction rays. (right) All rays from a unique object point that pass through the lens aperture meet at a unique image point.

✦ How many rays do we need? Just two. There is no rule stating which two of the three we need; however, the ray crossing the center of the lens (rule 3) appears to be the most convenient, in many cases. We can also draw all three rays to confirm that the point of intersection of the third ray is exactly the same as the point of intersecton of the first two. Just an ‘insurance policy’! ✥ What if the lens is not completely visible? Even if the lens is partially covered or semi-transparent, ray diagrams are valid and applicable. The image location and size are not dependent on the ‘ray population,’ i.e., how many rays pass through the lens. The image sharpness and brightness are affected by the ray population (§ 7.5).

Figure 4-16: Image location and shape are not affected by possible lens obscureness. Even a partial lens (but with the same optical power) will form the image at the same location.

✤ What about any other, random ray? Sure, we can draw any other ray (from the same object point), and we can apply the rules of refraction on the two surfaces of the thin lens. We may draw the path of any random ray through the lens (this ray, called the wild ray, is discussed § 4.5). By doing so, we will realize that 4-126

IMAGING WITH LENSES

any ray originating from the same object point meets the construction rays at the same image point (barring, of course, a drawing error). This is the simplicity offered by construction rays: We do not have to engage in challenging ray-tracing calculations! ❊ Are the rules applicable to the reverse propagation of light? Yes. The only difference is that for reverse light propagation (from right to left), the secondary focal point F΄ switches sides and is placed where the primary focal point once was. F΄ is located to the left of a positive lens and to the right of a negative lens. Likewise, the primary focal point

F is now placed to the right of a positive lens and to the left of a negative lens.

Figure 4-17: The construction rays rules also apply to the reverse propagation of light (right to left).

✦ I am not convinced about rule #3. Why doesn’t a ray directed toward the center of the lens change direction upon exiting the lens? This rule is a simplification, subject to certain conditions. The lens must be infinitesimally thin, and the medium must be the same on both sides of the lens. Then the emerging ray is only slightly displaced with respect to the incident ray. Its direction is the same as that of the incident ray, owing to the application of Snell’s law to a set of almost parallel surfaces. This resembles the parallel displacement that occurs in a flat slab. Naturally, the thinner the lens [i.e., the lesser its power (less curvature)], the more accurate the approximation. If the medium is not the same along the two lens surfaces, application of Snell’s law will produce the new (different) angle of refraction.

Figure 4-18: Ray-tracing rule ❸ assumes a thin lens and the same medium on both sides of the lens.

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GEOMETRICAL OPTICS

4.3.3 Ray-Tracing Examples with Plus Lens Imaging We can now re-evaluate the cases presented in Figure 4-9 through Figure 4-12 using ray tracing. A directional object that clearly rises above the optical axis is used. Select a point on the top of the object, well separated from the optical axis, and draw two construction rays. Example ☞: An object is placed 100 cm before (in front of) a lens with focal length +20 cm. Find the image location and magnification.

We use ray-tracing rules ❶ and ❷. The two rays intersect after the lens, at a short distance after the focal point. As calculated previously, the image is real (rays intersect), minified (image smaller than object), and inverted. The magnification is m = (+25 cm) / (−100 cm) = −0.25.

Note

: In this case, as in many examples, the illustration depicts only two (out of the three) ray-tracing

construction rays. It is suggested that the reader attempt to implement the missing rule for verification.

Example ☞: An object is placed 40 cm before (in front of) a lens of focal length +20 cm. Find the image location and magnification.

This is a special case of the previous example. The object is situated at twice the focal length in front of (before) the lens. We use ray-tracing rules ❶ and ❸. The two rays intersect after the lens, at a distance twice that of the focal length. The image is real (rays intersect), inverted, and the same size as the object. The magnification is m = (+40 cm) / (−40 cm) = −1.0.

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Example ☞: An object is placed 30 cm before (in front of) a lens of focal length +20 cm. Find the image location and magnification.

We use ray-tracing rules ❶ and ❷. The two rays intersect after the lens, a long distance after the focal point. As was found previously, the image is real (rays intersect), magnified (image larger than object), and inverted. The magnification is m = (+60 cm) / (−30 cm) = −2.0.

Example ☞: An object is placed 10 cm before (in front of) a lens of focal length +20 cm. Find the image location and magnification.

We use ray-tracing rules ❶ and ❸. The two image-forming rays do not intersect after the lens; therefore, the image is virtual. To locate the image, we extrapolate the image-forming rays. The extrapolations intersect at a distance x΄ = −20 cm, which is before the lens (hence, the negative sign). The magnification is m = (−20 cm) / (−10 cm) = +2.0. This suggests that the image is magnified and erect.

In all cases presented, it is clear that both the numerical solutions, based on imaging relationships presented in Eqs. (4.1) and (4.5), and the schematic solutions, based on ray tracing, have agreeing results.

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4.4 IMAGING BY A MINUS-POWERED LENS Imaging with a diverging, minus-powered lens (because it has a negative optical power, it is also called a negative lens) follows practically the same rules as imaging with a positive (pluspowered) lens. Because the focal length is negative, the secondary focal point F΄ is located to the left (in front) of the lens. Accordingly, the first rule states that a ray propagating parallel to the optical axis, after passing through the lens, diverges and appears as if it originated from the focal point F΄, which is situated in front of the lens. Example ☞: A beam collimated along the optical axis is incident on a lens of focal length −25 cm. Find the beam shape and image location after the lens.

Figure 4-19: Application of the optical power addition principle in a negative lens.

To visualize the vergence, we draw the first construction ray: A ray parallel to the optical axis meets the lens and diverges, deviating as if it originates from the focus. ✔ Principle of optical power addition: •

collimated beam vergence incident on the lens: L = 0 D,

•

lens optical power: F = 1/f΄ = 1/(−0.25 m) = −4 D,

•

vergence leaving the lens: L΄ = 0 D + (−4 D) = −4 D, suggesting a diverging beam.

Image location is the reciprocal of image vergence: x΄ = 1/L΄ = 1/(−4 D) = −0.25 m = −25 cm. ✔ In summary:

L + F = L΄  L΄ =

4-130

1

+

1

 −0.25 m

= 0 D + ( −4.0 D ) = − 4.0 D  x΄ =

1

L΄

=

1

−4.0 D

= − 0.25 m = − 25 cm

IMAGING WITH LENSES

An interesting case is when a converging beam (e.g., L = +4 D) is incident on a lens with an equal and opposite optical power (F = –4 D). This lens ‘neutralizes’ (collimates, or produces a zero vergence) the incident beam. This is an example of a virtual object being located after the lens (further discussed in § 4.7). Usually, the object is real and is located in front of the lens.

Figure 4-20: Neutralization of a converging beam by a negative lens.

4.4.1

Ray Diagrams for a Minus-Powered Lens

The ray-tracing rules for a minus lens are very similar to those for a plus lens. The key differences are that the rays diverge after the lens, and that the locations of the secondary and primary focal points are reversed. The image from a real object is erect, virtual, and minified.

Minus-lens imaging for an object placed to the left of the lens: Image is virtual, erect (m > 0), and formed to the left (in front) of the lens. Magnification is always less than unity (m < 1.0).

The three ray-diagram rules specific to a minus (negative) lens are as follows: A ray parallel to the optical axis (parallel ray) diverges after the lens, and its path appears as if it originates from the secondary focal point F΄.

A ray directed to the primary focal point F (focal ray) becomes parallel to the optical axis after the lens.

A ray directed to the center of the lens (nodal ray) does not change direction; after the lens, it continues along the same path with the same inclination (undeviated).

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GEOMETRICAL OPTICS

Figure 4-21: Ray diagrams for a diverging (minus-powered) lens. Example ☞: An object is placed 50 mm before (in front of) a lens of focal length −16.66 mm. Find the image location and magnification.

✔ Schematic solution: We apply the first and the third ray-tracing rules for a pair of rays that originate from the object. Ray ❶ is parallel to the optical axis and diverges after the lens, as if it originated from the secondary focal point F΄. Ray ❸ simply transcends the center of the lens undeviated. We observe that the rays diverge after the lens. The image is virtual, being located at the point of intersection of the ray extrapolations (Figure 4-22). This point is located before (in front of) the lens at ¼ of the object location. The magnification is positive (the image is erect) and is less than 1.0 (the image is minified).

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Figure 4-22: Virtual image from a negative lens. The magnification is positive but less than 1.0 (m = +0.25).

✔ Numerical solution: Object location is x = −0.05 m. Object vergence is L = 1/x = 1/(−0.05 m) = −20 D. Lens power is

F = 1 / (−0.0166 m) = −60 D. Image vergence is L΄ = (−20 D) + (−60 D) = −80 D. The image is therefore virtual, since it has a negative vergence. Image location is x΄ = 1/L΄ = 1/(−80 D) = −0.0125 m = −12.5 mm:

L + F = L΄  L΄ = ( −20 D ) + ( −60 D ) = − 80 D  x΄ =

1

L΄

=

1 −80 D

= − 0.0125 m = − 12.5 mm

The magnification is positive (erect image): m = (−12.5 mm) / (−50 mm) = +0.25 = ¼.

Figure 4-23: The same problem as in Figure 4-22 but managed with application of optical power and vergence addition.

When a (real) object is placed in front of the lens, the object vergence is negative. The lens power is also negative for a negative lens. This results in the image vergence being negative, too, and numerically larger than the object vergence. Thus, a real object forms an image that is always located to the left of (before) the negative lens and is virtual, erect, and minified.

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4.5 IMAGING BY AN SSRI Imaging by a single spherical refracting interface (SSRI) follows exactly the same principles as imaging by a lens. The sign rules follow the Cartesian convention; the object vergence is added to the SSRI power, which is calculated via Eq. (1.8) to produce the image vergence. Attention is drawn to the fact that, by rule, the SSRI has differing refractive indices across the interface, which is important when calculating vergence. Here are the imaging relationships: Vergence and Optical Power:

+

L object vergence

Object and Image Locations:

F

=

optical power

L΄

n

n΄

n΄

object space

image space

image space

x object location

+

f΄ focal length

(4.8)

image vergence

=

x΄

(4.9)

image location

The only difference between these relationships and the imaging relationships introduced for lenses [Eqs. (4.3) and (4.4)] is that here it is impossible for both n’s to be equal, or even for both to be 1.0 because, by rule, an SSRI separates two media with different refractive indices. Example ☞: A convex SSRI separates air from glass with refractive index 1.5. The radius of curvature is +0.10 m. An object is placed 50 cm in front of the interface. Identify the image location and magnification. The object space is filled with air, so n = 1.0, while the image space is glass of n΄ = 1.5. (This is essential in SSRI imaging because, by rule, the interface separates two media with different refractive indices.) The SSRI optical power is calculated using Eq. (1.8): F = (n΄– n)/r = (1.5 – 1.0)/(+0.1 m) = +5.0 D. The object is placed at a distance of –0.5 m, which is before the SSRI: x = – 0.5 m. Therefore, the object vergence is L = n / x = –2.0 D. We now add the object vergence to the SSRI power to obtain the image vergence and image location:

L + F = L΄ ⇒ L΄ = (–2.0 D) + (+5.0 D) = +3.0 D. L΄ = n΄ / x΄ ⇒ x΄ = 1.5/ (+3.0 D) = +0.5 m = 50 cm to the right of the interface. For the magnification, do not be tempted to use m = x΄/x. This relationship is not valid in this case because the two refractive indices are not the same. We use instead the more generally applicable relationship m = L/L΄ ⇒ m = L/L΄ = (–2.0 D)/(+3.0 D) = – 0.66. The image is real (positive vergence), inverted (negative magnification), and minified (magnification less than 1.0). 4-134

IMAGING WITH LENSES

Note

: A convex SSRI always adds convergence (or positive vergence) to the incident rays, rendering

the image vergence more converging (or less diverging); a concave SSRI always adds negative vergence to the incident rays, rendering the image vergence less converging (or more diverging).

Figure 4-24: Formation of a real image by a convex SSRI.

Alternatively, we can implement ray-tracing rules in a fashion very similar to that prescribed in § 4.3. These are the SSRI imaging ray diagrams. As in lens imaging, there are three construction rays that originate from (or pass through) the same object point. In the case of a convex SSRI, the ray-tracing rules can be stated as follows: Rule ❶: A ray propagating parallel to the optical axis. At the interface, the ray is deviated toward and crosses the secondary focal point F΄. Note that in a convex SSRI, F΄ is situated to the right of the interface. Rule ❷: A ray crossing the primary focal point F (which, in a convex SSRI, is situated to the left of the interface). At the interface, the ray is refracted and propagates parallel to the optical axis. Rule ❸: A ray directed toward and crossing the center of curvature (which serves as the nodal point N) continues to propagate undeviated.

Figure 4-25: Ray-tracing rules for a convex SSRI.

In the case of a concave SSRI, the ray-tracing rules can be stated as follows: 4-135

GEOMETRICAL OPTICS

Rule ❶: A ray propagating parallel to the optical axis. At the interface, the ray is deviated away from the secondary focal point F΄. Note that in a concave SSRI, F΄ is situated to the left of the interface, so the backward extrapolation of this ray crosses point F΄. Rule ❷: A ray directed toward the primary focal point F. In a concave SSRI, F is situated to the right of the interface. As this ray is intercepted at the interface, only the forward extrapolation of this ray crosses point F. The ray is then refracted and becomes parallel to the optical axis. Rule ❸: A ray that crosses the center of curvature (which serves as the nodal point N) continues to propagate undeviated.

Figure 4-26: Ray-tracing rules for a concave SSRI. Example ☞: A concave SSRI separates air from glass with refractive index 1.5 (as shown in Figure 4-26). The radius of curvature is r = – 0.10 m, and an object is placed 50 cm in front of the interface. Identify the image location, type, and magnification. The object space is filled with air, so n = 1.0, while the image space is filled with glass (n΄ = 1.5). The SSRI optical power is calculated using Eq. (1.8): F = (n΄– n)/r = (1.5 – 1.0)/( –0.1 m) = –5.0 D. The object is placed at a distance of –0.5 m to the left of the SSRI: x = –0.5 m. Thus, the object vergence is

L = n / x = 1.0/ (–0.5 m) = –2.0 D. The imaging relationship L + F = L΄ yields (–2.0 D) + (–5.0 D) = L΄ = –7.0 D. Then we calculate the image location: L΄ = n΄ / x΄ ⇒ x΄ = n΄ / L΄ = 1.5/ (–7.0 D) = –0.214 m. The image is located 21.4 cm before (in front of) the interface. Magnification: m = L/L΄ ⇒ m = (–2.0 D)/( –7.0 D) = +0.285. The image is virtual (negative vergence), erect (positive magnification), and minified (magnification less than 1.0).

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IMAGING WITH LENSES

4.6 NOTES ON IMAGING 4.6.1 The Optical Invariant The propagation of an arbitrary/random ray though an optical system provides some interesting insights. Figure 4-27 presents a ray whose origin is at the on-axis point (distance x) of an object, whose height is h. The ray terminates at the conjugate of the axis point (distance x΄) of the image, whose height is h΄. The ray with the maximum possible height at the lens is a marginal ray (defined in § 7.2.2). This ray forms an angle ϑ with the optical axis in object space and an angle ϑ΄ with the optical axis in image space, and intersects the lens at a height noted as s.

Figure 4-27: A random ray through an optical system.

We now apply some basic geometry for small angles to obtain (expressed in rad): ϑ = s / x and ϑ΄ = s / x΄, which is rearranged to: s = ϑ · x = ϑ΄ · x΄ Note that angle ϑ is computed here as the ratio of height to distance. Therefore, it is simply the slope of the ray, which approximates the angle for small angles. Then, we recall the definition of magnification [Eq. (3.1)]:

m = h΄/h, which expands to Eq. (4.6): m = h΄/h = n · x΄/ n΄ · x Combining the above, we obtain

x   = x΄  ΄ =

h΄ n΄ h΄ n΄   x  ΄   =   ΄  n   h = n΄  ΄  h΄ h n h n

(4.10)

This result is independent of the arbitrary choice of ray height s so is valid regardless of which ray is chosen (within the approximation of small angles, which is also called the paraxial approximation).

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GEOMETRICAL OPTICS

Additionally, this result is an expression of a fundamental law of optics, not just a restatement of the magnification formula. In any optical system, the product of the image size (height h΄) and ray angle ϑ΄ is a constant, or invariant, of the system, known as the optical invariant. The result is valid for any number of lenses, as could be verified by tracing the ray through a series of lenses. This is also called the Lagrange invariant or the Smith–Helmholtz invariant. In the case where the optical system is surrounded by the same medium, Optical Invariant:

ϑ · h = ϑ΄ · h΄

(4.11)

The optical invariant is a very useful concept in determining the maximum amount of light that a lens system can accept (see Numerical Aperture, presented in § 7.1.4), the maximum number of resolvable spots in the image (see Resolution, discussed in § 7.5), and the luminous flux collected by a lens system.19 In the case where an optical system consists of a number of elements (and/or surfaces), this relationship can be iterated at each element or surface:

n1 · ϑ1 · h1 = n2 · ϑ2 · h2 = … = n΄ · ϑ΄ · h΄ where n is the refractive index preceding the surface, ϑ is the angle of the ray, and h is the object or image height. Here, the numeric indices suggest ‘same side’ variables. Note : It is important to keep the correct algebraic signs. Recall (§ 3.2.1) that the height has a positive sign if it extends above the optical axis and a negative sign if it extends below it. The angle sign convention states that an angle is positive if it is counter-clockwise and negative if clockwise. Both height and angle are measured from the optical axis.

Example : Is the following optical arrangement possible?

Figure 4-28: Application of optical invariance in a black-box system: Example 1.

19

Introduction to Optics § 2.2.1 Photometric Quantities.

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IMAGING WITH LENSES

We first check the sign preservation. In object space, both the image height h and ray angle ϑ are positive; their product is positive. In image space, both the image height h΄ and ray angle ϑ΄ are negative; their product, however, is still positive. This is good news. A closer look also suggests that the object height is greater than the image height, and that the object ray angle is less than the image ray angle, keeping the numerical value of the optical invariant, well… invariant!

Example : Is the following optical arrangement possible?

Figure 4-29: Application of optical invariance in a black-box system: Example 2. This system fails the sign preservation, so the arrangement is impossible. In object space, both the image height h and ray angle ϑ are positive; their product is positive. In image space, the image height h΄ is positive and the ray angle ϑ΄ is negative; their product is negative.

Example : Is the following optical arrangement possible?

Figure 4-30: Application of optical invariance in a black-box system: Example 3. If the black box is surrounded by the same optical medium, this system fails magnitude preservation. In object space, the image height h and ray angle ϑ are greater than their counterparts in image space. It appears that the optical invariant in object space has a greater magnitude (although they are both positive) than in image space. The sign preservation is only possible on the condition that the refractive index in image space is greater than that in object space.

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GEOMETRICAL OPTICS

4.6.2 The Wild Ray The propagation path of an arbitrary/random ray (wild ray) through a thin lens can be traced by the oblique incidence or Listing method. The reasoning behind the method is that every ray parallel to any optical axis (including the secondary optical axes, not just the principal optical axis) that emerges from the lens must transcend the corresponding focal point, defined as the intersection of the specific axis with the focal plane. The steps are as follows: ⑴ Draw an auxiliary line parallel to the incident ray such that it crosses the focal point F and follows the focal ray (ray diagram rule ❷). ⑵ The ray exiting the lens transcends the point where the path of the auxiliary line crosses the focal plane at the other side.

Figure 4-31: Tracing the random ray path with the Listing method for a positive lens.

Alternatively, applying reverse light propagation, the steps can be as follows: ⑴ From the focal plane / ray intersection, draw an auxiliary line toward the lens center. ⑵ At the point of intersection of the ray with the lens, the direction of the exiting ray is parallel to the auxiliary line.

Figure 4-32: Alternative drawing of the random ray path with the oblique incidence method.

These rules are fully compatible with the ray diagrams (Figure 4-14 and Figure 4-21). In the case of a lens with a negative optical power, for the application of rule ❶ we draw the reverse ray extrapolation of the incident ray to find its intersection with the focal plane at F, 4-140

IMAGING WITH LENSES

which is now located after the lens, not before it, as in the case with a converging lens (positive optical power).

Figure 4-33: Tracing the random ray path with the Listing method for a diverging lens.

The procedure is applicable to any single lens element in a lens system (§ 6.4.3) and, with some variation, for a thick lens of a lens system, if the locations of the cardinal points (defined in § 6.2.1) are known.

4.6.3 Optical Infinity An interesting case exists when the object is placed exactly on the focal plane of the lens (x = f = –f΄ ). Then we can say that ‘there is no image’ because, after the lens, the rays are collimated—they do not converge! The image, however, does exist. It is ‘formed’ at optical infinity, x΄ = ∞. It appears as if the rays originate from a point so far away that all of the rays from this point, when drawn propagating toward the lens, are collimated.

Figure 4-34: Imaging from a positive lens to optical infinity. The object is placed at the primary focal point of the lens.

The mathematical analysis is as follows:

L + F = L΄

 L΄ =

1

− f΄

+

1

f΄

= 0D 

x΄ =

1

L΄

=

1 0D

=

4-141

GEOMETRICAL OPTICS

This image can be observed with another lens—one that can converge the collimated beam to its focal point. This is the case of observation with our naked eye: The optical system of the human eye images the collimated beam on the retina. An object may also, accordingly, be located at optical infinity (x = –∞). Then the rays originating from the object and propagating to the lens (or any optical system, such as an SSRI) are collimated, and the image is formed on the secondary focal point (x΄ = f΄):

L + F = L΄



L΄ =

1



+

1

f΄

=

1 f

 x΄ =

1

L΄

=

1 1

= f΄

f΄

Example ☞: A convex glass (n΄= 1.5) SSRI, preceded by air (n = 1.0), has a power F = +10 D. Where is the image of an object placed at optical infinity? By rule, we can simply answer this as ‘at the secondary focal point.’ The very definition of the secondary focal point is the image point of an object at optical infinity. Let us prove this, as well, using imaging relationships. This SSRI with F = +10 D separates air from glass. Therefore, n = 1.0 and n΄ = 1.5. The secondary focal length of the SSRI is f΄ = n΄ / F = 1.5 / (+10.0 D) = +0.15 m. Since the object is at optical infinity, the object vergence is zero. Therefore, L + F = L΄ ⇒ 0 + (+10 D) = L΄ ⇒ L΄= +10 D. The location of this image, x΄= n΄ / L΄= 1.5 / (+10.0 D) = +0.15 m = +15 cm, is to the right of this interface, coinciding with the secondary focal point.

Figure 4-35: The image of an object at optical infinity is formed at the secondary focal point.

Optical infinity is not exactly located at infinity, since this is a mathematical idealization. In reality, we may accept that an object (or image) is at optical infinity if it is located a great distance away in comparison to the focal length of the optical system. Then the corresponding object or image ray bundle is considered collimated. For example, for the optical system of the human eye (F = +60 D, f = 16.6 mm), any distance beyond 6 m may be considered as optical infinity. 4-142

IMAGING WITH LENSES

In this case, the image has, in principle, infinite size. The concept of lateral magnification, as defined by Eq. (4.5), is not applicable. A similar problem exists when the object is located at optical infinity (the object beam is collimated along the optical axis): The image is infinitesimally small. Then we apply angular magnification Mϑ (defined in § 3.3.1).

Optical Infinity

Object or image ray bundle is collimated.

Object or image may be far away (a star), or...

The corresponding ray bundle may be collimated by an optical element.

4.7 VIRTUAL OBJECT IMAGING Despite the fact that the majority of imaging places the object in front of a lens (or SSRI, or mirror), it is not rare for the object to be situated after the lens. This configuration can be created in lens systems: The object of the second lens is the image formed by the first lens. This is a virtual object. Recall that an object does not need to be an actual, physical object or a radiating light source—it is the rays that matter. An object is associated with rays incident on the optical system (§ 3.1.1) that are either diverging (real object) or converging (virtual object). Optics principles do not change and are not dependent on the object location and type (real or virtual). Using vergence, the object vergence is positive because the object location x is now positive. We simply add to this vergence the lens optical power; thus, the image-forming beam becomes accordingly more (or less) converging, depending on the algebraic sign of the lens optical power. A positive lens makes the image-forming beam more converging, and a negative lens makes it less converging, causing the image to be closer or farther away from the virtual object location, respectively.

Figure 4-36: A virtual object has positive vergence: The rays incident on the lens are converging.

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GEOMETRICAL OPTICS

If the object is virtual, we may notice some adaptations in the way the ray-tracing rules, as originally presented in Figure 4-14, are implemented for a real object. The difference is that the rays do not originate from a point on the object because it is not a real object, but their extrapolations pass through the selected object point.

4.7.1 Virtual Object Imaging with a Plus Lens The rules pertaining to imaging of a virtual object with a plus lens are as follows: A ray that follows rule ❶ is drawn parallel to the optical axis such that, if extrapolated, it would meet the object point of interest. A ray that follows rule ❷ first crosses the primary focal point F, aiming at the same object point as the extrapolation of rule 1 ray; it is then refracted to become parallel to the optical axis. A ray that follows rule ❸ intersects the lens center then continues undeviated to the same object point as the intersection of the extrapolation of the first two rays.

Figure 4-37: Ray-tracing diagrams for a converging (plus) lens in the presence of a virtual object. Example ☞: An object is located 120 cm after a positive lens of focal length 30 cm. Find the image location and magnification.

Figure 4-38: Imaging from a virtual object with a positive lens. The image location is determined by ray tracing. 4-144

IMAGING WITH LENSES

✔ Schematic solution: We place the object 120 cm after the lens. The image is formed at the point where the rays now converge (the image is real)! To apply rule ❸, we draw (and extend) a line that joins an object point and the center of the lens. To apply rule ❶, we draw a line parallel to the optical axis that would intersect the same object point; this ray is refracted by the lens toward the focal point F΄. These rays intersect at a point to the right of the lens. The image is therefore real and is also erect, as it has the same orientation as the object. ✔ Numerical solution: Object location is x = +120 cm.

L + F = L΄  L΄ =

1

+

1

+1.20 m +0.30 m

= +4.16 D  x΄ =

1

L΄

=

1

+4.16 D

= +0.24 m = +24 cm

The image location is therefore x΄ = +24 cm. The image location is positive (0.2× the object location), so the magnification (minification) is positive m = (+0.24)/(+1.20) = +0.2.

Figure 4-39: Imaging from a virtual object with a positive lens. The image location is determined by vergence and power addition.

The object is termed virtual because its formation is assumed. This object functions just like any other object and forms an image, which, in this case, is real and erect. This image formation should not be a surprise at all. As the rays were in a path to converge, the lens made them converge even more. The final ray convergence forms a real image. With a virtual object, the object vergence is positive. With a positive power lens, this results in the image vergence being positive, too, and numerically larger than the object vergence. Thus, the image located to the right of (after) the positive lens is always real and is shorter than the object. Positive Lens Imaging of a Virtual Object: The image formed by a positive lens of a virtual object is always real, minified, and erect.

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GEOMETRICAL OPTICS

Virtual Object

Object location: positive, after the optical element.

Object vergence: positive.

Usually the image is 'to be formed' by another optical element.

Plus lens (and convex SSRI) imaging of a virtual object can be described by an expansion of the top-right quadrant in Figure 4-13. The image is real and erect with a magnification less than unity, and is formed between the lens and the secondary focal point (see Figure 4-40).

Figure 4-40: Expanding Figure 4-13, the entire curve 1/x + 1/ f ΄ = 1/x΄ is shown in the object–image location space (left) and in the inverse space of object–image vergence (right). The top-right quadrants correspond to a virtual object.20 Example ☞: A convex SSRI of glass (n= 1.5), followed by air (n΄= 1.0), has power F = +10 D. If an object is located x = +0.09 m = +9 cm to the right of the interface, where is the image, and what type of image is it? First we clarify the two refractive indices: object space n = 1.5 and image-space n΄ = 1.0. (Recall that object space is associated with rays entering the interface, and image space with rays leaving the interface.) The object is located at x = +0.09 m = +9 cm to the right of the interface. It is a virtual object with a vergence

L = n / x = 1.5 / (+0.09 m) = +16.66 D.

The statements that ‘a negative lens forms only virtual images’ and ‘a real image is inverted’ apply to real objects. It is obvious, as we see here, that they do not apply to virtual objects! Yet, here is one more surprise: The negative lens now forms images with a magnification larger than 1.0! 20

4-146

IMAGING WITH LENSES

The imaging relationship yields: L + F = L΄ ⇒ L΄ = +16.66 D +10.0 D = +26.66 D. Thus, the image location is x΄= n΄/L΄= 1.0 / (+26.66 D) = +0.0375 m. The image is located 3.75 cm to the right of the surface. The magnification is m = L/L΄ = (+16.66 D)/(+26.66 D) = +0.625. The image is real (positive vergence), erect (positive magnification), and minified (magnification less than 1.0).

Figure 4-41: Real image produced by SSRI imaging of a virtual object.

4.7.2 Virtual Object Imaging with a Minus Lens Next we examine image formation for a virtual object with a negative lens. Now that the object is virtual, we may, again, notice some adaptations in the way the ray-tracing rules, as originally presented in Figure 4-21, are implemented for a real object. As in the case with a positive lens, the rays do not originate from an object point but, instead, their extrapolations project to a selected point. Rule ❶: A ray drawn parallel to the optical axis, to the height where its forward extrapolation meets a selected object point. At the lens, this ray is refracted as if it originates from the secondary focal point F΄. Rule ❷: A ray aiming at (appearing at, or extrapolating to) the primary focal point F and directed to the same object point as ray 1 (i.e., its extrapolation connects the primary focal point and the object point). At the lens, this ray is refracted to become parallel to the optical axis. Rule ❸: A ray intersecting the lens center and directed toward the same object point as rays 1 and 2. At the lens, this ray continues undeviated.

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Figure 4-42: Ray diagrams for a diverging (minus) lens in the presence of a virtual object. Example ☞: An object is located 10 cm after (to the right of) a negative lens of focal length −12.50 cm. Find the image location, type, and magnification.

✔ Schematic solution: The lens has a negative focal length: The focal point F΄ is at f΄ = −0.125 m. We place the object at a location x = +0.10 m after the lens, where it would be formed if there was no lens. To apply ray-tracing ray rule ❶, we draw a line parallel to the optical axis whose extrapolation crosses an object point (Figure 4-43). This ray, refracted by the lens, propagates as if it originated from the lens focal point F΄.

Figure 4-43: Image from a virtual object via a negative lens: application of ray tracing.

To apply ray-tracing rule ❷, we draw a ray that extrapolates (is aimed) to connect the same object point and the primary focal point F; at the lens, this ray is refracted to become parallel to the optical axis.

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IMAGING WITH LENSES

To apply ray-tracing rule ❸, we draw a line that joins the center of the lens and the same object point. The three rays intersect the same point that is past the lens, which indicates the location of the image. Since both the image and object locations are positive (after the lens), the image is erect—it has the same orientation (up) as the object. The fact that the image-forming rays intersect (i.e., the image is formed by converging rays) means that the image is real. ✔ Numerical solution: The object vergence is L = 1/x = 1/(+0.10 m) = +10 D. The lens optical power is

F = 1/(−0.125 m) = −8 D. Thus, the image vergence is +10 D+(−8 D) = +2 D. The positive sign of the image vergence indicates intersecting image-forming rays and therefore indicates a real image. The image location is x΄ = 1/(+2 D) = +0.5 m = +50 cm. This result indicates that the rays were coverging to form the object (+10 D), and the presence of the negative lens caused less (by the amount of lens power –8 D) vergence, which became +2 D. The rays then converged to a point farther away from the lens:

L + F = L΄  L΄ =

1

1

+

+0.10 m −0.125 m

Image location is x΄ = +0.50 m: x΄ =

1

L΄

=

1

+2 D

= (+10.0 D ) + ( −8.0 D ) = +2.0 D

= + 0.50 m = + 50 cm

The fact that the image location is 5× the object location means that the magnification is positive and larger than 1.0 (m =+5.0). The image is erect (+). It is also real (positive image vergence).

Figure 4-44: Image from a virtual object and a minus lens. The image location is calculated by applying vergence and lens power addition. 4-149

GEOMETRICAL OPTICS

Negative Lens Imaging with a Virtual Object: The negative lens may produce a variety of images with a virtual object, depending on the object 'location' with respect to the focal point.

Example ☞: An object is located 16.66 cm after (to the right of) a negative lens of focal length –12.50 cm. Find the image location, type, and magnification.

The object vergence is L = 1/x = 1/(+0.166 m) = +6 D. The lens optical power is F = 1/(−0.125 m) = −8 D. Thus, the image vergence is L΄ = (+6 D)+(−8 D) = −2 D. The negative sign indicates nonintersecting rays, or a virtual image. The image location is x΄ = 1/(−2 D) = −0.5 m = −50 cm. The magnification is m = x΄ / x = −3. The image is virtual, magnified, and inverted.

Example ☞: An object is located 25 cm after (to the right of) a negative lens of focal length −12.50 cm. Find the image location, type, and magnification.

The object vergence is L = 1/x = 1/(+0.25 m) = +4 D. The lens power is F = 1/(−0.125 m) = −8 D. The image vergence is L΄ = (+4 D)+(−8 D) = −4 D. This is a virtual image (negative vergence,

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IMAGING WITH LENSES

diverging rays) formed at x΄ = 1/(−4 D) = −0.25 m = −25 cm. The magnification is m = x΄ / x = (−0.25 m)/(+0.25 m) = −1.0. The image is virtual, the same size as the object, and inverted. Example ☞: An object is located 50 cm after (to the right of) a negative lens of focal length −12.50 cm. Find the image location, type, and magnification.

The object vergence is L = 1/x = 1/(+0.50 m) = +2 D. The lens power is F = 1/(−0.125 m) = −8 D. The image vergence is L΄ = (+2 D)+(−8 D) = −6 D. The negative vergence of nonintersecting rays indicates a virtual image. The image location is x΄ = 1/(−6 D) = −0.166 m = −16.6 cm. The magnification is m = x΄ / x = (−0.166 m)/(+0.50 m) = −⅓. The image is virtual, minified, and inverted. Example ☞: An object is located 12.50 cm after (to the right of) a negative lens of focal length −12.50 cm (at the primary focal point of the lens). Find the image location, type, and magnification.

✔ Ray-tracing schematic solution: To apply ray-tracing rule ❷, we draw a ray that extrapolates (is aimed) to connect the same object point and the primary focal point F; at the lens, this ray is refracted to become parallel to the optical axis. To apply ray-tracing rule ❸, we draw a line that joins the center of the lens and the same object point. The three rays intersect the same point, past the lens, which indicates the image location.

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✔ Numerical solution—vergence and power summation: The object vergence is L = 1/x = 1/(+0.125 m) = +8 D. The lens power is F = 1/(−0.125 m) = −8 D. The image vergence is L΄ = (+8 D) − (+8 D) = 0 D. A zero image vergence means that the wavefront is flat (collimated rays), so the image is formed at optical infinity. Note

: A virtual object is never ‘placed’ as it does not relate to a physical object or an actual light

source, but rather to a configuration of rays incident on the optical element in a converging fashion. The correct verb is ‘situated.’

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4.8 ADVANCED LENS-IMAGING EXAMPLES Example ☞ : A real object is placed 20 cm before (in front of) a lens. The image is formed 40 cm after the lens. What is the image type and magnification? What type of lens produces such an image?

The object location is x = −0.20 m, and the image location is x΄ = +0.40 m. The object vergence is L = 1/ x = 1/ (−0.20 m) = −5.0 D, and the image vergence is L΄ = 1/ x΄ = 1/ (+0.40 m) = +2.5 D. Because the image vergence is positive, the image is real. We calculate the magnification using Eq. (4.5): m = x΄ / x = (+0.40 m) / (−0.20 m) = −2.0. The image is inverted and magnified. This type of image can only be produced by a positive lens, particularly when the object is located such that 2 f > |x| > f, forming a real image at 2 f΄ < x΄ < ∞. To calculate the lens optical power, we use the imaging relationship given in Eq. (4.1): L + F = L΄. The only unknown is the lens power: −5 D + F = +2.5 D, which leads to F = +7.5 D, or focal length in air f΄ = 1/(+7.5 D) = +0.133 m = +13.3 cm. Note

: We assumed that the medium surrounding the lens is air. However, this should not be taken

for granted. In this example, the calculated values for the lens focal length would not be affected, as long as the medium surrounding the lens is the same on both sides.

Example ☞ : A symmetric biconcave lens surrounded by air is constructed of a material of refractive index n = 1.5 and has radii of curvature of 20 cm. When a real object is placed in front of the lens, an image of ⅕ the object size is produced. Where is the object located?

We calculate the power of the lens using the lens-maker’s formula [Eq. (2.4)]:

F =

1 1 1   1  −2  = ( n − 1)   −  = (1.5 − 1)   − = ( 0.5)     = − 5.0 D f r r − 0.2 m + 0.2 m 0.2 m      1 2

1

We note that the imaging relationship provided in Eq. (4.1) (L + F = L΄ ) has two unknowns, as we know neither the object location nor the image location.

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The second piece of information is that the magnification is +0.2. The algebraic sign is positive, since a negative lens can only produce an erect and minified image of a real object. We can therefore write the magnification in the form of Eq. (4.6):

m=

L = 0.2  L΄ = 5 L L΄

We can now combine the lens imaging equation by substituting accordingly:

L + F = L΄  L + F = 5 L  L =

F 4

=

−5 4

D  x = − 0.8 m = − 80 cm

The object is therefore located 80 cm in front (to the left) of the lens.

Example ☞ : A symmetric biconvex lens surrounded by air is constructed of a material of refractive index n = 1.5 and has radii of curvature of 20 cm. When a real object is placed in front of the lens, an image of ⅕ the size of the object is produced. Where is the object located?

We calculate the power of the lens using the lens-maker’s formula [Eq. (2.4)]:

F =

1 1 1   1  +2  = ( n − 1)   −  = (1.5 − 1)   − = ( 0.5)     = +5 D f  +0.2 m −0.2 m   0.2 m   r1 r2 

1

Again, the imaging relationship given in Eq (4.1) (L + F = L΄ ) has two unknowns, as we know neither the object location nor the image location. The second piece of information is that the magnification is −0.2. The algebraic sign is negative because, when a positive lens produces a minified image from a real object, this image is inverted. We can therefore write the magnification in the form of Eq. (4.6): 4-154

IMAGING WITH LENSES

m=

L = − 0.2  L΄ = − 5 L L΄

We now combine the lens imaging relationships by substituting accordingly:

L + F = L΄  L + F = − 5 L  L = −

F 6

5

= − D  x = − 1.2 m = − 120 cm 6

The object is therefore located 120 cm in front (to the left) of the lens.

Example ☞ : A lens has a focal length of +20 cm. When a real object is placed in front of the lens, an image of 5× the size of the object is produced. What type of lens produces such an image, what type of image is this, and where exactly is the object located?

We note that the image is magnified. This excludes a negative lens. It is therefore a positive lens with power F = +5 D. A positive lens can produce two types of magnified images. ✔ Case of a real image: This is an inverted, magnified image formed to the right of (after) the lens. The magnification is m = −5:

m=

L = − 5  L = − 5 L΄ . L΄

We now combine the lens imaging relationships:

L + F = L΄  − 5 L΄ + F = L΄  L΄ = To calculate the object location:

F 6

=

5 6

D  x΄ =

6 5

m = 1.2 m = 120 cm

5

6

6

25

L = − 5 L΄ = − 5 D  x = −

m = − 0.24 m = − 24 cm

✔ Case of a virtual image: This is an erect, magnified image formed to the left of (before) the lens. The magnification is m = +5:

m=

L = + 5  L = + 5 L΄ . L΄

We now combine the lens imaging relationships:

L + F = L΄  + 5 L΄ + F = L΄  L΄ = −

F 4

5

4

4

5

= − D  x΄ = − m = − 0.8 m = − 80 cm

Finally, to calculate object location, we take the reciprocal of the object vergence:

 5  4

L = + 5 L΄ = + 5  −  D  x = −

4 25

m = − 0.16 m = − 16 cm

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4.9 LENS IMAGING QUIZ Unless otherwise specified, the lenses in this quiz are surrounded by air. A ‘placed’ image is real; a ‘formed’ image is virtual. 1)

An object placed before a lens forms a minified image. What type of lens is it (two correct answers)? a) b) c) d)

5)

a plus lens, if the image is inverted a plus lens, if the image is erect a minus lens, if the image is inverted a minus lens, if the image is erect

An image formed by a lens is virtual and erect. What type of lens is it (two correct answers)? a) b) c) d)

6)

a plus lens a minus lens can be either a plus or a minus lens the information provided is not sufficient

a plus lens, if the image is magnified a minus lens, if the image is minified a plus lens, if the image is minified a minus lens, if the image is magnified

An image formed by a lens is real and erect. What type of lens is it (two correct answers)? a) b) c) d)

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a plus lens, if the image is magnified a minus lens, if the image is minified a plus lens, if the image is minified a minus lens, if the image is magnified

9)

at twice the focal length of a plus lens at the focal length of a plus lens at twice the focal length of a minus lens at the focal length of a minus lens

An object located 25 cm in front of (before) a lens forms an image 12.5 cm to the right of (after) the lens. What is the lens power? a) b) c) d) e) f)

a real image at +10 cm a real image at +5 cm a real image at –5 cm a virtual image at –5 cm a real image at object-space infinity

An object placed in front of (before) a lens forms an image that that is minified to 1/4 of the object size and inverted. What type of lens is it? a) b) c) d)

4)

8)

Collimated light is incident on a +20.0 D lens. Determine the type and location of the image. a) b) c) d) e)

3)

a plus lens a minus lens can be either a plus or a minus lens the information provided is not sufficient

I want to produce an image on a sheet of paper with a magnification equal to unity (regardless of being erect or inverted). I will place the object … a) b) c) d)

An object placed before a lens forms an image four times bigger in size. What is the lens type? a) b) c) d)

2)

7)

–6.0 D –4.0 D +4.0 D +6.0 D +8.0 D +12.0 D

An object located 10 cm in front of (before) a lens forms an image 8 cm in front of (before) the lens. What is the focal length of this lens? a) b) c) d) e) f) g)

+2.5 m +80 cm +12.5 m +40 cm –40 cm –12.5 cm –10 cm

10) An object placed 50 cm in front of (before) a lens forms an image 25 cm to the right of (after) the lens. What is the lens power? a) b) c) d) e) f)

–4.0 D –4.0 D –2.0 D +4.0 D +6.0 D +10.0 D

11) A beam of vergence +10.0 D is incident on a plus lens with power +8.0 D. What is the vergence of the beam leaving the lens? a) b) c) d) e)

+2.0 D +8.0 D +10.0 D +18.0 D +80.0 D

IMAGING WITH LENSES

12) A –6.0 D power lens forms an image 10 cm in front of (before) the lens. Where is the object located? a) b) c) d) e) f)

6.25 cm to the right of (after) the lens 6.25 cm to the left of (before) the lens 25 cm to the right of (after) the lens 25 cm to the left of (before) the lens 10 cm to the right of (after) the lens 10 cm to the left of (before) the lens

13) An object is located in front of (before) a lens such that the object vergence is –6.00 D. The lens power is +10.00 D. Where is the image located? a) b) c) d) e) f)

m = +0.5 m = –0.5 m = +1.0 m = –1.0 m = +2.0 m = –2.0

15) A beam of vergence +10.0 D is incident on a plus lens with optical power +20.0 D. What is the vergence of the beam leaving the lens? a) b) c) d)

–6.0 D –3.0 D –2.0 D +4.0 D +6.0 D

19) An object is located in front of a lens such that the object vergence is L = –2.00 D. The lens power is F = +7.00 D. Where is the image (x΄= ?)? a) b) c) d) e) f)

2 m in front of (before) the lens 4 m to the right of (after) the lens 20 cm in front of (before) the lens 20 cm to the right of (after) the lens 50 cm to the right of (after) the lens 50 cm in front of (before) the lens

20) An object is placed 10 cm in front of (before) a lens with power +20 D. What type of image is it, where is it formed, and what is its magnification? a) b) c) d)

real, 10 cm in front of the lens, m = +1.0 real, 5 cm to the right of the lens, m = –1.0 real, 10 cm to the right of the lens, m = –1.0 virtual, 10 cm to the right of the lens, m = +1.0

21) Which diagram represents a proper ray-tracing rule implementation?

+10.0 D +20.0 D +30.0 D +40.0 D

16) An object located 25 cm before (in front of) a lens forms an image 25 cm after (to the right of) the lens. What is the focal length of this lens? a) b) c) d) e)

a) b) c) d) e)

4 m in front of (before) the lens 4 m to the right of (after) the lens 25 cm in front of (before) the lens 25 cm to the right of (after) the lens 40 cm to the right of (after) the lens 40 cm in front of (before) the lens

14) When an object is placed 20 cm in front of (before) a +10 D lens, what is the magnification? a) b) c) d) e) f)

18) An object located 10 cm before (in front of) a lens forms an image 16.6 cm before (to the left of) the lens. What is the lens power?

+5 cm +10 cm +12.5 cm +25 cm +50 cm

a)

A

b)

B

c)

C

d)

D

22) Which diagram represents a proper ray-tracing rule implementation?

17) An object located 20 cm before (in front of) a lens forms an image with a magnification of –1.0. What is the focal length of this lens? a) b) c) d)

+5 cm +10 cm +20 cm +100 cm

a)

A

b)

B

c)

D

d)

D 4-157

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23) Which diagram represents a proper ray-tracing rule implementation?

c) d) e) f)

real, 60 cm to the left of the SSRI real, 40 cm to the left of the SSRI real, 60 cm to the right of the SSRI real, 40 cm to the right of the SSRI

28) Referring to Q 27, the (linear) magnification in this case is …

a)

A

b) B

c)

C

d)

D

24) Which diagram represents a proper ray-tracing rule implementation?

a) b) c) d) e) f)

m = –1.5 m = –1.0 m = –0.66 m = +1.5 m = +1.0 m = +0.66

29) An object situated 50 cm to the left of an air (n =1.0) – glass (n΄ =1.5) convex SSRI forms an image 50 cm to the right of the SSRI. The power of this SSRI is … a) b) c) d) e) f)

–6.0 D –5.0 D –4.0 D +6.0 D +5.0 D +4.0 D

30) Referring to Q 29, the (linear) magnification in this case is … a)

A

b)

B

c)

C

d)

D

25) An object is placed in front of a plus lens. Its location from the lens is greater than twice the lens focal length, in absolute terms (i.e., ignoring + or – signs). The image produced is … a) b) c) d) e)

real, inverted, and minified real, inverted, and magnified virtual, upright, and magnified virtual, upright, and minified virtual, inverted, and magnified

26) An object is placed in front (to the left) of a plus lens at a distance exactly equal to the focal length, in absolute terms (i.e., ignoring + or – signs). Where is the image? a) b) c) d)

at the secondary focal point at twice the focal distance, behind the lens at twice the focal distance, in front of the lens at optical infinity

27) A point object is placed 40 cm to the left of an air (n =1.0) – glass (n΄ =1.5) convex SSRI with +5.0 D power. The image type and location are … a) b) 4-158

virtual, 60 cm to the right of the SSRI virtual, 40 cm to the right of the SSRI

a) b) c) d) e) f)

m = –1.5 m = –1.0 m = –0.66 m = +1.5 m = +1.0 m = +0.66

31) An object is placed 10 cm to the left (the air side) of an air (n =1.0) – glass (n΄ =1.5) concave SSRI with –5.00 D power. Where is the image being formed, what type is it, and what is the (linear) magnification? a) b) c) d) e) f)

x΄ = –15 cm, virtual, m = +1.5 x΄ = –10 cm, virtual, m = +1.0 x΄ = –10 cm, virtual, m = +0.66 x΄ = –10 cm, real, m = –1.5 x΄ = +15 cm, real, m = +1.0 x΄ = +15 cm, real, m = –0.66

32) An insect is trapped in a large amber cube that forms a flat amber (n =1.5) – air (n΄ =1.0) SSRI. We observe its image appearing from 6.6 cm inside the cube. What is its actual (object) location inside the amber cube? a) b)

3.3 cm 6.6 cm

IMAGING WITH LENSES

c) d)

10.0 cm 13.3 cm

33) Referring to Q 32, for the sake of geometry, let us reduce the object to a point. What is the vergence of the rays entering (L) the amber–air interface from an object ‘point’ representing this insect, and what is the vergence of the rays leaving (L΄) that interface? a) b) c) d) e) f)

L = –6.66 D, L΄ = –10.0 D L = –10.0 D, L΄ = –10.0 D L = –15.0 D, L΄ = –10.0 D L = –15.0 D, L΄ = –15.0 D L = –22.5 D, L΄ = –15.0 D L = –22.5 D, L΄ = –22.5 D

37) Are you still here? OK, you are not quitting… (you must be enjoying this). This image is ___________ and its linear magnification is ____________. a) b) c) d) e) f)

real virtual real virtual real virtual

m = +1.5 m = +1.5 m = +1.0 m = +1.0 m = –1.5 m = –1.0

38) Yes, you must still be here. We cannot believe that this image is formed at the exact same location as the object, yet the image is greater in (linear) size. Which one of the following ray tracings is correct?

34) Ah, that insect, again. Now the piece of amber is a spherical (n =1.5) – air (n΄ =1.0) SSRI with a radius of curvature of –0.1 m, as shown in the figure. What is the power of this SSRI?

a) b) c) d) e) f)

–15.0 D –10.0 D –5.0 D +5.0 D +10.0 D +15.0 D

35) Let’s stay with this SSRI. We just calculated its power (hopefully, correctly). The SSRI has two focal lengths, the primary f (object space) and the secondary f΄ (image space). Their lengths are … a) b) c) d) e) f)

primary –30 cm, secondary +20 cm primary –30 cm, secondary +30 cm primary –20 cm, secondary +30 cm primary +20 cm, secondary +30 cm primary +20 cm, secondary +20 cm primary +30 cm, secondary +30 cm

36) The insect (yes, that, again), is still situated (physically) 10 cm inside the spherical amber SSRI (x = –10 cm). If the SSRI has +5.0 D power, the image formed by this SSRI is situated at (x΄ = ?) … a) b) c) d) e) f)

–30 cm –20 cm –10 cm +10 cm +20 cm +30 cm

a)

A

b)

B

c)

C

d)

D

39) In our diving experiment, we submerge into seawater (n = 1.33) a glass lens (n = 1.5), whose power is +20.0 D when surrounded by air. What is the focal length of this lens when submerged? a) b) c) d)

5 cm 10 cm 15 cm 20 cm

40) While it is still underwater, we use this lens (recall, power in air = +20.0 D) to image a hippocampus that is buoying just 10 cm in front (x = –10 cm) of the lens. Where is the image (x΄), and what is the magnification (m)? a) b) c) d) e) f)

x΄ = –20 cm, m = +2.0 x΄ = –20 cm, m = +1.0 x΄ = –20 cm, m = –1.0 x΄ = +10 cm, m = –1.0 x΄ = +10 cm, m = +2.0 x΄ = +10 cm, m = –2.0

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GEOMETRICAL OPTICS

41) What is a hippocampus? a) b) c) d) e)

a hippopotamus a seahorse a unicorn a hyperopic field Hippocrates

42) A glass (n=1.5) symmetric biconcave lens, whose surfaces fit on the edges of a 10 cm diameter circle, produces a 3× erect image (m = +3.0). What is the type and location of the object? a) b) c) d) e)

44) In the photograph below, multiple images are being formed from an object, a plane on its final approach. The images are formed in rain droplets on sunroof glass while the photographer is in the comfort of his car. What type of images are they? What is the magnification? (Hint: Start by determining if the image is erect or inverted.)

real, placed 10 cm to the left of the lens real, placed 3.33 cm to the left of the lens virtual, placed 10 cm to the right of the lens virtual, placed 3.33 cm to the left of the lens virtual, placed 3.33 cm to the right of the lens

43) A virtual object is on a blind date with a flat refractive interface. What is the outcome? a) b)

c) d) e)

social media status remains complicated depending on the effect the surface has on the incident vergence, it can be a real or a virtual image definitely a virtual image definitely a real image mission impossible! Virtual objects won’t date flat refractive interfaces!

Photograph of a Lufthansa Airlines Airbus A 321 by Konstantin von Wedelstaedt, used with permission. a) b) c) d)

the image is real and inverted; the magnification is less than 1.0 (negative) the image is virtual and real; the magnification is less than 1.0 (positive) the image is real and inverted; the magnification is less than 1.0 (positive) the image is virtual and inverted; the magnification is less than 1.0 (negative)

45) Back to this plane photograph (danke, Konstantin!). What type of lens can produce such an image? a) b) c) d) e)

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a biconvex lens of water a planoconvex lens of water, convex side up a planoconvex lens of water, convex side down a planoconcave lens of glass, concave side up a biconcave lens of water

IMAGING WITH LENSES

4.10 LENS IMAGING SUMMARY Lens Imaging Relationship In lens imaging we employ (assuming propagation in air) the following relationships: •

For an object placed at location x, the object vergence is L = 1/x.

•

For an image formed at location x΄, the image vergence is L΄ = 1/x΄.

•

For a lens or mirror with focal length f΄, the optical power is F = 1/f΄.

The imaging relationship states that the object vergence is added to the power of the imaging element to produce the image vergence: Vergence and Optical Power:

+

L object vergence

1 x

Object and Image Locations:

F

=

optical power

1 f΄

+

object location

L΄ image vergence

1 x΄

=

focal length

image location

The above assumes that the optical medium surrounding the lens (imaging element, in general) is air. If the refractive index in object space is n and in image space is n΄, the relationship takes the more generally applicable form: n x object location

Linear Magnification (using location):

n΄ f΄

+

focal length

m

=

n΄ x΄ image location

h΄ x΄ = h x

We use this relationship if the imaging is taking place in air, or if the medium surrounding the lens is the same on both sides of the lens. Otherwise, we use the more generally applicable relationship: Linear Magnification (using vergence):

m 

h΄ nx΄ n/ x L = = = h n΄x n΄ / x΄ L΄

Lens Imaging Ray-Tracing Diagrams In a plus-powered lens, 1: The ray parallel to the optical axis refracts at the lens to the secondary focal point. 2: The ray originating from (or crossing through) the primary focal point refracts at the lens to become parallel to the optical axis. 4-161

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3: The ray targeting the center of the lens refracts without any ray deviation.

Figure 4-45: Summary of ray-tracing rules in a plus lens.

In a minus-powered lens, 1: A ray parallel to the optical axis refracts at the lens as if it originated from the secondary focal point. 2: A ray originating from (or targeting) the primary focal point refracts at the lens to become parallel to the optical axis. 3: A ray targeting the center of the lens refracts without any ray deviation.

Figure 4-46: Summary of ray-tracing rules in a minus lens.

We summarize lens-imaging image formation configurations by grouping the plus lens and the minus lens variations that include a real object at various locations (including the object at the primary focal point) and a virtual object (Figure 4-47). •

There are three distinct outcomes with a plus lens and a real object: The image can be real and inverted (left column, top three cases), formed at infinity, and virtual and magnified. When the object is virtual, a plus lens can only produce a real, minified, and erect image.

•

There is only one outcome with a minus lens and a real object: The image can be virtual, minified, and erect (right column, last case). However, there are several outcomes with a minus lens and a virtual object: The image can be virtual and inverted (right column, top three cases), formed at infinity, or real and magnified.

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Figure 4-47: Summary of the image formation configurations in lens imaging. [Left column: Positive lenses, top (1) to bottom (6)] 1, 2, and 3: Real object, real image. 4: (Real) object at the primary focal point, image at optical infinity. 5: Real object, virtual image. 6: Virtual object, real image. [Right column: Negative lenses, top (1) to bottom (6)] 1, 2, and 3: Virtual object, virtual image. 4: (Virtual) object at the primary focal point, image at optical infinity. 5: Virtual object, real image. 6: Real object, virtual image. (Compare to Figure 5-63.) Note: For simplicity, the rays shown correspond to two of the three construction rays (rules 1 and 3). 4-163

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5 IMAGING WITH MIRRORS

5.1 PLANE MIRROR PRINCIPLES Consider an object placed in front of a flat, plane mirror. We follow any two rays from a given point on the object. Naturally, the rays are diverging, for the object is real. These rays then reflect off of the mirror, and they are still diverging—they do not intersect. However, their extrapolations meet at a point behind the mirror (Figure 5-1). The rays leaving the reflecting surface on their way to form the image appear to originate from that point.

Figure 5-1: When the object distance from the mirror increases, the corresponding image distance from the mirror increases. The image is always erect.

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For any object point placed at x to the left of the mirror, there is an image point formed at x΄, which is an equal distance to the right the mirror. The image separation from the mirror equals the object separation from the mirror, and this is valid for every object–image pair. When looking at a plane mirror, the world is not upside down; images maintain an up– down, vertical orientation. Thus, if the object is oriented upward, the image is also oriented upward. These images are upright or erect. However, we cannot read a line of text through a mirror. Image parity changes in reflection are associated with an apparent left–right reversal; this symmetry is known as enantiomorphism. We stress, however, that this ‘left–right’ reversal is misleading. In reality, what happens is just a front–back reversal. Parity is determined by looking back against the propagation direction toward the object or image in that space.

Figure 5-2: The image resulting from mirror reflection is erect and has the same size as the object, but has enantiomorphic symmetry. What is reversed is the front–back (red line) orientation.

Figure 5-3: The stereotypical mistake made with mirror reflection is due to the fact that we compare the object (to the image) as if the object appears reversed with respect to the image. The object, however, is not reversed! The image fomed by a plane mirror:

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• is virtual (formed behind the mirror), • has a size equal to that of the object (magnification = 1.0), and • maintains the same vertical orientation (erect +).

IMAGING WITH MIRRORS

The fact that the image is erect, in combination with the fact that, in reality, the ray bundle vergence does not change—just the path changes—results in the image always having the same size as the object, regardless of the object or observer locations. In a plane mirror, the magnification always equals +1.0. In general, just as in lenses [see Eq. (4.5)], the relationships expressing linear magnification in mirrors are Mirror Magnification (location): Mirror Magnification (vergence):

m

h΄ x΄ = h x

(5.1)

L L΄

(5.2)

m=

5.1.1 The Cartesian Convention in Mirrors Here is an important note. All mirror imaging relationships are the same as their counterparts in lens imaging. The Cartesian sign convention (§ 3.2) is followed; here, the difference is that, because of reflection, the directional distances that apply to image space are now associated with a reflected wave, which has a reversed direction of propagation with respect to the initial wave that is incident to the mirror and is associated with object space.

Figure 5-4: The Cartesian sign convention in mirrors.

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For example, the distance to the image location x΄ should be considered along the direction of light after reflection. Thus, for an image formed to the left (in front) of the mirror, the distance x΄ is positive (the image is real), while for an image formed to the right of the mirror (the image is virtual), the distance x΄ is negative. This is simply because light changes direction upon reflection: The positive algebraic sign for the image location applies to images located in front (to the left) of the mirror. This also applies to the radius of curvature and the focal length (as shown in § 5.2). Recall that, just as in lenses, in mirrors, the focal length is directionally drawn to the secondary focal point, which is situated in image space. Thus, in spherical mirrors, the focal length is positive if it points to the left (concave mirrors) and negative if it points to the right (convex mirrors).

5.1.2

Multiple Plane Mirror Surfaces

With two successive reflections (or generally, with an even number of reflections), the right–left appearance is restored to normal. An image seen by an even number of reflections maintains its parity so we have a congruent reversal. The image now is rotated by 180°. An odd number of reflections changes the parity.

Figure 5-5: Congruent image reversal from a double reflection.

Of course, with two mirrors there is more than one image! The number of images depends on the angle between the mirrors. The simplest case is when the mirrors are perpendicular, forming a 90° angle (right-angle mirrors). The additional image (#3 in Figure 5-6) is due to successive beam reflections off of both mirrors. This can be explained if we consider that each mirror is mirrored on the other mirror. It is as if we have two infinitely extending (to the virtual space behind the real mirrors) perpendicular mirrors. These, too, function as mirrors. Image 1 is formed by mirror 1, which now functions as an object for the ‘apparent mirror’ 2 forming image 3, which is also the image of image 2 for the apparent mirror 1. Thus, from two mirrors forming an angle of 90°, three reflection images are produced.

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Figure 5-6: Multiple images from two perpendicular (right-angle) mirrors.

Figure 5-7: (left) Apparent right–left reversal in simple reflection. (right) Four images from two perpendicular mirrors.

The number of these secondary images increases for a smaller angle between the reflecting surfaces. With this angle at ϑ = 60°, five of such images are formed, whereas with the angle at 45°, seven images are formed. The following simple relationship applies: Number of Images from an Angled Mirror:

N =

360o

  o 

−1

(5.3)

which provides integer results if the angle ϑ between the two mirrors is a divider of 360, such as 90°, 60°, 45°, 40°, and so on. Therefore, with a 2° angle, we expect 179 images (good luck counting!).

Figure 5-8: Multiple images from two mirrors at angles of 60° (left) and 45° (right).

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5.2 SPHERICAL MIRRORS 5.2.1 Geometry of a Spherical Mirror A mirror can have any shape. In practice, however, we study reflecting surfaces that are rotationally symmetrical with respect to an axis, the (principal) optical axis. The point where the optical axis intersects the reflecting surface is the vertex point V. The simplest reflecting surface other than flat, is a spherical surface. A spherical surface is similar to the single spherical refracting interface (§ 1.2) but instead of refraction, there is reflection! A spherical mirror can be part of—regardless of the size of the part—an ideal spherical surface, like a fragment of a Christmas ball ornament.

Figure 5-9: Rotationally symmetrical surfaces: parabola, ellipse, and circle. The parabolic shape (parabola) results in stigmatic imaging, the formation of a point in image space from a distant object (§ 2.2.3.). The spherical surface (circulus) is considered an approximation of a parabola. (Illustration reprinted from Athanasius Kircher’s 1645 massive treatise Ars Magna Lucis et Umbrae.)

The similarities to the spherical refracting surfaces also include the fact that the sphere center is the center of curvature, and its radius is the radius of curvature r. The reciprocal of the radius of curvature is the curvature C. Every line passing through the center of curvature (other than the principal optical axis) is a secondary optical axis.

Figure 5-10: The two types of spherical reflecting surfaces: convex (left) and concave (right). 5-170

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The exterior surface of a sphere (a bowl or a spoon, for example) forms a convex mirror, and its interior surface forms a concave mirror. We can consider a convex mirror as wrapping around ‘darkness,’ and a concave mirror as wrapping around ‘light.’

Figure 5-11: The interior (left) and the exterior (right) surface of a spherical bowl can form a concave and a convex mirror, respectively.

Figure 5-12: (left) The lighting of the Olympic flame (in a concave mirror). (right) “…one giant leap for mankind.” Neil Armstrong is reflected in Buzz Aldrin's faceplate (a convex mirror) during the 20 July 1969 Apollo 11 mission. (Left photo used with permission from the Hellenic Olympic Committee; right photo credit: NASA.)

The larger the size of the sphere, the larger the radius of curvature, and the smaller the curvature. A flat mirror has an infinite radius of curvature and a zero curvature.

Figure 5-13: Vertex V, center of curvature, and radius of curvature r in spherical mirrors.

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In a spherical mirror, r = 1/C, and, inversely, C = 1/r. The radius of curvature (and therefore the curvature) has an algebraic sign. The sign follows the Cartesian sign convention (presented in § 5.1.1). Attention must be drawn to the fact that, because of reflection, light reverses direction. Therefore, the directional distance for the radius of curvature, focal length, and image location is positive to the left and negative to the right. In Figure 5-13, the convex mirror has a radius of curvature that has a negative value and the concave mirror has a radius of curvature that is positive.

Figure 5-14: Satellite dishes are concave mirrors operating in the microwave part of the electromagnetic spectrum. In mirrors, light reverses direction after reflection. Therefore, any directional distance associated with reflected light (focal length, radius of curvature, or image location) is: * Positive, if it is in front (to the left) of the mirror, and * Negative, if it is after (to the right of) the mirror. The object location is unaffected, as it is associated with the initial light propagating to the right. The object location is: • Negative, if the object is in front (to the left) of the mirror, and • Positive, if the object is after (to the right of) the mirror.

Figure 5-15: Identification of reference geometrical aspects in a convex mirror.

A spherical geometry (circular in two dimensions) can be used to determine the ray propagation after reflection. The optical axis serves as the horizontal axis and is perpendicular to the reflecting surface at the vertex point [Figure 5-15 (left)]. The key to understanding 5-172

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spherical mirror geometry is knowing that any line segment that is normal to the surface lies along a radius and therefore passes through the center of curvature [Figure 5-15 (right)]. We now consider a ray reflected off of a convex mirror. This ray is incident on the mirror from the left and propagates parallel to the optical axis [Figure 5-16 (left)]. We draw the normal to the surface at the point of incidence. For a spherical surface, this line projects to the center of curvature and is, in other words, a radius of that circle. Because the ray is initially parallel to the optical axis, if the angle of incidence is ϑi, the normal to the surface at the point of incidence forms exactly the same angle as the angle formed by the optical axis at the center of curvature (these are alternate angles).

Figure 5-16: Reflection of a ray propagating parallel to the optical axis off of a convex mirror. (left) Incident ray and (right) reflected ray.

We then apply the law of reflection, which simply states that the angle of reflection ϑr equals the angle of incidence ϑi. We note in Figure 5-16 (right) that the extrapolation of the reflected ray forms an equal angle (opposing angles) with the normal to the surface. Then an isosceles triangle is formed, its sides being r/2.

Figure 5-17: Relationship between the radius of curvature and focal length in spherical mirrors.

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5.2.2

Focal Points and Focal Lengths in a Spherical Mirror

The above considerations suggest that the reflected ray appears to originate from a point on the optical axis that is situated at one-half of the radius of curvature r. This is true for any incident ray parallel to the optical axis. In other words, these paths repeat for all rays that were originally parallel to the optical axis, regardless of the ray height.21 Therefore, when a collimated pencil of rays is incident along the optical axis of a convex mirror, the reflected bundle diverges: All rays appear to originate from a unique on-axis point situated r/2 from the vertex [Figure 5-18 (left)]. Similarly, in a concave mirror [Figure 5-18 (right)], the reflected ray passes through a point at one-half of the radius of curvature. Therefore, when a collimated pencil of rays incident along the optical axis is reflected off of a concave mirror, the ray bundle converges: All rays pass through a point that is situated on the optical axis at r/2, or one-half of the radius of curvature.

Figure 5-18: Reflection and focal points. (left) Convex mirror and (right) concave mirror.

This unique point situated on the optical axis is the focal point of the mirror. Its distance from the vertex is the focal length f, which is one-half of the radius of curvature: Mirror Focal Length f:

f = r/2

(5.4)

Mirror Focal Length f :

always equals one-half of the radius of curvature, and

has the same algebraic sign as the radius of curvature,

is positive to the left and negative to the right.

Only within the limits of the paraxial approximation. In other words, this statement is true for relatively small angles, or for rays that are relatively close to the optical axis (small height). 21

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Note

: How many focal points / lengths are in a mirror? One or two?

In fact, there are two focal points and, consequently, two focal lengths, in a spherical mirror. For imaging purposes, the focal point in Figure 5-19 serves as the secondary focal point F΄, just as in lens imaging. This is the image point of an axially collimated ray bundle incident on the mirror. Optical infinity and the focal point are therefore optical conjugates. There is also a primary focal point F, situated at exactly the same location. If an object (virtual for convex mirrors or real for concave mirrors) is to be placed at this point, reflection will form a collimated beam parallel to the optical axis. Again, the focal point and optical infinity are optical conjugates. This is easily demonstrated by simply flipping the arrowheads in Figure 5-19. Because these two mirror focal points always coincide, they are simply called ‘the’ focal point. For the same reason, f and f΄ can be used interchangeably and … (almost without guilt) can be referred to as ‘the’ focal length, even dropping the prime!

Figure 5-19: A collimated beam propagating along the optical axis (left) diverges in a convex mirror (as if it originated from the focal point) and (right) converges to the focal point in a concave mirror.

In convex mirrors, the focal point is situated in the space past the mirror, and in concave mirrors, the focal point is situated in front of the mirror. The focal length, as well as the radius of curvature, is always drawn from the vertex toward the center of the spherical surface. Because of reflection, however, light changes direction, and we must be careful with the algebraic sign. The focal length has a negative value for a convex mirror and a positive value for a concave mirror (Figure 5-19).

5.2.3

Nodal Point in a Spherical Mirror

Consider a ray that is incident on a spherical mirror such that it is perpendicular to the mirror surface. Geometrically, this means that the ray is directed to (in a convex mirror), or it originates from (in a concave mirror), the center of curvature of the mirror. The angle of incidence is zero, since the ray is normal to the surface (a projection of a spoke); therefore, it forms an angle of 5-175

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zero degrees (ϑi = 0°) with the normal. As a result, the reflected ray also forms a zero degree angle (ϑr = 0°), which means that the ray is undeviated in its path following refraction (just reversing direction).

Figure 5-20: A ray incident on a spherical mirror with 0° angle of incidence (normal incidence) continues to propagate undeviated; either the ray or its extrapolation cross the nodal point, which is the center of curvature.

For this reason, the center of curvature in a spherical mirror is its nodal point. In general, the nodal point N is a point to which a ray is directed; the reflected ray appears as if it originates from the nodal point N΄ and continues to propagate undeviated. We realize that the two mirror nodal points N and N΄ are identical—the only difference is their function. Nodal point N relates to object space, associated with the ray incident on the mirror, while nodal point

N΄ relates to image space, associated with the ray reflected off of the mirror. We will further discuss nodal points as a concept in thick lenses and lens systems (§ 6.2.2).

5.2.4

Optical Power in a Spherical Mirror

The shorter the focal length, the more the beam converges or diverges. This property of a collimated beam converging from an optical element (as in an SSRI § 1.2.3 and a lens § 2.4) is the reflecting or mirror optical power F, reported in diopters (D), which is the reciprocal of a meter (m–1). In air, the optical power is simply the reciprocal of the focal length f. Thus, for all spherical mirrors (either convex or concave), the optical power is Mirror Optical Power (in air):

Fmirror = 1/f = 2/r

(5.5)

Example ☞: A plane (flat) mirror has ∞ focal length and zero optical power F = 1/(∞) = 0 D. Example ☞: A mirror with focal length f = −25 cm is convex. Its power is F = 1/(−0.25 m) = −4.0 D. Example ☞: A mirror with power F = +10 D is concave. It has focal length f = 1/(+10 D) = +0.1 m = +10 cm.

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Mirror Optical Power F:

Plane mirror, F is zero (collimated beam remains collimated)

Concave mirror, F is positive (collimated beam converges)

Convex mirror, F is negative (collimated beam diverges)

If the space in front of the mirror is filled with a medium with refractive index n instead of air, the mirror optical power is expressed as Mirror Optical Power (in medium n):

Fmirror = n/f = 2n/r

(5.6)

Figure 5-21: (left) Negative optical power (F = −5.0 D) in a convex mirror and (right) positive optical power (F = +20.0 D) in a concave mirror.

Focal length and radius of curvature are positive in a concave mirror and negative in a convex mirror. In both mirror types, Image location is positive if the image is formed before the mirror (real image) and negative if it is formed after the mirror (virtual image). Object location is negative if the object is placed before the mirror (real object) and positive if it is to be formed after the mirror (virtual object).

Figure 5-22: Algebraic signs and locations for a real object / image and a virtual object / image in mirror imaging.

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5.2.5 Vergence and Spherical Mirror Power We can use vergence arguments in order to offer another proof for Eq. (5.6): Mirror Optical Power (in medium with refractive index n): Fmirror = 2n/r where n is the refractive index in the medium preceding the mirror surface, and r is the radius of curvature of the surface, expressed in meters. It follows that the power is expressed in diopters. To find the mirror power, we specify that the vergence leaving the surface after reflection

L΄ is the sum of the power F and the vergence L reaching the surface (before reflection): Vergence before reflection (object) + Mirror Power = Vergence after reflection (image):

L + F = L΄ ⇒

F = L΄ – L

We consider a beam that is directed toward the center of curvature of a convex mirror. This means that all rays intersect the surface at normal incidence. Then both the angle of incidence ϑi and the angle of reflection ϑr equal 0° for all rays in this configuration. The rays therefore (would) converge to a point that is the center of curvature, also serving as the nodal point for this spherical interface (further discussed in § 5.2.3).

Figure 5-23: A beam incident on a convex spherical mirror and aiming at the center of curvature is subject to normal incidence, so the reflected rays follow the same path.

The refractive index in the space before the mirror is n. Regarding the incident rays, the ‘distance to convergence’ r is the mirror radius of curvature. The vergence is positive because the rays are converging. Because r is negative—according to the ‘mind the sign’ convention— the distance to convergence is –r, and the vergence for the rays incident on the surface is expressed as Vergence entering:

L = n/(–r)

which is positive because the radius of curvature in this convex mirror is negative.

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Immediately after reflection, the distance to convergence is the same, again equal to r, but this time with no change in sign, since the rays are now diverging: Vergence leaving:

L΄ = n/(r)

which is negative, since the radius of curvature in this convex mirror is negative. Therefore, F = L΄ – L ⇒ F = n/(r) – n/(–r) = +2n/(r), which is the mirror optical power. Consider a beam directed toward the center of curvature of a concave mirror (Figure 5-24). The rays first pass through the mirror’s center of curvature and then intersect the mirror surface at, again, normal incidence. The refractive index has value n. For the incident rays, the ‘distance to convergence’ r is the radius of curvature of the mirror. The vergence is negative because the rays are diverging. Because r is positive—according to the ‘mind the sign’ convention—the distance to convergence is –r, and the vergence for the rays incident on the surface is Vergence entering:

L = n/(–r) (negative)

Immediately after reflection, the distance to convergence is the same, again, equal to r, and, again, has no change in sign, since the rays are converging (r is positive in concave mirrors): Vergence leaving:

L΄ = n/(r)

(positive)

Therefore, F = L΄ – L ⇒ n/(r) – n/(–r) = +2n/(r), which is, again, the mirror optical power. The mirror power is positive, since the radius of curvature in this concave mirror is positive. In either a convex or concave mirror, vergence considerations provide a proof for the mirror–image relationship.

Figure 5-24: A beam incident on a concave spherical mirror and aiming at the center of curvature (left) is subject to normal incidence, so the reflected rays (right) follow the same path.

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5.3 IMAGING WITH A SPHERICAL MIRROR The ability of reflecting surfaces to transform a reflected beam is the reason that an image is formed. A simple case is shown in Figure 5-25, where a ray bundle from far away (optical infinity), collimated along the optical axis, is incident on a convex mirror. The vergence of the ray bundle leaving the mirror (image) is simply the sum of the incident object vergence and the mirror optical power. Specifically, • The collimated object beam vergence incident on the mirror is L = 0 D. • The convex mirror optical power is F = −10 D. Their sum is −10 D. • The image vergence leaving the mirror is L΄ = −10 D.

Figure 5-25: Imaging for a beam coming from optical infinity (object vergence = 0). Focal point F and optical infinity are optical conjugates.

5.3.1

Spherical Mirror Imaging Relationship

This addition rule is valid, in general, for any object location, not just infinity. For object location

x and image location x΄, image vergence L΄ = 1/x΄ is produced by adding the object vergence L = 1/x to the mirror optical power F. Hence, the mirror imaging relationship has the following expressions: Vergence and Optical Power:

L

+

object vergence

Object and Image Locations:

1 x object location

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F

=

optical power

+

1 f΄ focal length

L΄

(5.7)

image vergence

=

1 x΄ image location

(5.8)

IMAGING WITH MIRRORS

The above relationships are identical to those in lens imaging [Eqs. (4.1) and (4.2)] and follow the Cartesian sign convention (§ 5.1.1). Once again, we must be mindful that, because of reflection off of the mirror, the direction of light propagation is reversed. This implies that the image distance and focal length have positive values if they are directed to the left, and negative values if they are directed to the right. Object distance, however, is not affected, since it is associated with the incident light, not the reflected light. This simple formula is valid for every object location [i.e., in front of the mirror (real object) or after the mirror (virtual object)] and can be used in both concave and convex mirrors, and even for plane mirrors. For a plane mirror, we simply use zero optical power.

Figure 5-26: Optical power and vergence addition principle applied to mirror imaging. This principle is applicable to all mirrors: plane, convex, and concave.

Imaging relationship for reflecting surfaces Expression with optical power:

Expression with object–image location and focal length:

Object vergence + Mirror power = Image vergence

1/x +1/f΄ = 1/x΄ Object placed to the left, x is negative. Image formed to the left, x΄ is positive. In a convex mirror, f΄ is negative. In a concave mirror, f΄ is positive.

Note

Valid for both convex and concave surfaces

: While we assume that the medium in front of the mirror is air, this is not always the case!

The above expressions are valid in the (typical) case where the medium in which light propagates is air, so the value of the refractive index n is 1.0. If this is not the case, then for either the vergence or the power we use the expressions L = n / x, L΄ = n / x΄, and F = 2n / r = n / f΄, respectively, where n is the refractive index of both object space and image space, as is often the case in mirrors.

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5.3.2

Convex Mirror Imaging Examples

The object beam in Figure 5-25 is collimated along the optical axis, which means that the object is located at optical infinity. A more realistic case is an object (a real object, that is) placed at a specific point in front of a convex mirror. In Figure 5-27, the object is placed 25 cm in front of the mirror, which means that

x = −0.25 m. The vergence originating from this object point is L = 1/x = −4.0 D. The convex mirror has focal length −0.1 m and optical power F = −10.0 D. We apply the imaging Eq. (5.7): Image Vergence: L + F = L΄ ⇒ L΄ = (–4.0 D) + (–10.0 D) = –14.0 D. Image Location: x΄ = 1/L΄= 1/(–14.0 D) = –0.071 m = –7.1 cm.

Figure 5-27: Imaging for an object originating from a point located 25 cm in front of the mirror.

The negative result for image location x΄ means that the image is 7.1 cm after (to the right of) the mirror, opposite to the direction of light propagation after reflection. This space, of course, is in complete ‘darkness.’ It only exists in the visual perception of the observer (that’s us, folks!). The image is virtual because the image-forming rays are diverging. It is perceived to originate from point x΄, where the image ray extrapolations intersect when traced backward. Power Magnification Imaging Relationship Figure 5-28: The three important mirror imaging relationships. Mind that the algebraic sign for image location and focal length are reversed in relation to the initial light direction.

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5.3.3

Ray Diagrams for Convex Mirrors

Alternatively, to find the image location and size, we can apply three simple ray diagram rules similarly to the way we applied lens ray diagram rules (presented in § 4.3). For a convex reflecting surface, the ray-tracing rules are as follows:

A ray parallel to the principal optical axis (parallel ray) is reflected as if it originated from the focal point. A ray directed at (targeting) the focal point (focal ray) becomes parallel to the principal optical axis upon reflection. A ray directed at the center of curvature of the mirror (radial or nodal ray) is retroreflected (simply reverses direction).

Figure 5-29: Ray-tracing diagrams for a convex mirror.

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Example ☞: An object is placed 20 cm in front of a convex mirror (F = −10 D, f = −10 cm). Find the image location, type, and magnification.

Figure 5-30: Image location in a convex mirror determined using ray diagrams.

✔ Schematic solution: We place the object 20 cm before (in front of) the mirror. We apply the first and the third raytracing rules for a pair of rays that originate from the object. Ray ❶ is parallel to the optical axis and is reflected at the mirror, as if it originated from focal point F. Ray ❸ is directed toward the center of curvature of the mirror and is retro-reflected, reversing direction. We observe that the rays diverge after the lens. The image is virtual, since it is produced by diverging rays. Its location is ⅔ of the focal length, and its magnification is less than 1.0 (= +⅓). ✔ Numerical solution: The object location is x = −20 cm = −0.2 m. The object vergence is L = 1 / −0.2 m = −5.0 D. The mirror power is F = −10.0 D. We now apply the imaging relationship given in Eq. (5.7): Image vergence: L + F = L΄ ⇒ L΄ = (–5.0 D) + (–10.0 D) = –15.0 D. Image location: x΄ = 1/L΄= 1/(–15.0 D) = –0.066 m = –6.6 cm. The negative sign indicates that the image location is in the direction opposite to the direction of the light after reflection. Therefore, the image is formed to the right of the mirror. The value of 6.66 cm suggests that the image is at a shorter (in absolute value) distance than the object (20 cm), which means that the magnification is less than 1.0. We apply Eq. (5.1): Magnification:

m =

x΄ −6.66 cm 1 = = + x −20 cm 3

The image is erect because the magnification is positive. These results are in agreement with those derived with the schematic solution. 5-184

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Figure 5-31: The example in Figure 5-30 with the image location now determined by applying optical power and vergence addition. Example ☞: An object is placed 5 cm in front of a convex mirror (F = −10 D, f = −10 cm). Find the image location, type, and magnification.

Figure 5-32: Image location in a convex mirror determined using ray diagrams.

✔ Schematic solution: We place the object 5 cm before (in front of) the mirror. We apply the first and the third raytracing rules. Ray ❶ is parallel to the optical axis and is reflected at the mirror, as if it originated from focal point F. Ray ❸ is directed at the center of curvature of the mirror and is retroreflected, reversing direction. The image (virtual), as indicated in Figure 5-33, appears to be located past the mirror at ⅓ of the focal length, and its magnification is less than 1.0 (= ⅔).

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Figure 5-33: The example in Figure 5-32 with the image location now determined by applying optical power and vergence addition.

We observe in both Figure 5-31 and Figure 5-33 that the image is virtual, erect, and minified. This is the generalized imaging outcome of a convex mirror with an object placed in front of it. Because the object vergence and mirror power are both negative, the image vergence is negative and numerically larger than the object vergence. Thus, a real object forms an image that is always located to the right of a convex mirror and is virtual, erect, and minified. The farther away the object is from the mirror, the more the image approaches the focal point and the smaller the image appears. This is because, for a very distant object (|x| is very large), the vergence (L = 1/x) becomes small and tends to zero. Thus, the value of the image vergence L΄ tends to equal the mirror optical power.

Figure 5-34: A (real) object placed at any location in front of a convex mirror results in an image that is always minified, erect, and virtual.

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Example ☞: An object is placed in front of a convex mirror at one-half of the focal length. Find the image location, type, and magnification.

This is a generalized version of the example discussed previously (5 cm in front of a convex mirror, f = −10 cm). We are mindful that both the object location and the focal length are negative. Thus, x = f/2 ⇒ L = 1/x = 1/(f/2) = 2/f and F = 1/f: Image vergence: L + F = L΄  L΄ = L + F = Image location: x΄ =

2 1 3 + = + . f f f

1 1 f = = . 3 L΄ 3 f

f x΄ 2 = 3 = + . Magnification: m = f x 3 2

Figure 5-35: (left) The curvier (steeper) the convex mirror, the smaller the image. This may give the false impression that viewed images correspond to objects that are farther away than their actual location. (right) Optics professors, rise in protest: There are NO OBJECTS in the mirror! Only virtual images appearing through this convex mirror. A proper statement would read: “The objects relating to the images appearing in mirror are larger in size and closer in distance than their conjugate images.”

Image of a Real Object by a Convex Mirror

Is always virtual, formed in the space behind the mirror, and erect.

Is smaller than the object (m < 1.0).

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5.3.4

Concave Mirror Imaging Ray Diagrams and Examples

The main difference between a concave mirror and a convex mirror is that the focal point is situated in front of a concave mirror. The focal length is positive (for example, +10 cm), where the (+) indicates that the focus location is in front of the mirror. The optical power according to Eq. (5.5) is positive. In such a mirror, rays parallel to the optical axis converge to a point on the optical axis located at one-half of the radius of curvature r. Consider a collimated ray bundle that is propagating from far away and is incident on a concave mirror. Its vergence is zero: L = 0 D. The reflected vergence equals the mirror optical power. For example, in Figure 5-36, we have collimated beam vergence occurring before reflection L = 0 D, mirror optical power F = +10 D, and reflected (image) vergence L΄ = +10 D, which means that the reflected beam converges to focus at x΄ = +0.1 m.

Figure 5-36: Imaging from a concave mirror for an object at optical infinity.

Just as in convex mirrors, we can apply the mirror imaging relationship to calculate the image location in concave mirrors. Alternatively, to find the image location, we apply the raytracing rules, which in the case of concave reflecting surfaces are:

A ray parallel to the principal optical axis (parallel ray) is reflected toward the focal point.

A ray passing via the focal point (focal ray) becomes parallel to the optical axis upon reflection on the mirror.

A ray that crosses the mirror center of curvature (radial or nodal ray) is retro-reflected (simply reverses direction).

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Figure 5-37: Ray-tracing diagrams for a concave mirror.

A second difference between concave and convex mirrors is that the image from a concave mirror can be real: The image is formed at the point where the image-forming rays intersect following reflection. If the rays do not intersect, however, the image is virtual, and the image-forming rays appear to originate from a virtual point situated behind the mirror. Similar to positive lens imaging, concave mirror imaging has three distinct outcomes, depending on the object location in relation to the focal point. The object is, in all of these cases, real and placed to the left of the mirror. Recall that the focal length f in concave mirrors is positive, while the (real) object location x, placed in front of the mirror, is negative. ❶ The first case involves an object placed beyond the radius of curvature: |x |> 2f . Example ☞: An object is placed 40 cm in front of a concave mirror with focal length f = 10 cm.

✔ Schematic solution: Ray ❶, propagating parallel to the optical axis, is reflected toward the focal point. Ray ❷, passing through the focal point, is reflected parallel to the optical axis. Ray ❸ crosses the 5-189

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center of curvature and is retro-reflected. All rays intersect at the image point. The image is real because it is formed by intersecting (converging) rays, it has a negative ‘height’ (it is inverted with respect to the object), and its magnification is less than 1.0. It is located a distance (in front of the mirror) that is shorter than the object.

Figure 5-38: Determination of the image location in a concave mirror using ray diagrams. The object is real and located a long distance from the focal point.

✔ Numerical solution: The mirror power is positive: F = 1 / f = 1 / (+0.1 m) = +10 D. The object location is negative (in front of the mirror): x = −0.4 m; therefore, the object vergence is L = 1 / x = 1 / (−0.4 m)= −2.5 D. Image vergence: L + F = L΄ ⇒ L΄ = (–2.5 D) + (+10.0 D) = +7.5 D. Image location: x΄ = 1/L΄= 1/(+7.5 D) = +0.133 m = +13.3 cm.

Figure 5-39: The example in Figure 5-38 with the image location now determined by applying optical power and vergence addition.

The positive sign indicates that the image-forming rays are converging to a point along the direction of light after reflection. Therefore, the image is real, located in front (to the left) of the mirror. The value of 13.3 cm suggests that the image is at a shorter (in absolute value) distance than the object (20 cm), which means that its magnification is less than 1.0. We apply Eq. (5.1) to calculate the magnification: m = x΄/x = +13.3 cm/(−40 cm) = −⅓. 5-190

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❶a This case involves an object at a location exactly equaling the radius of curvature: |x | = 2f . Example ☞: An object is placed 20 cm in front of a concave mirror with focal length 10 cm.

Figure 5-40: Determination of the image location in a concave mirror using ray diagrams. The object is real and is placed at the mirror center of curvature.

✔ Schematic solution: We use ray-tracing rules ❶ and ❷. The image is real, since it is formed by intersecting (converging) rays, it has a negative ‘height’ (is inverted with respect to the object), and its magnification equals 1.0. It is located in front of the mirror, at a distance exactly equaling the radius of curvature (= 2× the focal length). ✔ Numerical solution: The mirror power is F = 1 / f = 1 / (+0.1m) = +10 D. The object location is negative (in front of the mirror): x = −0.2 m; therefore, the object vergence is L = 1 / x = −5.0 D. Image vergence: L + F = L΄ ⇒ L΄ = (–5.0 D) + (+10.0 D) = +5.0 D. Image location: x΄ = 1/L΄= 1/(+5.0 D) = +0.2 m = +20 cm. Magnification: m = x΄/x = +20 cm/(−20 cm) = −1.0. The generalized version of this example states that the object is at a distance equal to the mirror radius of curvature. We are mindful that, while the object location is negative, the focal length is positive. Thus, x = –2f ⇒ L = 1/x = 1/(−2f ) and F = 1 / f. Image vergence: Image location:

L + F = L΄  L΄ =

x΄ =

1 1 = = 2f 1 L΄ 2f

1 1 1 + = + −2f f 2f

Magnification: m =

x΄ 2f = = − 1.0 x −2 f

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Figure 5-41: (left) Imaging from a concave mirror for an object at |x| > 2f. The image is real, inverted, and smaller (|m| < 1) than the object. (right) The image and the object share the same location and have the same size (x = −2f , x΄= +2f, m = −1). ❷ The second case involves an object placed such that 2f >| x | > f ; i.e.,

x is between the radius

of curvature and the focal length. Example ☞: An object is placed 15 cm in front of a concave mirror with focal length 10 cm.

✔ Schematic solution: We use ray-tracing rules ❶ and ❷. The image is real, since it is formed by intersecting (converging) rays, it has a negative ‘height’ (inverted), and its magnification is larger than 1.0 (|m| > 1). It is located in front of the mirror, beyond the center of curvature (x΄ > r = 2 f ). ✔ Numerical solution: The mirror optical power is F = 1 / f = 1 / (+0.1 m) = +10.0 D. The object location is negative (in front of the mirror): x = −0.15 m; therefore, the object vergence is L = 1 / x = −6.66 D. Image vergence: L + F = L΄ ⇒ L΄ = (–6.66 D) + (+10.0 D) = +3.33 D. Image location: x΄ = 1/L΄= 1/(+3.33 D) = +0.3 m = +30 cm. Magnification: m = x΄/x = +30 cm/(−15 cm) = −2.0.

Figure 5-42: Determination of the image location in a concave mirror using ray diagrams. The object is real, being located between the center of curvature and the focal point.

If the object approaches the focal point (from the side of the center of curvature), the closer the object is to the focal point, the larger and the farther away the image is formed. 5-192

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❸ The third case involves an object that is even closer to the mirror, for which |x | < f . The

object is therefore ‘inside’ the focal point. Example ☞: An object is placed 5 cm in front of a concave mirror with focal length 10 cm.

✔ Schematic solution: Ray tracing (rules ❶ and ❸) indicates that the image is virtual, erect, and larger than the object.

Figure 5-43: Imaging from a concave mirror of an object placed at | x | < f. The image is virtual, erect, and larger than the object.

✔ Numerical solution: The mirror power is F = 1 / f = 1 / (+0.1 m) = +10.0 D. The object location is negative (in front of the mirror): x = −0.05 m; therefore, the object vergence is L = 1 / x = −20.0 D. Image vergence: L + F = L΄ ⇒ L΄ = (–20.0 D) + (+10.0 D) = –10.0 D (virtual image). Image location: x΄ = 1/L΄= 1/(–10.0 D) = –0.10 m = –10 cm. Magnification: m = x΄/x = (–10 cm)/(−5 cm) = +2.0.

Figure 5-44: The example in Figure 5-43 now managed with optical power and vergence addition.

If the object moves away from the focal point (but is still in front of the mirror), the closer the object is to the mirror, the closer to the mirror the virtual image is formed. This is the case 5-193

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of the vanity mirror. The generalized version of this example states that the object is at a distance equal to one-half of the focal length.

Example ☞: An object is placed in front of a concave mirror at location |x| = f/2. Find the image location and magnification.

We are mindful that, while the object location is negative, the focal length is positive (concave mirror), so |x| = f/2 means that x = −f/2. We apply the imaging relationship for x = −f/2 ⇒

L = 1/x = 1/(−f/2) = −2/f and F = 1/f :

 2 1 1 +  = − f  f f

Image vergence: L + F = L΄  L΄ =  − Image location:

x΄ =

1 1 = L΄ − 1

= −f

Magnification:

m=

f

x΄ −f = = + 2.0 − f x 2

It is noteworthy that we can find the exact results if we reverse the object–mirror–image arrangement. Now the object is placed in front of a convex mirror [Figure 5-45 (right)] at a location x = f (note that f is now a negative number). By applying ray-tracing diagrams and/or vergence and optical power summation rules, we find that the image is formed at the location

x΄ = f/2. To obtain this, we apply the imaging relationship for x = f ⇒ L = 1/x = 1/f and F = 1/f : Image vergence:

1 1 2 L + F = L΄  L΄ =   +   = f f f

1 1 f = = Image location: x΄ = 2 L΄ 2 f

x΄ = Magnification: m = x

f

2 = +1 f 2

Figure 5-45: The power and beauty of symmetry in the rules governing optics. The object becomes the image, and the image becomes the object!

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Figure 5-46: (left) When the (real) object is placed closer than the focal length, the (real) image is inverted and magnified. (right) When the object is farther than the focal length, the (real) image is inverted—this this is how ‘turning the world upside down’ got its name! (Left image © Anil Kapoor taken at Kensington Gardens, Westminster, UK; right image by E. N. Gall taken at Israel Museum, Jerusalem, from Wikimedia Commons under license CC BY-SA 4.0.)

Figure 5-47: The giant sculpture Cloud Gate in Millennium Park, Chicago, Illinois, affectionally known as "The Bean" because its legume-like shape forms (mostly) a convex mirror. Cloud Gate was created by the world-renowned sculptor Anish Mikhail Kapoor.

Concave Mirror, object at a distance longer than the focal length (|x| > f ): The image is real, inverted, and is located in front (to the left) of the mirror. The image and the object share the same location if |x| = 2f. Then |m| = 1.0.

Concave Mirror, object at a distance shorter than the focal length (|x| < f ):

The image is virtual, erect, and magnified (m >1), and is located behind (to the right of) the mirror.

If |x|> 2f , then |m| > 1.0 and if 2f >|x|> f, then |m| < 1.0.

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An interesting case involves an object placed at the focus, where |x| = f . This means that the object is placed exactly at the primary focal point of the mirror. Example ☞: An object is placed 10 cm in front of a concave mirror with focal length 10 cm.

✔ Schematic solution: We apply rules ❶ and ❸. The reflected rays form a collimated beam. The image is virtual, being formed at optical infinity (see also § 4.6.3). ✔ Numerical solution: The mirror optical power is F = 1 / f = 1 / (+0.1 m) = +10.0 D. The object location is negative (in front of the mirror): x = −0.10 m; therefore, the object vergence is L = 1 / x = −10.0 D. Image vergence: L + F = L΄ ⇒ L΄ = (–10.0 D) + (+10.0 D) = 0.0 D. Image location: x΄ = 1/L΄= 1/(0.0 D) = ∞.

Figure 5-48: Imaging from a concave mirror of an object at |x| = f. The image is formed at optical infinity.

We note that the image vergence is zero, so the image is formed at optical infinity. This is in agreement with the ray-tracing solution. The generalized version of this example states that the object is at a distance equal to the focal length.

5.3.5 The Virtual Object in Mirror Imaging The cases investigated so far share an object placed in front (to the left) of the mirror. This object is real, and its vergence is negative (Figure 5-49). A virtual object is ‘located’ after the mirror. This object can only be formed by a converging bundle of rays. Therefore, this object carries a positive vergence, and its location is positive, x > 0 (Figure 5-50). 5-196

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Figure 5-49: Real objects placed in front of spherical mirrors. The location and vergence are negative.

Figure 5-50: Virtual object in spherical mirrors. The object location and vergence are positive. Example ☞: An object is formed 20 cm after a flat mirror. Find the image location, type, and magnification.

✔ Schematic solution: The object location is 20 cm after the mirror, where it would be formed if the mirror was not present. The drawn rays obey the simple rules of reflection; i.e., the angle of reflection equals the angle of incidence. We therefore draw two rays directed at the same object point (object ray 1 and object ray 2, Figure 5-51). These rays are reflected and intersect in front of the mirror. The image is formed at the point where image ray 1 and image ray 2 now converge (yes, the image is real!). It is located 20 cm in front of the mirror.

Figure 5-51: Virtual object in a flat mirror. The image is real.

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✔ Numerical solution: The mirror optical power is F = 1 / f = 1 / (∞) = 0 D. The object location is positive (to the right of the mirror): x = +0.20 m; the object vergence is therefore L = 1 / x = +5.0 D. Image vergence: L + F = L΄ ⇒ L΄ = (+5.0 D) + (0.0 D) = +5.0 D. (The image is real.) Image location: x΄ = 1/L΄= 1/(+5.0 D) = +0.20 m = +20 cm. Magnification: m = x΄/x = (+20 cm)/(+20 cm) = +1.0. There are many similarities between the case of a virtual object in a flat mirror and a real object placed in front of a flat mirror (see also § 5.1). The virtual image and the real object are optical conjugates. Since the mirror optical power is zero, the image vergence (image distance) equals the object vergence (object distance), and the magnification equals +1.0. We now examine cases of virtual objects in spherical mirrors. Example ☞: An object is located 20 cm after a concave mirror with a focal length of 10 cm. Identify the image type and magnification.

✔ Schematic solution: To apply rule ❶, we draw a line parallel to the optical axis that would intersect an object point; this ray is reflected by the mirror toward focal point F. To apply rule ❸, we draw a line that joins the center of curvature and the same object point. These rays intersect at a point in front the mirror. The image is therefore real and is also erect, having the same orientation as the object.

Figure 5-52: Imaging of a virtual object from a concave mirror. The image is real, erect, and smaller than the object.

✔ Numerical solution: The mirror optical power is F = 1 / f = 1 / (+0.1 m) = +10.0 D. The object location is positive (after the mirror): x = +0.20; the object vergence is therefore L = 1 / x = +5.0 D.

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Image vergence: L + F = L΄ ⇒ L΄ = (+5.0 D) + (+10.0 D) = +15.0 D. Image location: x΄ = 1/L΄= 1/(+15.0 D) = +0.066 m = +6.66 cm. Magnification: m = x΄/x = (+6.66 cm)/(+20 cm) = +⅓. We note that the image vergence is positive, so the image is real. The image is minified and erect, as indicated by the magnification value of +⅓.

Figure 5-53: The example in Figure 5-52 now with application of optical power and vergence addition.

This is the general imaging outcome from a concave mirror and a virtual object. Because both the object vergence and the mirror power are positive, the image vergence is always positive and numerically larger than the object vergence. Thus, a virtual object forms an image that is always located to the left of a concave mirror and is real, erect, and minified. Example ☞: A virtual object is formed 40 cm after a convex mirror with a focal length f = 10 cm. Identify the image type and magnification.

Figure 5-54: Imaging of a virtual object from a convex mirror. The image is virtual, inverted, and smaller than the object. 5-199

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✔ Schematic solution: To apply rule ❶, we draw a line parallel to the optical axis that would intersect an object point; this ray is reflected by the mirror, diverging away from the focal point F. To apply rule ❷, we draw a ray that is directed at the focal point and the same object point. This ray is reflected parallel to the optical axis. To apply rule ❸, we draw a line that joins the center of curvature and the object point. These rays are diverging. Their extrapolations intersect at a point behind (to the right of) the mirror. The image is therefore virtual and is also inverted. ✔ Numerical solution: The mirror optical power is F = 1 / f = 1 / (−0.1 m) = −10 D. The object location is positive (after the mirror): x = +0.40 m; therefore, the object vergence is L = 1 / x = 1/ (+0.40 m) = +2.5 D. Image vergence: L + F = L΄ ⇒ L΄ = (+2.5 D) + (–10.0 D) = –7.5 D. Image location: x΄ = 1/L΄= 1/(–7.5 D) = –0.133 m = –13.3 cm. The negative image vergence suggests that the image is virtual. Its location, also negative, suggests that the image is after the mirror. Magnification: m = x΄/x = (–13.3 cm)/(+40 cm) = –⅓.

Example ☞: An object is formed 15 cm after a convex mirror with a focal length of 10 cm. Identify the image type and magnification.

✔ Schematic solution: We implement rules ❶, ❷, and ❸ as in the previous example. These rays are diverging. Their extrapolations intersect at a point behind (to the right of) the mirror. The image is therefore virtual and is also inverted. ✔ Numerical solution: The mirror optical power is F = 1 / f = 1 / (−0.1 m) = −10 D. The object location is positive (after the mirror): x = +0.15 m; therefore, the object vergence is L = 1 / x = 1/ (+0.15 m) = +6.66 D. Image vergence: L + F = L΄ ⇒ L΄ = (+6.66 D) + (–10.0 D) = –3.33 D. Image location: x΄ = 1/L΄= 1/(–3.33 D) = –0.3 m = –30 cm. Magnification: m = x΄/x = (–30 cm)/(+15 cm) = –2.0. The negative image vergence suggests that the image is virtual. Its location, also negative, suggests that the image is formed after the mirror. The difference between this image and the one in the previous example is that this image is magnified—the value of its magnification is greater than 1.0. 5-200

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Figure 5-55: Imaging of a virtual object from a convex mirror. The image is virtual, inverted, and larger than the object.

We thus observe that a convex mirror presents the same optical behavior as a negative lens, in practically all aspects. The image formed from a real object is always virtual, minified, and erect. The image formed from a virtual object can be virtual, magnified, and inverted. It can also be real and magnified. This occurs in the case where the (virtual) object is between the mirror and the focal point. Example ☞: An object is formed 5 cm after a convex mirror with focal length f = 10 cm. Identify the image type and magnification.

✔ Schematic solution (Figure 5-56): We implement rules ❶ and ❸. These rays are converging a point in front (to the left) of the mirror. The image is therefore real and is also erect. ✔ Numerical solution: The mirror optical power is F = 1 / f = 1 / (−0.1 m) = −10.0 D. The object location is positive (after the mirror): x = +0.05 m; therefore, the object vergence is L = 1 / x = 1/ (+0.05 m) = +20.0 D. Image vergence: L + F = L΄ ⇒ L΄ = (+20.0 D) + (–10.0 D) = +10.0 D. The image is real. Image location: x΄ = 1/L΄= 1/(+10.0 D) = +0.1 m = –10 cm. The image is in front of the mirror. Magnification: m = x΄/x = (+10 cm)/(+5 cm) = +2.0. The image is magnified and erect (m = +2.0).

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Figure 5-56: Imaging of a virtual object from a convex mirror. The image is real, erect, and larger than the object.

5.4 ADVANCED MIRROR IMAGING EXAMPLES Example ☞: An object is placed in front of a concave mirror at a location |x| = f. Find the image location.

The object location is negative, while the focal length is positive, so |x| = f means that x = −f. The imaging relationship for x = −f ⇒ L = 1/x = 1/(−f ) = −1/f and F = 1/f ΄ results in zero image vergence, so the image is formed at optical infinity:

 1  1  L + F = L΄  L΄ =  −  +   = 0  f   f΄  Example ☞: A spherical mirror that forms only virtual images from objects placed in front of it has a radius of curvature of half a meter. What is the mirror type, and what are the radius of curvature and the power?

The key words are ‘that form only virtual images.’ This has to be a convex mirror (cases as shown in Figure 5-34). Therefore, the radius of curvature and the focal length are negative:

r = −0.50 m and f = −0.25 m. We now calculate the optical power in air: Fmirror =

1

f

=

2

r

=

2

−0.50 m

= − 4.0 D

Example ☞: A mirror with a focal length of half a meter forms an erect image one-quarter the size of an object placed in front of it. What type of image is this, and what is the separation between the object and the image?

This must be a convex mirror because the image is minified and erect. The focal length is negative. We write ‘mirror of half a meter focal length’ as: f = −0.50 m and F = −2.0 D.

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Apparently, in the imaging relationship we have two unknowns, as we know neither the object vergence nor the image vergence (and, consequently, nor the location). All we know is that the object is real (it is placed in front of the mirror).

Figure 5-57: Mirror imaging resulting in an image of ¼× the object size.

The second piece of information is that the image height is ¼ of the object height. (This case is similar to those presented in Figure 5-34.) Therefore, we write m = +0.25 = +¼. We use Eq. (5.2) [we can also use Eq. (5.1) in the form of reciprocal locations]:

m=

L 1 L  =  L΄ = 4 L L΄ 4 L΄

Now, we have two equations with two unknowns, L and L΄. We can simply substitute the above into the imaging relationship provided in Eq. (5.7), transforming it into an equation with only one unknown:

L + F = L΄ ⇒ L + (–2 D) = 4 L ⇒ L = –2/3 D The object location is, therefore,

x =

1

L

1

= −

2 3

3

= − m = − 150 cm. D

2

It is a real object, placed in front (to the left) of the mirror. 2

8

Image vergence is L΄ = 4 L = 4   − D  = − D. 3  3  Image location is x΄ =

1

L΄

1

= −

8 3

3

= − m = − 37.5 cm. D

8

It is a virtual image, formed to the right (after) the mirror. The object is placed to the left of the mirror (x = −150 cm), while the image is formed to the right of the mirror (x΄ = −37.5 cm). The object–image separation in this case is the numerical sum (both having the same origin, but being oppositely directed) of these two length quantities; therefore, the separation = 187.5 cm. 5-203

GEOMETRICAL OPTICS

Note that both locations are actually negative! Indeed, the reversal of what is positive and what is negative in mirrors can be fun, right?

Example ☞: An image that is five times larger than the object is formed on a screen. Both the object and the image are located in front of a mirror. If the object is 3.6 m from the screen, what type of image is formed, and what is the type and focal length of the mirror?

Figure 5-58: Mirror imaging resulting in a real image of 5× the object size.

The key is that the image is formed on a screen. This means that the image is real. Also, the image is magnified. The combination of a real and a magnified image suggests that the mirror is concave. This corresponds to the example shown in Figure 5-42. The information ‘five times larger’ is written as m = −5.

m=

x΄ x΄  −5 =  − 5 x = x΄ x x

The second piece of information is that the image is spaced 3.6 m from the object. Note that x is negative and x΄ is positive (aren’t we having fun with positive and negative directional distances in mirrors?). Thus, the object–image separation is written as x΄ + x = 3.6 m. The two-equation system is now written as x΄ = –5 x and x΄ + x = 3.6 m. This is a system of two equations with two unknowns. The solutions are: x = –0.9 m and x΄ = +4.5 m. We now use the imaging relationship given in Eq. (5.8) to find the focal length of the mirror: 1 1 1 1 1 1 + =  + =  x f x΄ −0.9 m f +4.5 m

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1 1 1 6 = − =  f = 0.75 m f +4.5 m −0.9 m 4.5 m

IMAGING WITH MIRRORS

5.5 MIRROR IMAGING QUIZ Note: Unless specified otherwise in this quiz, the mirror is preceded by air. Object placed = real object. Object formed = virtual object. 1)

+20 D +10 D +5 D –5 D –10 D –20 D

a) b) c) d) e) 8)

A B C D They are all wrong!

Which of the following represents a proper raytracing rule implementation?

A concave mirror has a radius of curvature of 1 m. Its focal length is … a) b) c) d) e) f)

5)

Which of the following represents a proper raytracing rule implementation?

+20 D +10 D +5 D –5 D –10 D –20 D

A concave mirror with a focal length of 20 cm has a power of … a) b) c) d) e) f)

4)

7)

+3.33 D +2.66 D +2.0 D +1.5 D +1.33 D

A convex mirror with a focal length of 10 cm has a power of … a) b) c) d) e) f)

3)

–100 cm –50 cm –25 cm +25 cm +50 cm +100 cm

A concave mirror with a 1 m radius of curvature is submerged in water (n =1.33). Its power is … a) b) c) d) e)

A convex mirror has a radius of curvature of half a meter. Its focal length is … a) b) c) d) e) f)

2)

6)

+100 cm +50 cm +25 cm –100 cm –50 cm –25 cm

A concave mirror with a 1 m radius of curvature is submerged in water (n =1.33). Its focal length is … a) b) c) d) e)

+133.3 cm +100 cm +66.66 cm +50 cm +33.33 cm

a) b) c) d) e)

A B C D They are all wrong!

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GEOMETRICAL OPTICS

9)

An object ray intersects the surface of a concave mirror parallel to the optical axis. Which of the diagrams represents the reflected ray?

12) A ray propagating parallel to the optical axis intersects a convex mirror. After reflection, the ray … a) b) c) d) e)

crosses the center of curvature of the mirror crosses the focal point of the mirror remains parallel to the optical axis diverges as if its origin was the center of curvature of the mirror diverges as if its origin was the focal point of the mirror

13) Which two of the following diagrams represent a proper ray-tracing rule implementation?

a) b) c) d) e)

A B C D None is correct.

10) An object ray intersects the surface of a convex mirror parallel to the optical axis. Which of the diagrams represents the reflected ray?

a) b) c) d)

A B C D

14) Which one of the following diagrams represents a proper ray-tracing rule implementation?

a) b) c) d) e)

A B C D None is correct.

11) A ray propagating parallel to the optical axis intersects a concave mirror. After reflection the ray … a) b) c) d) e)

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crosses the center of curvature of the mirror crosses the focal point of the mirror remains parallel to the optical axis diverges as if its origin was the center of curvature of the mirror diverges as if its origin was the focal point of the mirror

a) b) c) d)

A B C D

IMAGING WITH MIRRORS

15) An object is located 40 cm in front of (before) a convex mirror whose focal length is 20 cm. The image magnification is … a) b) c) d) e)

+⅓ +⅔ –1 –2 +2

16) An object is located 20 cm in front of (before) a convex mirror whose focal length is 20 cm. The image magnification is … a) b) c) d) e)

+½ +⅓ +1 –2 infinity

17) Select the correct statement regarding the object and image location algebraic signs in the figure:

19) An object beam with –10 D vergence is incident on a mirror. The image-forming beam has –30 D vergence. The mirror is … a) b) c) d)

convex; power –10 D convex; power –20 D concave; power +10 D concave; power –20 D

20) An object beam with –20 D vergence is incident on a mirror. The image-forming beam has +30 D vergence. The mirror is … a) b) c) d)

convex; power –10 D convex; power –30 D concave; power +50 D concave; power +10 D

21) An object placed before a mirror forms a minified image. What type of mirror is it (two answers)? a) b) c) d)

concave, if the image is inverted concave, if the image is erect convex, if the image is inverted convex, if the image is erect

22) An image formed by a mirror is virtual and erect. What type of mirror is it (two correct answers)? a) b) c) d) a) b) c) d)

the object location is positive, the image location is positive the object location is positive, the image location is negative the object location is negative, the image location is positive the object location is negative, the image location is negative

18) Select the correct statement regarding object and image type in the figure:

23) An image formed by a mirror is real and erect. What type of mirror is it (two correct answers)? a) b) c) d)

the object is real, the image is real the object is real, the image is virtual the object is virtual, the image is real the object is virtual, the image is virtual

concave, if the image is magnified concave, if the image is minified convex, if the image is magnified convex, if the image is minified

24) I want to produce an image on a sheet of paper (hint: this means that the image is ...) with a magnification equal to unity (regardless of being erect or inverted). I will place the object at … a) b) c) d)

a) b) c) d)

concave, if the image is minified concave, if the image is magnified convex, if the image is magnified convex, if the image is minified

the radius of curvature of a concave mirror the focal length of a concave mirror twice the focal length of a convex mirror the focal length of a convex mirror

25) An object is placed in front of a concave mirror at a distance greater than the mirror radius of curvature (in absolute terms, disregarding possible + or –). The image produced by the mirror is … a) b)

real, inverted, and minified real, inverted, and magnified 5-207

GEOMETRICAL OPTICS

c) d) e)

virtual, upright, and magnified virtual, upright, and minified virtual, inverted, and magnified

26) An object is placed in front of a concave mirror at a distance that is less than the mirror focal length. The image produced by the lens is … a) b) c) d) e)

real, inverted, and minified real, inverted, and equal to the object virtual, upright, and magnified virtual, upright, and equal to the object virtual, inverted, and magnified

27) A +20 D object vergence is incident on a +20 D mirror power. The image vergence is … a) b) c) d) e)

0D +20 D –20 D –30 D +40 D

31) When a virtual object is formed at a distance greater than the radius of curvature of a convex mirror, the image is … a) b)

c) d)

32) An object is placed 6.66 cm in front of a concave mirror whose radius of curvature is 3.33 cm. Determine the image type, location, and magnification. a) b)

28) A +10 D object beam vergence is incident on a – 20 D mirror power. The image vergence is … a) b) c) d) e)

–10 D +10 D +20 D +30 D –30 D

29) I am looking at a convex mirror. As I approach the mirror, my image … a) b) c) d)

becomes larger but remains smaller than my ‘object’ suddenly becomes inverted and larger becomes larger and eventually greater than my ‘object’ suddenly becomes inverted and smaller

30) How can an inverted and magnified image be produced by a convex mirror? a) b)

c)

d)

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Piece of cake. Just present a real object at a large distance in front of the mirror. Easy. All we need is a real object placed a distance between the center of curvature and the focal point. Not so easy, but doable. We must form a virtual object between the focal point and the center of curvature. Impossible. No matter what we do, a convex mirror always produces a minified and erect image by any object.

erect, minified, real, and formed between the focal point and the center of curvature inverted, minified, virtual, and formed between the focal point and the center of curvature erect, minified, real, and formed between the focal point and the vertex of the mirror inverted, minified, virtual, and formed between the focal point and the vertex of the mirror

c) d)

a virtual, inverted, minified image, 2.22 cm in front of the mirror, m = –⅓ a real, inverted, magnified image, 6.66 cm in front of the mirror, m = –1 a virtual, erect, minified image, 2.22 cm in front of the mirror, m = +⅓ a real, inverted, minified image, 2.22 cm in front of the mirror, m = –⅓

33) An object is located 10 cm in front of a vanity mirror marked 2.5× (meaning m = +2.5). Determine the image type and location. a) b) c) d) e)

virtual, erect, 25 cm to the right of the mirror virtual, erect, 4 cm to the right of the mirror virtual, erect, 2.5 cm to the right of the mirror virtual, erect, 25 cm to the left of the mirror real, erect, 2.5 cm to the right of the mirror

34) Back to Q 33. What is the mirror type and what is its radius of curvature r? a) b) c) d) e) f)

a concave mirror, +13.33 cm a concave mirror, +16.66 cm a concave mirror, +26.66 cm a concave mirror, +33.33 cm a convex mirror, –16.66 cm a convex mirror, –8.33 cm

35) An object is placed 20 cm in front of a convex mirror with a half-meter radius of curvature. Find the image type, location, and magnification. a) b)

a virtual, minified, erect image, 11.1 cm to the left of the mirror, m = +0.555 a virtual, minified, erect image, 14.28 cm to the right of the mirror, m = +0.714

IMAGING WITH MIRRORS

c) d)

a real, minified, inverted image, 14.28 cm to the right of the mirror, m = –0.714 a virtual, minified, erect image, 11.1 cm to the right of the mirror, m = +0.555

36) I have a 10 cm radius of curvature concave mirror and want to produce an image on a white screen that is 6.66 cm to the left of the mirror. Where should I place the (real) object? a) b) c) d)

20 cm to the left of the mirror 5 cm to the left of the mirror 2.85 cm to the left of the mirror 2.85 to the right of the mirror

41) A converging set of rays strikes a flat mirror such that its vergence is +2.0 D. This corresponds to a … a) b) c) d)

real object 50 cm to the left of the mirror real object 50 cm to the right of the mirror virtual object 50 cm to the left of the mirror virtual object 50 cm to the right of the mirror

42) An object is placed 75 cm to the right of a concave mirror of 100 cm radius of curvature (see figure). What is the image type, location, and magnification? (Hint: Pay attention to the direction of light propagation!)

37) Back to Q 36. The image formed is ________ and the magnification is ___________. a) b) c) d) e)

real, inverted; real, inverted; virtual, erect; real, erect; real, erect;

m = +0.333 m = –0.333 m = +0.666 m = +0.666 m = +3.0

38) I want to create a magnified image on a screen using mirrors. My two options are … a) b) c) d)

to place a real object between the focal point and the vertex of a concave mirror to place a real object between the center of curvature and the focal point of a concave mirror to form a virtual object between the vertex and the focal point of a convex mirror to form a virtual object between the focal point and the center of curvature of a convex mirror

a) b) c) d)

real, inverted, x΄ = –1.00 m, m = +2.0 real, inverted, x΄ = +1.50 m, m = +2.0 real, inverted, x΄ = +1.50 m, m = –2.0 virtual, inverted, x΄ = +1.00 m, m = +2.0

43) As in Q 42, we now place a flat mirror 1.2 m to the right of the concave mirror. The (final) image formed by the flat mirror is …

39) Let’s take a trip to London, UK (Cartesian signal convention). In mirrors, the image location is positive if the image is to the (select two) … a) b) c) d)

right of a convex mirror left of a convex mirror right of a concave mirror left of a concave mirror

40) An object is placed to the left of a mirror, and its image is also formed to the left of that mirror. Mind the gap and select the proper statement: a) b) c) d)

object location negative, image location positive, magnification negative object location negative, image location negative, magnification positive object location negative, image location positive, magnification positive object location positive, image location positive, magnification positive

a) b) c) d)

real, 90 cm to the right of the concave mirror real, 150 cm to the right of the concave mirror virtual, 90 cm to the right of the concave mirror virtual, 120 cm to the right of the concave mirror

44) What is the vergence of the (final) image formed by the flat mirror in Q 43? a) b) c) d)

L΄ = –3.33 D L΄ = +3.33 D L΄ = +1.11 D L΄ = –1.11 D

5-209

GEOMETRICAL OPTICS

5.6 MIRROR IMAGING SUMMARY Mind the Sign! In mirror imaging, all of the relationships are the same as their counterparts in lens imaging, and the algebraic signs follow the Cartesian sign convention (§ 3.2). The notation for object location is identical to that applied to lenses: An object location to the left of the mirror has a negative sign and to the right of the mirror has a positive sign.

Figure 5-59: The Cartesian sign convention in mirrors for object location.

The directional distances that apply to image space are associated with a reflected wave that has a reversed direction of propagation with respect to the initial, incident-to-the-mirror wave. Thus, image location, radius of curvature, and focal length values to the left of the mirror have a positive sign and to the right of the mirror have a negative sign.

Figure 5-60: The Cartesian sign convention in mirrors for image location, radius of curvature, and focal length.

5-210

IMAGING WITH MIRRORS

Focal Length and Power in Spherical Mirrors Mirror Focal Length f:

fmirror = r/2

where r is the radius of curvature of the mirror. Equivalently: r = 2 · fmirror Mirror Optical Power (in air): Fmirror = 1/f = 2/r Mirror Optical Power (in medium with refractive index n):

Fmirror = n/f = 2n/r

Mirror Imaging Ray-Tracing Diagrams Imaging relationships are identical those in mirrors and in lenses. The object vergence is added to the power of the imaging element to produce the image vergence: Vergence and Optical Power:

+

L object vergence

1 x

Object and Image Locations:

F

L΄ image vergence

1 f΄

+

object location

=

optical power

=

focal length

1 x΄ image location

If the refractive index in object (and image) space is n, the more generally applicable form is n x object location

Linear Magnification (using location):

n f΄

+

focal length

m

=

n x΄ image location

h΄ x΄ = h x

Ray tracing in mirrors is very similar to ray tracing in lens imaging. In a concave mirror: 1: The ray parallel to the optical axis reflects off of the mirror at the focal point. 2: The ray originating from (or crossing through) the focal point reflects at the mirror to become parallel to the optical axis. 3: The ray targeting the center of curvature of the mirror reflects back without any ray deviation.

Figure 5-61: Summary of ray-tracing rules in a concave mirror. 5-211

GEOMETRICAL OPTICS

In a convex mirror: 1: The ray parallel to the optical axis reflects at the mirror as if it originated from the focal point. 2: The ray directed toward (or targeting) the focal point reflects at the mirror to become parallel to the optical axis. 3: The ray targeting the center of curvature of the mirror reflects back without any ray deviation.

Figure 5-62: Summary of ray-tracing rules in a convex mirror.

We summarize mirror imaging formation configurations by grouping the concave and convex mirror variations that include a real object at various locations (including the object at the primary focal point) and a virtual object (see Figure 5-63). •

There are three distinct outcomes with a concave mirror and a real object: The image can be real and inverted (left column, top three cases), formed at infinity, and virtual and magnified. When the object is virtual, a concave mirror can only produce a real, minified, and erect image.

•

There is only one outcome with a convex mirror and a real object: The image can be virtual, minified, and erect (right column, last case). However, there are several outcomes with a convex mirror and a virtual object: The image can be virtual and inverted (right column, top three cases), formed at infinity, or real and magnified.

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IMAGING WITH MIRRORS

Figure 5-63: Summary of image formation configurations. [Left column: Concave mirrors, top (1) to bottom (6)] 1, 2, and 3: Real object, real image. 4: (Real) object at the primary focal point, image at optical infinity. 5: Real object, virtual image. 6: Virtual object, real image. [Right column: Convex mirrors, top (1) to bottom (6)] 1, 2, and 3: Virtual object, virtual image. 4: (Virtual) object at the primary focal point, image at optical infinity. 5: Virtual object, real image. 6: Real object, virtual image. (Compare with Figure 4-47.)

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GEORGE ASIMELLIS

LECTURES IN OPTICS, VOL 2

6 THICK LENSES AND LENS SYSTEMS

6.1 THE THICK LENS Any real lens is a thick lens, meaning that it has a non-zero lens thickness t , which is the axial separation between the vertexes of the two refracting surfaces. Lens thickness is a nondirectional distance (therefore, is always positive), measured at the center of the lens along the optical axis (i.e., the lens is not tilted), regardless of the shape or power of the lens, and regardless of whether it is plus- or minus powered. The lens thickness corresponds to a greater surface separation in plus lenses and a lesser separation in minus lenses. When the lens thickness is comparable in magnitude to the radii of curvature r and/or the lens focal length f, the lens is no longer thin. We can assume that a physical lens is thin only if t < ⅒ r.

6.1.1 Equivalent Optical Power and Focal Length In a thick lens, we cannot ignore the effect of the lens thickness on the optical power. If the surfaces have refractive powers F1 and F2, the lens material has a refractive index nlens, and the thickness is t, then Eq. (2.3), which describes the equivalent power of a thick lens, becomes Equivalent Power in a Thick Lens:

Fe = F1 + F2 −

t nlens

F1  F2

(6.1)

Equation (6.1) is known as Gullstrand’s relationship, named after the Swedish ophthalmologist Allvar Gullstrand. In addition to its importance in thick lenses and lens systems 6-215

GEOMETRICAL OPTICS

(§ 6.5), this relationship is also very important in visual optics because the optical elements of the human eye, such as the cornea and the lens, are thick lenses. Allvar Gullstrand was awarded the Nobel Prize in 1911 for his work on the optics of the eye, which includes a schematic eye.22 Notes

: The physical thickness in most ophthalmic and laboratory lenses is in the range of a few

millimeters. However, the lens thickness must be converted to meters when used in this relationship.

The first lens surface has a radius of curvature r1; its power F1 is the front or base curve power. The second surface has a radius of curvature r2; its power F2 is the back surface power. These two surface powers can be calculated independently using the SSRI power relationship introduced by Eq. (1.8): F1 =

nlens − next n −1 = (lens preceded by air) r1 r1

F2 =

next − nlens 1− n = (lens followed by air) r2 r2

(6.2)

where next is the refractive index of the surrounding medium (which may be different on the two sides of the lens), and nlens, or simply n, is the lens material refractive index. The equivalent lens focal length (distance) fe [m] is the reciprocal of the equivalent lens power Fe [D]: Equivalent Focal Length (in air):

fe =

1

(6.3)

Fe

If the medium on either side of the lens is not air, the reciprocal relationship involves the refractive index next of that external medium in the numerator: fe = next/Fe. Example ☞: A thick biconvex lens of glass (nlens = 1.5) has two symmetric surfaces with radii of curvature 5 cm separated by t = 7.5 cm. What are the equivalent lens power and the equivalent focal length in air?

Figure 6-1: Profile of a real, thick lens showing surface separation and radii of curvature. We apply Eqs. (6.2) to calculate the optical power of each refracting surface of the lens:

F1 =

22

n −1 r1

=

1.5 − 1.0 0.05 m

=

+0.5 +0.05 m

Introduction to Optics § 4.2.2 Schematic Eyes.

6-216

= +10.0 D and F2 =

1− n

r2

=

1.0 − 1.5

−0.05 m

=

−0.5 −0.05 m

= +10.0 D

THICK LENSES AND LENS SYSTEMS

We now use Eq. (6.1) with t = 0.075 m and n = 1.5. The equivalent lens power is: Fe = F1 + F2 −

t nlens

F1  F2 = + 10 D + 10 D −

0.075 m 1.5

 (+10 D)  (+10 D) = + 15 D

and the focal length is fe = 1/ Fe = +0.066 m = +6.6 cm. If the lens is regarded as thin, we use Eq. (6.1) with t = 0.0 m and n = 1.5. The equivalent lens power is: Fe = F1 + F2 −

t nlens

F1  F2 = + 10 D + 10 D −

0 1.5

 (+10 D)  (+10 D) = + 20 D

and the focal length is fe = 1/ Fe = +0.05 m = +5.0 cm.

As can be seen from the example, the power of the biconvex lens, compared to the power of an ideally thin lens with the same radii of curvature (5 cm) and index (nlens = 1.5), becomes smaller as the lens thickness increases (Figure 6-2).

Figure 6-2: Equivalent optical power in a thick lens with increasing thicknesses.

6.1.1.1

Parametric Analysis of Thick Lens Power

We now realize that the equivalent lens optical power (and, accordingly, its equivalent focal length) of a real, thick lens depends on the lens thickness. Figure 6-3 illustrates the linear reduction of lens power with increasing lens thickness t, in agreement with Eq. (6.1). This example corresponds to a thick, symmetric biconvex lens with radii of curvature of 10 mm, surrounded by air, for two values of the lens refractive index, n lens = 1.5 and 1.8. Thickness t varies from zero (an ideally thin lens) to 70 mm. The maximum optical power, which in this case is the limit at which t diminishes, becomes zero. The optical power of a thin lens is a simple sum of the two surface powers: F = F1 + F2. In a thick biconvex lens, the product F1· F2 is positive; therefore, as the thickness increases, an additional (third) term subtracts the optical power from the corresponding ideal thin lens. There is a thickness value for which the lens obtains a zero power, and beyond that, as thickness further increases, the power becomes negative! This critical lens thickness is Critical Lens Thickness:

1 1 t = n lens   +   F1 F2 

(6.4)

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GEOMETRICAL OPTICS

Figure 6-3: Dependence of the biconvex lens equivalent power on the lens thickness t.

The third term –(t/nlens) · (F1· F2), which is a consequence of the lens having a non-zero thickness t, depends on the algebraic sign of the product of surface powers F1· F2 as follows: •

If the product is positive (the case in biconcave and biconvex lenses), the thick lens equivalent power decreases with the lens thickness (just as in Figure 6-3).

•

If the product is negative (the case in meniscus lenses), the thick lens equivalent power increases with the lens thickness (as in Figure 6-7).

•

If the product is zero (as in a planoconvex or planoconcave lens), the thick lens equivalent power is fixed, regardless of the lens thickness (for example, see Figure 6-73).

Figure 6-4: Lens type and algebraic sign of the surface power product. Example ☞ : In what type of lens does the equivalent power become even more positive as the thickness increases? The lens has to be positive, so we consider planoconvex, biconvex, and positive meniscus lenses. The effect of the lens thickness on the equivalent power is –(t/nlens)·(F1· F2), so the effect of the third term is to give an algebraic sign that is opposite to the sign of the product F1· F2.

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THICK LENSES AND LENS SYSTEMS

In a planoconvex lens, the product of the two surface powers is zero because one of the two surface powers equals zero. In this case, the equivalent lens power equals the thin lens power, which itself equals the convex power alone; the lens thickness has no effect on the lens power! In a biconvex lens, the product of the two surface powers is positive because both surface powers are positive (both are convex surfaces). This renders the term –(t/nlens)·(F1· F2) negative, meaning that the thicker the lens, the less positive the equivalent lens power. In a plus meniscus lens, the product of the two surface powers is negative because there is one convex and one concave surface. This renders the term –(t/nlens)·(F1· F2) positive, meaning that the thicker the lens, the more positive the equivalent power.

Example ☞ : The following thick lenses have the same center thickness t and the same numerical values for the radius of curvature (where applicable), are made of the same refractive index material, and are surrounded by air. Which lens is subject to the most reduction in power, in comparison to the thin lens power?

In planoconvex and planoconcave lenses, the lens thickness has zero effect on the equivalent power, which equals the thin lens power: Fe = F1+ F2 = either F1 or F2, depending on the lens orientation. In both biconcave and biconvex lenses, the lens thickness has a negative effect on the power by the amount of –(t/nlens)·(F1· F2). Since the radii of curvature have equal numerical values and the lenses are made of the same material, the surface powers have equal values: positive in the biconvex lens and negative in the biconcave lens. The term –(t/nlens)·(F1· F2) is, in both cases, negative, so these two (thick) lens types have less power in comparison to the simple, thin lens power (Fe < F1+ F2). Finally, in the case of the meniscus lens with r1 = r2 (therefore, F2 = –F1; this is the Höegh meniscus, § 6.6), the thin lens power (t = 0; Fe = F1 + F2) equals zero; however, the thick lens equivalent power [t ≠ 0; Fe =

F1+ F2 –( t/nlens)·F1· F2 = –( t/nlens)·F1· F2] is positive by the amount of –( t/nlens)·F1· F2, since F1· F2 is negative. For this reason, as the lens thickness increases, this meniscus power becomes more positive. Thus, the biconvex and the biconcave lens types are subject to the most reduction in equivalent power. As these lenses become thicker, the biconvex lens becomes less positive and the biconcave lens becomes more negative.

6-219

GEOMETRICAL OPTICS

The magnitude of the equivalent power in a thick lens also depends on the relationship between the refractive index of the surrounding medium and that of the material. Figure 6-5 presents the dependence of the lens equivalent power for the material refractive index n lens = 1.5 on the value of the surrounding medium refractive index next. Three cases are shown, an ideally thin lens (t = 0 mm) and two thick lenses with t = 40 mm and t = 80 mm.

Figure 6-5: Dependence of the lens optical power on the refractive index of the surrounding medium.

We note that the equivalent power for a converging thick lens is, in general, less than that of a thin lens with same material and radii of curvature. It diminishes when the surrounding space has the same refractive index as the lens optical medium; in Figure 6-5, this occurs at

next = nlens = 1.5, which is where (just as in thin lenses) the power becomes negative when the surrounding medium is optically more dense than the lens (see, for example, Figure 2-38). Note : Why is the lens equivalent power affected by the lens thickness? To calculate the power (in a thin lens), we simply add the optical powers of each surface. In essence, we ignore the separation between the two refracting surfaces. In the case of a non-zero lens thickness, as light wavefronts propagate into the lens medium, their curvature decreases; a wavefront near its point source has a greater curvature, while away from the source, it has an increasingly smaller curvature (Figure 3-19 § 3.4.2). Because of the inherent link between the optical power and the change it induces on the light vergence, the reduction in wavefront curvature within the volume of a thick lens affects the lens power, which therefore differs from the lens power of a thin lens with identical refracting surfaces.

The equivalent optical power of a real lens depends on: ✔ The radii of curvature r of the refracting surfaces, ✔ The lens refractive index in relation to the surrounding medium refractive index, ✔ The surface separation (lens thickness) t.

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6.1.2 Specialty Lenses The ball lens is a glass sphere. It is a simple lens, usually of small size, although it is now commonly available in diameters of several centimeters. Due to its flexibility, the ball lens has many applications, such as in optical fiber coupling or focusing sun rays on photovoltaic energy converters, and has recently found its niche as an element for artistic photography (Figure 6-6).

Figure 6-6: This 9 cm diameter ball lens is used to create real, minified, and inverted images of Roman-era antiquities at the city of Thessaloniki, Greece. (Photos by Eftratios I. Kapetanas used with permission.)

The size lens is another interesting lens: It is a meniscus lens with a concentric design23 [i.e., its centers of curvature are coincidental, with the radii of curvature differing slightly (Figure 6-7). A size lens can have a positive power, a zero power, or even a negative power. These lenses are prescribed for aniseikonia.24

Figure 6-7: The size lens is a thick lens with different radii of curvature but identical centers of curvature.

6.1.3 The Cornea Equivalent Power The cornea of the human eye has the shape of a meniscus lens. Because it is a strong positive lens, the cornea is the most powerful optical element of the human eye. The anterior corneal surface has an average radius of curvature of r1 = +7.7 mm, and the posterior surface has an 23

Visual Optics § 9.2.5.1 The True Plano Lens, the Equi-sided Lens, and the Concentric Lens.

24

Introduction to Optics § 5.2.7 Low-Vision Aid and Aniseikonia Telescopic Devices. 6-221

GEOMETRICAL OPTICS

average radius of curvature of r2 = +6.8 mm. To the left of the anterior surface is air (n air = 1.0); internally, to the right of the posterior surface, is the aqueous humor with a refractive index of

naqueous = 1.336. The corneal refractive index is ncornea = 1.376. The two refracting surfaces are separated by an average distance of t = 540 μm = 0.54 mm (Figure 6-8).

Figure 6-8: The cornea is a meniscus lens with different radii of curvature, surrounded by optical media with different refractive index values.

The optical powers of the first (anterior) surface and the second (posterior) surface are calculated via the single spherical refracting interface power relationships: F1 =

n΄ − n r

=

ncornea − nair r1

=

1.376 − 1.0 +0.0077 m

= + 48.83 D and F2 =

n΄ − n r

=

naqueous − ncornea r2

=

1.336 − 1.376 +0.0068 m

= − 5.88 D

The equivalent optical power of the cornea is Fcornea = F1 + F2 −

t F F = n 1 2

( +48.83 D ) + ( −5.88 D ) −

0.540×10 −3 m 1.376

( +48.83 D )  ( −5.88 D ) =

+ 43.06 D

The cornea is therefore a strong converging lens with optical power ≈+43 D. Food for thought

: What would the corneal power be if both sides of the cornea were surrounded by air?

There is no difference in the power of the anterior surface, which would be +48.83 D. However, the power of the posterior surface would be very different: −55.29 D instead of −5.88 D:

F2 = (n΄−n)/r = (nair − ncornea)/r2 = (1.0 − 1.376)/(+0.0068 m) = −55.29 D. Thus, this ‘lens’ would be a minus lens with a power of about (+48.83 D) +(−55.29 D) = −6.46 D. The fact that the cornea is not a negative lens but a strong positive lens is mainly due to the lens being surrounded by two different refractive indices. The anterior surface is preceded by air (refractive index = 1.0), while the posterior surface is followed by the aqueous, whose refractive index (n aqueous = 1.336) has nearly the same value as that of the ‘lens’ material (n cornea = 1.376).

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6.2 CARDINAL POINTS: CONCEPT AND APPLICATIONS 6.2.1

Principal Points and Principal Planes

The question that arises in a thick lens is: Where is the origin of the directional distance associated with the equivalent focal length? In a thin lens, the focal lengths are measured from a line bisecting this lens, which is considered to have a zero thickness. Conveniently, we assume that the lens refracting action occurs abruptly, exactly at the center of the lens. The hypothesis that the thin lens has a zero thickness helps reference object–image locations; however, this approximation is not valid in thick lenses. When a ray parallel to the optical axis is incident on a thick lens, it is refracted at both surfaces of the lens on its way to the secondary focal point F΄. To find the actual (true) ray path inside the lens, we apply the refraction laws on each surface (Figure 6-9).

Figure 6-9: Actual ray path in a thick lens.

We now extend (in a forward direction) (Figure 6-10) the path of the initial, incident ray (parallel to the optical axis) and extrapolate (backward) the exiting ray. These extensions define a conventional ray path inside the lens. At the intersection of these two extrapolated paths, the plane that is perpendicular to the optical axis is the principal plane H΄ corresponding to the image-space focal point F΄ (both with primed letters). Any ray incident on the lens parallel to the optical axis leaves the lens as if it was subject to refraction exactly on this surface. The principal plane is, in this case, the assumed surface on which a single, instant refraction takes place. The intersection of the principal plane with the optical axis is the principal point P΄. Its distance from the secondary focal point F΄ is the equivalent focal length f΄e. Because they relate to light leaving the lens, the focal point F΄, principal plane H΄, and principal point P΄ are also termed the image-space focal point, image-space principal plane, and image-space principal point, respectively.

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GEOMETRICAL OPTICS

Figure 6-10: Image-space principal point, principal plane, and equivalent focal length.

A ray that originates from, or crosses, the primary (object-space) focal point F (Figure 6-11) exits the lens parallel to the optical axis. The assumed refraction takes place on the object-space principal plane H, which again is found by the intersections of the projections of the incident and exiting rays. Thus, the focal point F, the principal plane H, and the principal point P are termed the object-space focal point, object-space principal plane, and object-space principal point, respectively.

Figure 6-11: Object-space principal point, principal plane, and equivalent focal length.

Thus, any thick lens has two principal planes (H, H΄). The object-space principal plane H and point P are associated with the object-space focal point F, and the image-space principal plane H΄ and point P΄ are associated with the image-space focal point F΄. Note

: Similar to the focal length in thin lenses (f or f΄), again, there are two equivalent focal lengths,

the object-space fe and the image-space f΄e (note the prime distinguishing the image space). The length associated with imaging is the image-space focal length f΄e, measured from the image-space principal plane H΄ to the secondary focal point F΄. Conversely, the object-space focal length f is measured from the object-space principal plane H to the primary focal point F. Because of the principle of reversibility, a ray (parallel to the optical axis) traveling from right to left is refracted on the principal plane H (which could be termed H΄ because in this configuration it would serve in image space), and then crosses the primary focal point F (which could be termed F΄). Similar to the thin lens, in principle, these two focal lengths are not equal in magnitude. They are equal only if the medium surrounding the lens has the same refractive index (e.g., air on both sides). 6-224

THICK LENSES AND LENS SYSTEMS

The surface termed the principal ‘plane’ is not exactly a plane, flat surface. It is, in fact, the surface defined by the assumed points of inflection of all of the conventional rays propagating inside the lens for an incident collimated ray bundle that travels along the optical axis. This surface can be spherical with a radius of curvature equal to the equivalent focal length (Figure 6-12).

Figure 6-12: The refracting locus that we call the principal ‘plane’ is not exactly plane. In reality, the principal plane is a curved surface.

For small angles with respect to the optical axis (paraxial approximation, presented in § 8.1.2), the principal plane is, to a good approximation, a flat surface. The significance of the principal planes is that the object / image distances and the focal lengths are measured from the principal planes H and H΄, respectively, or from the two principal points P and P΄, respectively, as opposed to from the lens center in a thin lens.

Figure 6-13: Principal planes (and points) and focal planes (and points) in a thin lens (top) and a thick lens (bottom). For the thin lens, the two principal planes coincide and the principal points are located at the lens center. 6-225

GEOMETRICAL OPTICS

The principal planes are optical conjugates: One is the equisized image of the other through the optical system.25 For this reason, any ray that crosses a principal plane at any height meets the other principal plane at exactly the same height; thus, between the two principal planes, rays are drawn parallel to the optical axis (they ‘teleport’). The principal planes in a thick lens can be located either inside or outside the lens, or in front of or after the lens. Their locations are determined by the two directional distances δ and

δ΄ from the vertex points V and V΄ to the principal planes H and H΄, respectively. Their expressions can be derived from simple ray-tracing calculations and have the following form: Location of Principal Planes: F1

΄ image-space principal plane

= − t  front surface power  Fe thick lens power

n΄ image-space refractive index

n lens

n

F2

 object-space principal plane

= t  back surface power  Fe

object-space refractive index

n lens

(6.5)

thick lens power

where F1 and F2 are the lens surface powers [front (1) and back (2)], Fe is the equivalent lens power, t is the lens thickness, n΄ and n are the refractive indices to the right (image space) and to the left of the lens (object space), respectively, and n lens is the lens material refractive index.

Figure 6-14: Principal plane locations in various lens forms (glass lens surrounded by air). In a planoconvex (or a planoconcave) lens, one of the two principal planes coincides with the convex (or concave) side. In a meniscus, one (or possibly two) principal planes can be located outside the lens.

Either of δ΄ or δ can be positive or negative, depending on the outcome of Eq. (6.5); the presence or absence of the minus (–) sign, itself, in the formula does not preclude the outcome. The algebraic sign on δ΄ or δ shows the direction and therefore the location of the principal plane in relation to the vertex point, according to the Cartesian convention. Thus, for light traveling from left to right, a positive value for δ΄ or δ suggests that the principal plane (H΄ or H) is to the right of (after) the corresponding vertex point (V΄ or V). Likewise, a negative value for

δ΄ or δ means that the principal plane is to the left of (before) its vertex point. The meaning of the plus (+) or minus (–) sign in the measure of the principal plane location is as follows (see the example in Figure 6-15): The location of an object placed at the principal point P is x = 0, and the vergence L = ∞, which yields the image vergence L΄ = ∞ regardless of the lens power. Thus, the image location is x΄ = 0, meaning that the image is formed exactly at the principal point P΄. 25

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THICK LENSES AND LENS SYSTEMS

•

If δ is positive, then H (and P) is located to the right of the front surface, object-space vertex

V, possibly inside the lens (if δ < t). If δ is negative, then H (and P) is located to the left of V. •

If δ΄ is positive, then H΄ (and P΄) is located to the right of the back surface, image-space vertex point V΄. If δ΄ is negative, then H΄ (and P΄) is located to the left of V΄, possibly inside the lens (if |δ΄| < t) or outside the lens (if δ΄ > t).

Figure 6-15: The locations of the principal planes are measured from the vertex of the refracting surfaces V and V΄. In this example, the distance δ is positive and the distance δ΄ is negative.

If the lens is symmetric (F1 = F2) and is surrounded by the same medium, such as air, then |δ| = |δ΄|; i.e., the principal planes are equidistant from their corresponding vertex points: Locations of Principal Planes (lens surrounded by air):

΄ = −t 

F1 1  Fe nlens

and

 = t

F2 1  Fe nlens

(6.6)

• A thick lens or lens system has two principal points, both situated on the optical axis. • The planes perpendicular to the optical axis at the principal points are the principal planes. • By utilizing these planes, we can treat any thick lens (or lens system) as a thin lens.

Principal Points and Planes

Example ☞: Locate the principal planes in a symmetric biconvex lens (t = 3 cm, n lens = 1.5) with radii of curvature r1 = −r2 = 10 cm. The lens is located in air. The optical power of each refracting surface is F1 = F2 = +5.0 D. The equivalent lens power, calculated using Eq. (6.1), is Fe = +9.5 D. To find the location of the principal planes, we apply Eq. (6.6) to determine the distances δ΄ from principal plane H΄ and δ from principal plane H.

΄ = −t 

F1



1

Fe nlens

= − (3 cm) 

5.0 D



1

9.5 D 1.5

 − 1 cm and

 = t

F2



1

Fe nlens

= (3 cm) 

5.0 D



1

9.5 D 1.5

 + 1 cm

These results suggest that both principal planes are inside the biconvex lens at equal distances from the respective vertex points by a distance of ≈ ⅓ of the lens thickness (Figure 6-16). 6-227

GEOMETRICAL OPTICS

Figure 6-16: Calculation of the principal plane locations (δ and δ΄) in a thick biconvex lens. •Principal planes are NOT real, hard surfaces; they are virtual surfaces.

Do principal

planes ‘exist’?

• Their locations can be calculated by the δ and δ΄ distances from the respective vertex points. • They assist in determining the image location if we follow simple ray-tracing rules and/or apply the imaging relationships. • They apply to a thick lens as well as any to optical system comprising more than one refractive element.

The order of appearance of the principal planes (i.e., which of H or H΄ is positioned first) depends on the lens type, the lens surface powers, and the refractive indices of the lens and the surrounding medium at either side of the lens. We can normally expect plane H to be positioned ahead of plane H΄ [Figure 6-17 (left)]. However, sometimes the principal planes are crossed, in which case plane H is situated after plane H΄ [Figure 6-17 (right)]. Example ☞: Locate the principal planes in minus meniscus lenses (n lens = 1.5) with surface powers F1 = +5.0 D and F2 = −7.0 D. The lenses are located in air: lens 1, t = 1.5 cm; lens 2, t = 4.0 cm. The only difference between these two lenses is their thickness, which affects the equivalent power. Lens 1,

Fe1 = −1.65 D, δ΄ = +3.03 cm, and δ = +4.24 cm. Lens 2, Fe2 = −1.0 D, δ΄ = +12.5 cm, δ = +17.5 cm. The principal planes are crossed in the second lens: Plane H΄ is closer to the lens than plane H.

Figure 6-17: Principal planes in a minus meniscus lens: (left) not crossed and (right) crossed. 6-228

THICK LENSES AND LENS SYSTEMS

6.2.2

Nodal Points

A thick lens has two nodal points. Their role is equivalent to that of the center of curvature in an SSRI (§ 1.2.6) or a mirror (§ 5.2.3), or the center of a thin lens. The undeviating ray (§ 4.3.1) crosses the center of a thin lens, maintaining its inclination with the optical axis. We note that there is no nodal plane; there are only nodal points. However, we use the notion of ‘nodal ray,’ which is a ray directed toward, or appearing to originate from, a nodal point. The nodal points function as follows: A ray directed at the object-space nodal point N emerges from the lens as if it originated from the image-space nodal point N΄ without a change in the angle formed with the optical axis (parallel to its original direction).

Figure 6-18: Nodal points in a thick lens and ray tilt preservation: Angle ϑ, which expresses the ray tilt, is equal along both sides of the lens.

The two focal points, the two principal points, and the two nodal points are the six cardinal points. These points are all situated on the optical axis.

Cardinal Points

• A thick lens or lens system has six (6) cardinal points: • These are the two focal points, two principal points, and two nodal points. • They are all situated on the optical axis.

If both sides of the lens are surrounded by the same medium (n = n΄), the nodal points coincide with the corresponding principal points: PN = P΄N΄ = 0. If not (n ≠ n΄), the nodal points (N,

N΄) and the principal points (P, P΄) are separated by Principal-to-Nodal Point Displacement:

Note

PN = P΄N΄ = f΄ + f =

n΄ n n΄ − n − = Fe Fe Fe

(6.7)

: There is no direct formula for determining the nodal point locations; they are, essentially,

referenced to their corresponding principal points. The good news is that Eq. (6.7) is not restricted to thick lenses but also applies to single refracting interfaces and reflecting surfaces. 6-229

GEOMETRICAL OPTICS

The symmetry relationships can be expressed as follows: •

Principal and nodal point separation: PP΄ = NN΄

•

Focal point F to principal point P = nodal point N΄ to focal point F΄: FP = N΄F΄

•

Focal point F΄ to principal point P΄ = nodal point N to focal point F : F΄P΄ = NF

Example ☞: A symmetrical glass (nglass = 1.5) biconvex lens has radii of curvature 5 m and thickness t = 5 m. What are some possible principal and nodal point locations? When the lens is surrounded (both sides) by air (n = n΄= nair = 1.0), the principal plane locations are P (δ = =+2.0 cm), i.e., to the right of V, and P΄ (δ΄ = –2.0 cm), i.e., to the left of V΄. The nodal points coincide with the principal points (PN = P΄N΄ = 0.0 cm). When the lens is surrounded (both sides) by water (n = n΄ = nwater = 1.33), the principal plane locations are

P (δ = =+2.353 cm), i.e., to the right of V, and P΄ (δ΄ = –2.353 cm), i.e., to the left of V΄. The nodal points coincide with the principal points (PN = P΄N΄ = 0.0 cm). The symmetry is preserved, but the locations of the two principal points are now different. When the lens is surrounded by different media [front by air (n = nair = 1.0), back by water (n΄ = nwater = 1.33)], the principal planes are P (δ = =+0.909 cm), i.e., to the right of V, and P΄ (δ΄ = –3.636 cm), i.e., to the left of V΄. The nodal points do not coincide with the principal points (PN = P΄N΄ = 2.727 cm). The nodal points are shifted to the right of the principal points (always in the direction of the greater refractive index).

Note

: When displaced from the principal points, the nodal points are closer to the side of the system

with the optically more dense medium, if the system has a net plus converging optical power (Fe > 0). Note

: If the equivalent focal lengths of the thick lens are known, the PN = P΄N΄ displacement can be

expressed by PN = P΄N΄= f΄+ f . This distance simply becomes zero if f΄ = – f, which is the case when both sides of the lens are surrounded by the same medium.

Table 6-1: Notation of principal points in object space and image space.

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Description

Object space

Image space

focal length

fe (primary)

f ΄e (secondary)

focal points

F (primary)

F΄ (secondary)

principal points

P

P΄

principal planes

H

H΄

nodal points

N

N΄

THICK LENSES AND LENS SYSTEMS

6.2.3 Cardinal Points in an SSRI and a Mirror 6.2.3.1 Cardinal Points in an SSRI A single spherical refracting interface (SSRI) has two distinct focal points, the object-space primary focal point F and the image-space secondary focal point F΄ (defined in § 1.2.4.1). Because of the two different refractive indices straddling the interface (n and n΄), the two focal lengths have different magnitudes as well: f΄ = n΄/FSSRI and f = –n/FSSRI. The SSRI has two principal points (P and P΄), both located at the same point, the vertex. The tangent(s) at the vertex is (are) the principal plane(s). In its dual role, this point (and this plane) serves as both the object-space principal point (rays entering the interface) and as the image-space (rays leaving) principal point (plane). A ray parallel to the optical axis leaves the principal plane H directed toward the secondary focal point F΄. A ray from the object-space primary focal point F, when refracted at the image-space principal plane H΄, leaves this plane parallel to the optical axis. Finally, the SSRI has two nodal points (N and N΄), both located at the same point, the center of curvature (§ 1.2.6). This point serves dual roles: It is both the object-space nodal point

N (for rays entering the interface) and the image-space nodal point N΄ (for rays leaving the interface). Any ray directed at the center of curvature (to nodal point N) leaves the interface in an undeviated direction away from the interface (from nodal point N΄). The formula to locate the nodal points with reference to the principal points is also valid in an SSRI. From Eq. (6.7), Principal Point – Nodal Point Separation (SSRI): PN = P΄N΄ =

n΄ − n = FSSRI

n΄ − n n΄ − n r

= r

(6.8)

where r is the SSRI radius of curvature, FSSRI is its power, n is the object-space refractive index, and

n΄ is the image-space refractive index. Indeed, as in § 1.2.6, the nodal point of the SSRI is its center of curvature.

Figure 6-19: Cardinal points in a single spherical refracting interface. 6-231

GEOMETRICAL OPTICS

Example : A convex SSRI separates air (n = 1.0) from glass (n΄ = 1.5). If the radius of curvature is r = +10 cm, what is the principal point – nodal point separation P΄N΄? The answer is simply the radius of curvature = +10 cm, the space between the principal point (vertex) and the nodal point (center of curvature). We can also use Eq. (6.8) to find that P΄N΄ = r = +10 cm. Note that we really do not need the values of the refractive indices!

6.2.3.2 Cardinal Points in a Spherical Mirror A mirror has a pair of focal points, a pair of principal points, and a pair of nodal points. Yet each of the two point pairs coincides at the same unique physical point. The object-space primary focal point F and the image-space secondary point F΄ coincide with the focal point halfway through the center of curvature. The principal points are both situated at the mirror vertex. The tangent(s) to the surface at the vertex is (are) the principal plane(s). In its dual role, this point (plane) serves as the object-space (relating to rays entering the mirror) principal point (plane) and the image-space (relating to rays reflected off the mirror) principal point (plane).

Figure 6-20: Cardinal points in a spherical mirror.

Finally, the object-space nodal point N and the image-space nodal point N΄ coincide with the center of curvature. The formula locating the nodal points with reference to the principal points (PN and P΄N΄) is also valid in a mirror. Now we implement Eq. (6.7): Principal Point – Nodal Point Separation (mirror): PN = P΄N΄ =

n΄ − n ( − n) − n = = −r 2n Fmirror r

(6.9)

where r is the mirror radius of curvature, Fmirror is its power, and n is the refractive index.26 This confirms that the nodal point in a mirror is always its center of curvature (§ 5.2.3).

Note that the refractive indices do not need to equal 1.0, but they are identical in a mirror; specifically, image-space n΄ = –n, where the minus sign corresponds to a folded space, not a physical space. 26

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THICK LENSES AND LENS SYSTEMS

6.3 VERTEX POWERS IN A THICK LENS 6.3.1 The Concept of Front and Back Vertex Power and Focal Length The two vertex focal lengths in a thick lens are the front focal length and the back focal length. The front focal length (FFL) or front focal distance (FFD) is the distance along the optical axis that separates the object-space focal point F from the vertex V of the first (front) lens surface. The back focal length (BFL) or back focal distance (BFD) is the distance along the optical axis that separates the image-space focal point F΄ from the vertex V΄ of the second (back) lens surface.

Figure 6-21: Front focal length (FFL) and back focal length (BFL) in a thick lens versus the equivalent focal length (fe). (left) FFL and BFL in a biconvex plus lens. (right) FFL and BFL in a minus meniscus lens. Note that in the minus lens, BFL points to a point to the left (‘front’) of the lens because the image-space focal point is to the left of the lens. Thus, the ‘F’ and the ‘B’ in FFL and BFL simply suggest the surface of the lens from which they are being referenced, not the physical space they are pointing at.

The associated vertex powers, reciprocally related to the front and back focal lengths, are the front vertex power FFVP and the back vertex power FBVP, respectively.

Focal lengths in a thick lens:

Equivalent Focal Length:

Front and Back Focal Lengths:

Measured from the principal plane / point.

Measured from the front / back vertex point.

Associated with the equivalent power.

Associated with the front / back vertex power.

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GEOMETRICAL OPTICS

The vertex powers and vertex lengths are very practical, since we can actually measure them in a laboratory setting (in clinical practice with a lensometer) without knowing the locations of the principal planes. The back vertex power is measured with the back surface of the lens against the lensometer stop (temples facing away from the observer); the front vertex power is measure with the front surface of the lens against the lensometer stop. The back vertex power provides a realistic value for the vergence of the light rays exiting the back surface of a thick lens (toward image space) if a collimated pencil of rays enters the thick lens from the front side (object space). In ophthalmic lenses, the back vertex power is also known as the prescription power of that lens. The front vertex power is also known as the ‘neutralizing,’ or ‘front neutralizing’ power, since it is the vergence (of opposite sign) of a beam incident on the lens, which, upon exiting the lens becomes collimated (zero optical power). What we measure when we neutralize the lens by hand is therefore its front vertex power.

6.3.2 The Measure of Vertex Powers Just as in the case of the equivalent optical power, the back and front surface powers are affected by the lens thickness. In a thin lens, all of these powers are identical; they are the simple summation of the powers of the two SSRIs that constitute the lens itself. But in a thick lens, we cannot simply add the two powers because they correspond to different locations, separated by the lens thickness t. The power of the front surface (i.e., the vergence of a collimated beam leaving the first SSRI) is different from that of the back surface due to the downstream vergence upon reaching the second, back interface. This is expressed by the effective vergence relationship [Eq. (3.7)]. The initial vergence L is the front surface power F1, adjusted to a downstream distance equal to the lens thickness t (always positive). The medium between is the lens material with refractive index n: Downstream-adjusted Front Surface Power F1 (lens thickness t, index n):

F΄1 =

F1 t 1− F nlens 1

Then, the back vertex power is the simple sum of the two powers exactly at the back vertex point: the downstream-adjusted, effective front surface power F΄1 and the back surface power F2: Back Vertex Power:

6-234

FBVP =

F1 + F2 t 1− F nlens 1

(6.10)

THICK LENSES AND LENS SYSTEMS

Back Vertex Power

• is the beam vergence leaving the image-space back surface of a thick lens if the beam entering the lens is collimated (relating to an object at infinity); • is the sum of the downstream-adjusted front surface power and the back surface power; • can be measured with the back surface of the lens against the lensometer stop; and • is the power reported in prescription spectacle lenses.

We can calculate the front vertex power following similar logic. The justification is simple. It is again, the sum of two powers: The front surface power F1 is not adjusted (it is already there!), but the back surface power F2 is now adjusted downstream. To calculate F΄2 we use Eq. (3.7), in which t is the lens thickness. F΄2 is again downstream (which practically means that t is positive) because for this calculation light travels from right to left—from the back lens surface to the front lens surface. Front Vertex Power:

Front Vertex Power

FFVP = F1 +

F2 t 1− F nlens 2

(6.11)

• is the—opposite sign—beam vergence entering the object-space front surface of a thick lens such that the beam leaving the lens is collimated (producing an image at infinity); • is the sum of the front surface power and the downstream-adjusted back surface power; • can be measured with the front surface of the lens against the lensometer stop; and • is the power typically measured by hand neutralization and is used for referencing added power in presbyopic prescription lenses.

Once the front vertex power FFVP and back vertex power FBVP are established, we can compute the front focal length (FFL) and the back focal length (BFL). These are reciprocally related to the refractive index in the front and back medium. Front Focal Length and Front Vertex Power:

Back Focal Length and Back Vertex Power:

FFVP = FBVP =

nF n  FFL = F FFL FFVP

(6.12)

nB n  BFL = B BFL FBVP

(6.13)

where nF (also n) and nB (also n΄) are the refractive indexes in front of the thick lens (before the first element, in general) and in back of the thick lens (after the last element, in general), respectively. Not surprisingly, the medium is most often air, so both nF and nB are simply 1.0.

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GEOMETRICAL OPTICS

Note

: Owing to the definition of the front vertex power (whose computation considers light traveling

from right to left), the corresponding front focal length is positive for a positive FFVP. This explains why

FFL = +n/FFVN.

Example ☞: A symmetric biconvex glass (n = 1.5) lens, 6 cm thick, has a front surface power F1 = +10 D and a back surface power F2 = +10 D. The lens is surrounded by air. Calculate the thick lens equivalent power, the front and back vertex powers, and the corresponding focal lengths. Thick lens equivalent power: Fe = F1 + F2 −

t 0.06 m F1  F2 = +10 D + 10D −  (+10 D)  (+10 D) = +20 D − 4 D = +16.0 D n 1.5

Front vertex power: F2

FFVP = F1+ 1−

t n

+10 D

= +10 D +  F2

1−

0.06 m 1.5

= +10 D +

 (+10 D)

+10 D 1 − 0.04  10

= +10 D +

+10 D 0.6

= +10 D + 16.66 D = +26.66 D

Back vertex power: F1

FBVP = 1−

t n

+10 D

+ F2 =  F1

1−

0.06 m 1.5

 (+10 D)

+10 D =

+10 D 1 − 0.04  10

+10 D =

+10 D 0.6

+10 D = + 16.66 D +10 D = +26.66 D

Figure 6-22: BFL, FFL, and equivalent focal length in a symmetric biconvex glass lens of 6 cm thickness. Equivalent focal length: f΄e = 1/Fe = 1/(+16 D) = +0.0625 m = +6.25 cm Front focal length: FFL = 1/FFVP = 1/(+26.66 D) = +0.0375 m = +3.75 cm Back focal length: BFL = 1/FBVP = 1/(+26.66 D) = +0.0375 m = +3.75 cm

Note

: In this symmetric biconvex lens, the two vertex powers are equal in magnitude but differ in thick

lens equivalent optical power.

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THICK LENSES AND LENS SYSTEMS

The difference between BFL and f΄e is none other than the spacing V΄P΄, which is the location δ΄ of the image-space principal plane H΄. Likewise, the difference between fe and FFL is the spacing VP, which is the location δ of the object-space principal plane H. Since BFL and

FFL can be measured on the optical bench, usually with a lensometer, the location of the principal planes in relation to BFL and FFL are as follows: δ΄ = BFL – f΄e and δ = fe – FFL

Principal Planes and Vertex Focal Length:

(6.14)

Example ☞: Where are the principal planes in the symmetric biconvex lens of the previous example? By Eq. (6.6): Principal plane H΄ location:

΄ = −t 

F1 1 +10 D 1  = − 0.06 m   = − 0.025 m = − 2.5 cm Fe nlens +16 D 1.5

We then compare this to the difference: δ΄ = BFL – f΄e = (+3.75 cm) – (6.25 cm) = –2.5 cm

By Eq. (6.6): Principal plane H location:

 = t

F2 1 +10 D 1  = 0.06 m   = + 0.025 m = + 2.5 cm Fe nlens +16 D 1.5

We then compare this to the difference: δ = fe – FFL = (+6.25 cm) – (3.75 cm) = +2.5 cm

Note

: The two vertex powers (and the back and front focal lengths) are not, in principle, of equal

magnitude. Even if the medium is the same around the lens, if the surfaces are not symmetric (r1 ≠ r2 ;

F1 ≠ F2), then, particularly in high-power thick lenses, these two vertex powers are not equal in magnitude. In ophthalmic lenses, which are mainly meniscus with a convex front surface (base curve), the difference between FBVP and FFVP is very small, especially in minus and low-power plus lenses because of the small lens thickness. Thus, despite different front and back surface powers, the vertex power difference is less than 0.15 D. This is not the case in high-power, thick plus lenses, in which this difference may exceed half a diopter.

Example ☞: A meniscus glass ophthalmic lens27 (n = 1.5), 3 mm thick, has a front surface power F1 = +10 D and a back surface power F2 = –5 D. Calculate the front and back vertex powers and compare them to the equivalent power (Figure 6-23). Thick lens equivalent power:

Fe = F1 + F2 −

t 0.003 m F  F = +10.0 D − 5.0 D −  (+10.0 D)  ( − 5.0 D) = +5.0 D + 0.1 D = +5.10 D n 1 2 1.5

Equivalent focal length: f΄e = 1/Fe = 1/(+5.10 D) = +0.196 m = +19.6 cm

27

Visual Optics § 9.4.4 Lens Thickness. 6-237

GEOMETRICAL OPTICS

Front vertex power: F2

FFVP = F1 + 1−

t n

−5 D

= +10 D +  F2

1−

0.003 m 1.5

= +10 D +

 ( − 5 D)

−5 D 1+ 0.002  5

= +10 D +

−5 D 1.01

= +10 D − 4.95 D = +5.05 D

Front focal length: FFL = 1/FFVP = 1/(+5.05 D) = +0.198 m = +19.8 cm Object-space principal plane location δ = fe – FFL = (+19.6 cm) – (19.8 cm) = –0.2 cm

Figure 6-23: BFL, FFL, and equivalent focal length in a plus meniscus ophthalmic lens of 3 mm thickness. Back vertex power: F1

FBVP = 1−

t n

+10 D

+ F2 =  F1

1−

0.003 m 1.5

−5 D =

 (+10 D)

+10 D 1 − 0.002  10

−5 D =

+10 D 0.98

− 5 D = 10.20 D − 5 D = +5.20 D

Back focal length: BFL = 1/FBVP = 1/(+5.20 D) = +0.192 m = +19.2 cm Image-space principal plane location δ΄ = BFL – f΄e = (+19.2 cm) – (+19.6 cm) = –0.4 cm

Note

: The two vertex powers are slightly different in magnitude and are also different from the thick

lens equivalent optical power.

Example ☞: Calculate the front and back vertex powers, focal lengths, and principal plane location for a hemispherical planoconvex lens. The lens has a radius of curvature r = +10 cm and a lens material refractive index n = 1.5. The lens is placed in air. The lens thickness equals the radius of curvature of the convex refracting surface: t = r = +10 cm. Surface powers:

6-238

F1 =

n −1 r1

=

n −1 r

=

1.5 − 1 +0.1 m

= +5.0 D

and

F2 =

1− n

r2

=

1− n



= 0

THICK LENSES AND LENS SYSTEMS

Equivalent power:

Fplanoconvex = F1 + F2 −

Equivalent focal length:

n

F1  F2 = +5 D + 0 −

t n

 (+5 D)  0 = +5.0 D

f΄e = 1/Fe = 1/(+5.0 D) = +0.20 m = +20 cm F2

Front vertex power: FFVP = F1 + 1−

Front focal length:

t

t n

0

= +5 D +  F2

1−

0.01 m 1.5

= +5.0 D  (0 D)

FFL = 1/FFVP = 1/(+5.0 D) = +0.20 m = +20 cm

Object-space principal plane location: δ = FFL – fe = +20 cm – (+20 cm) = 0 cm

Figure 6-24: BFL, FFL, and equivalent focal length in a hemispheric planoconvex glass lens of 10 cm thickness.

Back vertex power:

F1

FBVP = 1−

Back focal length:

t n

+5 D

+ F2 =  F1

1−

0.10 m 1.5

+0 D =

 (+5 D)

+5 D 1 − 0.333

=

+5 D 0.66

= + 7.5 D

BFL = 1/FΒVP = 1/(+7.5 D) = +0.133 m = +13.33 cm

Image-space principal plane location: δ΄ = BFL – f΄e = +13.33 cm – (+20 cm) = –6.66 cm

Note

: In this example, the principal plane location calculations have yielded the same results as the

ones obtained in § 6.2.1 using the Eq. (6.6) for δ and δ΄. Note

: In an SSRI, the principal planes are tangent to the vertex (or, equivalently, the principal planes

coincide with the vertex points, or δ = δ΄ = 0). For this reason, the vertex power equals the actual SSRI power, which, in turn, is the lens power, and FFL and BFL equal f and f΄, respectively.

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GEOMETRICAL OPTICS

6.4 IMAGING WITH A THICK LENS 6.4.1

Ray Propagation in a Thick Lens

The cardinal points, principal planes, and auxiliary rays are ‘construction materials,’ in the sense that they help draw a simplified model of ray propagation in a thick lens or, as we will soon see, in a complex optical system comprising more than one lens. In reality, all rays are subject to (1) refraction by the first surface, (2) propagation inside the thick lens medium, and (3) refraction by the second surface. The only laws that apply are the laws of refraction and propagation. The rays are not ‘aware’ of the existence of any cardinal points and do not follow the conventional paths, but instead follow actual paths, as in Figure 6-25.

Figure 6-25: Cardinal points and ray paths in a thick lens surrounded by different optical media (here, the medium to the right of the lens is optically more dense than the medium to the left, n΄ > n). Note that the nodal points are separated from the principal points.

However, the consideration of actual ray paths can be quite complicated in the case of a thick lens (or a lens system). To facilitate image location in a thick lens (or a lens system), assuming that the cardinal points are known, we use conventional (construction) rays in a technique similar to ray tracing in thin lenses, as presented in § 4.3. These ray-tracing rules significantly simplify finding the location of the image. Here are some rules: ✘ Rays do not refract at the surfaces. We ignore their interaction with any lens surface. Instead, ✔ Rays parallel to the optical axis continue (teleport) inside the lens, maintaining ray height: ✏ A ray traveling from the left continues until it reaches the principal plane H΄. There, the ray is subjected to refraction and bends, and it exits the lens directed toward the secondary (image plane) focal point F΄. ✏ A ray traveling from the right continues until it reaches the principal plane H. There, it is subjected to refraction and bends, and it exits the lens directed toward the primary focal point F. 6-240

THICK LENSES AND LENS SYSTEMS

Figure 6-26: Ray-tracing rule 1 employing principal plane H΄ and focal point F΄. Note that between principal planes H and H΄ the ray propagates parallel to the optical axis.

If we use the principle of reversibility (which can be visualized by simply flipping the arrow tip, as shown in Figure 6-27), in essence, we produce a version of ray-tracing rule 2. For (the rarely used case of) light propagating from right to left, a ray propagating parallel to the optical axis refracts at the principal plane H and then crosses the primary focal point F.28

Figure 6-27: Ray tracing employing principal plane H for light propagating from right to left.

Back to the usual case of light propagating from left to right, a ray crossing the primary focal point F propagates until it reaches the principal plane H where it refracts parallel to the optical axis (Figure 6-28). Recall that then the primary focal point is the object point that produces a collimated beam. The ray-tracing rule now states that a ray crossing the primary focal point F refracts at the principal plane H and exits parallel to the optical axis.

Figure 6-28: Ray-tracing rule 2 employing the principal plane H and focal point F.

Your optics professor rises in protest, stating that now what we call F is indeed a secondary focal point and should be primed. We can promptly agree (there is no other choice, is there?). For propagation from right to left, image space is to the left of the system, which suggests that the case in Figure 6-27 is simply the same as in Figure 6-26, considering the reversed path of light propagation. 28

6-241

GEOMETRICAL OPTICS

Note

: Because the principal planes are optical conjugates, a construction ray crossing a principal

plane at any height meets the other principal plane at exactly the same height; between the two principal planes, rays are drawn parallel to the optical axis.

✔ A ray directed at the object-space nodal point N, forming an angle ϑ with respect to the optical axis, appears to exit as if it originated from the image-space nodal point N΄ with the same angle ϑ. There is a virtual parallel displacement (teleporting) over the optical axis from N to N΄.

Figure 6-29: Ray-tracing rule 3 employing nodal points N and N΄.

6.4.2

Ray Diagrams in a Thick Lens

The ray diagrams in a thick lens (and in any lens system) follow the same rules as in thin lenses, if we acknowledge the axial displacement between the two principal planes and between the nodal points. The ray diagram rules for a positive (plus) lens are as follows: ① parallel: A ray propagating parallel to the optical axis continues until it reaches the imagespace principal plane H΄; at that point the ray bends toward the secondary focal point F΄. ② focal: A ray that crosses the primary focal point F continues until it reaches the objectspace principal plane H, where it is bent, becoming parallel to the optical axis. ③ radial / nodal: A ray directed at the object-space nodal point N exits the lens originating from the image-space nodal point N΄ with the same angle (with respect to the optical axis) as the angle of entry (with respect to the optical axis).

Figure 6-30: Ray diagrams in a thick converging (plus) lens. Compare with Figure 4-14. 6-242

THICK LENSES AND LENS SYSTEMS

Actual rays

Conventional rays

• are subject to two refractions, one at each bounding (refracting) lens surface. • change path only when they encounter the principal planes (parallel and focal ray) or nodal points (the nodal / radial ray).

Figure 6-31: Ray diagrams in a thick converging (plus) lens. The difference between this figure and Figure 6-30 is that, due to a different location of the object (in relation to the lens primary focal point—here the object location is at a shorter distance than the lens focal length), the image-forming rays are diverging, as opposed to the configuration in Figure 6-30 (where the object is located at a longer distance than the lens focal length), in which the image-forming rays are converging.

Imaging with negative lenses follows the same rules as are used with positive lenses, with some minor differences. Because the focal length is negative, the primary focal point F is now after the lens and the secondary focal point F΄ is before the lens. Thus, rule ➀ states that a ray parallel to the optical axis, after reaching principal plane H΄, diverges and appears as if it originated from the secondary focal point F΄, whereas rule ② states that a ray directed to the primary focal point F diverges at the principal plane H and then continues on its path parallel to the optical axis.

Figure 6-32: Ray diagrams in a thick diverging (minus) lens. Compare with Figure 4-21. Note that, although the principal points (in this figure as well as in Figure 6-30 and Figure 6-31) coincide with their respective nodal points, this is not a general property; it results from the fact that the surrounding medium is the same around both lens surfaces (n΄ = n).

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GEOMETRICAL OPTICS

Ray tracing in a thick lens: • In a thick lens, the real rays do not all pass by the cardinal points. • However, it is very useful to apply the rules of ray tracing to find the image location. • In object space, locations are measured from the object-space principal plane Η (such as x). • In image space, locations are measured from the image-space principal plane Η΄ (such as image location x΄ and focal length f΄ ).

6.4.3

The Wild Ray in a Thick Lens

The Listing method (presented in § 4.5) facilitates ray tracing of any random ray through a thin lens. Extending the application to a thick lens, we follow these steps: ⑴ We draw an auxiliary line parallel to the incident ray crossing the primary focal point F. ⑵ The incident and the auxiliary rays become parallel to the optical axis just after intersecting principal plane H.

Figure 6-33: Wild ray in a thick lens rules: steps 1 and 2.

⑶ The point at which the extension of the auxiliary line intersects the image-space focal plane is also the point where the actual, exiting ray intersects the image-space focal plane. ⑷ We connect this point on the principal plane H΄ to the point that meets the parallel displacement of the incident beam over the optical axis and between the principal planes. Thus, we have the path of the exiting ray.

Figure 6-34: Wild ray in a thick lens: steps 3 and 4. 6-244

THICK LENSES AND LENS SYSTEMS

6.4.4

Thick Lens Imaging Relationship

In addition to ray tracing, we can use numerical, analytical imaging relationships in a similar fashion to the way these relationships are used for a thin lens. The conventions are: light travels from left to right; object-space notations are non-primed; image-space notations are primed; the object location x is measured from the principal plane H; and the image location x΄ is measured from the principal plane H΄. We use the equivalent focal length f΄e, which in air is simply the reciprocal of the equivalent optical power. This length is measured from the principal plane H΄. Then, the thick lens imaging relationship takes the form Thick Lens Imaging Relationship:

1

x object location

+

1

f΄

=

equivalent focal length

1

x΄

(6.15)

image location

Figure 6-35: Imaging in a thick positive biconvex lens, forming a real image to the right of the lens. The object and image distances x and x΄ are measured from the principal planes H and H΄, respectively.

• The imaging relationships in a thick lens are similar to those in a thin lens. • The equivalent (total) optical power is calculated from Gullstrand's formula. • The object location x is measured from the object-space principal plane Η. • The focal length f΄e and image location x΄ are measured from the image-space principal plane Η΄.

Thick Lens Imaging

Example ☞: Identify the image location and type if an object is placed 15 cm in front of (a) a thin lens and (b) a thick lens. Both lenses are symmetric biconvex with a radius of curvature r = 10 cm and a lens material refractive index n = 1.5, and are placed in air. The thick lens has central thickness t = 3 cm. Surface powers (common to both lenses):

F1 =

n −1 1.5 − 1 1− n 1− 1.5 = = +5.0 D and F2 = = = +5.0 D r1 +0.1 m r2 −0.1 m 6-245

GEOMETRICAL OPTICS

Thin lens power: Fthin = F1 + F2 = +5 D +5 D = +10.0 D Thin lens focal length: f΄ = 1/F = 1/(+10.0 D) = +0.10 m = +10 cm Thick lens power: Fe = F1 + F2 −

t 0.03 m F1  F2 = +5 D + 5D −  (+5 D)  (+5 D) = +10 D − 0.5 D = +9.5 D n 1.5

Thick lens equivalent focal length: f΄e = 1/F e = 1/(+9.5 D) = +0.105 m = +10.5 cm Location of principal planes, δ΄ for principal plane H΄ and δ for principal plane H:

΄ = −t 

F1 1 5D 1  = − (3 cm)    − 1 cm Fe n lens 9.5 D 1.5

and

 = t

1 5D 1  = (3 cm)    + 1 cm Fe n lens 9.5 D 1.5

F2

Thin lens imaging (Figure 6-36): For an object at x = –0.15 m (L = –6.66 D) and a lens with F = +10.0 D, the image is produced with L΄ = +3.33 D, located at x΄ = +0.3 m. It is a real and inverted image with magnification m = x΄ / x = (+0.3 m) / (–0.15 m) = –2.0.

Figure 6-36: Thin lens imaging example with the object placed 15 cm in front (to the left) of the lens. Thick lens imaging (Figure 6-37): We assume that the object is the same distance in front of the lens, so the object distance from the front vertex is xv = –0.15 m. To use the lens equation for the thick lens, we must find the object distance to the principal plane H, which is δ = +0.01 m. Therefore, x = –0.16 m. For an object at x = –0.16 m (L = –6.25 D) and a lens with Fe = +9.5 D, the image is produced with L΄ = +3.25 D, located at x΄ = +0.307 m. (This is now measured from the principal plane H΄, which means that the image distance from the back vertex is x΄v = 0.297 m.) The image is real and inverted with magnification m = x΄ / x = (+0.307 m) / (–0.16 m) = –1.92.

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THICK LENSES AND LENS SYSTEMS

Figure 6-37: Thick lens imaging example with the object placed 15 cm in front (to the left) of the lens.

Note

: Despite the significant lens thickness in relation to its focal length, we notice that the location

as well as the magnification of the image obtained with the precise thick lens formulae differs by less than 5% compared to that obtained using the simple thin lens formula.

Example ☞: For the arrangement that corresponds to Figure 6-38, determine the lens thickness, lens power, lens surface powers, and object and image location, and verify the imaging relationship. The lens is symmetric, thick, and biconvex with a material refractive index n = 1.5, placed in air.

Figure 6-38: Imaging with a thick lens: determination of object and image locations. The following can be obtained from the information presented by the ruler: The object is located 8 cm to the left of the object-space principal plane H (PO = –0.08 m), and the image is located 10 cm to the right of the image-space principal plane H΄ (P΄I = +0.10 m). This is adequate information to enumerate the focal length / equivalent power of the lens by means of the imaging relationship given in Eq. (6.15): 1

− 0.08 m object location

1

+

f΄ equivalent focal length

=

1 +0.10 m



1

f΄

= Fe = +22.5 D  f΄ = +4.44 cm

image location

This result is in accordance with the information that can be obtained by the ruler (P΄F΄ = 4.44 cm). Alternatively, one can use the information on the equivalent focal length to extract the lens power and verify the imaging relationship.

6-247

GEOMETRICAL OPTICS

In addition, the lens thickness (VV΄) appears to be 5 cm. We can apply the thick lens equivalent power relationship [Eq. (6.1)], using the information that the lens is symmetric, meaning that the two surface powers have the same magnitude. This leads to F1 = F2 = +15 D.

Figure 6-39: Object, image, focal length, and lens thickness determination for the case in Figure 6-38.

In minus thick lens imaging, we follow the same rules that stipulate the origin of the directional distances for the object (measured from the object-space principal plane), the image, and the focal length (from the image-space principal plane). The only apparent difference is that the (secondary) image-space focal point is situated to the left of the lens.

Figure 6-40: Imaging with a thick, negative (minus-powered) meniscus, forming a virtual image to the left of the lens.

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THICK LENSES AND LENS SYSTEMS

6.5 LENS SYSTEMS A lens system (the notion of a system can be extended to include not just lenses, but single refracting surfaces, reflecting surfaces, etc.) can be modeled as a thick lens. Here, too, we employ the concepts of equivalent optical power (now also called total optical power), equivalent focal length, principal plane, and so on. We assume that all lenses (and refracting surfaces or mirrors) share the same optical axis. Such a system is called uniaxial.

6.5.1

Optical Power in a Lens System

We can apply Eq. (6.1) in a system of two thin lenses separated by d; if the medium between the lenses is air (the simplest case), then for nbetween we use 1.0. If the (thin) lenses are in contact, then d = 0, and we simply add their optical powers. The system optical power is now called total optical power FTOT, which is essentially the equivalent optical power concept in the thick lens. In the imaging formulas, as with the thick lens, the equivalent optical power is now simply and interchangeably called the total optical power. General relationship (Gullstrand's relationship) Thin lenses in contact Lenses separated by distance d (in air)

6.5.1.1

• F = F1 + F2 − (d/n) · (F1· F2) TOT

• F = F1 + F2 TOT

• F TOT = F1 + F2 − d · F1· F2

Thin Lenses in Contact

The simplest case of a lens system involves two thin lenses in contact. Their separation d is therefore zero. Because the lenses are thin, the optical power of the system is simply the sum of the two individual optical powers.

Figure 6-41: Optical power in a system of two thin lenses in contact.

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GEOMETRICAL OPTICS

In Figure 6-41, the two thin planoconvex lenses (F = +5 D)—with their flat sides in contact—form a system that is equivalent to a thin biconvex lens. This lens has 2× the power (FTOT = 5 + 5 = +10 D) and half the focal length (fTOT = 1/FTOT = +0.1 m) of each component.

6.5.1.2 Equivalent (Total) Power of Two Thin Lenses in Air Now, the lens system involves two thin lenses separated by air. Their separation d is therefore not zero, but the refractive index of the medium between them is n = 1.0. Because of this separation, the total optical power of the system is not the sum of the two individual optical powers. Example ☞: A two-lens optical system consists of lens 1 with power +10.00 D and lens 2 with power – 10.00 D, situated 0.15 m after lens 1. What is the total system optical power and focal length? If we consider each lens as a thin lens with power F1 and F2, and we simply add the power of each, the sum is +10 D + (–10) D = 0.0 D, which means that the system power is zero! However, the lenses are not in contact. They are separated by a thickness d = 15 cm in air (refractive index

nbetween = 1.0). So the lens system (total) equivalent optical power is FTOT = F1 + F2 −

d 0.15 m F1  F2 = (+10 D) + ( − 10 D) −  (+10 D)  ( − 10 D) = + 15 D nbetween 1.0

Thus, the system has a net positive optical power. The focal length (in air) is simply the reciprocal of the total power: fe = 1/FTOT = 1/(+15.0D) = +0.066 m = +6.6 cm.

Figure 6-42: System of a thin plus lens (+10 D) and a thin minus lens (–10 D) in air, separated by 15 cm. Example ☞: A two-lens optical system consists of lens 1 with power +4.00 D and lens 2 with power – 4.00 D, situated 25 cm after lens 1. What is the total system optical power and focal length? Again, if we simply add the two powers F1 and F2, we produce a zero power system. The lenses are separated by d = 0.25 m in air (refractive index nbetween = 1.0). So the total equivalent optical power is

FTOT = F1 + F2 −

6-250

d 0.25 m F1  F2 = (+4 D) + ( − 4 D) −  (+4 D)  ( − 4 D) = +4.0 D nbetween 1.0

THICK LENSES AND LENS SYSTEMS

Thus, the system has a net positive optical power of +4.0 D. The focal length (in air) is the reciprocal of the total power: fe = 1/FTOT = 1/(+4.0D) = +0.25 m = +25 cm.

Figure 6-43: System of a thin plus lens (+4.0 D) and a thin minus lens (–4.0 D) in air, separated by 25 cm.

Example ☞: A two-lens optical system consists of lens 1 with focal length 40 cm and lens 2 with focal length 30 cm, situated 80 cm after lens 1. What is the total system optical power and focal length?

Figure 6-44: System of a two thin plus lenses (f1 = 40 cm and f2 = 30 cm) in air, separated by 80 cm. The lenses have powers F1 = 2.5 D and F2 = 3.3 D. If we simply add the powers, a plus-powered system (+5.83 D) would be produced. The lenses are separated by a thickness d = 0.80 m in air (refractive index

nbetween = 1.0). So the equivalent optical power is –0.833 D:

FTOT = F1 + F2 −

Note

d nbetween

F1  F2 = (+2.5D) + (+3.33 D) −

0.80 m 1.0

 (+2.5 D)  (+3.33 D) = − 0.833 D

: Surprisingly, a system of two plus lenses produces a system whose power is negative. The

reason for this is that the corrective third term in Gullstrand’s formula exceeds the simple sum of the two powers. Since the two individual powers have the same algebraic sign, the term –d·F1·F2 becomes negative, which is particularly dominating when the separation d between the two lenses is significant.

6-251

GEOMETRICAL OPTICS

6.5.2

Cardinal Points in a Thin Lens System

If the lenses are surrounded by air and separated by a distance d, to calculate the system optical power we apply Eq. (6.1) with n = 1.0 and use t = d. The vertex points (noted as V) from which we measure the separation d and the distances δ and δ΄ are the optical centers of the two thin lenses. The distances from the principal points P and P΄ to the vertex points are calculated via a version of Eq. (6.6), in which

nlens is simply nmedium between, which most likely equals 1.0 if the medium between the points is air: ΄ = −d

F1 1  Fe n medium between

and

 = d

F2 1  Fe n medium between

(6.16)

Finally, because (if) the surrounding medium is the same on either side of the system, the principal points coincide with their respective nodal points.

Figure 6-45: System of two thin lenses in air. The lenses each have an optical power of F = 5.0 D and are separated by d = 0.1 m. The principal planes are crossed (plane H is situated after plane H΄). Example ☞ : Two lenses (F1 = F2 = +5.0 D) are separated in air by a distance d = 0.1 m (Figure 6-45). What is the equivalent power, and where are the principal points located? The equivalent optical power is FTOT = 5.0 + 5.0 – 0.1· 5.0·5.0 = 7.5 D, and the effective focal length is f΄e = 1/FTOT = 0.133 m = 13.3 cm. The principal points P΄ and P are separated from the corresponding vertices by distances δ΄ and δ:

 ΄ = − (0.1 m) 

Note

+5.0 D 1



+7.5 D 1

= − 0.066 m

 = (0.1 m) 

+5.0 D 1



+7.5 D 1

= + 0.066 m

: In this example, δ > d/2. This means that the two principal planes are crossed, as we first

encounter plane H΄ and then plane H.

6-252

and

THICK LENSES AND LENS SYSTEMS

For two thin lenses in air: The spacing d is the separation between the two lenses. The medium between them has a refractive index of 1.0. The vertex points correspond to the locations of the centers of the thin lenses. Equivalent focal length and object /image locations are measured from the corresponding principal planes. Back focal length (BFL) and front focal length (FFL) are measured from their corresponding vertex points.

Example ☞: Determine the cardinal points in a system comprising a positive lens 1 with F1 = +10.0 D (f1 = +10 cm) and a negative lens 2 with F2 = –10.0 D (f2 = –10 cm) lens 2. The lenses are separated by 15 cm in air. This two-lens system has a total power of FTOT = +15.0 D and an effective focal length in air fe = +6.6 cm. To find the principal plane locations, we note that the medium between them is air (nbetween = 1.0), so

΄ = − d   = d

F1 1 +10 D 1  = − (0.15 m)   = − 0.1 m and FTOT nbetween +15 D 1

F2 1 −10 D 1  = (0.15 m)   = − 0.1 m FTOT nbetween +15 D 1

The principal planes are located such that δ΄ = −10 cm and δ = −10 cm. The negative sign means that both principal planes are located to the left of their respective vertex points (which coincide with the positions of each individual lens), as shown in Figure 6-46.

Figure 6-46: Locations of the principal planes for a two-lens system. To find the nodal points, we note that the refractive indices before (to the left of) and after (to the right of) the two-lens system are equal (n = n΄ = 1.0). Therefore, the nodal points coincide with the corresponding principal points. 6-253

GEOMETRICAL OPTICS

Example ☞:: Find the image formed by this system if an object is placed 20 cm in front of (before) lens 1. In object space we measure locations from the object-space principal plane H, while in image space we measure from the image-space principal plane H΄. The object location is therefore measured from the principal plane H; since the object in this case is 20 cm before lens 1, the object location is x = −10 cm (still to the left of H). We now apply the imaging relationship for f΄e = +6.6 cm, measured from principal plane H΄. The calculations show that the image location is x΄ = +20 cm, which is also to be measured from principal plane

H΄. The plus sign indicates that the image is formed to the right of H΄, which itself is 10 cm to the left of lens 2, so the image is formed 10 cm to the right of lens 2. The magnification is m = x΄/x = −2.0. The image is real because it is produced by intersecting rays (a converging ray bundle), is magnified (×2), and is inverted (note the − sign).

Figure 6-47: The system is now treated using cardinal points (reduced system, § 6.5.4). Shown are the three ray-tracing rules used to locate the image. Note

: The system acts like a positive lens. The object is placed at a distance longer than the

equivalent focal length and forms a real, magnified, and inverted image.

6.5.3 The Afocal System Consider a system of two plus lenses, a +2.0 D and a +10.0 D lens, separated by a distance of +0.6 m in air. One might expect this lens system to produce a positive total power FTOT as follows:

FTOT = F1 + F2 − d· F1· F2 = (+2.0 D)+(+10.0 D) – (0.6 m)·(+2.0 D)·(+10.0 D) = +12.0 D – 0.6·20.0 = 0.0 D.

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THICK LENSES AND LENS SYSTEMS

Wait a minute—this must be a mistake. Two plus lenses produce … a zero-power system? After re-checking the math a few times in vain, we realize that this system has indeed a net optical power of zero diopters; there is no numerical error. This is an afocal system, examples of which are telescopes,29 beam expanders, and teleside converters. In an afocal system, the image-space focal point of the first system is coincident with the object-space (front) focal point of the second system. In an afocal system, the equivalent optical power FTOT is always zero regardless of the actual focal length values; we can generalize the previous result by using d = f1 + f2 : Afocal System Equivalent Power FTOT:

FTOT

n ( F2 + F1 ) n n + F F2 F1  F2 d = F1 + F2 − F1  F2 = F1 + F2 − 1 F1  F2 = F1 + F2 − F1  F2 = 0 nbetween n n

(6.17)

There are several ramifications to this arrangement. Essentially, because of the zero optical power, an afocal system acts according to ‘collimated in – collimated out.’ This means that if a collimated ray bundle (zero object vergence) is incident on the first lens, light leaving the system is also collimated (zero image vergence).

Figure 6-48: An afocal system comprising two lenses separated by the sum of their focal lengths.

This also means that in an afocal system both principal planes lie at infinity on opposite sides of the system. Indeed, the locations (δ and δ΄) of the principal planes, derived using the relationships in Eq. (6.16), are

΄ = − d 

F1 1 F  = −d 1 =  FTOT 1 0

and

 = d

F2 1 F  = d 2 =  FTOT 1 0

(6.18)

Two other properties of afocal systems are that the transverse and longitudinal magnifications are constant, and that equispaced planes map into equispaced planes. The angular magnification achieved with an afocal system is the ratio of the two focal lengths of the two lenses comprising the system.

29

Introduction to Optics § 5.2.2 Telescope Principle of Operation. 6-255

GEOMETRICAL OPTICS

Example ☞ : Two lenses with F1 = +1.0 D and F2 = +5.0 D are used to form an afocal system in air. What would be their separation d ? There are two ways to obtain this. One is to compute and add the focal lengths:

f1 = 1/F1 = 1.0 m and f1 = 1/F2 = 0.2 m, so d = f1 + f2 = 1.2 m. The second is to determine d from the total power relationship:

FTOT = F1 + F2 −

d 1

F1  F2 = 0  + 1.0 + 5.0 − d 1.0  5.0 = 0  d =

6.0 5.0

m = 1.2 m

6.5.3.1 Telescope Types The telescope is an example of an afocal optical system. In its simplest form, it comprises two plus lenses separated by the sum of their focal lengths.30 The system presented in Figure 6-48 has two plus lenses. This is a form of the astronomical (Keplerian) telescope. The observed object (such as a star) is located far away, typically at optical infinity. The lens closer to the object is called the objective lens, while the lens closer to the observing eye is called the eyepiece (Figure 6-49). The objective lens forms a real and inverted image at the secondary focal point of the objective, which is also the primary focal point of the eyepiece; thus, the telescope length is d = fo΄ + fe. This is a mathematical expression of the fact that the back focus (secondary focal point) of the objective coincides with the front focus (primary focal point) of the eyepiece.

Figure 6-49: An astronomical telescope consists of two plus lenses spaced by the sum of the their focal lengths.

The Galilean telescope (Figure 6-50) is another type in which the objective lens is, as always, a plus-powered lens that forms a real and inverted image at the primary focal point of the eyepiece. The eyepiece, however, is a minus-powered lens. The two lenses are separated not by the sum but by the difference of their focal lengths. Numerically, this separation is the algebraic sum of the two focal lengths since the eyepiece focal length is negative. 30

Introduction to Optics § 5.2.4 Telescope Types and Designs.

6-256

THICK LENSES AND LENS SYSTEMS

Figure 6-50: A Galilean telescope consists of a plus lens (objective) and a minus lens (eyepiece) spaced by the ‘sum’ of the their focal lengths (their numerical difference because the eyepiece is a minus lens).

6.5.4

Cardinal Points in a Thick Lens System

When dealing with a complex system that involves many thick lenses, we want to replace the system with just six cardinal points: a pair of principal planes/points, a pair of nodal points, and a pair of focal points. Thus, we replace the complex system of a number of thick lenses with a reduced system. The generally applicable rules for a system with real, thick lenses is presented in the case of two thick lenses (Figure 6-51), which is the simplest case; the model can be extended by adding more thick lenses one at a time. The cardinal points for each thick lens are noted with the corresponding subscripts 1 and 2, while in the reduced system they have no subscripts. The principal points P1 of the first lens and P΄2 of the second lens function as vertex points (V ≡ P1) and V΄ ≡ P΄2) for the reduced system.

Figure 6-51: A system of two actual (thick) lenses can be replaced by six cardinal points.

To calculate the equivalent system power (FTOT), we use Eq. (6.1). Now d is the distance from the object-space principal plane of the first lens (H1) to the image-space principal plane (H΄2) of the second lens. For F1 and F2, we use the optical powers of the two individual (thick) lenses. Typically, there is air between the thick lenses and on either side of the lenses. In this simplified case, n, n΄, and n lens all equal 1.0. 6-257

GEOMETRICAL OPTICS

Then we calculate locations δ and δ΄ of the principal points P and P΄, employing Eq. (6.5). The origin for δ is the vertex point V = P1, and the origin for δ΄ is the vertex point V΄ = P΄2. Finally, we calculate the locations of the nodal points from the locations of the principal points by employing Eq. (6.7).

Figure 6-52: Equivalence of a system with many optical elements to a single reduced system, employing a pair of principal planes/points, a pair of nodal points, and a pair of focal points.

6.5.5

Intermediate Image Technique in Two or More Lenses

There are two ways to model a system having two or more lenses: The first employs the principal planes and ray diagrams as if there is only one set of cardinal points—in other words, the reduced system. The challenge is that we have to find the location of the principal (and/or) nodal points. We then determine the final image location and size from the initial object in relation to the cardinal points by applying either thick ray diagrams (Figure 6-35), or the analytical imaging relationship given in Eq. (6.15), or by using a computer-based ray-tracing program. The numerical approach to this type of solution employs the analytical imaging expressions of Eq. (6.15). Essentially, the steps are just as are implemented in the thin lens summation of incident vergence and lens power. The difference is in the details of the location calculation of these vergence values. Specifically, •

The object vergence L is calculated from the object location x with respect to the principal plane H.

•

The equivalent (total) optical power FTOT is added to this value to provide the image vergence L΄.

•

The reciprocal of the image vergence is the image location x΄ with respect to the principal plane H΄.

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THICK LENSES AND LENS SYSTEMS

1. Find the reduced-system cardinal points.

2. Locate the image by application of the thick ray diagrams,

(or) employ the analytical imaging expressions [Eq. (6.18)].

The second type of solution employs the notion of successive, or intermediate imaging. We apply the simple, one-thin-lens imaging rules (as shown in § 4.3) to locate the image formed by the first lens, ignoring the presence of the second lens. This is the intermediate image. The second lens now treats the intermediate image as an object at the same physical location and of the size as the image formed by the first lens (with a change in the point of origin, which now lies on the second lens). We then proceed to find the secondary image from this object (by applying ray diagrams or lens imaging rules), ignoring the presence of the first lens, even if it intersects the actual ray paths, and even if the auxiliary ray diagrams do not exist in reality; i.e., the rays in the diagrams correspond to non-permitted paths through the optical system. In other words, we simply disregard the presence of the first lens. Its usefulness ‘perishes’ after the formation of the intermediate image. The total, overall magnification is the product of the individual magnifications. This process can be repeated indefinitely until it is applied to the last lens (or optical element). Example ☞: A lens system is formed by two positive lenses with focal lengths f1 = 15 cm and f2 = 20 cm, separated in air by d = 56 cm. An object is placed at location x1 = –40 cm in front of (before) lens ❶. Find the final image location and magnification through the two-lens system. Step 1: Determination of the intermediate image (Figure 6-53): For an object at x1 = –40 cm (L1 = –2.50 D) and a lens with F1 = +6.66 D, the image is produced with L1΄ = +4.166 D, located at x1΄ = +0.24 m. The image is therefore formed at +24 cm after lens ❶. It is real and has a (primary) magnification m1 = (+24)/(−40).

Figure 6-53: Step 1: determination of the intermediate image. 6-259

GEOMETRICAL OPTICS

Step 2: Determination of the new object (Figure 6-54): This initial image is the object for the second lens. Because this object will be imaged by lens ❷, its location must be referenced in relation to lens ❷, which is

d = 56 cm to the right of lens ❶. Therefore, this intermediate object is located x2 = −32 cm from lens ❷. Step 3: Determination of the final image: We implement the imaging relationship for x2 = –32 cm (L2 = –3.125 D) and F2 = +5.0 D. The image is produced with L2΄ = +1.875 D, located at x2΄ = +0.533 m. The image is therefore formed +53.3 cm after lens ❷. It is a real image and has a (secondary) magnification of

m2 = (+53.3)/(−32) = −1.66.

Figure 6-54: Steps 2 and 3: determination of the new object location and final image location. The final image is erect (a reversal of the inverted image). The total magnification is the product of the two individual magnifications. The final image size equals the size of the initial object, since the magnification is

m = (−24/+40) · (−53.33/+32) = +1.0.

Figure 6-55: Imaging in a two-lens system using the intermediate image method.

Note

: This system has a total (equivalent) optical power of FTOT = –7.00 D. The object-space principal

plane H is situated at δ = –40 cm, and the image-space principal plane H is situated at H΄ = +53.33 cm. The optics professor is elated to discover that this case illustrates a previous statement that the principal planes are optical conjugates. When the object is placed at the object-space principal point, its image is 6-260

THICK LENSES AND LENS SYSTEMS

formed at the image-space principal point with unity magnification. Think of it in terms of the imaging relationship if x = 0: This relationship can only be satisfied if, for any system power, x΄ = 0 as well.

1. We find the location and size of the primary image by applying a summation of vergence and optical power (analytical solution);

(or) we locate the primary

2. This is the object for the

image with ray diagrams

second lens. The procedure

applicable for a single, thin

can be repeated for any

lens (schematic solution).

additional optical element.

Example ☞: A system consists of two thin lenses, a positive lens with focal length f1 = +20 cm and a negative lens with f2 = –20 cm. The lenses are separated by d = 30 cm. We place an object at a distance

x1 = – 30 cm in front of (before) the first lens. Find the location and determine the properties of the final image through this two-lens system. Step 1: Determination of the intermediate image: According to the imaging relationships, the initial image from lens ❶ is formed at the location x΄1 = +60 cm, which is validated by construction ray diagrams [Figure 6-56 (left)]. The magnification is m1 = −2.0.

Figure 6-56: A lens system with a plus lens and a minus lens: (left) the initial image and (right) final image. Step 2: Determination of the new object: This initial image is the new object for lens ❷. The object location is now measured from lens ❷, where the new coordinate origin (0, 0) is placed. Therefore, the new object location is x2 = +30 cm (the object is now virtual). Step 3: Determination of the final image: Given the new object for lens ❷, we schematically find the final image location, ignoring the fact that the rays have to go through lens ❶. The final image is virtual, located at x΄2 = −60 cm. The magnification due to lens ❷ is m2 = −2.0. The final image is formed at the 6-261

GEOMETRICAL OPTICS

exact location of the initial object, with a total magnification of m = (−2.0)·(−2.0) = +4.0, the product of the two individual magnifications.

Note

: The final image is magnified (4×) compared to the object, despite the fact that the system is

composed of a converging and a diverging lens. This is similar to the operation of the telephoto lens, which is a combination of converging and diverging lens groups that produces large magnifications.

Even if the second lens is between, or appears

To find the final image employing the

to block, the construction rays, we simply

intermediate image, we now ignore the

ignore its presence and consider only the primary object with respect to the first lens and the optical power (focal length) of the

presence of the first lens, and we draw the rays and calculate the location and size of the

first lens.

final image as if the first lens does not exist.

Example ☞: Find the equivalent optical system of the previously described lens system. We consider each lens as thin, with powers F1 and F2, separated by a ‘medium’ of air of thickness d = 30 cm and refractive index n between = 1.0. If we simply add the optical power of each lens, the sum is 5 D + (–5) D = 0 D, which means that the optical power is zero! However, the lenses are not in contact. ✏ Optical power:

FTOT = F1 + F2 −

d 0.30 m F1  F2 = (+5 D) + ( − 5 D) −  (+5 D)  ( − 5 D) = +7.5 D n 1.0

✏ Focal length (in air): fe = 1/FTOT = 1/(+7.5 D) = +0.133 m = +13.3 cm ✏ Principal plane locations. The medium between the lenses is actually air, so n between = 1.0.

΄ = −d  = d

F1 1 +5.0 D 1  = − (0.3 m)   = − 0.2 m and FTOT nbetween +7.5 D 1

F2 1 −5.0 D 1  = (0.3 m)   = − 0.2 m FTOT nbetween +7.5 D 1

✏ Nodal points: These coincide with the corresponding principal points because the refractive indices before and after the two lens system are equal (n = n΄ = 1.0). The principal planes are located at δ΄ = –20 cm and δ = –20 cm. The negative sign in both means that the two principal planes are located to the left of their respective vertex points, which in this case coincide with the positions of each thick lens. After locating the cardinal points (i.e., defining the reduced system), we need the object location with respect to the principal plane. The object location is referenced from the object principal plane H, so x = −10 cm. We now apply the imaging relationship given in Eq. (6.15) for Fe = +7.5 D. The image location is x΄ = −40 cm, referenced from 6-262

THICK LENSES AND LENS SYSTEMS

the principal plane H΄. The negative sign indicates that the image is to the left of H΄. The magnification is

m = x΄ / x = +4.0. The image is virtual because it is produced by ray extrapolation in a direction opposite to the direction of ray propagation.

Figure 6-57: Reduced system showing the construction rays that help locate the image. Note

: This lens system acts as a magnifying lens. The object is located (from the corresponding

object-space principal plane) at a distance that is shorter than the equivalent focal length. We note the complete agreement between the results by either technique.

6.5.6 The Thick Lens as a Two-Element System The technique of identifying the image formed by two distinct thin lenses can be expanded to include any two optical elements. A thick lens is formed by two of such elements, the front and back surface, which form two independent SSRIs separated by the lens thickness t in a medium, the lens material having refractive index n. For simplicity, we will assume that the lens is surrounded by air. This technique enables the calculation of the front and back vertex powers as well as the corresponding focal lengths of the lens. To calculate BFL, we locate the image formed by the two SSRIs from an object at optical infinity and relate the location of this object from the back surface vertex. To calculate FFL, we locate the object location that will produce an image at optical infinity. Alternatively, we can reverse the direction of light propagation, place an object at optical infinity to the right of the system, and identify the object location. Example ☞: A symmetric biconvex glass (n = 1.5) lens that is 6 cm thick and surrounded by air has a front surface power F1 = +10 D and back surface power F2 = +10 D. Where is the image of an object at optical infinity? 6-263

GEOMETRICAL OPTICS

The first step is to locate the intermediate image formed by the front surface, which alone acts as the first imaging SSRI; this will be the object for the back surface, which acts as the second imaging SSRI; the final image is formed at the secondary focal point of the system, which leads to the back vertex focal length. We use subscript 1 for the quantities relating to the front surface. The first SSRI with F1 = +10 D separates air from glass. For this SSRI, object-space n1 = 1.0 and image-space n1΄ = 1.5. The second SSRI with F2 = +10 D separates glass from glass. For this SSRI, object-space n2 = 1.5 and image-space n2΄ = 1.0.

Figure 6-58: BFL calculation using the two-image, element-by-element method produced by the two SSRIs of the front and the back lens surfaces. We implement the imaging relationship, considering that object vergence is zero (the object is at optical infinity): L1 + F1 = L1΄ ⇒ 0 + (+10 D) = L1΄ ⇒ L1΄ = +10 D. The location of this initial image, x1΄= n1΄ / L1΄= 1.5 / (+10.0 D) = +0.15 m = +15 cm, is to the right of this refracting element (the front lens surface). This intermediate image is now the object for the second surface. Considering a lens thickness of 6 cm, this point is located x2 = +0.09 m = +9 cm to the right of the lens back surface. It is a virtual object with L2 = n2 / x2 = 1.5 / (+0.09 m) = +16.66 D. It is important to note that the vergence and the distance are now measured from the second element, the lens back surface (vertex point V΄). We implement the imaging relationship for SSRI 2: L2 + F2 = L2΄ ⇒ L2΄ = +16.66 D +10.0 D = +26.66 D. This brings the second, final image to x2΄= n2΄ / L2΄= 1.0 / (+26.66 D) = +0.0375 m. This is the back focal length (BFL) of the thick lens. The result is identical to the result in an earlier example (see Figure 6-22). Therefore, Back vertex power: FBVP = L2΄ = +26.66 D. Back focal length: BFL = 1.0/FBVP = 1.0/(+26.66 D) = +0.0375 m = +3.75 cm.

Note

6-264

: Because this specific lens is symmetric (F1 = F2 = +10 D), it follows that BFL = FFL.

THICK LENSES AND LENS SYSTEMS

6.6 ADVANCED PRACTICE EXAMPLES Equivalent Power in a Ball Lens Because the ball lens has a perfectly circular profile, it is a biconvex lens of a thickness equal to the diameter d = 2r of the sphere, and therefore equal to twice the radius r. The two radii of curvature r1 and

r2 have the same magnitude but opposite algebraic signs: r1 = r and r2 = –r (Figure 6-59). If the lens material has a refractive index n, the lens equivalent optical power Fe is 2

Fe, ball lens

( n − 1) − 2 r ( n − 1) n − 1 1− n 2 r  n − 1   1− n  = + −  = 2    r −r n  r   −r  r n r2

=

2 ( n − 1)

nr

(6.19)

The equivalent focal length fe of this lens in air is the reciprocal of the equivalent optical power:

fe, ball lens =

1

Fe, ball lens

=

nr 2 ( n − 1)

For a ball lens in air, made of glass with n = 1.5, the equivalent focal length is fe ball lens = 1.5 · r.

Figure 6-59: The ball lens is a thick, converging lens.

Principal Planes in a Ball Lens Once we identify the principal planes in a glass ball lens surrounded by air (Figure 6-59), we can answer the question, ‘From where do we measure the focal length 1.5 · r in the ball lens?’ The answer is simply, ‘from the principal plane H΄.’

n −1 1− n , F2 = = F1 r −r

F1 =

According to Eq. (6.6):

n −1 ( n − 1)  n  r = − r F1 1 1 ΄ = −t   = − ( 2r )  r  = − ( 2r )  n −1 n Fe nlens 2 ( n − 1)  n  r 2 nr

and

Fe = 2

n −1 nr

Optical powers:

Principal point P΄ is located at the center of this lens, which has a thickness of 2r. Principal point P is also located at the center of this lens (δ = +r) (Figure 6-60).

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GEOMETRICAL OPTICS

Figure 6-60: The principal points in a ball lens are located exactly at the center of the lens. Recall that the principal planes are not flat surfaces, but are actually curved.

Summation Power Property in a Hemispheric Planoconvex Lens A ball lens can be regarded as the simple sum of two back-to-back hemispheric planoconvex lenses (Figure 6-61). Geometrically, it is easy to accept this schematic summation property. The questions now are, how does the summation property apply to the optical power, and from exactly where do we measure the focal length, etc., if the separation between the two individual, thick lenses d is not zero?

Figure 6-61: The ball lens as a superposition of two hemispheric planoconvex lenses. A particular advantage of a two-thin-lens system is the fact that the vertex V coincides with the center of its respective thin lens. Despite the fact that a planoconvex lens is a thick lens, its principal plane H is directly on the refracting (curved) surface vertex (P coincides with V). Therefore, the lens separation d can be the distance from the first (front) principal plane of the first lens (H1) to the second (back) principal plane (H΄2) of the second lens. Recall the following: Ball lens power [Eq (6.19)]: F ball lens = 2

( n − 1) nr

Planoconvex lens power [Eq. (6.20)]: Fplanoconvex =

n −1 r

We now apply Gullstrand’s formula to the two planoconvex lens optical powers, assuming that d = 2R. The medium between has a refractive index of n. Therefore, the optical power of the ball lens equals the thicklens sum of the two planoconvex lens optical powers:

Fd=2R =

F1 planoconvex 1

6-266

+

F2 planoconvex 2

−

2 ( n − 1) d n − 1 n − 1 2r n − 1 n − 1  F1  F2 = + −   = n r r n r r nr

THICK LENSES AND LENS SYSTEMS

Principal Planes in a Symmetric Biconvex Lens In a symmetric biconvex lens, the object-space principal plane H is positioned to the right of the objectspace vertex point V because δ is always positive. The image-space principal plane H΄ is positioned to the left of the image-space vertex point V΄ because δ΄ is always negative. Example ☞: Locate the principal planes in the following cases of a symmetric biconvex lens located in air:

n lens = 1.5, surface powers F1 = F2 = +10.0 D.

Case 1: Lens thickness t = 5.0 cm. The first step is to calculate the equivalent lens power. Using Eq. (6.1), Fe = +16.66 D. Then we calculate the location of the principal plane. Using Eq. (6.6), δ = +2.0 cm and δ΄ = –2.0 cm. Both principal planes are inside the biconvex lens at equal distances from their respective vertex points [Figure 6-62 (left)]. Because δ < t/2, we first encounter plane H and then plane H΄.

Figure 6-62: Principal planes in a symmetric biconvex lens: (left) t = 5 cm and (right) t = 15 cm. Case 2: Lens thickness t = 15.0 cm. Following the same steps as in case 1, we find that Fe = +10.0 D, δ = +10.0 cm, and δ΄ = –10.0 cm. Both principal planes are inside the biconvex lens at equal distances from their respective vertex points. In this example, δ = |δ΄| = 10 cm > t/2 = 7.5 cm. This means that in this case the tip of the directional distance for

δ is situated after δ΄—the two principal planes are crossed, as we first encounter plane H΄ and then plane H [Figure 6-62 (right)].

The transition from the non-crossed principle planes to the crossed principal planes occurs when the lens thickness increases to the point at which the lens becomes a ball lens, meaning that t = 2r. In this case, the object-space principal plane H is positioned at the same spot as the image-space principal plane H΄. When the lens thickness further increases to the point where t > 2r, the directional distances will begin to show that δ is situated after δ΄, and the principal planes will be crossed (Figure 6-63).

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Figure 6-63: Parametric analysis of the locations of the principal planes in a biconvex lens (with F1 = +10.0 D and F2 = +10.0 D) as a function of lens thickness.

Principal Planes in a Symmetric Biconcave Lens In a symmetric biconcave lens, the object-space principal plane H is positioned to the right of the object-space vertex point V because δ is always positive. The image-space principal plane

H΄ is positioned to the left of the image-space vertex point V΄ because δ΄ is always negative. The principal planes never cross in a symmetric biconcave lens because the object-space principal plane H is always positioned before the image-space principal plane H΄. The sum of the absolute values of δ and δ΄ is always less than the lens thickness t. Example ☞: Locate the principal planes in the following cases of a symmetric biconcave lens located in air:

n lens = 1.5, surface powers F1 = F2 = –10.0 D. Case 1 [Figure 6-64 (left)]: Lens thickness t = 5.0 cm: Fe = –23.33 D, δ = +1.4 cm, and δ΄ = –1.4 cm. Case 2 [Figure 6-64 (right)]: Lens thickness t = 15.0 cm: Fe = –30.00 D, δ = +3.3 cm, and δ΄ = –3.3 cm.

Figure 6-64: Principal planes in a symmetric biconcave lens: (left) t = 5 cm and (right) t = 15 cm.

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THICK LENSES AND LENS SYSTEMS

Figure 6-65: Parametric analysis of the locations of the principal planes in a biconcave lens (with F1 = – 10.0 D and F2 = –10.0 D) as a function of lens thickness.

Principal Planes in a Plus Meniscus Lens The locations of the principal planes are influenced by the various shapes that a meniscus lens may have. These locations are determined by the positive and negative surface powers, which, in a meniscus lens, are different. A plus meniscus lens can have a front surface power of F1 = +10.0 D (like many ophthalmic lenses) and a back surface power of F2 = –5.0 D, or a front surface power of F1 = –5.0 D and a back surface power of F2 = +10.0 D; the latter case is a flipped version of the former case. In a plus meniscus lens, the directional distances for the principal plane locations δ and

δ΄ have an algebraic sign that is opposite to the sign of the front surface power F1. If the front surface power is positive, then both δ and δ΄ are negative, and the object-space principal plane H is closer to its vertex point V than the image-space principal plane H΄ is to its vertex point V΄. Example ☞: Locate the principal planes in the following cases of a plus meniscus lens located in air:

n lens = 1.5, surface powers F1 = +10.0 D and F2 = –5.0 D. Case 1 [Figure 6-66 (left)]: Lens thickness t = 1.0 cm: Fe = +5.33 D, δ = –0.63 cm, and δ΄ = –1.25 cm. Case 2 [Figure 6-66 (center)]: Lens thickness t = 5.0 cm: Fe = +6.66 D, δ = –2.5 cm, and δ΄ = –5.0 cm. Case 3 [Figure 6-66 (right)]: Lens thickness t = 15.0 cm: Fe = +10.0 D, δ = –5.0 cm, and δ΄ = –10.0 cm.

Figure 6-66: Principal planes in plus meniscus lens: (left) t = 1 cm, (center) t = 5 cm, and (right) t = 15 cm. 6-269

GEOMETRICAL OPTICS

Figure 6-67: Parametric analysis of the locations of the principal planes in a plus meniscus lens (with F1 = +10.0 D and F2 = –5.0 D) as a function of lens thickness.

Principal Planes in a Minus Meniscus Lens A minus meniscus lens can have a front surface power of F1 = +10.0 D and a back surface power of F2 = –20.0 D, or a front surface power of F1 = –20.0 D and a back surface power of F2 = +10.0 D, the latter case being a flipped version of the former case. In a minus meniscus lens, the directional distances for the principal plane locations δ and

δ΄ have the same algebraic sign as the front surface power. In other words, if the front surface power is positive (as in most ophthalmic lenses, being convex), then both δ and δ΄ are positive, and the object-space principal plane H is farther from its vertex point V than the image-space principal plane H΄ is to its vertex point V΄. Example ☞: Locate the principal planes in the following cases of a minus meniscus lens located in air.

n lens = 1.5, surface powers F1 = +10.0 D and F2 = –20.0 D. Case 1 [Figure 6-68 (left)]: Lens thickness t = 1.25 cm: Fe = –8.33 D, δ = +2.0 cm, and δ΄ = +1.00 cm. Case 2 [Figure 6-68 (center)]: Lens thickness t = 2.5 cm: Fe = –6.66 D, δ = +5.0 cm, and δ΄ = +2.50 cm. Note that in this case the principal planes coincide! Case 3 [Figure 6-68 (right)]: Lens thickness t = 3.75 cm: Fe = –5.0 D, δ = +10.0 cm, and δ΄ = +5.00 cm. Note that in this case the principal planes are crossed!

Figure 6-68: Principal planes in a minus meniscus lens: (left) t = 1.25 cm, (center) t = 2.5 cm, and (right) t = 3.75 cm. Note that in the right case the principal planes are crossed. 6-270

THICK LENSES AND LENS SYSTEMS

The Minus Meniscus as an Afocal System The minus meniscus lens becomes an afocal system when the lens thickness is t = 7.50 cm. If we use the surface powers F1 = +10.0 D and F2 = –20.0 D, we will find that Fe = 0.0 D. This is an afocal system (§ 6.5.3), in which case the principal planes are at infinity: δ = –∞ and δ΄ = –∞. In this afocal system, the image-space focal point of the front surface is coincident with the object-space (front) focal point of the second surface.

The Minus Meniscus as a Positive Power Lens For any other thickness greater than t = 7.50 cm, the equivalent lens power is positive—yes a minus meniscus lens can be positive; for example, for a lens of thickness t = 15 cm, Fe = +10.0 D, δ = –20.0 cm, and δ΄ = –10.00 cm. The principal planes swing back to being to the left of their respective vertex points.

Figure 6-69: Parametric analysis of the locations of the principal planes in a minus meniscus lens (with F1 = +10 D and F2 = –20 D) as a function of lens thickness.

The Höegh Meniscus and the Size Lens The Höegh meniscus, named after the German optical lens designer Emil von Höegh, is a thick lens whose refracting surfaces have the same (magnitude and sign) radii of curvature (r1 = r2 = R), but whose centers of curvature are slightly displaced. Because of this, each surface power is equal in size, with opposite signs:

F1 =

n −1 n' − n = lens r R

and

F2 =

n΄ − n 1− nlens = = − F1 r R

If the lens is regarded as thin (t = 0), its optical power (the sum of the two surface powers) would be zero. However, the equivalent optical power is not zero because the lens is thick (t ≠ 0). The power is positive, which means that the lens is a converging lens:

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n − 1 1− n lens t t Fe, Höegh = F1 + F2 − F1  F2 = lens + − n R R nlens

t  n lens − 1   nlens − 1   1− n lens   R  R  = n  R      lens  

2

Figure 6-70: The Höegh meniscus is a thick lens in which the refracting surfaces have the same radii of curvature, but the centers of curvature are slightly displaced. Example ☞: Calculate the principal plane locations for the Höegh meniscus when surrounded by air.

n −1 n −1 = , Surface powers: F1 = r1 R

F2 =

1− n

r2

1− n

t  n −1 = = − F1 , and Fe = FHoegh =  R n  R 

2

Locations of principal planes: n −1

F 1 1 R R ΄ = −t  1  = −t   = − 2 Fe nlens n −1 t  n −1 n   n R 

1− n

and

F 1 1 R R  = t 2  = t  = − 2 Fe nlens n −1 t  n −1 n   n R 

We note that if the Höegh meniscus is surrounded by the same medium, then δ΄ = δ; i.e., the principal points P and P΄ (and the corresponding principal planes H and H΄) are separated from their respective vertex points by equal distances, or d = t. There are some even more interesting findings. For example, the principal points are located (in the typical case where the lens is optically more dense than the surrounding medium such as glass lens in air) on the side that is opposite to the side of the centers of curvature because both δ values have algebraic signs that are opposite to the algebraic sign of the radii of curvature. For example, if n lens = 1.5, δ΄ = δ = –2R. Finally, at least one, if not both, principal plane (more likely the object-space H) is located outside the lens, since

δ < 0 (Figure 6-71).

Figure 6-71: Principal planes of the Höegh meniscus located outside and to the left of the lens. 6-272

THICK LENSES AND LENS SYSTEMS

Equivalent Power and Principal Planes in a Meniscus Lens Example ☞: Calculate the equivalent optical power, equivalent focal length, and principal plane locations for a meniscus lens in air, whose front surface power is F1 = +10.0 D and back surface power is F2 = –15.0 D. The lens is made of glass of refractive index nlens = 1.5 and has a central thickness of t = 1.5 cm. Equivalent power:

t 0.015 Fe = F1 + F2 − F1  F2 = (+10)+( − 15) − (+10)  ( − 15) = (+10)+( − 15) + 1.5 = − 3.5 D n 1.5 Focal length: fe = 1 =

Fe

1 = − 0.0285 m −3.5 D

This distance is measured from the image-space principal point P΄. The location of the image-space principal point P΄ from the image-space vertex point V΄, which is δ΄, is calculated by application of Eq. (6.6):

΄ = − t 

F1



1

Fe n lens

= − (0.015 m) 

+10 D



1

−3.5 D 1.5

= +0.0285 m = +2.857 cm

The finding that δ΄ is positive suggests that principal point P΄ (and principal plane H΄) is to the right of the image-space vertex point V΄, outside of the lens at a distance of 2.85 cm. Likewise, the location of the object-space principal point P (and principal plane H) from the object-space vertex point V, which is δ, is

 = t

F2 1 −15 D 1  = (0.015 m)   = +0.0428 m = +4.28 cm Fe nlens −3.5 D 1.5

The finding that δ is positive suggests that P (and principal plane H) is 4.28 cm to the right of the objectspace vertex point V. Considering that the lens thickness is 1.5 cm, P (and principal plane H) is also to the right and outside of the lens proper at a distance of about 2.78 cm (Figure 6-72).

Figure 6-72: Equivalent focal length and principal plane locations in a meniscus lens with F1 = +10 D, F2 = – 15 D, and lens t = 1.5 cm. 6-273

GEOMETRICAL OPTICS

Equivalent Power and Principal Planes in a Planoconvex Lens Example ☞: Calculate the equivalent optical power, equivalent focal length, and principal plane locations for a hemispherical planoconvex lens. Assume that the lens has a radius of curvature r and a lens material refractive index n = 1.5. The lens is surrounded by air.

The thickness of the hemispherical lens equals the radius of curvature of the convex surface: t = r. Optical Powers:

F1 =

n −1 n −1 = r1 r

Equivalent Power: Fplanoconvex = F1 + F2 − Focal Length:

t n

and

F1  F2 =

fe planoconvex =

n −1 r

F2 =

+ 0 −

1− n

r2

=

1− n



= 0

t n −1 n −1 0.5  0 = = n r r r

1 1 = = 2r 0.5 Fplanoconvex r

(6.20) (6.21)

This distance is measured from the image-space principal point P΄. Its location δ΄ is

F 1 ΄ = − t  1  Fe nlens

n −1 1 r r = −r  r  = − = − = − 0.66r n −1 n n 1.5 r

The finding that δ΄ is negative and less than t (δ΄ = r) suggests that P΄ is inside the lens, toward the refracting surface, at ⅔ the thickness of the hemispherical planoconvex lens (t = r). We now calculate δ:

 = t

F2 1 0 1  = r  = 0 n −1 n Fe nlens r

The finding that δ = 0 suggests that the object-space principal point P coincides with the refracting surface vertex point V. Point P (and plane H) is located exactly on the front surface of this lens (Figure 6-73).

Figure 6-73: Equivalent focal length and principal planes in a planoconvex hemispherical lens.

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Imaging in a Thick Meniscus Lens Example ☞: Identify the image location and type if an object is placed 24 cm in front of a thick meniscus lens with a front surface power (base curve) of F1 = +10.0 D, a back surface power of F2 = –8.0 D, a center thickness to t = 2.5 cm, and a lens material refractive index of n = 1.5, placed in air (Figure 6-74). Lens power: Fe = F1 + F2 −

t n

F1  F2 = +10 D − 8 D −

0.025 m 1.5

 (+10 D)  ( − 8 D) = +2 D + 1.33 D = +3.33 D

Thick lens equivalent focal length in air: f΄e = 1/F e = 1/(+3.33 D) = +0.30 m = +30 cm Location of principal planes, δ΄ for plane H΄ and δ for plane H:

΄ = −t 

F1



1

Fe n lens

= − (2.5 cm) 

10 D



1

3.33 D 1.5

= − 5 cm

and

 = t

1 −8 D 1  = (2.5 cm)   = − 4 cm Fe n lens 3.33 D 1.5

F2

Because the object is placed 24 cm in front of the lens, which means in front of the object-space vertex point, the object location is x = –20 cm, referenced to the object-space principal plane, which is 4 cm to the left of the lens.

Figure 6-74: Thick lens imaging example with the object placed 24 cm in front (to the left) of the lens. We can now implement the imaging relationship given in Eq. (6.15) for the known object position (position

x = –0.2 m, or vergence L = –5.0 D) and lens equivalent power (F = +3.33 D). This results in an image vergence of L΄ = –1.66 D, or an image location of x΄ = –0.6 m. The negative vergence suggests that the image is virtual. The magnification is m = +3.0, which suggests a magnified and erect image. The image location x΄ = –0.6 m is referenced to the image-space principal plane H΄, which lies 5 cm to the left of the back surface, or 2.5 cm to the left of the front surface. Therefore, the image is 62.5 cm to the left of the lens.

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Two-Lens System Imaging Example ☞: A lens system is formed by two positive lenses with focal lengths f1 = 40 cm and f2 = 30 cm, separated by d = 80 cm. An object is placed at a location x1 = –50 cm in front of (before) lens ❶. Find the final image location and magnification through the two-lens system.

Figure 6-75: Imaging in a two-lens system using the intermediate image. Step 1: Determination of the intermediate image: For an object at x1 = –50 cm (L1 = –2.0 D) and a lens with F1 = +2.50 D, the image is produced with L1΄ = +0.50 D, formed at x1΄ = +2.0 m after lens ❶. It is real and has a (primary) magnification of m1 = (+2.00)/(−0.50) = −4.0. Step 2: Determination of the new object: The initial image is the object for the second lens. Because this object will be imaged by lens ❷, its location must be referenced in relation to lens ❷, which is 80 cm to the right of lens ❶. Therefore, this intermediate object is located x2 = +1.20 m from lens ❷.

Figure 6-76: Determination of the new object location and final image location. Step 3: Determination of the final image: Object location x2 = +1.2 m (L2 = +0.833 D) and F2 = +3.33 D. The image vergence is L2΄ = +4.166 D, located at x2΄ = +0.24 m. The image is therefore x΄2 = +24 cm after lens ❷. It is a real image and has a (secondary) magnification of m2 = (+24)/(+120) = +0.2. The final image is inverted with respect to the original object. The total magnification is m = m1 · m2 = (−4.0)·(+0.2) = −0.8. The equivalent optical power of this system is FTOT = −0.833 D. The principal planes are located at δ = –3.2 m and δ΄ = +2.4 m. Therefore, the object location from the object-space principal plane is x = +2.7 m (L = +0.37 D). Combined with the equivalent power FTOT = −0.833 D, the image vergence is L΄ = −0.463 D, or x΄ = −2.16 m from the image-space principal plane H΄, which is +0.24 m from the image-space vertex point (lens ❷). 6-276

THICK LENSES AND LENS SYSTEMS

Example ☞: A lens system is formed by a positive lens with focal length f1 = +40 cm and a negative lens with focal length f2 = –60 cm, separated by d = 20 cm. An object is placed at a location x1 = –60 cm in front of (before) lens ❶. Find the final image location and magnification through the two-lens system.

Figure 6-77: Imaging in a two-lens system using the intermediate image. Step 1: Determination of the intermediate image: For an object at x1 = –60 cm (L1 = –1.66 D) and a lens with F1 = +2.50 D, the image is produced with L1΄ = +0.833 D, formed at x1΄ = +1.20 m after lens ❶. It is real and inverted, and has a (primary) magnification of m1 = (+1.20)/(−0.60) = −2.0.

Figure 6-78: Determination of the new object location and final image location. Step 2: Determination of the new object: This initial image is the new object to be imaged by lens ❷. Its location is now referenced in relation to lens ❷, which is 20 cm to the right of lens ❶. Therefore, this intermediate object is located x2 = +1.0 m from lens ❷.

Step 3: Determination of the final image: We implement the imaging relationship for x2 = +1.0 m (L2 = +1.0 D) and F2 = −1.66 D. The image is produced with L2΄ = −0.66 D, located at x2΄ = −1.5 m. The image is therefore formed x΄2 = −150 cm to the left of lens ❷. It is a virtual image and has a (secondary) magnification of m2 = (−150)/(+100) = −1.5. The final image is erect. The total magnification is m = m1 · m2 = (−2.0)·(−1.5) = +3.0.

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6.7 THICK LENS AND CARDINAL POINTS QUIZ Thick Lens Power Unless otherwise stated, the lenses in this quiz are surrounded by air. 1)

a) b) c) d) e) f)

The following diagram represents two single spherical refracting interfaces A and B made of glass (n = 1.6). The radius of curvature is 10 cm in each. What are their powers? 5)

a) b) c) d) 2)

b) c)

F = +0.0 D F = +12.0 D F = +24.0 D F = +0.0 D

Twist to Q 2. The lens produced in Q 2 is not thin; its surfaces are separated by t = 4.44 cm. The equivalent power of the thick lens is … a) b) c) d) e) f)

4)

biconvex biconvex biconvex meniscus

d) e)

6)

Fe = –0.1 D Fe = +0.1 D Fe = +11.0 D Fe = +12.1 D Fe = +12.9 D Fe = +13.0 D

7)

this is a mistake; both lenses should have the same power the front surface power in lens B has changed to +2.5 D the back surface power in lens B has changed to +2.5 D the contribution of lens thickness in lens B, which equals –2.5 D the contribution of lens thickness in lens B, which equals +2.5 D

Two lenses have the same surface powers (as shown). Lens A is thin, while lens B is thick. Their equivalent powers are …

a) b) c) d) e)

Given the surface powers and lens thickness shown in the figure, the equivalent power of this lens is (lens material n = 1.5) …

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Lenses A and B have the same radii of curvature and material. Thin lens A power FeA is +10.0 D, and thick lens B power FeB is +7.5 D. The reason is …

a)

We now proceed to make a thin lens (thickness = 0 cm) using the SSRIs presented in Q 1. The lens produced is ___________ and has power ______. a) b) c) d)

3)

FA = +6.0 D, FB = –6.0 D FA = +6.0 D, FB = +6.0 D FA = +12.0 D, FB = +12.0 D FA = +3.0 D, FB = –3.0 D

Fe = +24.0 D Fe = +20.0 D Fe = +16.0 D Fe = +6.0 D Fe = –4.0 D Fe = –6.0 D

Lens A: FeA = –10.0 D, Lens B: FeB = –12.5 D Lens A: FeA = –10.0 D, Lens B: FeB = –10.0 D Lens A: FeA = –10.0 D, Lens B: FeB = –7.5 D Lens A: FeA = 0.0 D, Lens B: FeB = –7.5 D Lens A: FeA = +25.0 D, Lens B: FeB = –12.5 D

Five thick lenses with the same center thickness and same refractive index material are placed in air.

THICK LENSES AND LENS SYSTEMS

Select the two correct statements for the thick lens equivalent power (Fe) compared to the thin lens power (the sum of front and back surface powers).

a) b) c) d) e) 8)

What is the thin lens power (F) in the lenses in Q 7 if their spherical surfaces each have 5 D of power (three correct answers)? a) b) c) d) e)

9)

A: FeA is less positive than the thin lens power B: FeB is less negative than the thin lens power C: FeC is equal to the thin lens power D: FeD is more positive than the thin lens power E: FeE is more positive than the thin lens power

A: FA = +5.0 D B: FB = –10.0 D C: FC = +5.0 D D: FD = –10.0 D E: FE = 0.0 D

11) The following thick lenses are made of the same glass material, have the same surface curvatures, and are placed in air. Which lens has the most positive equivalent power?

a)

A

b) B

c)

C

d) D

12) The following thick lenses are made of the same glass material, have the same surface curvatures, and are placed in air. Which lens has the most negative equivalent power?

a)

A

b) B

c)

C

d) D

13) Given the surface powers and the lens thickness as shown, the equivalent thick lens power (Fe) is (nlens= 1.5, lens placed in air) …

The following thick lenses have the same center thickness t, are made from the same refractive index n material, and are placed in air. Select the two correct answers pertaining to the contribution of the third term in Gullstrand’s equation: –t/n · (F1·F2).

a) b) c) d) e) f) g)

positive in lenses A and D positive in lenses A and B zero in lenses D and E zero in lenses C and D zero in lenses B and E negative in lenses A and C negative in lenses C and D

10) Back to Q 9. The lenses, classified by their thick lens power, are (select two answers) … a) b) c) d)

plus lenses: A, B, and C plus lenses: A, B, and D minus lenses: D and E minus lenses: C and E

a) b) c) d) e) f)

+20.0 D +1.5 D +1.0 D 0D –1.0 D –1.5 D

14) Select 2 correct statements regarding the cornea of the human eye … a) It is a positive (plus) lens, despite the fact that the cross-section shape resembles a minus (negative) meniscus lens. b) It is a negative (minus) lens because the crosssection resembles a minus (negative) meniscus. c) The anterior radius of curvature significantly affects the optical power of the cornea. d) The cornea thickness value significantly affects the optical power of the cornea. e) The posterior radius of curvature significantly affects the optical power of the cornea. 6-279

GEORGE ASIMELLIS

LECTURES IN OPTICS, VOL 2

Vertex Powers 15) The back vertex power for a lens (t = 4 mm, n = 1.6) with front surface (base curve) power F1 = +10.00 D, back surface power F2 = – 5.00 D is … a) b) c) d) e)

+4.76 D +4.94 D +5.06 D +5.26 D –14.76 D

16) Calculate the front vertex power of the lens in Q 15. a) b) c) d) e)

+4.76 D +4.94 D +5.06 D +5.26 D –14.76 D

17) In a meniscus lens in air (n = 1.0) with front radius of curvature +5 cm and back radius curvature +10 cm (t = 1.5 cm, n = 1.5), (three correct) … a) b) c) d) e) f)

the first (front) surface power is +5.0 D the first (front) surface power is +10.0 D the second (back) surface power is –5.0 D the second (back) surface power is –10.0 D the equivalent thick lens power is +6.0 D the equivalent thick lens power is +5.5 D

18) In a meniscus lens in air (n = 1.0) with front radius of curvature +10 cm and back radius of curvature +5 cm (t = 1.5 cm, n = 1.5), (three correct statements) … a) b) c) d) e) f)

the first (front) surface power is +5.0 D the first (front) surface power is +10.0 D the second (back) surface power is –5.0 D the second (back) surface power is –10.0 D the equivalent thick lens power is –5.5 D the equivalent thick lens power is –4.5 D

19) What is the neutralizing (front vertex) power of a lens (t = 2.5 cm, n = 1.5) with front surface power +20.0 D and back surface power +10.0 D?

a) b) c) d)

+25.0 D +30.0 D +32.0 D +40.0 D

20) What is the prescription (back vertex) power of a lens (t = 3 cm, n = 1.5) with front surface power +10.00 D and back surface power –5.00 D? a) b) c) d)

+7.50 D +6.00 D +5.45 D +5.00 D

21) What is the prescription (back vertex) power of a lens (t = 2.5 cm, n = 1.5) with front surface power +20.0 D and back surface power +10.0 D? a) b) c) d)

+25.0 D +30.0 D +32.0 D +40.0 D

22) An ophthalmic lens has a base curve (front power) +8.00 D and a back surface power –4.00 D (t = 2.5 mm, n = 1.6). Select three correct statements regarding the equivalent, back, and front vertex powers. a) b) c) d) e) f)

The equivalent thick lens power is +4.05 D. The equivalent thick lens power is +3.95 D. The back vertex power is + 4.05 D. The back vertex power is + 4.10 D. The front vertex power is +4.10 D. The front vertex power is +4.02 D.

23) A lens has back focal length +3.75 cm and effective focal length +6.25 cm. Where is the principal plane located in relation to the vertex point? (Do not use formulas, just draw a sketch!) a) b) c) d)

P΄ is 2.5 cm to the right of V΄ P΄ is 2.5 cm to the left of V΄ P΄ is 3.75 cm to the right of V΄ P΄ is 3.75 cm to the left of V΄

c) d) e)

all situated on the optical axis all situated on the principal plane(s) all situated on the focal plane(s)

Cardinal Points 24) The cardinal points are (two correct answers) … a) b)

three pairs: two focal, two principal, two vertex three pairs: two focal, two principal, two nodal

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THICK LENSES AND LENS SYSTEMS

25) When the principal planes are crossed ... a) b) c) d) e) f) g)

30) Back to Q 29. What is the lens equivalent power?

principal plane H΄ is in object space principal plane H is in image space principal plane H is after nodal point N΄ principal plane H is ahead of focal point F΄ principal plane H is after focal point F΄ principal plane H is ahead of principal plane H΄ principal plane H is after principal plane H΄

26) In a thick lens, the lens thickness is the separation between … a) b) c) d)

vertex point V and vertex point V΄ principal point P and principal point P΄ nodal point N and nodal point N΄ focal point F and focal point F΄

27) In the optical system of the eye, the object-space nodal point N is 5.73 mm to the right of objectspace principal point P (PN = 5.73 mm). What is the separation between image-space nodal point N΄ and image-space principal point P΄ (P΄N΄)? a) b) c) d)

11.46 mm 5.73 mm 2.865 mm 0 mm

a) b) c) d)

Fe = +10.0 D Fe = –2.75 D Fe = –3.5 D Fe = –6.5 D

31) Back to Q 29. Where is the object-space principal plane H in relation to the vertex point V? a) b) c) d)

H is 2.3 cm to the right of V H is 1.54 cm to the right of V H is 2.3 cm to the left of V H is 4.28 cm to the right of V

32) A glass (n = 1.5, t = 2.5 cm) planoconcave lens with back surface power F2 = –10.0 D is surrounded by air. Where are the principal planes H and H΄? a) b) c) d)

H is at V (δ = 0); H΄ is at V΄ (δ΄ = 0) H is at V (δ = 0); H΄ is to the right of V΄ (δ΄ = +1.66 cm) H is to the right of V (δ = 1.66 cm); H΄ is at V΄ (δ΄ = 0) H is at V (δ = 0); H΄ is to the left of V΄ (δ΄ = –1.66 cm)

33) Back to Q 32. What is the lens equivalent power?

28) Select the correct description of the effective (fe΄ and fe) and vertex focal lengths (BFL and FFL).

a) b) c) d)

Fe = +10.0 D Fe = 0.0 D Fe = –7.5 D Fe = –10.0 D

34) A glass (n = 1.5, t = 1.5 cm) planoconvex lens with front surface power F1 =+10.0 D is surrounded by air. Where are the object-space principal plane H and the image-space principal plane H΄? a) b) a)

A

b)

B

c)

C

d)

D

29) A meniscus glass lens (nlens = 1.5, t = 1.5 cm) has front surface power F1 = +10.0 D and back surface power F2 = –15.0 D. Where is the image-space principal plane H΄ in relation to the vertex point V΄? a) b) c) d)

H΄ is 2.85 cm to the right of V΄ H΄ is 1.54 cm to the right of V΄ H΄ is 2.85 cm to the left of V΄ H΄ is 4.28 cm to the right of V΄

c) d)

H is located at V (δ = 0); H΄ is located at V΄ (δ΄ = 0) H is located at V (δ = 0); H΄ is located to the right of V΄ (δ΄ = +1.0 cm) H is located to the right of V (δ = +1.0 cm); H΄ is located to the right of V΄ (δ΄ = +1.0 cm) H is located at V (δ = 0); H΄ is located to the left of V΄ (δ΄ = – 1.0 cm)

35) Back to Q 34. What is the lens equivalent power? a) b) c) d)

Fe = +10.0 D Fe = 0.0 D Fe = –7.5 D Fe = –10.0 D 6-281

GEOMETRICAL OPTICS

36) A twist to Q 34. The lens is followed by water (image space n΄ = 1.333), while the object space is still air (n = 1.0). What is different (three correct answers)? a) b) c) d) e)

the lens equivalent power Fe the object-space principal plane location the image-space principal plane location the object-space nodal plane location the image-space nodal plane location

42) In a concave mirror … a) b) c)

d) 37) Back to Q 36. The lens equivalent power Fe is … a) b) c) d)

+7.50 D +10.00 D +13.33 D +15.00 D

38) Back to Q 36. The image-space principal plane is … a) b) c) d)

at the image-space vertex V΄ (δ΄ = 0.0 cm) 1.0 cm to the left of V΄ (δ΄ = –1.0 cm) 1.3 cm to the left of V΄ (δ΄ = –1.3 cm) 1.3 cm to the right of V΄ (δ΄ = +1.3 cm)

39) Back to Q 36. The image-space nodal plane N΄ is … a) b) c) d)

at the image-space principal point P΄ (P΄N΄= 0) 1.0 cm to the right of P΄ (P΄N΄= 1.0 cm) 1.0 cm to the left of P΄ (P΄N΄= –1.0 cm) 3.3 cm to the right of P΄ (P΄N΄= 3.3 cm)

40) When a lens is surrounded in object space by air (n = 1.0) and in image space by water (n΄ = 1.333), … a) b) c) d) e)

the nodal points are coincident with the principal points only the object-space nodal point shifts to the right of the principal point only the image-space nodal point shifts to the right of the principal point both nodal points shift to the right of the principal points by equal amounts both nodal points shift to the left of the principal points by equal amounts

41) In a convex mirror … a) b) c)

d)

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the principal point is at the vertex; the nodal point is at the focal point the principal point is at the vertex; the nodal point is at the center of curvature the principal point is halfway through the center of curvature; the nodal point is at the center of curvature the principal point is at the center of curvature; the nodal point is at the vertex

the principal point is at the vertex; the nodal point is halfway to the center of curvature the principal point is at the vertex; the nodal point is at the center of curvature the principal point is halfway to the center of curvature; the nodal point is at the center of curvature the principal point is at the focal point; the nodal point is at the vertex

43) An air (n = 1.0) to glass (n΄ = 1.5) convex SSRI has a radius of curvature of +10 cm. What is the principal point P΄ to nodal point N΄ separation? a) b) c) d) e)

0 cm +5 cm +10 cm +15 cm +30 cm

44) A twist to Q 43. The SSRI now separates water (n = 1.33) from glass (n΄ = 1.5). The distance between principal point P΄ and nodal point N΄ is … a) b) c) d) e)

0 cm +5 cm +10 cm +15 cm +30 cm

45) A refractive element has an object-space focal length of f = –0.250 m and an image-space focal length of f΄ = +0.375 m. The nodal points with reference to the principal points are … a) b) c) d) e) f)

0.125 to the right (N to the right of P) 0.125 to the left (N to the left of P) 0.500 to the right (N to the right of P) 0.500 to the left (N to the left of P) 0.625 to the right (N to the right of P) 0.625 to the left (N to the left of P)

46) The refractive indices in image space (n΄) and in object space (n) in Q 45 have a ratio of … a) b) c) d)

n΄ = –1.5 · n n΄ = +1.5 · n n΄ = –0.66 · n n΄ = +0.66 · n

THICK LENSES AND LENS SYSTEMS

Ray Tracing with Principal Points 47) A ray propagating toward the front (objectspace) nodal point Ν exits the optical system … a) b) c) d) e)

parallel to the optical axis intersecting the image-space focal point F΄ as if it originated from the image-space nodal point Ν΄ as if it originated from the image-space principal point P΄ as if it originated from the image-space focal point F΄

a) b) c) d) e)

continues parallel to the optical axis until it reaches the secondary focal point F΄ continues parallel to the optical axis until it reaches the object-space principal plane H bends toward the image-space nodal point Ν΄ bends toward the image-space principal point P΄ bends toward the image-space focal point F΄

51) Select the correct ray-tracing implementation:

48) A ray propagates parallel to the optical axis from the left until it reaches the object-space principal plane H. Then, the ray … a) b) c) d) e)

continues parallel to the optical axis until it reaches the secondary focal point F΄ continues parallel to the optical axis until it reaches the image-space principal plane H΄ bends toward the image-space nodal point Ν΄ bends toward the image-space principal point P΄ bends toward the image-space focal point F΄

49) Select the correct ray-tracing implementation: a)

A

b)

B

c)

C

d) D

52) Select the correct ray-tracing implementation:

a)

A

b)

B

c)

C

d)

D

50) A ray coming from the left (as usual) propagates parallel to the optical axis until it reaches the image-space principal plane H΄. Then it …

a)

A

b)

B

c)

C

d) D

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GEOMETRICAL OPTICS

53) Select the correct ray-tracing implementation:

a)

A

b)

B

c)

C

d) D

54) Select the correct ray-tracing implementation:

55) Select the correct ray-tracing implementation:

a)

A

b)

B

c)

C

d) D

56) Which setup correctly denotes the directional distances for the object, image, and focal length?

a)

A

b)

B

c)

C

d) D

57) Which setup correctly denotes the directional distances for the object, image, and focal length?

a)

A

b)

B

c)

C

d) D

a) 6-284

A

b)

B

c)

C

d) D

THICK LENSES AND LENS SYSTEMS

58) Use the ruler to identify the object distance, image distance, and focal length.

a) b) c) d)

a) b) c) d) e)

object distance = –9 cm, image distance = +8 cm, focal length = +4 cm object distance = –8 cm, image distance = +9 cm, focal length = +4 cm object distance = –8 cm, image distance = +8 cm, focal length = +5 cm object distance = –8 cm, image distance = +8 cm, focal length = +4 cm object distance = –10 cm, image distance = +7 cm, focal length = +3 cm

59) Use the ruler to identify the object distance, image distance, and focal length.

e)

object distance = –17.66 cm, image distance = –6.25 cm, focal length = –10 cm object distance = –16.66 cm, image distance = –6.25 cm, focal length = –10 cm object distance = –12.66 cm, image distance = –1.25 cm, focal length = –5 cm object distance = –16.66 cm, image distance = –6.25 cm, focal length = –5 cm object distance = –12.66 cm, image distance = –1.0 cm, focal length = –10 cm

60) Use the ruler to identify the object distance, image distance, and focal length.

a) b) c) d) e)

x = –8 cm, x΄ =+10 cm, f΄ = +4.44 cm x = –5.75 cm, x΄ =+10 cm, f΄ = +2.24 cm x = –5.75 cm, x΄ =+10.5 cm, f΄ = +2.24 cm x = –8 cm, x΄ =+10.5 cm, f΄ = +2.24 cm x = –5.75 cm, x΄ =+10 cm, f΄ = +4.44 cm

d) e)

F1 = +25.0 D, F2 = –25.0 D F1 = +50.0 D, F2 = –50.0 D

Lens Systems 61) A system of two lenses, plus (F1) and minus (F2), has zero power when the lenses are in contact. The lenses are separated in air. The new, equivalent (total) power equals that of the plus lens (F1). What is the separation d of the two lenses? a) b) c) d) e)

d = +1/(F2 · F1) d = –2/F2 d = +2/F1 d = +1/F1 d = +1/(F1+F2)

62) A system of a plus lens (F1) and a minus lens (F2) has zero power when the lenses are in contact. When the lenses are separated by 1 m in air, the equivalent (total) power becomes +25.0 D. What is the power of each element lens? a) b) c)

F1 = +1.25 D, F2 = –1.25 D F1 = +5.0 D, F2 = –5.0 D F1 = +12.5 D, F2 = –12.5 D

63) A system comprises a plus lens (F1 = +10.0 D) and a minus lens (F2 = –25.0 D), separated by d = 6 cm in air. The equivalent power of this system is … a) b) c) d)

0.0 D –15.0 D –21.0 D –30.0 D

64) Back to Q 63. This system makes a … a) b) c) d)

Galilean telescope astronomical (Keplerian) telescope microscope magnifying lens

65) Back to Q 63. The principal planes of this system are situated at (two correct answers) … 6-285

GEOMETRICAL OPTICS

a) b) c) d) e) f)

object-space Η: at infinity (δ = ∞) object-space Η: at lens 1 (δ = 0) object-space Η: at lens 2 (δ = d) image-space Η΄: at infinity (δ΄ = ∞) image-space Η΄: at lens 1 (δ΄ = –d) image-space Η΄: at lens 2 (δ΄ = 0)

66) Two thin plus lenses F1 = +10.0 D and F2 = +50.0 D, separated by d = 12 cm in air, form a system whose total (equivalent) power is … a) b) c) d)

+120.0 D +60.0 D +48.0 D 0.0 D

67) The system described in Q 66 is a … a) b) c) d)

Galilean telescope astronomical (Keplerian) telescope microscope magnifying lens

68) What is the total power of a system comprising a thin plus lens (F1 = +10.0 D) and a thin minus lens (F2 = – 25.0 D), separated by d = 12 cm, in air? a) b) c) d) e)

+15.0 D +5.0 D –27.0 D –35.0 D –45.0 D

69) A thin plus lens (F1 = +10.0 D) and a thin minus lens (F2 = –25.0 D) are separated by d = 12 cm in a medium with refractive index n = 1.5. The equivalent (total) power of this system is … a)

+15.0 D

b) c) d) e)

+5.0 D –27.0 D –35.0 D –45.0 D

70) Two thin plus lenses (F1 = +10.0 D and F2 = +10.0 D) are separated by d = 12 cm in a medium with refractive index n = 1.5. The equivalent (total) power of this system is … a) b) c) d) e)

+8.0 D +10.0 D +12.0 D +20.0 D +28.0 D

71) Two thin planoconvex lenses L1 and L2 in contact at their flat surfaces produce a +10.0 D lens. What is the power if the lenses are separate by 8 cm in air? a) b) c) d)

+12.0 D +10.0 D +8.0 D +5.0 D

72) Back to Q 71. Where are the principal planes (two correct answers)? a) b) c) d)

Η: 5 cm to the right of L1 (δ = +5 cm) Η΄: 5 cm to the right of L1 (δ΄ = +5 cm) Η: 5 cm to the left of L2 (δ = +5 cm) Η΄: 5 cm to the left of L2 (δ΄ = –5 cm)

73) The back focal length of the system in Q 71 is … a) 12.5 cm b) 10.0 cm c) 7.5 cm d) 5.0 cm

Lens System Imaging 74) In the system shown in the drawing, identify the type and location of the object for lens 2.

a) b) c) d) e)

a real object located 1.0 m to the right of lens 2 a real object located 1.2 m to the right of lens 1 a virtual object located 1.2 m to the right of lens 2 a virtual object located 1.0 m to the right of lens 2 a real object located 1.0 m to the right of lens 1

75) In the system shown in the drawing, identify the type and location of the object for lens 2.

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THICK LENSES AND LENS SYSTEMS

c) d)

the image from lens 2 the image from lens 3

81) In a two-lens system, the object is at the objectspace principal plane H. Where is the final image? a) b) c) d) e)

a real object located 2.0 m to the right of lens 2 a real object located 1.2 m to the right of lens 1 a virtual object located 1.2 m to the right of lens 2 a virtual object located 2.0 m to the right of lens 2 a real object located 1.2 m to the right of lens 2

76) The intermediate image shown in the illustration is a (two correct answers) …

a) b) c) d)

at the image-space focal point F΄ at the image-space principal plane H΄ at the image-space nodal point N΄ at optical infinity

82) Back to Q 81. What is the (total) magnification? a) b) c) d)

+1.0 –1.0 it cannot be determined infinity

83) A lens system comprises a thin plus lens 1 (F1 = +10.0 D) and a thin minus lens 2 (F2 = –10.0 D), separated by 5 cm, in air. The total power is … a) b) c) d)

real image via lens 1 virtual image via lens 1 real object for lens 2 virtual object for lens 2

77) Back to Q 76. The final image is (two answers) … a) b) c) d)

the optical conjugate of the initial object the optical conjugate of the intermediate image a virtual object for lens 2 a virtual image via lens 2

78) What point serves as the object for lens 1 (dashed lines indicate ray extrapolation)?

a)

A

b)

B

c)

C

d)

A

b)

B

c)

C

d)

D

80) Back to Q 78. Point C functions as (select two) … a) b)

–10.0 D –5.0 D +5.0 D +10.0 D

84) Back to Q 83. The object-space principal plane Η in this lens system is situated (δ = ?) … a) b) c) d) e) f)

at optical infinity to the left –10 cm to the left of lens 1 –5 cm to the left of lens 1 +5 cm to the right of lens 1 +10 cm to the right of lens 1 at optical infinity to the right

85) Back to Q 83. The image-space principal plane Η΄ in this lens system is situated (δ΄ = ?) …

D

79) Back to Q 78. What point functions as the image for lens 1? a)

a) b) c) d)

a) b) c) d) e) f)

at optical infinity to the left –10 cm to the left of lens 2 –5 cm to the left of lens 2 +5 cm to the right of lens 2 +10 cm to the right of lens 2 at optical infinity to the right

the object for lens 3 the object for lens 2 6-287

GEOMETRICAL OPTICS

86) Which illustration correctly depicts the locations of the principal planes for the system in Q 83?

91) Comparing the answers obtained in Q 87 and Q 90, the answers … a)

b) c) d) a)

A

b)

B

c)

C

d)

D

87) Back to Q 83. A real object is placed –10 cm to the left of lens 1. The image of this object via the reduced system is situated … a) b) c) d) e) f)

at optical infinity to the left –10 cm to the left of lens 2 –5 cm to the left of lens 2 +5 cm to the right of lens 2 +10 cm to the right of lens 2 at optical infinity to the right

88) Back to Q 83. The image of this object via lens 1 (successive imaging intermediate image) is … a) b) c) d) e) f)

at optical infinity to the left –10 cm to the left of lens 2 –5 cm to the left of lens 2 +5 cm to the right of lens 2 +10 cm to the right of lens 2 at optical infinity to the right

89) Back to Q 88. Where is the object for lens 2 (use successive imaging)? a) b) c) d) e) f)

at optical infinity to the left –10 cm to the left of lens 2 –5 cm to the left of lens 2 +5 cm to the right of lens 2 +10 cm to the right of lens 2 at optical infinity to the right

90) Back to Q 88. Where is the (final) image created by lens 2 (use successive imaging)? a) b) c) d) e) f) 6-288

at optical infinity to the left –10 cm to the left of lens 2 –5 cm to the left of lens 2 +5 cm to the right of lens 2 +10 cm to the right of lens 2 at optical infinity to the right

are identical because the imaging technique used in Q 87 (using the principal planes) and the one used in Q 90 (successive imaging) were properly executed are identical because I got lucky—I’m buying a lottery ticket are escaping me—I have no idea what I’m doing are different. I need to solve by carefully drawing ray tracing and executing the calculations.

92) A thin minus lens 1 (F1 = – 2.0 D) and a thin minus lens 2 (F2 = –1.5 D) are separated by 33 cm in air. An object is placed 1 m to the left of lens 1. Where is the final image (use successive imaging)? a) b) c) d) e)

100 cm to the left of lens 1 at the plane of lens 1 at the plane of lens 2 33 cm to the left of lens 1 33 cm to the right of lens 2

93) Back to Q 92. What is the overall magnification? a) b) c) d) e) f)

m = +½ m = +⅓ m = +⅙ m = –⅙ m = –⅓ m = –½

94) Two identical plus lenses (lens 1 and lens 2, each with F = +4.0 D) are separated by 1.0 m in air. An object is placed half a meter to the left of lens 1. Where is the final image (use successive imaging)? a) b) c) d) e) f)

100 cm to the left of lens 1 50 cm to the left of lens 1 at the plane of lens 1 100 cm to the right of lens 2 50 cm to the right of lens 2 at the plane of lens 2

95) Back to Q 94. What is the overall magnification? a) b) c) d)

m = +2.0 m = +1.0 m = –1.0 m = –2.0

THICK LENSES AND LENS SYSTEMS

6.8 THICK LENS AND CARDINAL POINTS SUMMARY Thick Lens Powers: Summary A lens has two distinct surface powers. In a lens of refractive index nlens and radius of curvature

r, surrounded by a medium with refractive index next, the surface powers are described by the single surface refractive interface relationships (SSRI power formula): F1 =

nlens − next and r1

F2 =

next − nlens r2

If the lens is surrounded by air, the surface powers are F1 =

nlens − 1 r1

and

F2 =

1− nlens

r2

The two surface powers determine the lens power. In a thin lens (nominal lens formula), the simple addition of the front and back surface powers produces the lens refractive power. Thin Lens Power:

F = F 1 + F2

In a thick lens, the equivalent power is described by Gullstrand’s relationship: Thick Lens Equivalent Power:

Fe = F1 + F2 −

t F1  F2 nlens

where nlens is the refractive index, t is the lens thickness, and F1 and F2 are the lens surface powers. The reciprocal of the equivalent power (in air) is the equivalent focal length f΄e measured from the principal plane H΄. The thick lens equivalent power is different from the thin lens power by an amount determined via the third term in Gullstrand’s relationship: Third term in Gullstrand’s relationship:

−

t nlens

F1  F2

This term is positive if the two surface powers have different signs (as in a meniscus lens), is negative if the two surface powers have the same signs (as in a biconvex or biconcave lens), and is zero if any of the two powers is zero (as in a planoconvex or planoconcave lens). In a thick lens, there are two vertex powers and two vertex focal lengths: Back Vertex Power: FBVP =

F1 + F2 t 1− F nlens 1

Front Vertex Power: FFVP = F1+

F2 t 1− F nlens 2

6-289

GEOMETRICAL OPTICS

The reciprocal of the back vertex power (in air) is the back focal length BFL or fBFL measured from the back vertex point V΄; likewise, the reciprocal of the front vertex power is the front focal length FFL or fFFL measured from the front vertex point V. Front Surface Power F1 • Is the beam vergence that leaves the front (first) lens surface if a collimated beam (object at infinity) is incident on the lens. • Is calculated using the SSRI power formula. • Is referenced at the front (object-space) vertex point V. Equivalent Power Fe • Is the beam vergence that leaves the image-space principal plane H΄ of a thick lens if a collimated beam (object at infinity) is incident on the lens. • Is the sum of the front surface power F1 , the back surface power F2, and the third term introduced by Gullstrand's formula. • Is referenced at the back (image-space) principal plane H΄. Front Vertex Power FFVP (also known as the neutralizing power) • Is the beam vergence leaving the object-space front surface of a thick lens if a collimated beam is incident on the lens from the back side. • Is the sum of the front surface power F1 and the downstream-adjusted back surface power F΄2. • Is referenced at the front (object-space) vertex point V. Back Vertex Power FBVP (also known as the prescription power) • Is the beam vergence leaving the image-space back surface of a thick lens if a collimated beam (object at infinity) is incident on the lens. • Is the sum of the downstream-adjusted front surface power F΄1 and the back surface power F2. • Is referenced at the back (image-space) vertex point V΄.

Figure 6-79: Summary of power concepts and formulas in thick lenses. 6-290

THICK LENSES AND LENS SYSTEMS

Equivalent Focal Length:

Measured from the principal plane / point Associated with the equivalent power

Focal lengths in a thick lens: Front and Back Focal Lengths:

Measured from the front / back vertex point Associated with the front / back vertex power

The Cardinal Points: Summary An optical system can be described by a set of six cardinal points: the two focal points, the two principal points, and the two nodal points, all of which are situated on the optical axis. The primary focal point F is the unique object point that produces a collimated pencil of rays in image space. The secondary focal point F΄ is the unique image point to which a collimated pencil of rays incident to the lens converges (is imaged). In imaging relationships we use the secondary focal length, which is referenced from the principal point P΄ to the secondary focal point F΄.

Figure 6-80: Focal points: secondary (F΄) and primary (F).

The principal points are the intersections of the principal planes H and H΄, defined as the virtual plane on which a ‘sudden’ and ‘complete’ refraction (i.e., application of Snell’s law) occurs. The intersection of plane H with the optical axis is the principal point P. In a thick lens or a lens system, the focal length is termed the equivalent focal length and represents the distance separating the principal points from the focal points, i.e., P΄F΄.

Figure 6-81: Principal points (P΄ and P) and principal planes (H΄ and H).

6-291

GEOMETRICAL OPTICS

The nodal points are defined by the angular subtense property: A beam entering the optical system directed at the anterior nodal point N exits the system as if it originated from the posterior nodal point N΄ with the same inclination ϑ with respect to the optical axis.

Figure 6-82: Nodal points (N΄ and N). Function of Focal points • To collimate a beam originating from the primary focal point or to converge a collimated beam to the secondary focal point.

Function of Principal points • To preserve ray displacement. Between principal planes, there is unity magnification and no change in ray height.

Function of Nodal points • To preserve angle with the optical axis. The rays are undeviated with no change in direction.

The locations of the principal planes depend on the lens equivalent power and surface powers, as well as the lens shape and refractive index. They can be to the right, to the left, inside, or outside the lens. The locations of principal planes (lens surrounded by air) are

΄ = −t 

F1 1  Fe n lens

and

 = t

F2 1  Fe n lens

Figure 6-83: Principal planes in (left) a minus meniscus lens and (right) a plus biconvex lens.

If both sides of the lens are surrounded by the same medium (n object space = n΄ image space), the nodal points coincide with the corresponding principal points: PN = P΄N΄ = 0. If not (n ≠ n΄), the nodal (N, N΄) points and the principal (P, P΄) points are non-coincidental, separated by Principal-to-Nodal Point Displacement:

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PN = P΄N΄ = f΄ + f =

n΄ − n Fe

THICK LENSES AND LENS SYSTEMS

Imaging with Thick Lenses An optical system can be described by a set of six cardinal points: the two focal points, the two principal points, and the two nodal points, all of which are situated on the optical axis. •

Ray-tracing rule 1 employs principal plane H΄ and focal point F΄. Note that between principal planes H and H΄ the ray propagates parallel to the optical axis.

•

Ray-tracing rule 2 employs principal plane H and focal point F.

•

Ray-tracing rule 3 employs nodal points N and N΄.

Figure 6-84: Ray-tracing rule 1, employing principal plane H΄ and focal point F΄.

Figure 6-85: Ray-tracing rule 2, employing principal plane H and focal point F.

Figure 6-86: Ray-tracing rule 3, employing nodal points N and N΄.

When imaging with either thick lenses or lens systems, we use the equivalent focal length f΄e measured from the principal plane H΄. Now the imaging relationship takes the form Imaging Relationship:

1

x object location

+

1

f΄ equivalent focal length

=

1

x΄ image location

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Figure 6-87: Imaging with a thick positive (plus) biconvex lens, forming a real image to the right of the lens. The object and image distances x and x΄ are measured from the principal planes H and H΄, respectively.

• Object location x is measured from the object-space principal plane Η. • Focal length f΄e and image location x΄ are measured from the image-space principal plane Η΄. Optical Power in a Lens System In a lens system, the optical power is called the total optical power FTOT and is essentially equivalent to the optical power that applies in the thick lens. Here, F1 and F2 are the individual thin lens powers comprising the system, d is the separation between the two lenses, and n is the refractive index of the medium between the two lenses. General relationship (Gullstrand's relationship) Thin lenses in contact Lenses separated by distance d (in air)

• F = F1 + F2 − (d/n) · (F1· F2) TOT

• F = F1 + F2 TOT

• F TOT = F1 + F2 − d · F1· F2

Lens System Imaging There are two ways to solve an imaging problem. The first involves the use of the principal planes and ray diagrams (reduced system). Locations are measured from the principal planes, and the (final) image location and size derived from the initial object can be determined by applying either ray diagrams (Figures 6-84 to 6-86) or the analytical imaging relationship. An alternative is successive, or intermediate, imaging. We apply the simple, one-thinlens imaging rules to locate the image formed by the first lens, ignoring the presence of the second lens. This is the intermediate image. The second lens treats the intermediate image as its object, which is at the same location and of the same size as the object formed by the first lens (with a change in the point of origin, which now lies on the second lens). We then proceed to find the secondary image from this object. The total, overall magnification is the product of the individual magnifications. This process can be repeated indefinitely until it is applied to the last lens (in general, an optical element). 6-294

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7 FINITE TRANSVERSE OPTICS

In the preceding chapter, we discussed the effects associated with the thickness t of a lens or the separation d between two lenses. Because thickness/separation are distances along the optical axis, the finite axial (longitudinal) extent of an optical system is considered. This invokes the concepts involved in a thick lens and a reduced system, using the cardinal points. With this information, we can identify the image location (where it is formed) and size (how big it is). However, up to this point, the actual size of the optical elements perpendicular to the optical axis has not been discussed. We dealt with either infinitesimally thin rays or arbitrarily large cross-sectional optics. For example, the lens diameter has not been an issue so far: This is the finite transverse (lateral) extent. The actual lens diameter does not affect the image location or size, but it does affect the quality of the formed image, such as its sharpness and brightness.

Figure 7-1: A camera lens often involves a large number of lens elements with different diameters. This is a cutout of an AF-S Nikkor 14-24mm f/2.8G ED photography lens (cropped from the original photo taken by Morio from Wikimedia Commons under license CC SA-BY 4.0). 7-295

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7.1 APERTURE STOP AND PUPILS Consider two cases involving a lens of +20 D power. In case 1 the lens diameter is 1 inch (2.54 mm), while in case 2 the lens diameter is 2 inches (5.08 mm). If an object is placed 12.5 cm in front of (either) lens, the rules of imaging suggest that the image is formed 8.33 cm after (to the right of) this lens (Figure 7-2). The location of the image is independent of the finite size of the lens: It is the same with a 1 inch or a 2 inch diameter lens (2.54 mm or 5.08 mm diameter).

Figure 7-2: Image location for the same object location but different lens diameter. The transverse dimension of a lens does not affect the location of the image; the ray-tracing rules are identical. However, the lens diameter affects several of its qualitative aspects, such as brightness and resolution. (Cartoon vector images used in this chapter have been adapted from freepik www.freepik.com.)

Despite the fact that these images are formed at exactly the same location and therefore have the same magnification, there are some qualitative aspects that are affected by the lens transverse dimensions. Any point at a light source or an object can emit rays in all directions. However, only some of these rays enter the optical system and contribute to image formation. Why some and not all? To form the image, rays must pass through a lens (or a mirror). A lens with an infinite lateral extent is an impossibility: A lens extends up to a specific diameter (assuming a circular cross-section) and therefore subtends a finite solid angle from the object. Increasing the lens transverse size affects the image brightness because more light is collected by the larger lens, so more light eventually reaches the image. Other aspects affected by the lens size are the depth of field (§ 7.4), resolution, blur, aberrations, and contrast (presented in § 7.5).

7.1.1

The Aperture Stop

The extent of the rays from an object that can enter the optical system and eventually form the image is spatially limited by a light cone, which can be represented by a three-dimensional solid angle (angular aperture). Often, however, we draw two-dimensional representations and use simple angles. 7-296

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Figure 7-3: Angular subtense of the light cone entering the system for the two lenses presented in Figure 7-2. While the location of the image is not affected, the light cone is dependent on the lens diameter. As will be introduced in § 7.2.2, the outermost ray that delimits the light cone is called the marginal ray.

A stop, despite its name, is a clear opening of a measurable size in an opaque screen. We often assume circular cross-sections, in which case a stop is like an iris circular diaphragm. Now, the size of the stop is measured by the diameter of its clear opening. The physical element that limits the area over which light is collected from an axial object point is called the aperture stop (AS). The aperture stop can be the perimeter / edge / rim of a lens, a mirror, or an aperture (iris diaphragm, a physical opening). In rotationally symmetrical systems, the center of the aperture stop is on the optical axis.

Figure 7-4: A circular aperture between the object and the lens serves as an aperture stop.

The aperture stop (AS):

Determines the lightgathering capacity of the optical system.

Selection of the element to serve as the aperture stop depends on the specific object location.

To identify the aperture stop, we have to determine which element limits light acceptance by the system from a specific axial object point. Before we rush to say ‘the one with the narrower aperture,’ we should be mindful that this choice might be dependent on the object location and the size and arrangement of the other optical elements. Several techniques (recipes) are used for identifying the aperture stop. Invariably, these techniques require the comparison of some geometrical parameters among the potential candidate elements that might be the aperture stop. Often, many similar-triangle geometrical considerations are involved; this is a good time to refresh your geometry skills! 7-297

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The geometrical parameters used in the identification techniques are the following: •

The semi-diameter: the half-diameter of the clear aperture of a stop or a lens, or any other optical surface. We use the concept of semi-diameter (which is the radius of the clear opening) in order to avoid confusion between the rays and the radius of curvature, which are so often used in optics. Think of it as a metric for the caliper of this element.

•

The ray height: the lateral (transverse) distance of the ray from the optical axis. The farther the ray is from the axis, the greater the ray height, e.g., the altitude of an airplane in flight.

•

The height to semi-diameter ratio: the ratio of the ray height of an element to the semidiameter of that element. If the ray passes by the edge/rim of the element, the height to semi-diameter ratio is 1.0, which is the maximum attainable value.

•

The aspect or aspect ratio: the ratio of the semi-diameter to the distance separating the element from the object. Essentially, the aspect defines the angular subtense, expressed in radians: The greater (lesser) the aspect ratio, the greater (lesser) the angular subtense.

•

The angular subtense with respect to the optical axis, calculated by the inverse tangent of the aspect ratio. If an element has a semi-diameter of 3.0 mm and is separated from the object by 30 mm, the aspect ratio is 3.0/30 = 0.1, which is also the angular subtense expressed in radians. This can be converted to degrees using tan−1(3.0/30) = 5.7°.

7.1.1.1 Aperture Stop with No Lens or when the Lens is the Last Element The simplest aperture stop lacks a physical diaphragm and consists of just a single lens. This aperture stop is the edge of the lens; rays outside the lens edge do not enter the system. If the object is at optical infinity and there are no lenses or the lens is the last element, the aperture stop is the element with the smallest diameter (assuming that all elements are collinear, i.e., their centers line up with the optical axis). Example ☞: Collimated light illuminates an array of three circular apertures spaced apart by 30 mm. The aperture diameters are 6, 8, and 15 mm. Which one is the aperture stop in this arrangement?

Figure 7-5: Three equispaced circular diaphragms illuminated by collimated light. The semi-diameters are 3.0 mm, 4.0 mm, and 7.5 mm. The maximum height of an unobstructed ray through the system is 3.0 mm. 7-298

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Because the object is at optical infinity, we draw rays parallel to the optical axis that aim at or pass by the edge of each aperture (Figure 7-5). Here, the ray height is simply the semi-diameter of each element: •

for the ray passing by the edge of aperture ❶, 3.0 mm,

•

for the ray aiming at the edge of aperture ❷, 4.0 mm (blocked by aperture 1), and

•

for the ray aiming at the edge of aperture ❸, 7.5 mm (blocked by aperture 1).

Of the above, the ray height corresponding to aperture ❶ is the smallest, so aperture ❶ is the aperture stop for this configuration. It is noteworthy that the order of the apertures does not affect which becomes the aperture stop if the object is still at optical infinity.

If the object is not at infinity (i.e., the incident light is not collimated) but is a finite distance in front of the optical system, we consider the angular subtense formed from the axial object point. The element that forms the smallest angular subtense is the aperture stop.

A simple recipe to find the aperture stop (with no lens, or when a lens is the last element):

1. Launch a marginal ray from the onaxis object point that passes through the edges of all elements and is not blocked by any element or surface.

2. The element (or surface) that forms the smallest angular subtense is the aperture stop.

Example ☞: An object is placed 30 mm in front (to the left) of the aperture array (Figure 7-6). Which element is the aperture stop in this arrangement? We launch rays from the on-axis object point, each of which runs by the edge of each aperture.

Figure 7-6: Three equispaced circular diaphragms in front of a near object. In this technique we use the marginal ray (§ 7.2), which is the ray that originates from the on-axis point and traverses the entire optical system (not blocked by any element) through the edge of the aperture stop. The ray aiming at the edge of element ❷ is the marginal ray of the system. The angular subtense formed is: •

for the ray aiming at the edge of aperture ❶, tan−1(3.0/30) = 5.7° (eventually blocked by 2),

•

for the ray aiming at the edge of aperture ❷, tan−1(4/60) = 3.8°, and

•

for the ray aiming at the edge of aperture ❸, tan−1(7.5/90) = 4.8° (also blocked by 2). 7-299

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The smallest angular subtense is formed by the ray aiming at the edge of aperture ❷. This aperture serves as the aperture stop in this case. It is noteworthy that the order of the apertures can affect which one becomes the aperture stop if the object is still at a fixed position before the first aperture.

Figure 7-7: In order for an optical element to serve as the aperture stop, two parameters are considered: the element’s location with respect to the object and its aperture (diameter) size. A different element might be the aperture stop, depending on the object location.

The aspect ratio and the angular subtense (the inverse tangent of the aspect ratio) of the element that is the aperture stop are the smallest compared to those parameters of the other elements. The ray height to semi-diameter ratio is the largest: Practically, as the ray passes by the edge of the element that is the AS, the ratio is 1.0, which is the largest possible value. This is the basis of a more general recipe to identify the aperture stop:

A general recipe to identify the aperture stop (AS):

Draw a marginal ray (from an on-axis object point) that passes through all, or is not blocked by any, element(s) or surface(s).

Mark the height at which the ray intersects the optical elements or surfaces. These heights are at most the semidiameter of the element.

The optical element (or surface) with the largest height to semi-diameter ratio is the aperture stop.

Example ☞: An object is placed 60 mm to the left (in front) of a lens whose diameter is 36 mm. We place a circular aperture of 10 mm diameter a distance of 20 mm to the left of the lens. Find the aperture stop. The candidates are the diaphragm aperture and the lens edge, the lens being the last element. The diaphragm has the smallest aspect ratio (and consequently, the smallest angular subtense) and the largest height to semi-diameter ratio, so it is the aperture stop. Note that the height of the unobstructed marginal ray at the lens (7.5 mm) can be calculated using similar triangles. Aspect: at the diaphragm 5 mm/40 mm = 0.125; at the lens 18 mm/60 mm = 0.3. Angular subtense: at the diaphragm tan−1(0.125) = 7.125°; at the lens tan−1(0.3) = 16.7°. The smallest aspect and the smallest angular subtense determine the aperture stop, which is the diaphragm. 7-300

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Height to semi-diameter ratio: at the diaphragm 5 mm/5 mm = 1.0; at the lens 7.5 mm/18 mm = 0.42. The largest height to semi-diameter ratio determines the aperture stop, which is the diaphragm.

Figure 7-8: Determination of the aperture stop for an object 60 mm from the lens. Example ☞: An object is placed 25 mm to the left (in front) of lens whose diameter is 36 mm. We place a circular aperture of 10 mm diameter at a distance of 20 mm to the left of the lens. Find the aperture stop. Aspect: at the diaphragm 5 mm/5 mm = 1.0; at the lens 18 mm/25 mm = 0.72. Angular subtense: at the diaphragm tan−1(1.0) = 45°; at the lens tan−1(0.72) = 35.75°. The smallest aspect and the smallest angular subtense determine the aperture stop, which is the lens. Height to semi-diameter ratio: at the diaphragm 3.6 mm/5 mm = 0.72; at the lens 18 mm/18 mm = 1.0. The largest height to semi-diameter ratio determines the aperture stop, which is the lens.

Figure 7-9: When the object is placed closer to the lens, the lens edge becomes the aperture stop. The ray passing the diaphragm edge (the orange ray) misses the lens. This ray is lost, and for all purposes can be classified as ‘obstructed.’ The ray passing by the lens edge (the blue ray) has, at the lens plane, a height of 18 mm, equal to the lens semi-diameter, so the ratio height to semi-diameter is 18 mm/18 mm = 1.0. This ray at the diaphragm has a height of 3.6 mm (calculated using similar triangles) and therefore has a ratio of 3.6 mm/5 mm = 0.72.

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7.1.1.2 Aperture Stop when an Element is Placed After a Lens If an aperture is placed after the lens, we must acknowledge the refractive effect; as the rays bend on the lens plane, there are no longer any simple, straight-line rays. Therefore, we can only work with the height to semi-diameter ratio and not the angular subtense. The steps are as follows: First we identify the image point using vergence imaging relationships or ray tracing. Then we draw straight-line marginal rays from the on-axis object point to the lens, and from that lens–ray intersection to the on-axis image point. These rays are directed at (pass by) the edges of the candidate aperture stop elements. The element with the largest height to semi-diameter ratio is the aperture stop.

A recipe to identify the aperture stop if an aperture is located after a lens:

Identify the image point. Then draw marginal rays from the on-axis object point to the lens, and from that lens to the on-axis image point.

Identify the marginal ray that passes unobstructed by the edge of a lens(es) and/or diaphragm(s).

The element with the largest height to semidiameter ratio is the aperture stop.

Example ☞: A circular aperture of 10 mm diameter is placed 12.5 mm to the right of (after) a lens. The lens diameter is 36 mm and the lens power is +40.0 D. Identify the aperture stop for an object at optical infinity. First, we acknowledge that, because the object is at optical infinity, the image point is none other than the secondary focal point, situated 25 mm to the right of the lens (power = +40.0 D). Then, we draw rays from the object (drawn simply parallel to the optical axis); these rays, after refraction by the lens, are directed toward the secondary focal point. The two rays we drew reach the edge of each element separately; the orange ray arrives at the lens, and the blue ray arrives at the aperture.

Figure 7-10: Aperture placed after the lens; incident light is collimated.

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•

The ray that passes by the edge of the lens (the orange ray) strikes the aperture at 9 mm on its way to the focal point and is blocked.

•

The ray that passes by the edge of the diaphragm (blue, 5 mm height) continues unobstructed toward the focal point. The ray meets the lens at 10 mm height (use similar triangles). For this ray, the height to semi-diameter ratio is 5 mm/5 mm = 1.0 at the diaphragm and 10 mm/18 mm = 0.55 at the lens.

The diaphragm is the AS because it corresponds to the largest height to semi-diameter ratio. Example ☞: In the same setup as above, the object is placed 50 mm to the left (in front) of the lens. Identify the aperture stop. We first need to determine the image location. Our object is placed at x = –50 mm (to the left of the lens), so the object vergence is L = –20 D. Using the imaging relationship given in Eq. (4.2), we add the object vergence to the lens power of +40 D and find that the image vergence is L΄ = +20 D. The image is formed +50 mm after (to the right of) the lens (this is a simple imaging case in which the object is placed at twice the focal length).

Figure 7-11: Aperture placed after the lens. The object is placed 50 mm before the lens. We now launch the two marginal rays from the on-axis object point that pass by the edges of the lens (the orange ray) and the diaphragm (the blue ray). These rays connect the on-axis object point with the on-axis image point following refraction by the lens. We examine the ray heights as they cross the two aperture stop candidates, the lens and the circular diaphragm aperture. •

The ray that passes by the edge of the lens (the orange ray) meets the aperture at 13.5 mm on its way to the image point and is obstructed.

•

The ray that passes by the edge of the circular diaphragm (the blue ray) continues propagating toward the image point unobstructed. For this ray, the height to semi-diameter ratio is 6.6 mm/18 mm = 0.366 at the lens and 5 mm/5 mm = 1.0 at the diaphragm.

The diaphragm is the aperture stop because it corresponds to the largest ratio.

7.1.1.3 Aperture Stop when There are Only Lenses It is possible for a simple optical system to have no dedicated physical diaphragm at all; the system can be comprised only of lenses. The technique to determine the aperture stop is the same: We identify the image point (including possible intermediate images). We then launch 7-303

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the marginal rays from the on-axis object point toward the lens(es) that terminate at the on-axis image point. The aperture stop is determined by the highest height to semi-diameter ratio. Example ☞: Consider two identical lenses of +2.0 D power and 50 mm diameter spaced apart by 100 cm. What element is the aperture stop for an object that is placed 66 cm in front of the first lens? First, we must identify the image formed by the lens system (Figure 7-12). Here we use the intermediate image technique presented in § 6.5.5. The final image is real, inverted [m = m1 · m2 = (−3.0)·(+⅓) = −1.0], and formed 33 cm to the right of lens ❷. (Working out the details of this two-lens imaging as an exercise is strongly recommended.)

Figure 7-12: Determination of the aperture stop in a two-lens system. The object is 66 cm in front of lens ❶. One of the two lenses is the aperture stop. We launch the marginal ray from the on-axis object point. The ray reaches the edge of lens ❶, then is refracted toward the intermediate image. This ray is subsequently intercepted at lens ❷ and is steered toward the final image. This ray intersects lens ❷ at a height of 12.5 mm (use similar triangles for verification). The height to semidiameter ratios are 1.0 for lens ❶ and 0.5 for lens ❷. Therefore, the edge of lens ❶ is the aperture stop.

7.1.2 Significance of the Aperture Stop The aperture stop is a real element that physically restricts the transverse extent (cross-sectional area) of a ray bundle that propagates through an imaging system. Its center intersects the optical axis. There is always an aperture stop in any imaging system, regardless of how simple or complicated it may be. On the other hand, even if there are several apertures or stops in an imaging system, only one serves as the aperture stop for a given object location. Because the aperture stop limits the amount of light that eventually forms the image, its size affects the image brightness (defined in § 7.5). Assuming a uniform object luminance, the larger the area of the aperture stop, the brighter the image (illuminance). Therefore, an increase

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in the aperture stop diameter by a factor of 2 (and in turn, a corresponding increase in the aperture stop area by 4) results in an increase in the image brightness by a factor of 4.

Figure 7-13: (left) In a photography camera lens, the aperture stop is usually the adjustable leaf diaphragm placed after the front lens elements. (right) In the eye, the aperture stop is the periphery of the anatomical iris.

In addition, the aperture stop affects the depth of field (§ 7.4); a larger aperture stop leads to a shallower depth of field. We can view this in association with the geometrical blur; a larger aperture stop leads to increased blur away from the image point (§ 7.6). Conversely, a smaller aperture stop leads to an increased depth of field and less geometrical blur away from the image. Resolution is affected via the relationship that provides the diffraction-associated minimum angle of resolution (Rayleigh criterion, discussed in § 7.5). A larger aperture stop leads to finer resolution / smaller minimum angle of resolution (MAR). Conversely, a smaller aperture stop leads to a larger minimum angle of resolution, or a lesser resolution ability. Finally, the aperture stop affects aberrations (discussed in § 8.3). Since high-order monochromatic aberrations such as spherical aberration and coma affect the image quality significantly more when the imaging rays wander away from the optical axis, a larger-diameter aperture stop leads to an increased contribution of aberrations, and a smaller-diameter aperture

Aperture Stop

stop leads to a reduced impact of aberrations on image quality. ☞ is the element that physically restricts the transverse extent (cross-section) of a ray bundle through an optical system. ☞ is a real, physical opening. It may be the lens ring or the edge of a physical diaphragm inside or outside the optical system. ☞ the size of the aperture stop affects the image brightness, the extent of aberrations, the depth of field, and the resolving power of the optical system.

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7.1.3

Entrance and Exit Pupil

In any optical system, for a given object location there is only one aperture stop. It is a real, physical element. The aperture stop forms two pupils. The object-space image of the aperture stop, formed by the optical elements preceding it (if any), is the entrance pupil. The entrance pupil is the aperture stop as seen from an on-axis object point. It confines the light cone (angular aperture) entering the optical system. This light cone has its apex at (originates from) the on-axis object point and fills the entrance pupil.

Figure 7-14: The aperture stop is the element that physically limits the light acceptance in an optical system; the entrance pupil confines the light cone entering the optical system.

The image-space image of the aperture stop, formed by the optical elements succeeding it (if any), is the exit pupil. The exit pupil is the aperture stop as seen from an axial point on the image. It confines the light cone emerging from the optical system. This light cone has its apex at (originates from) the on-axis image point and fills the exit pupil.

Figure 7-15: The aperture stop is the element that physically limits light acceptance in an optical system; the exit pupil confines the light cone leaving the optical system.

We have stated that, in rotationally symmetrical systems, the center of the AS is on the optical axis; the same is true for the centers of the entrance and exit pupils, which, being images 7-306

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of the aperture stop, are also situated on the optical axis. In addition, in aberration-free systems, the shape of the pupils is a magnified (or minified) version of the aperture stop; i.e., if

Pupils

the aperture stop is circular, the pupils are circular, too (and not elliptical, for instance). ☞ A pupil determines the viewable angular subtense through an optical system. ☞ Through the front lens(es), we see the entrance pupil; through the back lens(es), we see the exit pupil. ☞ The size, shape, and location of the aperture stop (in relation to the imaging lens) determine the size, shape, and locations of the entrance and exit pupils.

Figure 7-16: Aperture stop, entrance pupil, and exit pupil in a two-lens optical system. Here, the entrance pupil and exit pupil are real images of the aperture stop.

By virtue of their definitions, the entrance and exit pupils are optical conjugates: They are both optical conjugates to (images of) the aperture stop. If the entrance pupil is an image of the aperture stop (for example, through lens 1, Figure 7-17) and, additionally, the exit pupil is an image of the aperture stop (for example, through lens 2), then the exit pupil is an image of the entrance pupil as well (through the system of lenses 1 and 2).

Figure 7-17: Reverse configuration of the optical setup presented in Figure 7-16. 7-307

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It also follows that, if we reverse light propagation, the aperture stop is an image of the entrance pupil as well. If we replace the image with the object, what functioned as the pupil now functions as the entrance pupil, and what functioned as the entrance pupil now functions as the exit pupil (Figure 7-17). What does not change, however, is the aperture stop!

Figure 7-18: Summary of the basic properties of the pupils and the angular aperture in relation to the object and image.

To maximize light admission in imaging systems when, for example, two optical instruments (or, better yet, an instrument and the human eye) are cascaded, the exit pupil of the first instrument must match (in location and diameter size) the entrance pupil of the second instrument. This is called pupil matching, which aims to ensure that all of the light from the observing instrument (such as a telescope) passes to the observing eye. The distance from the last optical surface to the (usually external) exit pupil is called the eye relief. Eye relief can be considered as the clearance between the instrument and the eye.

7.1.3.1 Simple Pupil Locations Depending on both the aperture stop location with respect to the lens focal point(s) and the configuration of the optics, the pupils may be real or virtual images of the aperture stop. The size of the aperture stop determines the size (diameter) of the pupils by virtue of the linear magnification relationship: If the magnification corresponding to the image of the aperture stop is m, the pupil diameter = m × aperture stop diameter. The pupils may be located anywhere on the optical axis, but they may not coincide with either the object or the image. Their order of appearance is dependent on the optics; for example, in Figure 7-16, the entrance pupil is before the exit pupil, but in Figure 7-25 (right) the entrance pupil is after the exit pupil (and the last lens!). •

If the optical system is only a single, thin lens without any other diaphragm, the aperture stop (AS) is the lens edge; the AS, as well as the two pupils, coincides with the lens edge.

•

If the AS is the first element (before the first lens), it is also the entrance pupil. Now the exit pupil is the image of the AS via the lens in image space (e.g., Figure 7-21).

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•

If the AS is the last element (after the last lens), the AS is also the exit pupil. Now the entrance pupil is the image of the AS via the lens in object space (e.g., Figure 7-24 and Figure 7-26).

•

If the AS is situated at the image-space (secondary) focal point of the part of the system that precedes it, then the exit pupil is at the AS and the entrance pupil is at object-space infinity [Figure 7-19 (right)]. The principal rays (§ 7.2.1) in object space are collimated and propagate parallel to the axis; the system is telecentric in object space.

•

If the AS is situated at the object-space (principal) focal point of the part of the system that follows it, then the entrance pupil is at the AS and the exit pupil is at image-space infinity [Figure 7-19 (left)]. The principal rays in image space are collimated and propagate parallel to the optical axis; the system is telecentric in image space.

Figure 7-19: A telecentric system has the aperture stop placed at a focal point of the lens: (left) A telecentric system in image space and (right) in object space.

7.1.3.2 Locating the Exit Pupil: the Aperture Stop Image in Image Space To determine the exit pupil, the aperture stop functions as a real object. Therefore, to find the image of the aperture stop in image space, we implement ray tracing by drawing construction rays directed from left to right; the positive directional distance is left to right, as usual. We can also implement the numerical solution using the object vergence and lens power relationships. For example, if the aperture stop is located at a distance greater than the focal length (i.e., to the left of the primary focal point, as in Figure 7-20), the exit pupil is a real image situated to the right of the secondary focal point of the lens. As the aperture stop gets closer to the focal point, the exit pupil moves away from the lens and increases in size (diameter)—note the increases in pupil size for pupils ② and ③ in Figure 7-20. The pupil diameter can be calculated using the magnification; for example, if m = –2, the pupil is an inverted image of the AS, with ×2 the diameter of the AS. Since the stops are usually circular apertures centered on the optical axis, their erect or inverted nature does not affect the pupil orientation: An inverted circle is the same as an erect circle. Magnification is thus used to calculate the pupil diameter in relation to the diameter of the AS.

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Figure 7-20: The exit pupil is the image of the AS in image space, formed by the optical elements succeeding it. Here, the exit pupil is a real image of the AS, and the magnification is negative. When the AS is placed in a different location, the exit pupil’s size and location both change . Example ☞: An iris diaphragm of 10 mm diameter is placed 40 cm in front of a +4.0 D lens of diameter 35 mm. Determine the aperture stop, entrance pupil, and exit pupil for collimated illumination. The aperture stop is the diaphragm, since it precedes the lens and has the smallest semi-diameter (note that collimated illumination means that the object is at infinity). The diaphragm is also the entrance pupil, since there are no imaging elements (i.e., lenses) before it. The exit pupil is the image of the aperture stop formed by the lens in image space. To calculate its location and size, we use the imaging relationship given in Eq. (4.2). The (real) object is the aperture stop, situated at x = –0.4 m; the object vergence is L = –2.5 D. The lens power is +4.0 D, so the image vergence is L΄ = +1.5 D and the image location is x΄ = +0.66 m = +66.6 cm. Therefore, the exit pupil is a real image of the aperture stop and is located 66.6 cm to the right of the lens. The magnification is m = –1.66. Therefore, the exit pupil diameter is 16.66 mm. Alternatively, we can use simple ray-tracing rules (Figure 7-21) to identify the image of the aperture stop. It is convenient to image the edge of the aperture stop via the lens.

Figure 7-21: Example of exit pupil determination. The exit pupil is a real image of the aperture stop.

What about other aperture stop locations? If the aperture stop is situated exactly at the lens primary focal point, the exit pupil is at optical infinity [creating a telecentric system, see 7-310

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Figure 7-19 (left)]. If the AS is situated at a distance less than the focal length (i.e., between the primary focal point and the lens), the exit pupil is a virtual image of the AS. Example ☞: A diaphragm of 10 mm diameter is placed 15 cm in front of a +4.0 D lens of diameter 35 mm. Determine the AS, entrance pupil, and exit pupil for collimated illumination. The AS is the diaphragm, since it precedes the lens and has the smallest semi-diameter. The AS is also the entrance pupil, since there are no imaging elements (lenses) before it. The exit pupil is the image of the AS formed by the lens in image space. To calculate the location and size of the exit pupil, we use the imaging relationship given in Eq. (4.2). The exit pupil is a virtual image, located at −37.5 cm (i.e., to the left of the lens), and the magnification is m = +2.5. Therefore, the diameter of the exit pupil is 25 mm (Figure 7-22).

Figure 7-22: Example of exit pupil determination. The exit pupil is a virtual image of the aperture stop.

7.1.3.3 Locating the Entrance Pupil: the Aperture Stop Image in Object Space The entrance pupil is the image of the AS in object space. This statement signifies a drastic departure from imaging as applied so far: Imaging of the AS takes place in object space (sounds odd!). Therefore, we consider construction rays directed from the right of the lens: the space that functions for this purpose as object space. These rays are directed to the left of the lens: the space that functions for this purpose as image space. What is normally object space becomes the ad hoc image space for the purpose of determining the entrance pupil. This is the meaning of the statement ‘the entrance pupil is the aperture stop as seen from the object.’ For this reason, the positive directional distance is reversed, i.e., light travels right to left, as seen in Figures 7-27 to 7-30. Therefore, to determine the entrance pupil (the image of the aperture stop in object space), we draw construction rays directed from right to left, and the positive directional distance is right to left. Another way to explain this oddity is that the aperture stop is always a real object; i.e., it cannot be treated as a virtual object. Therefore, its imaging assumes that it is situated before the lens (as far as light travels) as a real object for the formation of the entrance pupil (and also of the exit pupil). The system is now ‘turned around.’ 7-311

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Figure 7-23: To determine the entrance pupil location, we consider light traveling from right to left, i.e., in the direction opposite to that used to determine the exit pupil location.

The reversal of light propagation aside (and the consequent reversal of positive and negative direction signs), all imaging relationships are applied as usual. For example, if the aperture stop is located at a distance greater than the focal length, the entrance pupil is a real image of the aperture stop (Figure 7-24). If the aperture stop is located at a distance less than the focal length, the entrance pupil is a virtual image of the aperture stop [Figure 7-25 (right)].

Figure 7-24: The entrance pupil is the image of the aperture stop in object space, formed by the optical elements preceding it. Here, the entrance pupil (in all three cases) is a real image.

Figure 7-25: Aperture stop, entrance pupil, and exit pupil in a one-lens system. Here the exit pupil (in the setup on the left) and the entrance pupil (in the setup on the right) are virtual images of the aperture stop. Note that in order to find the entrance pupil, imaging is conducted in object space. 7-312

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Example ☞: An iris diaphragm of 8 mm diameter is placed 10 cm to the right of a +5.0 D lens of diameter 25.4 mm. Determine the aperture stop, entrance pupil, and exit pupil for an object at infinity.

Figure 7-26: Example of entrance pupil determination. Here the entrance pupil is a magnified virtual image of the aperture stop. The diaphragm is the aperture stop as well as the exit pupil, since there are no imaging elements after it. The entrance pupil is the image of the aperture stop formed by the lens. We locate the entrance pupil by drawing the construction rays from right to left. The image is virtual, at −20 cm (i.e., to the right of the lens), and the magnification is m = +2.0. Therefore, the diameter of the entrance pupil is 2.0 · 8 mm = 16 mm.

Example ☞: An iris diaphragm of 8 mm diameter is placed 10 cm after a –5.0 D lens of diameter 25.4 mm. Determine the aperture stop, entrance pupil, and exit pupil for collimated illumination.

Figure 7-27: Example of entrance pupil determination. Here the entrance pupil is a minified virtual image of the aperture stop. The diaphragm is the AS as well as the exit pupil, since there are no lenses following it. The entrance pupil is the image of the AS formed by the lens that precedes it, the negative lens. We draw the construction rays from right to left. The image is virtual, located between the lens and the stop. We can also use the vergence imaging relationships: The object is located at −10 cm = −0.1 m, so the object vergence is −10.0 D. We add the –5.0 D lens power, which gives an image vergence of −15.0 D. Therefore, the image is located at −0.066 m = −6.66 cm (i.e., to the right of the lens). The magnification is m = +⅔. Therefore, the diameter of the entrance pupil is (⅔) · 8 mm = 5.3 mm.

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☞ is the image of the aperture stop formed by the optical elements preceding it. ☞ determines the angular breadth of the rays that enter the system.

Exit Pupil

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☞ is associated with object space.

☞ is the image of the aperture stop formed by the optical elements succeding it. ☞ determines the angular breadth of the rays that exit the system. ☞ is associated with image space.

7.1.3.4 Locating the Pupils if the Aperture Stop is Unknown If the aperture stop is unknown, to find the pupils (and the aperture stop), we follow these steps: 1. Form the images of any possible aperture stop (AS) in object space.

2. Identify the angle subtended from the on-axis object point to the edge of each AS image.

3. The image of the element with the smallest angle is the entrance pupil.

4. The element producing this image is the aperture stop.

5. The image of the aperture stop in image space is the exit pupil.

7.1.3.5 The Entrance Pupil in the Human Eye The aperture stop of the eye is the anatomical iris. The entrance pupil is the object-space image of the iris, formed by the cornea, which is the ‘lens’ that is preceding it. For this imaging, the positive direction is from right to left. The object of the imaging is the iris opening, situated in the aqueous, which is the ad hoc object space with naqueous = 1.336. To determine the entrance pupil of the human eye, we assume the following values: average corneal power +42 D; anatomical iris situated 3.6 mm to the right of the cornea;31 eye filled with aqueous with refractive index naqueous = 1.336, and air with refractive index nair=1.0.

Figure 7-28: Entrance pupil of the human eye. This is the apparent pupil, which is situated slightly closer to the cornea and is about 12.7% larger in diameter than the anatomical iris, which is the aperture stop. 31

This is the anterior chamber depth. See Visual Optics § 2.4.2 Iris and Pupil.

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To calculate the entrance pupil location, we begin with the aperture stop (the iris) location, which is –3.6 mm = –0.0036 m before the ‘lens’ (the cornea). The object vergence is

L = 1.336/(–0.0036 m) = –371.11 D. The lens power is F = +42.00 D. We add the two such that the image vergence is L΄ = –371.11 + 42 = –329.11 D. The image space for this calculation is air, so the image location is x΄ = 1/(–329.11 D) = –0.0030 m = –3.0 mm. The entrance pupil is situated 3 mm to the left of the cornea. The magnification is calculated by the vergence ratio: n = L / L΄ = (–371.11 D)/(–329.11 D) = +1.1276. The eye’s entrance pupil is about 12.7% larger in diameter than the anatomical iris. The exit pupil, which relates the cone of rays to the retinal image, is the image of the anatomical iris via the crystalline lens. Its calculation is more complicated because of the proximity of the iris to the lens and the fact that the crystalline lens is a thick lens. Note

: The term ‘pupil’ is casually referred to as the clear aperture formed by the anatomical iris. In

optics terms, the iris is the aperture stop, not the pupil. The cornea forms an image of this aperture stop, which is the entrance pupil of the eye. In fact, the entrance pupil is what we ‘see’ and measure with pupilometry techniques.32, 33 The exit pupil of the eye is the image of the iris formed by the crystalline lens as ‘seen’ by the retina. Its calculation is a bit more complicated, since the crystalline lens can only be treated as a thick lens.

7.1.4 Numerical Aperture and F-number The entrance pupil determines the angular aperture φen of the object-originating ray bundle entering the system, while the exit pupil determines the angular aperture φex of the imageforming rays exiting the system. The half-angle (φ/2) is formed between a marginal ray [a ray passing by the edge of the pupil (§ 7.2.2)] and the optical axis. The numerical aperture NA is associated with these variables and expresses the ray angular breadth. It is defined as

Numerical Aperture:

( 2)

NA  n  sin 



D

2 f

(7.1)

where n is the refractive index of the medium, φ is the angular aperture (subtended by either the entrance or the exit pupil), f is the focal length, and D is the pupil diameter (for a system with just a simple lens, this is its diameter). The approximation (≈) is used to denote that this relationship applies to paraxial rays, which form a small angle with the optical axis. The larger 32

Spring KH, Stiles WS. Apparent shape and size of the pupil viewed obliquely. Br J Ophthalmol. 1948; 32(6): 347-54.

Kanellopoulos AJ, Asimellis G. Clear-cornea cataract surgery: pupil size and shape changes, along with anterior chamber volume and depth changes. A Scheimpflug imaging study. Clin Ophthalmol. 2014; 8: 2141-50. 33

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the numerical aperture, the greater the capacity of the lens to gather light, which is proportional to the square of the numerical aperture (NA)2. An equivalent quantity is the F-number or f/# (f-number, f-ratio, f-stop), defined as

f /# 

F-number (f/#):

1

2 (NA )



f D

(7.2)

Figure 7-29: F-number and numerical aperture for a simple lens.

According to Eq. (7.2), a larger lens diameter for the same focal length results in a greater NA and a smaller f/#. Both the NA and the f/# can be used to describe the beam spreading. A collimated ray bundle has zero numerical aperture, a positive diverging bundle, and a negative converging bundle (in a way, the numerical aperture is related to convergence with opposite algebraic signs). If a collimated pencil of rays completely illuminates a lens, the emerging bundle has the lens f/#. The smaller the ratio f/D, the larger the numerical aperture, the smaller the f/#, and the larger the collection angle of the lens—the cone of light that can be collected by an optical instrument. This lens forms brighter images of finer detail (brightness and resolution, § 7.5)

and f/#:

aperture

Numerical

compared to a lens with a larger f/#. ☞ NA describes the angular breath of a cone of light rays, in other words, the angular ray acceptance of an optical instrument. ☞ NA values range from 0 to n, the refractive index of the medium. ☞ Fast optical systems have large NA values and small f/# values.

Example ☞: The human eye’s pupil diameter D varies from 2 to 8 mm (daylight to nightlight). Its medium (aqueous/vitreous) has refractive index n = 1.33. The focal length is ≈ 23 mm. What is the eye’s f/# ? For the maximum pupil diameter, the semi-diameter is 4 mm. The half-angle angular aperture is φ/2, approximated by the ratio of semi-diameter to focal length: φ/2 = (4 mm)/(23 mm) [rad], and also,

sin(φ/2) ≈ 4/23. Then, NA = n · sin(φ/2) ≈ 1.33 · 4/23 = 0.23, or f/# = 1/ (2 · NA) ≈ 1/ (2 · 0.2) ≈ 2.5.

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For the minimum pupil diameter (semi-diameter 1 mm, φ/2 = 1 mm/23 mm), NA ≈ 1.33 · 1/23 = 0.058, or f/# = 1/ (2 NA) ≈ 1/ 2 · 0.058 ≈ 8.6.

The numerical aperture is very helpful in optical instruments with a short working distance, such as in fiber optics and microscopes, while the f/# is very helpful for distant objects, such as in telescopes, binoculars, and camera lenses.

Figure 7-30: (left) Photography lenses with different f-numbers. The first lens has f-numbers from 1.8 to 22, and the second lens has f-numbers from 1.2 to 16. The lens can be closed or opened such that its f/# can be varied from the largest value, indicating the minimum lens aperture, to the smallest value, indicating the maximum lens aperture. (right) Apertures (left to right) with f/# = 11, 5.6, and 1.2. As the f/# decreases, the lens aperture area increases, leading to a faster system with more light throughput. (Lens images on left from www.nikonusa.com.)

7.2 PRINCIPAL / CHIEF AND MARGINAL RAYS 7.2.1 The Principal / Chief Ray The principal or chief ray is an oblique ray that originates from a non-axial object point (height ≠ 0) and passes through the center of the aperture stop. Thus, the principal ray has three known points: the origin, which is the off-axis object point, a mid-point, which crosses the optical axis at the center of the aperture stop, and the off-axis conjugate image point.

Figure 7-31: Example of a principal ray in a system comprising just one lens and an aperture stop. 7-317

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Because the pupils are optical conjugates of the aperture stop, the principal ray (or its extrapolation) enters the optical system along a line directed toward the center of the entrance pupil and exits the system along a line passing through the center of the exit pupil. If a pupil is a real image of the aperture stop, the principal ray intersects the center of that pupil. If a pupil is a virtual image of the aperture stop, the extrapolation of the principal ray (not the ray) intersects the center of that pupil. We can determine the principal/chief ray by following these steps:

1. Launch a ray from an offaxis object point, directed toward the center of the entrance pupil.

Note

2. If there is a lens before the aperture stop, the ray bends at the lens and passes through the center of the aperture stop.

3. If there is lens after the aperture stop, the ray bends at that lens and passes through the center of the exit pupil.

: If the entrance pupil is after the lens (i.e., is a virtual image of the AS), the ray extrapolation passes

through the center of the entrance pupil. If the exit pupil is to the left of the lens (i.e., is a virtual image of the AS; Figure 7-31), the (virtual) ray extrapolation passes through the center of the exit pupil.

7.2.2 The Marginal Rays The marginal rays originate from the on-axis (paraxial) object point (height = 0), pass by the periphery (edge) of the aperture stop, and terminate at the on-axis image point.

Figure 7-32: Marginal rays. In this example, the exit pupil is a virtual image, so the extrapolations of the marginal ray pass by the edge of the exit pupil.

Both the principal rays and the marginal rays are real rays (i.e., they correspond to actual light propagation) that traverse the optical system unobstructed. They both connect an object point to its conjugate image point. 7-318

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•

The principal ray originates from an off-axis object point (e.g., the head of the farm animal in Figure 7-31) and travels toward its conjugate off-axis image point.

•

The marginal rays originate from the on-axis object point (e.g., the nose of the farm animal in Figure 7-32) and travel toward its conjugate on-axis image point.

•

The principal ray passes through the center of the aperture stop.

•

The marginal rays pass through the edge of the aperture stop. They are therefore the most extreme rays, forming the largest possible angle with the optical axis as they leave the object.

We can determine the marginal rays by following these steps:

1. Launch a ray from the on-axis object point, directed toward the edge of the entrance pupil.

Note

2. If there is a lens before the aperture stop, the ray bends at the lens and passes through the edge of the aperture stop.

3. If there is a lens after the aperture stop, the ray bends at that lens and passes through the edge of the exit pupil.

: Some variations of this rule:

If there is no physical diaphragm but just a lens, the lens edge is the aperture stop, the principal ray passes by the center of the lens, and the marginal rays pass by the edge of the lens. If the aperture stop is the entrance pupil as well, the marginal ray is directed to the edge of the aperture stop. If the aperture stop is also the exit pupil, the marginal ray leaves the system passing by the edge of the aperture stop. If the entrance pupil is after the lens (i.e., is a virtual image of the aperture stop; see Figure 7-35), a virtual extrapolation of the marginal ray passes through the edge of the entrance pupil. If the exit pupil is to the left of the lens (i.e., is a virtual image of the aperture stop; see Figure 7-32), the extrapolation of the virtual ray passes through the edge of the exit pupil.

Because the edge of the aperture stop is not a single point (unlike its center) but rather a geometrical shape (such as a circular rim), there is an infinite number of marginal rays (explaining the use of the plural in marginal ‘rays’). The marginal rays confine the light cone from the on-axis object; this is the light cone entering the system, as introduced in § 7.1.1. The edges of the entering and the exiting light cones that relate the object and image to the entrance pupil and exit pupil, respectively, are the marginal rays (see Figure 7-16) or extrapolations of the marginal rays (see Figure 7-35).

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7.2.2.1 Strategies for Drawing Principal and Marginal Rays The proper drawing of the principal and marginal rays can be facilitated by following some strategic steps. First and foremost is to identify the elements that comprise the system (such as the lenses and diaphragms) and the object location; then we determine: •

the (final) image location (including possible intermediate images), using either ray tracing or numerical solution / imaging relationships, along the lines of many of the examples presented in Chapter 4 and § 6.5.6, and

•

the aperture stop, and then its images in object space (entrance pupil) and in image space (exit pupil), using the techniques presented in § 7.1.1 and § 7.1.3.

Figure 7-33: System comprising two lenses and an aperture stop. Here the final image is real and erect. The entrance pupil is a virtual image of the AS, and the exit pupil is a real image of the AS.

To draw the marginal ray, we need the landmarks for this ray, which are the on-axis object point, the edge of the aperture stop, and the on-axis image point. The ray (or its extrapolation) enters via the edge of the entrance pupil and leaves via the edge of the exit pupil. We now launch a ray from the on-axis object point toward the edge of the entrance pupil (Figure 7-34).

Figure 7-34: First attempt to draw the marginal ray. The ray is aimed at the edge of the entrance pupil.

This is problematic! On its way to enter the system via the edge of the entrance pupil, this ray strikes a lens, so it bends. Any ray, in general, is refracted by a lens; the marginal ray is 7-320

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no exception. Because the ray is refracted, its extrapolation remains targeted at the edge of the entrance pupil (here the entrance pupil is a virtual image of the AS), and the new direction of the marginal ray is toward the edge of the AS, and then the on-axis point of the intermediate image.

Figure 7-35: The marginal ray strikes the lens, is refracted, and then passes by the edge of the aperture stop.

Τhe strategy to draw the marginal ray can be summarized so far as follows: • • • •

•

•

Launch a ray from the on-axis object point aiming toward the edge of the entrance pupil. The marginal ray is refracted by any lens that it encounters. The marginal ray always passes by the edge of the aperture stop. The marginal ray enters the system via the edge of the entrance pupil. If the entrance pupil is a real image of the aperture stop, the ray passes by its edge (see Figure 7-37). If the entrance pupil is a virtual image of the aperture stop, the extrapolation of the marginal ray passes by its edge (see Figure 7-35). The marginal ray crosses the optical axis at the intermediate image (if it exists). If the intermediate image is real, the ray passes by it; otherwise, the extrapolation of the ray passes by the virtual intermediate image. The marginal ray leaves the system by the edge of the exit pupil. If this is a real image of the aperture stop, the ray passes by its edge (Figure 7-36 and Figure 7-37). If the exit pupil is a virtual image, the extrapolation of the marginal ray passes by its edge.

Figure 7-36: The marginal ray passes through the edge of the exit pupil and the on-axis point of the image.

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•

The marginal ray passes by the on-axis point of the (final) image. If this image is a real image of the object, the ray passes by its on-axis point. If this image is a virtual image of the object, the extrapolation of the marginal ray passes by its on-axis point. Figure 7-37 presents a simpler example. It is simpler because the two images of the

aperture stop (i.e., the entrance pupil and the exit pupil) are real.

Figure 7-37: Determination of marginal rays when all pupils and images are real images. The marginal rays pass by the edges of the aperture stop, enter the system by the edge of the entrance pupil, and leave the system by the edge of the exit pupil.

For this reason, the marginal rays pass by the edges of the entrance pupil as well as by the edges of the exit pupil, in addition to passing by the edges of the aperture stop. The marginal rays originate from the on-axis object point and terminate at the on-axis image point; they also pass by the on-axis intermediate image point, the latter two of which are both real images. If the object or any of the image(s) are virtual, the extrapolation of the marginal ray crosses the optical axis at the location of the virtual object or image, including intermediate images. Τhe strategy to draw the principal/chief ray follows a similar set of rules. The key difference is that the principal ray originates from an off-axis point and always crosses the center of the aperture stop. Thus, it crosses the optical axis at least once, i.e., at the aperture stop. •

The first step is to launch the ray from an off-axis object point aiming toward the center of the entrance pupil. If the entrance pupil is a real image of the aperture stop, the ray passes by the center of the entrance pupil; if it is a virtual image of the aperture stop, the extrapolation of the principal ray passes by the center of the entrance pupil.

•

The principal ray is refracted by any lens that it encounters.

•

The principal ray always passes by the center of the aperture stop.

•

The principal ray leaves the system by the center of the exit pupil. If the exit pupil is a real image of the aperture stop, the ray passes by the center of the exit pupil; if the exit pupil is a virtual image, the extrapolation of the principal ray passes by the center of the exit pupil.

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Figure 7-38 presents the principal ray drawn for the system introduced in Figure 7-37. The principal ray crosses the optical axis three times: at the center of the aperture stop, at the center of the entrance pupil, and at the center of the exit pupil, both pupils being real images of the aperture stop.

Figure 7-38: Determination of the principal ray when all pupils and images are real images.

When both the principal ray and the marginal rays are shown (Figure 7-39), we can clearly note the major differences between these rays: The principal ray enters the system by the center of the entrance pupil, while the marginal rays enter by the edge of the entrance pupil. The principal ray leaves the system by the center of the exit pupil, while the marginal rays leave by the edge of the exit pupil.

Figure 7-39: Principal / chief ray and marginal rays in a two-lens system.

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Principal / Chief Ray • There is only one for any given off-axis object point. • It intersects the optical axis at the aperture stop center and the two pupil centers if the pupils are real images; otherwise, its extrapolation does so. • It passes by the edge of the field stop and the two windows / ports (§ 7.3.3). • Because it originates from an off-axis object point, it depends on the object height.

Marginal Rays • There is an infinite number of marginal rays for any on-axis object point. • They intersect the optical axis at the object and image on-axis points. • They pass by the edge of the aperture stop and the two pupils if the pupils are real images; otherwise, the marginal ray extrapolations pass by the edges of the pupils. • Because they originate from an on-axis object point, they do not depend on the object height.

Path of the Principal / Chief Ray: The ray intersects the optical axis at the center of the aperture stop. The principal ray corresponding to the field height passes by the edge of the field stop. It intersects the optical axis at either pupil center if the pupils are real images of the aperture stop. Its extrapolation intersects the optical axis at either pupil center if the pupils are virtual images of the aperture stop. Then: • The exiting pricipal ray extrapolation intersects the exit pupil. • The entering principal ray extrapolation intersects the entrance pupil.

Path of the Marginal Rays: These rays intersect the optical axis at the object and image points. They pass by either pupil edge if the pupils are real images of the aperture stop. Their extrapolations pass by either pupil edge if the pupils are virtual images of the aperture stop. Then: • The exiting marginal ray extrapolation passes by the edge of the exit pupil. • The entering marginal ray extrapolation passes by the edge of the entrance pupil.

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7.3 FIELDS, STOPS, AND RELATED EFFECTS 7.3.1 Field of View The field of view (FoV) is the lateral extent of the object-space area that can be imaged by an optical system. It can be expressed either angularly or linearly. The angle by which an object (in general, the observable extent of an object) is viewed is the angular field of view (aFoV) or angle of view (AoV). The angle of view is reported as either the full aFoV or the half-angle, semi-aFoV = aFoV/2. The latter is formed by the principal / chief ray and the optical axis intersecting at the entrance pupil of the optical instrument. The linear field of view (linear FoV) can be used to describe the field height at a given distance. Similar to the ray height, the field height is the (greatest value of the) lateral distance of the object that can be imaged. The linear FoV can also express the chord contained by the angle at an indicated range. In this concept, the linear field of view is an angular quantity that simply reports lengths and is used in estimating the suitability of an instrument for tasks in observation or photography.

Figure 7-40: Field of view: angular and linear.

Terrestrial telescopes and binoculars express the linear FoV in feet at a distance of 1000 yards (420 ft). When using the metric system, the linear FoV can be reported in a length ratio (y m over 1000 m). For example, a linear FoV of 0.5 m at 50 m can be converted to 10 m at 1000 m. In other cases, such as when using a hand-held magnifier, the linear field of view is expressed as the maximum viewable linear dimension of the object without referring to the viewing distance. Conversions between the linear FoV and the aFoV are facilitated by using geometry. For example, the ratio of the linear FoV (shown as y in Figure 7-41) to the observation length (shown as z) is a good approximation of the aFoV expressed in radians: ϑ = aFov = y/z [rad]. We can then convert this to degrees (1 rad = 57.3°), or use trigonometry: ϑ[°]= tan–1(y/z). If the 7-325

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observation distance is 1000 m, the radian angles can be converted to degrees by dividing the linear FoV by 1000 · 2π /360° = 17.45°. The linear FoV y can be computed by the observation distance z and the aFov as follows:

y = z · aFoV [rad] = z · tan(aFoV [°]).

Figure 7-41: Geometry for conversions between linear FoV and angular FoV. Examples ☞: A linear FoV of 5 m at a distance of 50 m corresponds to an aFoV of 5/50 = 0.1 rad. We can convert this to degrees by using the conversion 1 rad = 57 °, so aFov = 5.7°, or by using trigonometry: tan–1(5 / 50) = 5.7 °. An aFoV of 30° at 1000 m corresponds to a linear FoV of 1000 · tan(30°) = 577.35 m. A linear FoV of 87.2 m at 1000 m corresponds to aFoV [rad] ≈ 87.2/1000 = 0.0872 rad ≈ (87.2 / 17.45)° = 5°. A terrestrial telescope has an aFoV = 30’ = 0.5°. At 1000 m, this telescope may see in its entirety an object whose linear dimension is 1000 · tan(0.5°) = 1000 · 0.00872 = 8.72 m wide. This is the linear FoV at 1000 m.

Figure 7-42: Imaging with a decreasing field of view from the same observation point. The leftmost image corresponds to an aFoV of 50°. The rightmost image corresponds to an aFoV of 3.15°.

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Figure 7-43: (left) Aerial image showing the location of the images in Figure 7-42. (right) This IMACS (Inamori-Magellan Areal Camera and Spectrograph) picture is a composite image of the galaxy NGC 253 (used with permission from Alan Dressler).34 The dashed lines indicate the fully illuminated 15:40΄ × 15:40΄ field. The solid circumference line shows the projection of the CCD mosaic, which measures 28΄ across.

Astronomical telescopes have a particularly small aFoV. With an aFoV of 0.25º = 16΄, we can observe part of the sky around a celestial body such as the Silver Coin or Silver Dollar Galaxy NGC 253, which is an intermediate spiral galaxy in the constellation Sculptor. Wide-angle lenses have a large aFoV: A fish-eye lens AFoV can be close to 180°. In the human eye, the monocular horizontal aFoV is about 165°, and the vertical aFoV is 130°.35 In space, the field of view is referred to as a pyramid with its apex at the center of the entrance pupil. The horizontal FoV corresponds to the horizontal plane and the vertical FoV to the vertical plane. The diagonal FoV is measured between two diagonally opposite edges.

Figure 7-44: Horizontal, vertical, and diagonal field of view (aFoV).

7.3.2

The Field Stop

The field of view is limited by a physical element, the field stop (FS), which imposes a hard limit on the principal/chief rays. We observe in Figure 7-45 that the principal/chief ray that Dressler A, Bigelow B, Hare T, Sutin B, Thompson I, Burley G, Epps H, Oemler Jr A, Bagish A, Birk C, Clardy K. IMACS: The InamoriMagellan areal camera and spectrograph on Magellan-Baade. Publications of the Astronomical Society of the Pacific. 2011; 123(901): 288. 34

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corresponds to the maximum field height passes by the edge of the field stop. Thus, the field stop (its size and its separation from the entrance pupil) determines the lateral spatial extent (for example, how many surface points) of the object that can be imaged.

Figure 7-45: The field stop is the physical element that determines the spatial extent (points) of the object field that can be imaged. A larger field stop allows a larger field to be imaged.

The field stop is a physical aperture. It may be the edge/rim of a lens, such as the eyepiece in systems with at least two lenses, or an actual aperture or frame inside or outside the optical system. The field stop is often located at the image plane. In the eye, the periphery of the macula can be considered the field stop. This is because the responsiveness to light is significantly less in the area outside the macula. The most popular optical instrument, the photographic camera, also has a field stop, which is the frame (edge) of the sensor/film.

Food for Thought

: Have you ever wondered why photographs (digital or print) are rectangular?

The lens cross-section is circular; so is the aperture stop (almost). Yet, photographs are rectangular in shape. This is because the frame of the sensor (which used to be film in the old days) is the field stop. This frame is rectangular in shape.

7.3.2.1 Finding the Field Stop To determine which element functions as the field stop, we seek the answer to this question: What element limits the field of view? One element is definitely out: The aperture stop can never be the field stop because the principal/chief ray always passes through its center. At the field stop, the principal/chief ray has the largest possible height to semi-diameter ratio because it passes by the field stop edge.

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Figure 7-46: The principal ray that corresponds to the field height passes by the edge of the field stop.

This observation leads to the description of a simple method to locate the field stop in an optical system using the height to semi-diameter ratio:

A recipe to identify the field stop:

1. Launch a principal/chief ray from the object that passes through—is not blocked by— any element or surface.

2. Mark the height at which the ray intersects the optical elements or surfaces.

3. The optical element (or surface) with the largest height to semi-diameter ratio is the field stop.

In the example shown in Figure 7-47, the elements that can serve as the field stop are the two lenses (the aperture stop can never be field stop). The second lens, which is called the eyepiece because it is closer to the eye, serves as the field stop: The height to semi-diameter ratio (= 1.0) of the eyepiece is the largest of the two possible elements.

Figure 7-47: Field stop in a two-lens system: The eyepiece (lens 2) is the field stop in this arrangement.

In telescopes, for example, the field stop is a lens, typically, the eyepiece, which is usually the one with the shortest focal length and smallest semi-diameter.29 7-329

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In the example shown in Figure 7-48, the elements that can serve as the field stop are the two lenses; once again, the aperture stop can never be a field stop. The first lens (called the objective because it is closer to the object) serves as the field stop: The height to semidiameter ratio of the objective is the largest among possible elements (in this case = 1.0).

Figure 7-48: In this two-lens-imaging system example the field stop is lens 1 (the objective).

7.3.2.2 Simple Field Stop Cases The field stop is always the ‘other’ element, once the aperture stop is identified. This means that in systems with just two separate elements, the field stop is the element that is not the aperture stop! Therefore, in any system with just two elements, once we identify the aperture stop, we have also identified the field stop, which is the other element. In general, the field stop is the element that subtends the smallest angle from the center of the entrance pupil. It can be placed after the entrance pupil (e.g., Figure 7-49) or before the entrance pupil (e.g., Figure 7-55).

Figure 7-49: Determination of the field stop using the smallest angular subtense from the center of the entrance pupil.

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A simple recipe to find

1. Launch a principal ray from the center

2. The element that

the field stop (with no

of the entrance pupil that passes through

forms the smallest

lens, or when the lens

the edges of all possible field stops and is

angular subtense is the

is the last element):

not blocked by any element or surface.

field stop.

When viewing an opening at some distance, say a porthole window, the system comprises just two apertures: the porthole and the pupil of the viewing eye, which is much smaller in diameter (can be a few millimeters) than the porthole. For a distant object, the system’s entrance pupil is the entrance pupil of the eye; the remaining element, the porthole, is the field stop.

7.3.3 Entrance and Exit Windows / Ports The object-space image of the field stop, formed by the preceding optical elements (for example, any and all lenses before the field stop), is the entrance window or entrance port. Its edge subtends the smallest angle with the center of the entrance pupil. The image-space image of the field stop, formed by the succeeding optical elements (for example, any or all lenses after the field stop), is the exit window or exit port. Its edge subtends the smallest angle with the center of the exit pupil. In the example presented in Figure 7-50, because the field stop is (the edge of) the first lens, there is no other optical element to the left of it. Therefore, the entrance port coincides with the field stop. The exit port is the image of the field stop formed by the lens after it, which is the second lens. It is a real image of the field stop.

Figure 7-50: Field stop and its (real) images: entrance window / port and exit window / port. Also shown are the aperture stop and its images: the entrance pupil and exit pupil. The principal ray passes by the edge of the field stop (and its images) and the center of the aperture stop (and its images). 7-331

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By virtue of their definitions, the entrance and exit ports are optical conjugates, in addition to being optical conjugates of the field stop. In the case where the field stop is situated at the image plane, the entrance port (i.e., the field stop image through the lenses before it) is located in object space exactly at the object plane (Figure 7-51).

Figure 7-51: Field stop at the image-plane entrance port on the object plane. The exit port coincides with the field stop; the entrance port is the (real) image of the field stop formed by the lens before it.

The field stop (if there is one) is always a real object; i.e., it cannot be treated as a virtual object. Therefore, its imaging, for the purpose of determining the ports, reckons it as if it is located before the lens. Determining the exit port is easy: To find the image of the field stop in image space, we draw construction rays from object space to image space, as usual. The positive directional distance is forward, i.e., from left to right, as shown in Figure 7-50. Using a method similar to the way we find the entrance pupil, to determine the entrance port (the image of the field stop in object space), we draw construction rays from image space to object space. This is because the entrance port is the field stop as seen from the object. The positive directional distance direction is now reversed, i.e., from right to left, as in Figure 7-51. The object space becomes the ad hoc image space for the purpose of determining the entrance port.

Figure 7-52: To determine the entrance port location, we consider light traveling from right to left, i.e., in a direction opposite to the direction used for determining the exit port location. 7-332

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If the field stop is not known, a method to determine the stop and the entrance and exit windows/port can be articulated. This method is similar to the one used for determining the aperture stop; the difference is that the angle now has its apex at the center of the entrance pupil instead of at the on-axis object point.

1. Form the images of any possible field stop in object space.

2. Identify the angle subtended by the edge of each image from the center of the entrance pupil.

3. The image of the element forming the smallest angle is the entrance window / port.

4. The element producing this image is the field stop.

5. The image of the field stop in image space is the exit window / port.

A corollary to this is that the aperture stop can never be the field stop. Because its image in object space is the entrance pupil (so the subtended angle is 180°), there is no chance that the aperture stop is the smaller stop. Therefore, in this method, it is implied that there is no

☞ is the field stop image formed by the optical elements before it in object space. ☞ forms the smallest angle with the center of the entrance pupil. ☞ is associated with object space.

Exit Window / Port

Entrance Window / Port

need to image the aperture stop in object space as a candidate for determining the field stop.

☞ is the field stop image formed by the optical elements after it in image space. ☞ forms the smallest angle with the center of the exit pupil. ☞ is associated with image space.

Figure 7-53: Field stop situated at the lens plane (left) and at the image plane (right). 7-333

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Table 7-1: Comparison of aperture stop and field stop properties. Aperture stop

Can be situated before or after the lens, but is always before the image plane. Can be the edge of a lens, typically the objective. Confines peripheral rays (and associated aberrations, such as spherical aberration). Affects image brightness. Affects depth of field and geometrical blur (defocus outside the image plane). Defines the diffraction-limited optical resolution (image sharpness at the image plane).

Field stop

Can be situated exactly on the image plane (not a rare case). Can be the edge of a lens, typically the eyepiece. Confines field aberrations such as distortion and curvature. Defines the angle formed by the principal/chief ray at the image plane. Its size and its location with respect to the aperture stop determine the field of view. Confines the viewable object extent and restricts vignetting.

Figure 7-54: Relationship between the entrance pupil and entrance port (left) and the exit pupil and exit port (right).

7.3.4 Size of the Field of View One of the simplest cases of field of view involves the model of a long cardboard box with a narrow window opening at one of its sides. The box is viewed from the opposite side by the observer’s eye. The window opening is the field stop. If you’re still wondering about what role each element plays in this cardboard box example, just think about this: Squinting your eyes effectively reduces the eye pupil. It might make the image appear dimmer, but you’re still viewing the same field—the same number of cars across the street. Thus, your pupil is the aperture stop (and entrance pupil), not the field stop. If you close down the window or select a smaller porthole while standing at the same place, you’ll be viewing a smaller, but not dimmer, field. This is why the window is the field stop. 7-334

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Figure 7-55: In this cardboard box, the window opening is the field stop.

This example can help us quantify the aFoV, which is the angular subtense of the window opening from the viewing eye. For the same tube length (window–aperture-stop separation), the aFoV widens if the window is larger (wider). If we keep the window size fixed and move away from the window (longer tube length), the aFoV narrows; if we move closer to the window (shorter tube length), the field of view widens. Therefore, the size of the (angular) field of view depends on the ratio of two parameters: the size (diameter) of the field stop and its separation from the aperture stop. A simple way to express this is the ratio of the window size to the length of the tube; this works well for small angles. A more accurate way is to consider the right-angle triangle tangent defined by the ratio of the field stop semi-diameter (size a) to its separation from the aperture stop (distance dp). This is an exact derivation of the semi-aFoV (i.e., aFoV/2). Example ☞: A cardboard box that is 10 cm long has a window of 2 cm in diameter. What is the aFoV? The window opening is the field stop and the entrance port, while the viewer’s eye serves as the entrance pupil. The aFoV is the angular subtense of the field stop (semi-diameter a = 1 cm) from the center of the aperture stop (dp = 10 cm): aFoV = 2 · tan−1(1 / 10) = 11.4 °.

Figure 7-56: Calculating the aFoV using the angular subtense of the window opening from the aperture stop.

7.3.4.1 Definition and Computation of the aFoV The simple carboard box case helped us understand that the (angular) size of the field of view depends on two parameters: the size (half-diameter) of the field stop and its separation from the aperture stop.

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In systems with lenses, it is proper to determine the aFoV using the ratio of the entrance port diameter to its distance from the entrance pupil. Since the aFoV is the angular subtense of the principal/chief rays, all we have to do is launch a ray from the edge of the entrance port that passes by the center of the entrance pupil or from the center of the entrance pupil to the edge of the entrance port (whichever is positioned first). The angle formed by the principal ray and the optical axis is the semi- (half of the) aFoV. Implementing simple geometry in Figure 7-57, if a is the semi-diameter of the entrance port, and the entrance port is separated from the entrance pupil by a distance dp, then Field of View:

a  aFoV  tan  =   2  dp

a  aFoV [] = 2  tan −1    dp   

(7.3)

Figure 7-57: The aFoV is bound by the principal rays passing by the edge of the entrance port and the center of the entrance pupil.

The size of the entrance port semi-diameter a is critical: Doubling a doubles the aFoV. The separation dp of the entrance port from the entrance pupil is also important: An entrance port closer to the entrance pupil (smaller dp) forms a larger aFoV. The two stop images that determine the aFoV can be present in either order, e.g., the entrance pupil may be in front of (Figure 7-57) or behind the entrance port (Figure 7-58). This affects neither the definition nor the magnitude of the aFoV; the formulas are exactly the same.

Figure 7-58: Determination of the aFoV when the entrance port is in front of the entrance pupil.

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7.3.4.2 Object-Space aFoV versus Image-Space aFoV There are two distinct aFoVs: the object-space aFoV (more often used) and the image-space aFoV. To compute the object-space aFoV, we use the ratio of the entrance port semi-diameter a to its distance from the entrance pupil dp [Eq. (7.3)].

Figure 7-59: The object-space aFoV is calculated using the ratio of the entrance port semi-diameter (a) to its distance from the entrance pupil (dp).

To compute the image-space aFoV, we use the ratio of the exit port semi-diameter a to its distance from the exit pupil dp [Eq. (7.3)]. Essentially, this is the angular subtense of the exit port from the center of the exit pupil. The formulas for the image-space aFoV are therefore identical to those for the object-space aFov, differing only in the fact that we use the semidiameter of the exit port and its separation from the exit pupil in the former. In general, the object-space and image-space aFoVs are not equal in magnitude.36 A comparison between Figure 7-59 and Figure 7-60 shows that, indeed, the object-space aFoV can be quite different from the image-space aFoV.

Figure 7-60: The image-space aFoV is calculated using the ratio of the exit port semi-diameter (a) to its distance from the exit pupil (dp). Compare with Figure 7-59.

7.3.4.3 Field of View in the Presence of a Lens In a system comprising a thin lens (of focal length f) with an object placed at infinity and a field stop semi-size h (which would be the sensor’s half-height in modern cameras) placed at the image (sensor) plane, the entrance port is at optical infinity. They are only equal if the entrance and exit pupils are situated on their respective nodal points. This is the case of a thin lens surrounded by the same medium with the aperture stop at the lens. 36

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We can compute the image-space aFoV considering the angular subtense of the exit port from the exit pupil. In the case where the object is at infinity, the image is formed at the focal plane of the lens. Now, the separation of the exit pupil from the exit port (distance dp) equals the focal length of the lens f; therefore, Field of View (lens focused at infinity):

h  f

aFoV = 2  tan −1 

(7.4)

Figure 7-61: The aFoV when the lens is focused at infinity and the field stop is at the sensor (image) plane.

In many devices such as the photography camera, the sensor (field stop) has a fixed size, so the aFoV is inversely proportional to the focal length: The shorter the focal length, the larger the aFoV. This is why short-focal-length lenses (e.g., 35 mm or less) are considered to be wide field, while long-focal-length lenses (e.g., 125 mm or more) are considered to be narrow field.

Figure 7-62: Two photographs with a large and a small field of view, taken from the same spot (Molyvos, Lesvos Island, Greece) with different focal length lenses: (left) short-focal-length (35 mm) wide-angle lens with a large field of view and (right) long-focal-length (200 mm) telephoto lens with a small field of view.

When a lens is used as a collimating magnifier, the FoV that matters is the linear field of view, and specifically, just the linear extent of the viewable object. Consider a lens of focal length f (power F = 1/f) with a semi-diameter h. The lens is held at a distance d. We seek the size of the linear field through this lens, which is simply the linear length of the object. The object is situated at the primary focal point of the lens; therefore, the rays leave the lens collimated. If the lens diameter 2h is sufficiently larger than the eye’s pupil diameter, the 7-338

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lens rim is the field stop and the eye’s pupil is the aperture stop and exit pupil. A principal ray that leaves the topmost clearly visible object point (this is the field semi-diameter with height y) crosses the center of the aperture stop, forming an angle φ (= aFoV/2) with the optical axis. We now apply some basic geometry to obtain (for small angles [rad]): aFoV/2 = φ = h / d [rad] and φ = y / f, which is rearranged to: y = f · h/d = h/d · F

Figure 7-63: The linear FoV when the lens is used as a collimating magnifier.

Note that angle φ is computed as the ratio of height to distance; therefore, it is simply the ray slope, which approximates the angle for small angles, which, however, is not quite true for such short focal lengths. The above relationship is re-arranged (multiply both sides ×2) as Linear Field of View:

lens diameter observation distance  lens power

(7.5)

The above relationship expresses the common knowledge that when a more powerful plus lens is used as a hand-held magnifier, the FoV is smaller. In a hand-held magnifying lens, the maximum magnification is inversely proportional to the lens focal length (or proportional to the power):37 Lens Magnifying Power:

M ANG =

25 cm

f

+1

(7.6)

The combination of Eqs. (7.5) and (7.6) indicates that a large-magnification, low-vision aid produces a narrow linear field of view.

7.3.4.4 Field of View when a Lens is Fitted on a Frame Lenses in front of an optical device (such as a prescription lens in front of the eye) alter the aFoV. In a simple approach, we may state that a plus lens decreases the aFoV, while a minus

37

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lens increases the FoV. This is because we understand that with a plus lens the principal rays converge toward the axis, while with a minus lens they diverge away from the optical axis. The working principle is that when we view a distant object via a lens held at some distance from the eye, the rim of the lens plays the role of the field stop (FS), while eye’s pupil is the aperture stop (AS). The refractive effect of the lens alters the size and, more importantly, the location, of the entrance pupil, so the aFoV is affected. Consider a 2 inch diameter lens (semidiameter a = 1.27 cm) held at a distance dp = 10 cm from the observer’s eye. No lens, just a frame The simplest case is when there is no lens at all, just the lens rim or a plano lens fitted on the frame. This case is handled just like the window in a simple cardboard box. Example ☞: A plano lens (or a clear frame) of 1 inch diameter is held at a distance of 10 cm. What is the aFoV when the lens is viewed via the observer’s eye? The frame is the field stop and the entrance port, while the viewer’s eye serves as the aperture stop and entrance pupil. The aFoV is the angular subtense of the FS with semi-diameter a = 1.27 cm (= 0.5 inch), axially separated from the center of the entrance pupil by dp = 10 cm: aFoV = 2 · tan−1(1.27 / 10) = 14.5 °.

Figure 7-64: The aFoV when viewing a lens with no refractive effect (just the frame).

Minus lens We fit the frame with a minus lens. The entrance pupil is now a virtual image of the aperture stop, located closer to the lens, whose frame is still the field stop and the entrance port. The aFoV increases. Example ☞: A –5.00 D lens of 1 inch diameter is held at a distance of 10 cm. What is the aFoV when the lens is viewed via the observer’s eye? The lens rim is the field stop and the entrance port. The viewer’s eye is the aperture stop. Its image via the lens placed at the frame forms the entrance pupil. It is a minified (m = +⅔), virtual image of the AS, located closer to the lens than the AS (6.66 cm). The aFoV is the angular subtense of the FS (a = 1.27 cm semi-diameter) separated from the center of the entrance pupil by dp = 6.66 cm: aFoV = 2 · tan−1(1.27 / 6.66) = 21.6 °. 7-340

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Figure 7-65: The aFoV when viewing a minus lens.

Plus lens Next, we fit a plus lens. The entrance pupil is now an image of the aperture stop. Depending on the focal length and viewing distance, the entrance pupil may be located farther or closer to the lens whose rim is the field stop and the entrance port. Typically, with a low-power plus lens, the aFoV decreases; however, when a high-power plus lens is used, the aFoV may increase as well and appear inverted! Example ☞: A +5.00 D lens of 1 inch diameter is held at a distance of 10 cm. What is the aFoV?

Figure 7-66: The aFoV when viewing a plus lens at a distance shorter than its focal length.

In this case, the entrance pupil is a magnified (m = +2) virtual image of the AS, located farther from the lens than the ΑS (20 cm to the right). The aFoV is the angular subtense of the FS (a = 1.27 cm semidiameter) separated from the center of the entrance pupil by dp = 20 cm: aFoV = 2 · tan−1(1.27 / 20) = 7.3 °.

Example ☞: A +30.00 D lens of 1 inch diameter is held at a distance of 10 cm. What is the aFoV?

Figure 7-67: The aFoV when viewing a plus lens at a distance greater that its focal length.

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In this case, the entrance pupil is a minified (m = –0.5) real image of the AS, located closer to the lens than the AS (5 cm). The aFoV is the angular subtense of the FS (a = 1.27 cm semi-diameter) separated from the center of the entrance pupil by dp = 5 cm: aFoV = 2 · tan−1(1.27 / 5) = 28.5 °. The field here is inverted!

Figure 7-68: The FoV when viewing the same scene via a minus lens (left) and a plus lens (right).

7.3.5 Fields of Half and Full Illumination The full or total field of view is an extended concept of the aFoV that includes all possible points of the object field that can be imaged, even with reduced illumination. Rays that pass through the entrance port may partially fill the entrance pupil (Figure 7-69) and therefore illuminate the image. The rays that cross the edge of the entrance port and the opposite edge of the entrance pupil define the total or extended field of view, in which, however, illumination may drop gradually toward the periphery. Implementing simple geometry in Figure 7-69, if a is the semi-diameter of the entrance port, and b is the semi-diameter of the entrance pupil, which is separated from the entrance port by a distance of dp, then Total Field of View (tFoV):

a+b a+b  tFoV  tan  =  tFoV = 2  tan −1    dp  dp  2   

Figure 7-69: Total (or extended) field of view. 7-342

(7.7)

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While the extended field is viewable, the illumination is not necessarily uniform over its extent. This is only true at the maximum illuminated field of uniform illumination, which is defined as the geometrical confinement of object points that, when imaged, correspond to the same (uniform) illumination through the system. This field can be produced by the rays that pass by the edge of the entrance port and the (same side) edge of the entrance pupil. Implementing simple geometry in Figure 7-70, if a is the semi-diameter of the entrance port, and b is the semi-diameter of the entrance pupil, which is separated from the entrance port by a distance of dp, then Field of Uniform Illumination (FUI):

a−b  FUI  a − b tan   =  FUI = 2  tan −1   dp  dp  2   

(7.8)

Figure 7-70: The field of uniform illumination.

There is also a ‘soft’ definition of the field of at least one-half illumination that represents a compromise between the restricted field (field of uniform illumination) and the extended (total) field, which is commonly used as a working approximation for the aFoV. Note ❶

: It is getting confusing. Which one is the field of view?

The aFoV is the optically sensible field of view. It is defined as the angular subtense of the principal

rays—those that graze the field stop (entrance port) and pass through the center of the aperture stop (entrance pupil). ❷

The total field of view is the extended field that includes object points that form partially illuminated

images. It is defined by rays crossing the opposite-side edges of the entrance point and entrance pupil. ❸

The field of uniform illumination is the most restrictive of all. It requires that all light from an object

point that is admitted by the entrance pupil reaches the entrance port. It is defined by rays crossing the same-side edges of the entrance point and the entrance pupil. ❹

The field of (at least one) one half illumination is a soft compromise between the total FoV and the

field of uniform illumination. It is commonly accepted as the working aFoV.

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Example ☞: Determine the aFoV, the total FoV, and the field of uniform illumination in a system with an entrance port 20 mm in diameter and an entrance pupil 10 mm in diameter, separated by 4 cm. The aFoV is the angular subtense of the entrance port (semi-diameter a = 1 cm) from the center of the entrance pupil (dp = 4 cm): aFoV = 2 · tan−1(1.0 / 4) = 28 °. The total FoV is the angular subtense of the entrance port (semi-diameter a = 1 cm) plus the entrance pupil (b = 0.5 cm) from the center of the entrance pupil (dp = 4 cm): aFoV = 2 · tan−1(1.5 / 4) = 41 °. The field of uniform illumination is the angular subtense of the entrance port ( a = 1 cm) minus the entrance pupil (b = 0.5 cm) from the center of the entrance pupil (dp = 4 cm): aFoV = 2 · tan−1(0.5 / 4) = 14.25 °.

Figure 7-71: Total field of view, angular field of view (aFoV), and field of uniform illumination. Note

: In this example, the three angular fields have notably different values. By rule, the larger field is

the total FoV, and the narrower field is the field of uniform illumination. The closer the separation between the entrance port and the entrance pupil, and the larger the entrance pupil in relation to the entrance port diameter, the greater the difference between the three fields. However, if the entrance pupil is much smaller than the entrance port (for example, when viewing a porthole window), or the port–pupil separation is rather large, then the three fields are almost identical.

7.3.6 Vignetting and Glare Consider an on-axis object point. The beam of light accepted into the optical system is shaped like a cone that is limited by the aperture stop. For an off-axis object point, the beam through the system is shaped like a section of a skewed cone, as long as no aperture other than the aperture stop limits the beam. If the beam is partially intercepted by one or more additional apertures, the imageforming cone of light is trimmed. Often, a bundle of rays from the off-axis tip of the object that is smaller than a bundle from the on-axis base propagates into the system; thus, the image irradiance from the off-axis tip is smaller than that of the base. 7-344

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The result is vignetting, a gradual darkening of the image toward the corners that is noticeable in poorly designed optical imaging instruments (projectors, microscopes, etc.). Vignetting is perhaps the only optics terminology that is not directly related to any light or geometrical property. It is derived from the French vignette for vineyard. The vine leaf coloration resembles the gradual edge fading seen in old-time photos. The loss of light at the peripheral marginal rays might help to reduce aberrations, which is sometimes desirable.

Figure 7-72: Vignetting setup in a two-lens system. The peripheral cone of light (shown in red) is partially blocked by the aperture, resulting in reduced image brightness at the periphery.

Being a peripheral effect, vignetting strongly depends on the object field height. No vignetting occurs when the entire ray bundle from a given object point passes through all of the apertures non-intercepted. Each aperture radius must be equal to or larger than the maximum height of the ray bundle at that aperture.

Figure 7-73: Cross-section through the optical system at the plane of an aperture. The bundle of rays that forms an off-axis object cone of light may or may not get through this aperture in its entirety. Vignetting Results from partial blocking or attenuation of peripheral object rays by the optical system. Occurs when other apertures in the system, such as a lens clear aperture, block all or part of an off-axis ray bundle. Manifests as a gradual image fading at the periphery.

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We’ve been complaining about light being undesirably eliminated. Now, it’s time to complain about light being undesirably added. Often, particularly in complicated optical instruments involving many elements, there is unwanted light, called veiling glare or stray light.

Figure 7-74: An image with evident sun glare manifesting as stray light imposed on a fishing boat.

Sunlight, and perhaps some indirect reflections originating from outside the observation field, may be glare sources. Once it enters the optical instrument, such light may reach the image plane via multiple reflections on internal surfaces, sharp corners, and other mechanical elements inside the optical tube. This light is superimposed on the image in the form of a veil, which leads to image deterioration and contrast reduction. For this reason, such light is unwanted, and we wish to restrict it. The glare stop is an aperture used to restrict glare. It is typically placed at a location corresponding to an internal image of the aperture stop (or the entrance pupil). This ensures that all useful illumination participates in image formation, while unwanted light is cut off.

Figure 7-75: The glare stop is designed to prevent unwanted light from reaching the image plane.

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7.4 DEPTH OF FIELD AND DEPTH OF FOCUS Objects in the field of view may be located at quite different distances from the imaging lens, which can be a photography camera lens, a microscope objective, or the eye. If we apply the imaging relationship, we will find that for a different object location there is a different image location. For example, with a positive lens, the farther away the object, the closer to the lens focal point the (real) image is formed. Therefore, each object forms a clear, sharp image (the commonly used term is that the lens ‘focuses’) at different distances from the lens plane. Think of a specific location in image space. There is a unique point in object space, which is its optical conjugate. An object placed at this conjugate point forms a sharp image in the image-forming plane. An object displaced axially from this conjugate point forms an image whose projection on the set image plane is blurry, with less resolution and contrast. Thus, if the image plane is fixed, there is a specific object distance at which the formed image exactly falls for a given lens power. But wait! Without much effort, we can clearly see almost everything from a desk in front of us to a faraway building, and even farther! Some readers might object, rightfully. Is there only one ‘specific’ object distance? Enter the concepts of depth of field and depth of focus. Photographers know quite well that there is an acceptable image quality for objects if they are within a certain range, but not for objects that are outside this range. This object-space range is the depth of field. We can therefore define the depth of field as the axial range in object space within which objects may form reasonably sharp images. This range can be reported in length, in which case it is called the linear depth of field. The linear depth of field is the difference between the ‘endpoints,’ called the distal and the proximal limiting positions. Between these endpoints is the conjugate object point—the point that is optically conjugate to the image-forming plane.

Figure 7-76: (left) The range of object locations for which a sharp image is formed is quite large; i.e., the depth of field is large. (right) The same photograph with a much shallower depth of field.

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Figure 7-77: Objects at the limiting points of the depth of field project images of marginally perceivable equal blur. Objects within the range of the depth of field form reasonably sharp images.

The dioptric depth of field, reported in diopters, is the difference in object vergence between the two limiting positions. Due to the linear nature of the imaging relationship, the dioptric depth of field is symmetrical with respect to the conjugate object point. For example, if the distal point has a +1.0 D vergence difference compared to the conjugate object point, the proximal point has a –1.0 D vergence difference. Then the extent of the depth of field can be reported as either 2.0 D or ± 1.0 D. Note that the linear depth of field is not symmetrical with respect to the conjugate point, since distances are reciprocal to the vergence. Example ☞: Estimate the dioptric and the linear depth of field when the distal limiting position (farthest point) has −4.0 D vergence and the proximal limiting position (nearest point) has –6.0 D vergence. Where is the conjugate point dioptrically and linearly? The dioptric depth of field is 2.0 D or ±1.0 D. The conjugate point in object space is at −5.0 D vergence, dioptrically half way between the farthest and the nearest points. Its location is therefore 20 cm in front of the lens. The farthest point, with –4.0 D vergence, is 25 cm in front of the lens. The nearest point, with −6.0 D vergence, is 16.66 cm in front of the lens. The linear depth of field, the separation between the 25 cm and 16.66 cm points, is 8.33 cm.

Example ☞: A clear image is formed if an object is placed 50 cm in front (to the left) of the lens. If the dioptric depth of field is ±0.50 D, what are the proximal and distal points of the depth of field? When the object is placed 50 cm in front (to the left) of the lens, the object vergence is –2.0 D. The distal (farthest) point has vergence –1.5 D, and the proximal (nearest) has vergence –2.5 D. Therefore, the distal point is 66.6 cm in front of the lens, and the proximal point is 40 cm in front of the lens. The linear depth of field is 26.6 cm.

Objects outside the depth of field form blurred images such that their distal and proximal points are indeed perceived as blurred by at least some recognizable image blur, a circle of least confusion (Figure 8-3). There is no ‘hard’ transition but, instead, a gradual loss of 7-348

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sharpness. Objects just in front of or behind the conjugate object distance38 begin to lose sharpness, even if this may not be perceived by our eyes or on a printed picture. Therefore, the depth of field is determined by how the images of these two limiting points are projected. Each object at either the proximal point or the distal point projects a defocused image at the conjugate plane. These defocused images have a blur of at least a perceived circle of confusion. Thus, the circle of confusion is a quantitative criterion of how much a point can be blurred in order to be perceived as no longer acceptably sharp. The parameter ‘perceived’ can be subjective. Differences in perception exist when simply viewing with the naked eye versus when viewing through a magnifying lens, or when enlarging a picture to a large print format.

Figure 7-78: Nature photograph with a very large depth of field, taken with an f/22 setting.

An acceptably sharp circle of confusion that is commonly in use is one that is unnoticed when a 35 mm negative film is enlarged to a standard 8×10 inch print and viewed from 1 ft. At this distance and print size, it is assumed that the smallest blur spot is unnoticeable if it is smaller than a circle of confusion of no more than 0.033 mm (0.01 inch). This 0.01 inch standard is used by lens manufacturers when providing depth of field markers on photography lenses. The extent of the depth of field is dependent on optical parameters such as the pupil size and focal length, and is influenced by optical aberrations.39 Assuming ‘perfect’ optics, i.e., no aberrations, the depth of field is shorter (shallower) with a larger pupil diameter and a shorter working object distance, which is typically determined by the lens focal length. In photography lenses, the depth of field can be very short when the object is close to the lens and if the lens is ‘fast,’ for example, f/# = 1.8 (the F-number).40 The depth of field can

This point is commonly referred to as the nominal focus or simply the focus. It stems from the use of the verb ‘to focus,’ as employed in photography lenses. We avoid using the term ‘the focus’ to denote this point in an effort to avoid confusion with the focal point, which is different. An object placed at the primary focal point F will form a virtual image at optical infinity (§ 4.6.3). 38

For example, sphericity, a parameter that can compensate for spherical aberration (§ 8.3.1.1), can be used to improve (enlarge) the depth of field. 39

40

Introduction to Optics § 6.2.3 Light Control in a Photography Camera. 7-349

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be very large if the lens is ‘stopped down’ to f/22, meaning that the aperture stop diameter is very small. The depth of field increases with an increased f/# (reduced aperture stop diameter), an increased object distance, or a shorter lens focal length.

Figure 7-79: Dependence of the depth of field on the aperture stop. The depth of field increases with a smaller aperture stop (right).

Figure 7-80: Photographs with a short (left, using f/22) and a long (right, using f/2.8, i.e., a large aperture) depth of field. In the right sundial image, the foreground is sharp, while the background is blurred.

The depth of field applies in object space. In image space, the corresponding concept is the depth of focus, the stretch in image space within which acceptably sharp images are formed. In photography, the depth of focus is the range in which the sensor (or the film) can be moved back and forth with respect to the lens with no manifest change in image sharpness. In the eye, the depth of focus is the perceptual tolerance of the retinal defocus; possible sharpness differences for images formed within the depth of focus are not noticeable. Although, theoretically, the macula forms a perfectly sharp retinal image only for a precise optical conjugate point, slightly defocused images formed in the depth of focus are also perceived as sharp. Due to similarities in name and nature, the depth of field and the depth of focus are often confused. Some similarities are as follows: Any object within the depth of field forms its 7-350

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image within the depth of focus; both the depth of field and the depth of focus depend on the aperture stop diameter (for example, they both increase with a smaller diameter). By virtue of the vergence imaging relationship, the vergence difference in object space between the two limiting positions of the depth of field equals the vergence difference in image space between the image points corresponding to these positions. Therefore, the dioptric

☞ an interval in object space (e.g., where the object can be located and be in focus) ☞ depends on the aperture stop diameter (the larger the diameter, the smaller the depth of field)

Depth of Focus

Depth of Field

depth of focus equals the dioptric depth of field. ☞ an interval in image space (e.g., inside the eye) ☞ depends on the depth of field and magnification

Now, some differences: The depth of field refers to a range of object locations, while the depth of focus refers to a range of image locations in relation to a sensor in the instrument (the digital sensor or film in cameras, or the retina in the eye). In photography and vision, in which objects are typically far away from the (relatively small) focal point, the depth of field can often be several meters long, while the depth of focus might typically be fractions of a millimeter. The linear depth of focus depends on the depth of field and on the magnification, which is dependent on the lens–object distance and the focal length of the instrument. In vision, the depth of field is the separation between the proximal and distal limiting positions for objects that can be perceived with the same clarity at the same time.41 The depth of field of the eye has clinical relevance, since it relates to the perceived refractive status and to presbyopia. The depth of field is affected by neural and psychophysical factors. Because of the variable pupil size, the eye's depth of field is variable: The depth of field increases with a smaller pupil, affording a larger range of acceptable vision, which often helps myopes and presbyopes.42 For example, to compensate for accommodation loss, we might surgically reduce the pupil diameter. The technique, which employs a dark ring inserted in the cornea (Kamra corneal inlays), reduces the effective pupil.43 The depth of field increases, which relieves presbyopic vision to a degree.

This only applies without a change in accommodation—an increase in the optical power of the crystalline lens. Therefore, we clarify that these points are not the same as the near and far points. The near point involves maximum accommodation, and the far point involves zero accommodation, so these two points could never be perceived at the same time. 41

42

Green DG, Powers MK, Banks MS. Depth of focus, eye size and visual acuity. Vision Res. 1980; 20(10): 827-35.

Jalali S, Aus der Au W, Shaarawy T. AcuFocus corneal inlay to correct presbyopia using femto-LASIK. One year results of a prospective cohort study. Klin Monbl Augenheilkd. 2016; 233(4): 360-4. 43

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Figure 7-81: Depth of field and depth of focus in the human eye.

Because the depth of field defines the range of object locations for which clear images are perceived, often, in optics (and in photography) we do not need to focus (i.e., to fixate) on the background, which is often at infinity. Focusing at infinity does not make good use of at least half of the available range of object locations, which extends beyond infinity. It is preferable to focus closer than infinity, at the dioptric mid-point of the depth of field. This is the hyperfocal distance: the fixation distance (casually termed focus) that places the distal limiting position, which is the farthest edge of a depth of field at the scene background. Then, the proximal edge of the useful range of objects expands to the full depth of field.

Figure 7-82: Hyperfocal distance.

When we place our focus at the hyperfocal distance, both the foreground and the horizon or distant background (which is often at infinity) are imaged with acceptable sharpness. If the depth of field is large, for example, with a wide-angle, short-focal-length lens, we may need to focus just a short distance from the lens to get the horizon acceptably sharp. The hyperfocal distance dependencies are identical to those of the depth of field, which are the aperture diameter and focal length. Example ☞: If the hyperfocal distance is 10 m, the depth of field extends from 5 m to infinity. Objects placed between 5 m and the horizon appear sharp (they are in the depth of field).

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Example ☞: If the background is at infinity and the depth of field is ±0.50 D, where is the hyperfocal distance? When considering the hyperfocal distance, infinity (vergence 0.00 D) is the distal point of the depth of field. We therefore need to locate the mid-point of the dioptric field, which has vergence –0.50 D (by the way, the proximal point has vergence –1.00 D). The hyperfocal distance is the linear distance from the system to

☞ A lens is focused at infinity if the image plane coincides with the secondary focal point. Then, the object is at infinity.

infinity?

Focused at

the –0.50 D vergence point, which is at 2 m.

☞ The eye is focused at infinity if the fixation point is the far point. If the eye is emmetropic, the far point is at infinity.

7.5 BRIGHTNESS, CONTRAST, AND RESOLUTION We can determine the quality of a formed image based on how bright, crisp, and clear it is. Image brightness refers to the amount of light reaching the image. Ranging from dark to bright, brightness is related to luminance on the image surface. Of course, sometimes, we can have too much light if the image is ‘washed out.’ If an image looks too white, it has too much

Brightness

Image

brightess, and if is too dark, it does not have enough brightness. ☞ governed by the light-gathering power of the lens, which is: ☞

proportional to the square of the numerical aperture and

☞

inversely proportional to the square of the f/#.

If the distinction among the gradients of gray tone is unsatisfactory, the contrast is low. Contrast describes the difference in brightness between the darkest and brightest areas of the image. Good contrast facilitates the distinction of features relating to brightness variations. All of the parameters involved in the brightness variations are affected by many factors, some of which include the spatial extent of the optics involved in image formation. Can we see infinitely small detail? The answer is no. Many concepts presented in the previous chapters involve certain approximations that ignore optical aberrations, considering only perfect wavefronts. Chapter 8 is devoted to optical aberrations. It discusses how ‘perfect’ an illumination or imaging system can or cannot be. In order to proceed, we assume ‘perfect’ 7-353

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optics, in other words, no aberrations at all, and perfectly uniform and plane illumination. Can we now image a point from an object to a single image point? The answer is still no.

Figure 7-83: The Molyvos harbor, Greece, imaged with (top left) good resolution, brightness, and contrast. (top right) The same image with poor resolution, (bottom left) poor brightness, and (bottom right) poor contrast.

Geometrical optics practically fails to describe contrast. While there is blur outside the image plane (geometrical image blur is discussed in § 7.6), there is nothing in geometry that explains why an image formed exactly on the image plane by such an ideal, aberration-free optical system should not be infinitely sharp and crisp. Enter wave optics. A key wave effect, diffraction, describes what happens when rays pass through a circular aperture, such as that of a lens.44 In this case, the image from an object point formed by an ideal optical system has a specific limit to how small it can be. This limit is a spot, called the Airy disk. Therefore, if an optical system is at least as good as diffraction allows it to be (in other words, its point image is more or less the size of the corresponding Airy disk), it is called diffraction limited. The Airy disk has a radius of Airy Disk Radius (spatial extent): Airy Disk Radius (angular subtense):

1.22 · λ · (f/#) = 1.22 · λ · (f/D) 1.22 · λ /D

(7.9) (7.10)

where λ is the wavelength, f/# is the F-number, f is the focal length, and D is the pupil diameter.

44

Wave Optics § 5.4 Circular Aperture Diffraction.

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The angular subtense is computed in units of radians. If the medium is not air, the wavelength is substituted by λ/n, where n is the refractive index of the medium. Resolution describes the ability of an imaging system to discern (resolve) detail in the object being imaged. We want to distinguish two very small and closely spaced independent objects as separate entities. Thus, resolution refers to the minimum feature size of the object, or the imaging system's ability to reproduce object detail. The least separation (angular or spatial) between the closest distinguishable points is the resolution limit. We observe two black dots and see them as two. If located farther away, the two black dots form a smaller angle. At some point, we can barely tell if there is one dot or two. This angle is the resolution limit. Naturally, the smaller this smallest angle, the better. The lens (pupil) diameter D relates diffraction (wave optics) to resolution (imaging optics). The larger the D, the smaller is the extent of the Airy disk. Now, imagine two Airy disks that are image spots corresponding to two distinct object points. To be seen as two, their image spots must be separated such that the maximum (brightest) of one coincides with the minimum (darkest) of the other; in other words, their separation must be at least the radius of the Airy disk. This is therefore the smallest resolvable separation between two points, according to the Rayleigh criterion, which determines resolution.

Figure 7-84: (left) A single object imaged to an Airy disk, (center) two unresolvable spots, and (right) two marginally distinguishable spots. The right image shows the Rayleigh criterion.

The minimum angle of resolution ϑoMIN expresses how small the angular separation of two object points can be such that they are barely resolved:

oMIN =

1.22  wavelength lens diameter

=

1.22   / n

D

(7.11)

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The above equation applies to an ideal (diffraction-limited) system with uniform, plane wave illumination. The angle is expressed in radians (rad). Therefore, the resolution limit is proportional to the f/#, inversely proportional to the exit pupil diameter D, and proportional to the wavelength λ. The f/# in a lens (an optical setup) relates to its resolution. A lens with a small f/# (large aperture), e.g., f/1.2, can form a finer-detail, sharper image than a lens with f/1.8.

Resolution of the human eye The resolution limit of the human eye derives from ϑMIN = 1.22 (λ/n)/D. We use λ = 0.55 μm, n = 1.336, and 2 mm for daylight pupil diameter D. Based on the above values, the human eye can discern objects separated by a minimum angle: ϑMIN= (1.22 ∙ λ/n)/D = 1.22 ∙ (0.55 μm/1.336 ) /2 mm) = 0.25 mrad ≈ 0.86΄ (arcmin). The corresponding spatial separation of 0.25 mrad is ≈ 0.25 mm at 1 m away, ≈ 0.0625 mm at 25 cm (the typical near-vision distance), and ≈ 1.5 mm at 6 m (the typical testing distance for vision charts).

Often enough, however, we would like to express the quantity represented by ‘the better’ (in “the smaller this smallest angle, the better”) with a larger number. Therefore, we use the reciprocal of the resolution limit, which is the resolving power or ability: The larger the resolving power, the better. Resolution Limit • The separation (either angular or spatial) between the closest distinguishable points imaged through an optical instrument. • The smaller, the better: Higher resolution means smaller separation. • Is expressed in angle, e.g., mrad, arcmin (degree minutes), arcsec (degree seconds), or in length, e.g., mm, μm, nm.

Resolving Power (Ability) • The reciprocal of the separation between the closest distinguishable points imaged through an optical instrument. • The larger, the better. • Is expressed in inverse angle, e.g., arcmin–1, arcsec–1, or in inverse length, e.g., dots-per-inch (DPI), line pairs/mm.

It should be noted that a loss of resolution is accompanied by a loss of contrast because the blurring results in some gray spillover, which causes the light areas to become less white and the dark areas to become less black. See, for example, the ideal contrast in Figure 7-85 (left) compared to the images in the center and right, which are reduced-contrast versions. 7-356

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Figure 7-85: Image with gradually reduced contrast.

The examples in Figure 7-85 are not true imaging examples, as there is no loss of resolution: The landmarks between the black and the intermediate white/gray stripes are clearly seen. When an image is formed, there is always some inevitable loss of resolution; in addition, there is also a loss of contrast. The images in Figure 7-86 show slight (left) and gradually increasing (center and right) blur. It is obvious that the blurred images have lost contrast as well as resolution.

Figure 7-86: Image with gradually decreased resolution, which is accompanied by a loss of contrast.

Figure 7-87: Details of the photographs from Figure 7-80. (left) With f/ 32, we can see less detail on the sundial, while (right) with f/ 2.8, we can see finer detail on the sundial. The detail (at-focus resolution) in these examples is governed by diffraction.

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7.6 GEOMETRICAL IMAGE BLUR It is desirable to have objects placed within the depth of field such that their image is formed on the proper conjugate image location on the image-receiving plane, situated within the depth of focus. Images formed on this plane can be as sharp as the resolution limit allows.

Figure 7-88: Blur caused by defocus: Away from the image plane a defocused (blurred) projection of the image is formed.

However, away from the image-receiving plane, a blurry image projection is formed. This is geometric image blur. The farther away from the image plane the image is formed (the larger the defocus), the larger the blur. In other words, the amount of geometric blur is proportional to the amount of defocus. Also, blur is proportional to the pupil diameter: A greater pupil diameter (expressed by a small f/#) correlates to a set of marginal rays that diverge sharply away from the focal plane, causing a greater geometric blur (Figure 7-92 and Figure 7-95).

Figure 7-89: In a blurry image, there is a loss of sharpness and edge definition.

Geometric blur is produced by a geometric transformation of the sharply formed image as it is projected away from the image-conjugate plane. It re-distributes light intensity in a bell shape that spreads along the lateral cross-section of a sharply formed shape. Thus, edges lose clarity and certain discrete fine image features are gradually lost. While the linear dimensions of the image can be well defined, this is not true for the blurred version. The blurred version is 7-358

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slightly enlarged due to the ‘fudging’ of the edges. This is exactly the case of imaging an object that is outside the depth of field.

Figure 7-90: The larger the deviation from the image plane (defocus), the larger the geometric blur. Blur also results from simply moving the image-receiving plane away from the image plane location or increasing the aperture stop diameter.

Figure 7-91: With the same optical setup as in Figure 7-88, exactly on the image-receiving plane of the original object (the sea lion), an object at a different location (the penguin) will project a blurred (defocused) image.

Because the blur increases farther away from the image-receiving plane (in proportion to the defocus), we can use this property to determine the depth of field by considering that the largest acceptable blur is at the limiting positions of the depth of focus. For this purpose, we estimate the size (diameter d) of the geometric blur circle. This is the diameter of a blurred circle corresponding to a point projected on the image-conjugate plane, at the edge of the depth of focus Δ that we perceive as ‘acceptable.’ Consider an aperture (exit pupil) diameter D. By geometry, according to Figure 7-92,

D d =  f 2



D circle of confusion d =   =   NA 2 f

(7.12)

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Figure 7-92: Calculation of the geometric blur circle diameter. This diameter can help to estimate the circle of confusion.

Figure 7-93: Of the three objects, only one (the seal) forms a sharp image on the image-receiving plane corresponding to the seal. The walrus forms an image slightly in front of the image plane, while the penguin forms an image slightly behind the image plane. On this plane, the walrus and the penguin are projected as blurry images.

This relationship is used so that the largest allowable blur circle diameter d is the maximum acceptable spot size in the system’s image. For example, for digital cameras, the maximum d should be no larger than the size of a single detector element (pixel), and preferably smaller. We realize that the geometric blur is proportional to the depth of focus (and, by extension, to the depth of field) and to the aperture diameter D, or the numerical aperture NA (and, by extension, inversely proportional to the f/#). A small f/# (large aperture) creates a very large geometric blur. In photography, objects outside the depth of field form a defocused background known as bokeh, from the Japanese bo-ké (ボケ). Usually employing a large NA (small f/#) to produce a shallow depth of field, bokeh results in aesthetically pleasing photographic compositions with a background blur or haze (Figure 7-94).

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Figure 7-94: Portrait with a shallow depth of field and a strong background blur (an example of bokeh) shot with f/ 3, using a 100 mm lens. (Image taken on Mykonos Island, Greece.)

Bokeh can be achieved with a fast lens and a long focal length or a fast macro lens, which usually has a shallow depth of field. Yes, such lenses can be expensive; do not expect to achieve this with the lenses installed in mobile phones! The subject of the photograph, particularly in a portrait shot, is enhanced by bringing more attention to it because its perceived sharpness contrasts with the strongly defocused backdrop.

Figure 7-95: Nature photograph with (left) moderate and (right) strong background blur. The same lens (100 mm) was used in both photos. In the left photograph, the setting was f/18, while in the right photograph, it was f/ 3.

When using a photography lens with mechanical overlapping blades, the bokeh defocus takes the physical shape of the lens. Therefore, the ‘circle’ of confusion is not a circle, but is only approximated as such when very small. When the ‘circle’ of confusion is large, most lenses will render holiday background lights to an attention-grabbing polygon shape (Figure 7-96).

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Figure 7-96: Artistic bokeh effects in photographs of (left) a New Orleans horse hitching-post and (right) holiday lights in the Big Apple (New York City).

We realize by now that there are two different kinds of blur. The first relates to the sharpness of the image formed on the image plane. This blur is governed by diffraction and defines resolution. A larger NA or a smaller f/# (larger aperture) results in less image blur exactly on the image-receiving (focal) plane and therefore also results in better resolution. Conversely, a smaller NA or a larger f/# (smaller aperture) results in more image blur exactly on the image-receiving (focal) plane, or, equivalently, less resolution. Away from the image-forming plane, or for objects outside the depth of field, we have geometric image blur, which is produced by defocus and governed by geometrical optics. A larger NA or a smaller f/# (larger aperture) results in increased geometric image blur away from the image-forming plane—more defocus. Conversely, a smaller NA or a larger f/# (smaller aperture) results in reduced geometric image blur away from the image-receiving (focal) plane—less defocus (Figure 7-97).

Figure 7-97: Resolution, field of view, and depth of field.

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7.7 ADVANCED PRACTICE EXAMPLES Identification of the Aperture Stop Example ☞: An object is placed 25 mm to the left (in front) of a +40 D lens whose diameter is 36 mm. A circular aperture of 10 mm diameter is placed 12.5 mm to the right of the lens. Identify the aperture stop.

This object (located at the primary focal point) forms an image at optical infinity. Therefore, rays leaving the lens propagate parallel to the optical axis.

Figure 7-98: Aperture placed after the lens, and object placed 25 mm in front of the lens. We draw the two marginal rays from the on-axis object point that pass by the edges of the lens (the orange ray) and the diaphragm (the blue ray). •

The ray that passes by the edge of the lens strikes the diaphragm at 18 mm and is obstructed.

•

The ray that passes by the edge of the diaphragm (blue, 5 mm height) continues unobstructed.

The height to semi-diameter ratios are 5 mm/18 mm = 0.277 at the lens and 5 mm/5 mm = 1.0 at the diaphragm. The diaphragm is the aperture stop because it corresponds to the largest ratio.

Example ☞: Consider a lens of +40 D power and 24.5 mm diameter. We use a set of two diaphragms, 2.5 mm and 15 mm in diameter, spaced apart by 25 mm, the first of which is located 25 mm to the right of the lens (Figure 7-99). Identify the aperture stop for collimated illumination (object at infinity).

Because the object is at optical infinity, the image point is none other than the secondary focal point, situated 25 mm to the right of the lens (power = +40 D). Therefore, collimated rays traverse the system toward the secondary focal point. We draw rays parallel to the optical axis that pass by the edges of the lens (the orange ray) and diaphragm ❷ (the blue ray). The rays are directed toward the secondary focal 7-363

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point, which is situated exactly at the center of diaphragm ❶. Therefore, despite being just 1.25 mm in semi-diameter, no ray is obstructed by diaphragm ❶, as all rays pass through its center. Diaphragm ❶ cannot be the aperture stop in this case!

Figure 7-99: Set of two diaphragms placed after a lens. •

The (orange) ray that passes by the edge of the lens is obstructed at diaphragm ❷ (height = – 12.25 mm, which is larger than the diaphragm semi-diameter).

•

The (blue) ray that passes by the edge of diaphragm ❷ has a height of –7.5 mm and a height to semidiameter ratio = 7.5 mm/7.5 mm = 1.0 (here we ignore the – signs for the purpose of calculating the ratios). At the lens, this ray also has a height of +7.5 mm, but the height to semi-diameter ratio at this point is 7.5 mm/12.25 mm = 0.61.

The largest ratio corresponds to diaphragm ❷, which serves as the aperture stop in this case.

Note

: One may wonder what the role of diaphragm ❶ might be. Placed at the focal point of the lens,

the diaphragm is usually called the confocal aperture or pinhole. Its role is to eliminate light that does not correspond to the intended origin (for example, to block stray light).

Example ☞: A clear aperture of 30 mm diameter is placed 10 mm after a minus lens of –40.0 D power and 26 mm diameter (Figure 7-100). What element is the aperture stop for an object at infinity? We might be tempted to say that the lens is the aperture stop because it has the smallest opening. Caution! This may not be true. We start by recognizing that, because the object is at optical infinity, the image point is the secondary focal point, situated 25 mm to the left of the lens (power = –40.0 D). Next, we draw rays parallel to the optical axis, which, after refraction by the lens, are directed away from the secondary focal point.

•

The (orange) ray passing by the edge of the lens (semi-diameter 13.0 mm) strikes the aperture at 18.2 mm (use similar triangles and note that the aperture is 35 mm from F΄) and is blocked.

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Figure 7-100: Aperture placed after a minus lens and object at infinity. •

The (blue) ray passing by the edge of the aperture (15.0 mm) first meets the lens at 10.7 mm height (use similar triangles) and propagates through the system. For this ray, the height to semi-diameter ratios are 15 mm/15 mm = 1.0 at the aperture and 10.7 mm/13.0 mm = 0.82 at the lens.

The aperture is the aperture stop because it corresponds to the largest height to semi-diameter ratio. This is a bit of a surprise because the aperture stop has the larger opening of the two elements.

Example ☞: In the same setup as before, we now place the clear aperture of 10 mm before the lens (Figure 7-101). What element is the aperture stop for an object at infinity? We draw rays parallel to the optical axis, which, after refraction by the lens, are directed away from the secondary focal point. •

The (orange) ray passing by the edge of the aperture (semi-diameter 15.0 mm) is lost (not collected) by the lens and therefore does not traverse the system.

Figure 7-101: Aperture placed before a minus lens and object at infinity. •

The (blue) ray passing by the edge of the lens meets the aperture at 13.0 mm. For this ray, the height to semi-diameter ratios are 13.0 mm/15.0 mm = 0.866 at the aperture and 13.0 mm/13.0 mm = 1.0 at the lens. The lens is the aperture stop because it corresponds to the largest height to semi-diameter ratio.

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Example ☞: Consider two identical lenses of +2.0 D power and 50 mm diameter, spaced apart by 100 cm. What element is the aperture stop for an object that is placed 25 cm in front of the first lens? The first step is to identify the two images formed in this system, the intermediate image and the final image (the intermediate image technique is presented in § 6.5.5). The (final) image formed by the lens system is real, inverted [m = m1 · m2 = (+2.0) · (−⅓) = −1.0], and located 75 cm to the right of lens ❷.

Figure 7-102: Determining the aperture stop in a two-lens system when the object is 25 cm in front of lens ❶. One of the two lenses has to be the aperture stop. We launch two rays that reach the edges of each lens. •

The marginal ray from the on-axis object point directed toward the edge of lens ❶ is lost (not collected by lens ❷) and therefore does not traverse the system.

•

The marginal ray from the on-axis object point directed toward the edge of lens ❷ follows a path from the intermediate image formed by lens ❶ and intersects lens ❶ at height = 8.33 mm.

For this ray, the height to semi-diameter ratios are 0.33 for lens 1 and 1.0 for lens 2. Therefore, the edge of lens ❷ is the aperture stop.

Locating Pupils when the Aperture Stop is Unknown Example ☞: In a two-lens system, lens ❶ has focal length 6 cm and diameter 4 cm, and lens ❷ has focal length 2 cm and diameter 3 cm. Their separation is 4 cm. A circular diaphragm (D) of diameter 1 cm is placed 3 cm after the first lens (Figure 7-103). An object is placed 9 cm in front of the system. Find the aperture stop and the two pupils. The three aperture stop candidates are: lens ❶, diaphragm D, and lens ❷. The image of lens ❶ in object space is lens ❶ itself. The angle subtended by the lens edge with the on-axis object point is the ratio of the semi-diameter to the distance from the object 2/9 = 0.22 rad ≈ 12.5°.

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The image of D formed by lens ❶ is located 6 cm after the lens and has a diameter of 2 cm (m = +2). Therefore, its distance from the on-axis object point is 15 cm, and its semi-diameter is 1 cm. The angle subtended by the edge of this aperture image from the optical axis is 1/15 = 0.066 rad = 3.8°.

Figure 7-103: Image of diaphragm D in object space formed by lens ❶. The image of lens ❷ formed by lens ❶ is located 12 cm after lens ❶ and has a diameter of 9 cm (m = +3). Therefore, its semi-diameter is 4.5 cm, and its distance from the object is 21 cm. The angular subtense of this aperture image with the on-axis object point is 4.5/21 = 0.214 rad = 12°.

Figure 7-104: Image of lens ❷ in object space formed by lens ❶. The entrance pupil is therefore the image of D, since it forms the smallest angle (3.8°). Thus, D is the aperture stop. The exit pupil is the image of D formed by lens ❷ (Figure 7-105). It is located 2 cm to the left of the lens and has a diameter of 2 cm (m = +2).

Figure 7-105: Image of diaphragm D in image space formed by lens ❷. 7-367

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The system, along with the two pupils, is presented in Figure 7-106. The rays that pass by the edge of the aperture stop are marginal rays (§ 7.2). Both pupils are virtual images of the aperture stop.

Figure 7-106: Entrance and exit pupils in the two-lens system with diaphragm D serving as the aperture stop.

Determination of Field Stop and Ports Example ☞: In a two-lens system, both having power 100 D, lens ❶ has diameter 30 mm, and lens ❷ has diameter 45 mm. Their separation is 40 mm. A circular diaphragm (D) of diameter 15 mm is placed 25 mm after lens ❶. An object is placed 30 mm in front of lens ❶. Find the field stop and the two ports.

The aperture stop is diaphragm D. This can be verified using any of the methods discussed in § 7.1. The entrance pupil is its image via lens ❶, located 16.66 mm in front of lens ❶ with diameter 10 mm (m = −0.66).

Figure 7-107: Seeking the field stop in an example of a two-lens system. The next task is to image the possible field stops in object space. These are either lens ❶ or lens ❷, since the aperture stop is never a field stop. The image of lens ❶ is lens ❶ itself: There are no lenses preceding it. The image of lens ❷ is produced by lens ❶: It is located 13.33 mm in front of lens ❶ with diameter 15 mm (m = – 0.33); therefore, it is located 3.33 mm in front of the entrance pupil. 7-368

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The angle subtended by the edge of the two images from the entrance pupil center are:

tan−1(15/16.66) ≈ 42° for the image of lens ❶ (the lens itself) and tan−1(7.5/3.33) ≈ 66° for the image of lens ❷ (these are large angles, so we do not use the radian approximation but instead use the exact inverse tangent). The smallest angle is formed by the image of lens ❶, so this is the entrance port. It is also the field stop, since there are no lenses preceding it.

Figure 7-108: Object-space images of lens ❶ and lens ❷ in relation to the entrance pupil. The exit port is the image of lens ❶ formed in image space by lens ❷. It is situated 13.3 mm after lens ❷ and has diameter 5 mm (m = −0.33). For completion, the exit pupil is the image of the aperture stop formed in image space by lens ❷. It is situated 30 mm after lens ❷ and has diameter 30 mm (m = –2.0).

Figure 7-109: Stops, pupils, ports, marginal ray, and principal ray for the two-lens system.

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Principal Ray and Marginal Rays in a Three-Lens System Example ☞: The system presented in Figure 7-110 comprises three lenses and one aperture stop (AS). The object and image locations are indicated, as well as the entrance pupil and exit pupil. Draw the marginal ray and the principal rays.

Figure 7-110: Three-lens system. Note the two intermediate images and the entrance and exit pupil. To draw the marginal ray and the principal ray, we follow the strategy outlined in § 7.2.2.1. The marginal ray originates at the on-axis object point and is initially directed toward the edge of the entrance pupil. It bends (refracts) at each lens, aiming at the on-axis image point of that lens upon refraction. On its way, it passes by the edge of the aperture stop and leaves the system by the edge of the exit pupil. We note in Figure 7-111 that the marginal ray does not actually cross the edges of either the entrance pupil or the exit pupil; it is the extrapolation of the marginal ray that does so. This is because both the entrance pupil and the exit pupil are virtual images of the aperture stop.

Figure 7-111: The marginal ray in the three-lens system. 7-370

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The marginal ray originates at the off-axis object point and is initially directed toward the center of the entrance pupil. It bends (refracts) at each lens, aiming toward the off-axis image point of that lens upon refraction. On its way, it passes by the center of the aperture stop and leaves the system by the center of the exit pupil.

Figure 7-112: The principal ray in the three-lens system. As seen from Figure 7-112, the principal/chief ray intersects the optical axis only once, at the center of the aperture stop. It does not cross the centers of the entrance or the exit pupil because both of these pupils are virtual images of the aperture stop. Here, the ray extrapolation, not the actual chief ray, passes through the corresponding pupil centers. The principal/chief ray intersects the optical axis at the center of the aperture stop in any case, regardless of the type of images the pupils may be.

Figure 7-113: Principal/chief and marginal rays in a three-lens system. In this case, the entrance pupil and the exit pupil are virtual images of the aperture stop.

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Principal Ray, Marginal Rays, and Pupils in a Compound Microscope Example ☞: Determine the principal ray, marginal ray, and pupils in a compound microscope.45 The objective lens is +80 D, diameter 25 mm; the eyepiece (ocular) lens is +80 D, diameter 40 mm, separated from the objective by d = 4.5 cm; the space is filled with air; the object is placed 2 cm to the left of the objective. The object is at x1 = –20.0 mm to the left of the objective lens. This forms an intermediate image at x΄1 = +33.33 mm to the right of the objective. This image is real, inverted, and magnified by m1 = –1.66. The intermediate image becomes the object for the eyepiece lens. The object location is x2 = –11.66 mm to the left of the eyepiece. The final image is produced at x΄2 = –175.0 mm to the left of the eyepiece (not shown in Figure 7-114) and is virtual and erect (in relation to the intermediate image), with a magnification of

m2 = +15. Therefore, the final image is formed 130 mm to the left of the objective, with a magnification of m = m1 · m2 = –25.

Figure 7-114: Exit pupil in a compound microscope arrangement. Pupils: The objective lens is the aperture stop and the entrance pupil. The exit pupil is the image of the objective lens rim via the eyepiece lens. It is situated +17.3 mm to the right of the eyepiece lens. This can be verified using an imaging relationship: Now the object is the objective lens, located –45 mm to the left of the eyepiece (x = – 0.045 m). L = – 22.22 D, and F = +80.0 D, so L΄ = +57.77 D, x΄ = +17.3 mm, and m = – 0.384. Given that the objective lens diameter is 25 mm, the exit pupil diameter is –9.61 mm (the negative sign simply indicates that the image is inverted; given the circular aperture shape, it has no other significance). Principal and Marginal Rays: Having located the image(s) and the pupils, we can draw the marginal rays and the principal ray. The marginal rays, originating from the on-axis object point, pass by the edge of the entrance pupil/aperture stop (the objective). There, they bend, crossing the optical axis at the intermediate image. 45

Introduction to Optics § 5.1 Microscopes.

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Upon reaching the eyepiece, they refract toward the edge of the exit pupil. Their extrapolations point toward the on-axis point of the final image (not shown, virtual, 175.0 mm to the left of the eyepiece). The principal ray, originating from an off-axis object point, crosses the center of the aperture stop (the objective lens) and reaches the eyepiece lens. There it is refracted toward the optical axis, crossing it at the center of the exit pupil. Its extrapolation points toward the off-axis point of the final image (not shown).

Principal Ray, Marginal Rays, and Pupils in a Galilean Telescope Example ☞: Determine the principal ray, marginal rays, pupils, and ports in a Galilean telescope.46 The objective lens is +10.0 D, diameter 30.0 mm; the eyepiece (ocular) lens is –50.0 D, diameter 25.0 mm. The two lenses are separated by d = 8 cm (the difference between the two focal lengths); the space between is filled with air; the object is placed at infinity. The object is situated at infinity. The objective lens produces an intermediate image at location x΄1 = +fo = +10.0 cm. This intermediate image becomes the object for the eyepiece lens. In relation to this lens, the object location is x1 = –fe = +2.0 cm, which is to the right of the negative-power eyepiece lens (it is a virtual object placed at the lens primary focal point). Therefore, the final image is produced at infinity as well. Thus, there are two collimated ray bundles that enter and exit the telescope (an afocal system).

Figure 7-115: Exit pupil in a Galilean telescope. Pupils: The objective lens is the aperture stop. Because there are no lenses preceding it, the objective is also the entrance pupil. The exit pupil is the image of the objective lens rim formed by the eyepiece lens. The object is situated – 8.0 cm to the left of the eyepiece, so the object vergence is –12.5 D. This is added to the –50.0 D of the eyepiece lens power to produce an image vergence of –62.5 D, or an image location of –1.66 cm. Based on the above, the exit pupil magnification is m = +0.2 (the positive sign indicates that the image is erect; given the circular aperture shape, it has no other significance).

46

Introduction to Optics § 5.2.4.4 The Galilei-type Telescope. 7-373

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Because the objective lens diameter is 30 mm, the exit pupil diameter is +0.2 · 30 mm = 6.0 mm. The ratio of the pupil diameters is proportional to the ratio of the two lens focal lengths, which is 10:2. The fact that the exit pupil is only 6.0 mm in diameter means that a ray bundle that enters the telescope filling the entire objective lens diameter (entrance pupil of 30 mm) covers an area of only 6 mm in diameter upon leaving the telescope. Principal and Marginal Rays: Because the object is at optical infinity, the principal ray is the optical axis. The marginal rays cross the edge of the aperture stop (i.e., the objective lens) and propagate to the eyepiece lens. There, the marginal rays are bent (refracted) parallel to the optical axis; their extrapolation crosses the edge of the exit pupil. Field Stop and Ports: There are only two elements in this system, the objective and the eyepiece lens. We have determined that the objective lens is the aperture stop. Whatever remains (however improbable, per the Sherlock quote) must be the field stop. Therefore, the eyepiece lens is the field stop. It is also the exit port, since there are no more elements following the field stop. The entrance port is the image of the eyepiece formed by the objective lens in object space. The calculations remain to be done as an exercise; the entrance port is a virtual image of the eyepiece, placed 40 cm to the right of the objective lens (which is 32 cm to the right of the eyepiece lens); the magnification is m = +5.0. The size (diameter) of the entrance port is 12.5 cm.

Determination of Object-Space aFoV and Image-Space aFoV Example ☞: Consider a system of two lenses with an aperture stop placed between them (presented in Figure 7-36). Determine the object-space and image-space FoV using ray tracing.

Figure 7-116: System comprising two lenses and an aperture stop placed between them. The first step is to determine the locations of the intermediate image and the final image. This is left as a ray-tracing exercise, which requires implementation of the techniques described in § 6.5.5.

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Pupils: The entrance pupil is the image of the aperture stop in object space, formed by lens ❶ when the aperture stop is the real object. Ray tracing reveals that this is a virtual image located to the right of lens ❶ (and also to the right of lens ❷).

The exit pupil is the image of the aperture stop in image space, formed by lens ❷ when the aperture stop is the real object. Ray tracing reveals that this is a real image located to the right of lens ❷.

Figure 7-117: Determination of the entrance and exit pupils using ray tracing. The next step is to determine the field stop. There are two possible elements that could be field stop: lens ❶ and lens ❷, as we know that the aperture stop can never be a field stop!

We implement the recipe prescribed in § 7.3.3: We form the images of these two candidates in object space and identify the angle subtended by the edge of each image from the center of the entrance pupil. The image of the element forming the smallest angle is the entrance window/port, and the element producing this image is the field stop. We therefore find the image of lens ❶ (the lens itself, as no lenses precede it) and the image of lens ❷ via lens ❶ (this is the lens preceding it). The angular subtense from the center of the entrance pupil of the image of lens ❷ is smaller than the angular subtense of lens ❶. Therefore, the image of lens ❶ is the entrance port, and the element forming lens ❷ is the field stop. Since there are no more lenses after lens ❷, this is also the exit port.

Figure 7-118: Determination of the entrance and exit ports using ray tracing. 7-375

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The final step is to determine the object-space aFoV and image-space aFoV. For the object-space aFoV, we need the entrance port and the entrance pupil. The cone formed by the entering part of the principal ray determines the aFoV, which can be calculated by the ratio of the entrance port semi-diameter to the separation between the entrance port and the entrance pupil.

Figure 7-119: Determination of the object-space aFoV. For the image-space aFoV, we need the exit port and the exit pupil. The cone formed by the exiting part of the principal ray determines the aFoV, which can be calculated by the ratio of the exit port semi-diameter to the separation between the exit port and the exit pupil.

Figure 7-120: Determination of the image-space aFoV.

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7.8 TRANSVERSE OPTICS QUIZ Unless otherwise stated, the lenses and openings in these quiz questions are surrounded by air.

Aperture Stop and Pupils / Principal Ray and Marginal Rays 1)

In a system consisting of two collinear circular apertures of different sizes (diameters), the aperture stop for collimated illumination is … a) b) c) d)

6)

a) b) c) d)

the one with the smaller diameter the first the last the one with the greater diameter 7)

2)

A light source is placed 5 cm in front of a 10 mm diameter aperture. The angular subtense (with respect to the optical axis) formed by the edge of this aperture is … a) b) c) d) e)

3)

4)

5)

2 rad 1 rad 0.2 rad 0.1 rad 0.05 rad

Which of the following apertures forms the smallest angular subtense? a) b) c) d)

The object is a light source placed 30 mm in front of the 1st aperture. The apertures are spaced 20 mm apart. Diameters: 1st 8 mm, 2nd 10 mm, and 3rd 16 mm. The aperture stop is …

a) b) c) d) 8)

the object aperture 1 aperture 2 aperture 3

The object is a light source placed 40 mm in front of the 1st aperture. The apertures are spaced 15 mm apart. Diameters: 1st 10 mm, 2nd 16 mm, and 3rd 14 mm. The aperture stop is …

aperture diameter 6 mm, source 5 cm away aperture diameter 10 mm, source 10 cm away aperture diameter 13 mm, source 15 cm away aperture diameter 16 mm, source 20 cm away

A light source is placed 5 cm away from a 6 mm diameter aperture. Every 5 cm, we place more apertures with diameters 10 mm, 13 mm, and 16 mm. Which aperture is the aperture stop? a) b) c) d)

aperture of 6 mm diameter aperture of 10 mm diameter aperture of 13 mm diameter aperture of 16 mm diameter

2 rad 1 rad 0.2 rad 0.1 rad 0.05 rad

Back to Q 2. We move the aperture 10 cm from the light source. The angular subtense is now … a) b) c) d) e)

A twist to Q 5: Now the light source is placed 10 cm away. Which aperture is the aperture stop?

aperture of 6 mm diameter aperture of 10 mm diameter aperture of 13 mm diameter aperture of 16 mm diameter

a) b) c) d)

the object aperture 1 aperture 2 aperture 3

A 1 inch diameter lens of +20.0 D power and half-inch clear apertures are used in the following questions (9– 11). You are being asked to determine the aperture stop. Illumination is collimated (object at infinity). 7-377

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9)

Aperture 1 is 10 mm in front of the lens, and aperture 2 is 25 mm to the right of the lens.

a) b) c) d)

aperture 1 aperture 2 aperture 3 aperture 4

13) Back to Q 12. Without changing anything on the apertures, light now travels from right to left as shown. Which aperture is the aperture stop? a) b) c) d)

lens aperture 1 aperture 2 optical infinity

10) Aperture 1 is 20 mm to the right of the lens, and aperture 2 is 30 mm to the right of the lens.

a) b) c) d)

aperture 1 aperture 2 aperture 3 aperture 4

14) Which of the following illustrations correctly describes the aperture stop and its images? a) b) c) d)

lens aperture 1 aperture 2 optical infinity

11) Aperture 1 is 40 mm to the right of the lens, and aperture 2 is 50 mm to the right of the lens.

a)

A

b) B

c) C

d) D

15) The following are images of the aperture stop: a) b) c) d)

lens aperture 1 aperture 2 optical infinity

12) The following illustration shows collimated light traveling from left to right on a set of collinear apertures as shown. Which is the aperture stop?

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a) b) c) d) e)

entrance pupil; exit pupil entrance port; exit port entrance pupil; entrance port entrance pupil; exit port entrance port; field stop

16) Given the following: collimated light entering the system; lens +25.0 D and 20 mm in diameter placed in air; aperture stop size (diameter) 5 mm, placed 10 mm behind (to the right of) the lens, the location and size of the entrance pupil are …

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21) The entrance pupil is the … a) b) c) a) b) c) d) e) f) g) h)

at the lens; 5 mm in size at the lens; 20 mm in size 10 mm to the right of the lens; 5 mm in size 13.3 mm to the left of the lens; 5 mm in size 13.3 mm to the right of the lens; 6.66 mm in size 6.66 mm to the right of the lens; 6.66 mm in size 6.66 mm to the left of the lens; 3.33 mm in size 40 mm to the right of the lens; 20 mm in size

17) Back to Q 16. The entrance pupil is … a) b) c) d)

at the lens; 5 mm in size at the lens; 20 mm in size 10 mm to the right of the lens; 5 mm in size 13.3 mm to the right of the lens; 5 mm in size 13.3 mm to the left of the lens; 6.66 mm in size 6.66 mm to the right of the lens; 6.66 mm in size 6.66 mm to the left of the lens; 7.5 mm in size 40 mm to the right of the lens; 20 mm in size

19) Back to Q 16. The exit pupil is … a) b) c) d)

a virtual image of the aperture stop a real image of the aperture stop the aperture stop itself the lens itself

20) The exit pupil is the … a) b) c) d) e)

e)

22) Given the following: collimated light entering the system; lens +25.0 D and 20 mm in diameter located in air; aperture stop size (diameter) 5 mm, placed 20 mm before (to the left of) the lens, the location and size of the entrance pupil are …

a virtual image of the aperture stop a real image of the aperture stop the aperture stop itself the lens itself

18) Back to Q 16. What are the size and location of the exit pupil? a) b) c) d) e) f) g) h)

d)

image of the aperture stop formed by any lenses in front of (preceding) it image of the aperture stop formed by any lenses succeeding (following) it image of the field stop formed by any lenses in front of (preceding) it image of the field stop formed by any lenses succeeding (following) it first aperture in the optical system through which light enters

image of the aperture stop formed by any lenses in front of (preceding) it image of the aperture stop formed by any lenses succeeding (following) it image of the field stop formed by any lenses in front of (preceding) it image of the field stop formed by any lenses succeeding (following) it last aperture in the optical system through which light exits

a) b) c) d) e) f)

at the lens; 20 mm in size at the lens; 5 mm in size 20 mm to the right of the lens; 5 mm in size 20 mm to the left of the lens; 5 mm in size 40 mm to the right of the lens; 0 mm in size 40 mm to the left of the lens; 10 mm in size

23) Back to Q 22. The entrance pupil is … a) b) c) d)

a virtual image of the aperture stop a real image of the aperture stop the aperture stop itself the lens itself

24) Back to Q 22. What are the location and size of the exit pupil? a) b) c) d) e) f)

at the lens; 20 mm in size at the lens; 5 mm in size 20 mm to the right of the lens; 5 mm in size 20 mm to the left of the lens; 5 mm in size 40 mm to the right of the lens; 10 mm in size 40 mm to the left of the lens; 10 mm in size

25) Back to Q 22. The exit pupil is … a) b) c) d)

a virtual image of the aperture stop a real image of the aperture stop the aperture stop itself the lens itself 7-379

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26) The drawing presents a lens and an aperture stop placed to its left. The element marked as A is the …

c) d) e)

field stop entrance port exit port

31) Back to Q 29. Here, the entrance pupil is a … a) b) c) d) a) b) c) d) e)

entrance pupil exit pupil field stop entrance port exit port

27) Back to Q 26. The aperture stop is also the … a) b) c) d) e)

entrance pupil exit pupil field stop entrance port exit port

28) Back to Q 26. Here, the exit pupil is a … a) b) c) d) e)

real image of the aperture stop in image space real image of the lens in image space virtual image of the aperture stop in image space virtual image of the lens in image space real image of the entrance pupil in object space

29) The drawing presents a lens and an aperture stop placed to its right. The element marked as B is the …

e)

real image of the aperture stop in object space real image of the exit pupil in image space real image of the lens in object space virtual image of the aperture stop in object space virtual image of the lens in object space

32) In an imaging system, which of the following are optical conjugates (three correct answers)? a) b) c) d) e) f)

entrance pupil and exit pupil aperture stop and exit pupil field stop and entrance pupil object and exit pupil image and entrance pupil entrance pupil and aperture stop

33) In a system comprising two lenses and an aperture stop between them, identify the proper conjugate pairs among the following (select two): a) b) c) d)

on-axis object point and on-axis image point on-axis object point and center of the aperture stop center of the aperture stop and center of the entrance pupil center of the exit pupil and on-axis image point

34) The exit pupil is … a) b) c) a) b) c) d) e)

entrance pupil exit pupil field stop entrance port exit port

30) Back to Q 29. Here, the aperture stop is also the … a) entrance pupil b) exit pupil

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d) e)

always a real image of the aperture stop in image space always a virtual image of the aperture stop in image space either a real or a virtual image of the aperture stop possibly a real image of the aperture stop in object space always a virtual image of the aperture stop in object space

35) The entrance pupil is … a)

always a real image of the aperture stop in image space

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b) c) d) e)

possibly a virtual image of the aperture stop in image space either a real or a virtual image of the aperture stop always a real image of the aperture stop in object space always a virtual image of the aperture stop in object space

36) Which of the following are physical elements, if they actually exist (two correct answers)? a) b) c) d) e)

field stop aperture stop exit pupil entrance pupil Santa Claus

37) You reduce the diameter (size) of the aperture stop by half. What happens to the image surface area? a) it is reduced to ¼ b) it is reduced to ½ c) it remains the same d) it doubles 38) You reduce the diameter (size) of the aperture stop by half. What happens to the amount of light passing through the optical system? a) b) c) d)

it is reduced by ¼ it is reduced by ½ it remains the same it doubles

39) The pupil of the eye is a …

40) In the human eye, the ‘pupil’ we see and measure with pupilometry is (in terms of optics) the … a) b) c) d) e)

aperture stop, the anatomical iris itself entrance pupil, the image of the anatomical iris formed by the cornea entrance pupil, the image of the anatomical iris formed by the crystalline lens entrance port, the image of the anatomical iris formed by the cornea exit port, the image of the anatomical iris formed by the cornea

41) The exit pupil of a microscope is a … a) b) c) d)

virtual image of the objective lens real image of the objective lens virtual image of the ocular lens real image of the ocular lens

42) Which lens is the most likely choice for the eyepiece in an astronomical telescope? a) b) c) d)

a biconvex lens with power +10 D a piece of glass with zero power a biconcave lens with power –10 D a planoconcave lens with power –20 D

43) An astronomical telescope is constructed using an objective (F1 = +5.0 D, diameter 10 cm) and an eyepiece (F2 = +50.0 D, diameter 3 cm). The separation between the lenses is … a) b) c) d)

d = 22 cm d = 20 cm d = 18 cm d = 13 cm

44) Back to Q 43. The roles of the objective and the eyepiece lens are (two correct answers) …

a) b) c) d)

magnified virtual image of the iris, functioning as the entrance pupil minified real image of the iris, functioning as the aperture stop physical element, located ahead of the iris magnified virtual image of the iris, functioning as the exit pupil

a) b) c) d) e) f)

objective: aperture stop and entrance pupil objective: aperture stop and exit pupil objective: field stop and exit pupil eyepiece: aperture stop and entrance pupil eyepiece: field stop and exit port eyepiece: field stop and entrance port

45) Back to Q 43. The size of the entrance pupil is … a) b) c) d) e)

1 cm 6 cm 10 cm 20 cm 30 cm 7-381

GEOMETRICAL OPTICS

46) Back to Q 43. The size of the exit pupil is … a) b) c) d) e)

1 cm 6 cm 10 cm 20 cm 30 cm

47) Back to Q 43. The exit pupil is a … a) b) c) d)

real image of the eyepiece via the objective real image of the objective via the eyepiece virtual image of the eyepiece via the objective virtual image of the objective via the eyepiece

48) Back to Q 43. The exit pupil is located … a) b) c) d)

2.2 cm to the left of the eyepiece 2.2 cm to the right of the eyepiece 2.2 m to the left of the objective 2.2 m to the right of the objective

49) In an astronomical telescope, the following are true regarding aperture stop, entrance pupil, and exit pupil (three correct answers): a) b) c) d) e) f) g)

the objective is the aperture stop the objective is the entrance pupil the objective is the exit pupil the eyepiece is the exit pupil the eyepiece is the aperture stop the exit pupil is a real image of the objective formed by the eyepiece the exit pupil is a virtual image of the objective formed by the eyepiece

50) You are given two lenses: Lens 1 has diameter 8 cm, +4.0 D power, and lens 2 has diameter 1.5 cm, –20.0 D power. You can construct a … a) Keplerian telescope; lens 1 objective, lens 2 eyepiece b) Keplerian telescope; lens 1 eyepiece, lens 2 objective c) Galilean telecope; lens 1 objective, lens 2 eyepiece d) Galilean telescope; lens 1 eyepiece, lens 2 objective 51) A Galilean telescope is constructed with an objective (F1 = +5.0 D, diameter 10 cm) and an eyepiece (F2 = –50.0 D, diameter 3 cm). The separation between the lenses is … a) b) 7-382

d = 22 cm d = 20 cm

c) d) e)

d = 18 cm d = 13 cm d = 10 cm

52) Back to Q 51. The roles of the objective and the eyepiece lens are (two correct answers) … a) b) c) d) e) f)

objective: aperture stop and entrance pupil objective: aperture stop and exit pupil objective: field stop and exit pupil eyepiece: aperture stop and entrance pupil eyepiece: field stop and exit port eyepiece: field stop and entrance port

53) Back to Q 51. The size of the entrance pupil is … a) b) c) d) e)

1 cm 6 cm 10 cm 20 cm 30 cm

54) Back to Q 51. The size of the exit pupil is … a) b) c) d) e)

1 cm 6 cm 10 cm 20 cm 30 cm

55) Back to Q 51. The exit pupil is a … a) b) c) d)

real image of the eyepiece via the objective real image of the objective via the eyepiece virtual image of the eyepiece via the objective virtual image of the objective via the eyepiece

56) Back to Q 51. The location of the exit pupil is … a) b) c) d)

1.8 cm to the left of the eyepiece 1.8 cm to the right of the eyepiece 1.8 m to the left of the objective 1.8 m to the right of the objective

57) If an aperture stop is placed at the primary focal point of a plus lens, then (select two correct) … a) b) c) d) e) f)

the entrance pupil is at the primary focal point the exit pupil is at the primary focal point the entrance pupil is at the secondary focal point the exit pupil is at the secondary focal point the entrance pupil is at the optical infinity the exit pupil is at the optical infinity

FINITE TRANSVERSE OPTICS

58) An aperture stop is placed at the secondary focal point of a plus lens. Then (select two correct) … a) b) c) d) e) f)

the entrance pupil is at the primary focal point the exit pupil is at the primary focal point the entrance pupil is at the secondary focal point the exit pupil is at the secondary focal point the entrance pupil is at the optical infinity the exit pupil is at the optical infinity

a) b) c) d) e)

59) In the drawing, the principal ray is wrong. The mistake is that the principal ray …

an off-axis object point passing through the center of the field stop an off-axis object point passing through the center of the lens the on-axis object point passing via the edge of the entrance pupil an off-axis object point passing through the secondary focal point an off-axis object point passing through / pointing to the center of the entrance pupil

63) In the drawing, the marginal ray is wrong. The two mistakes are that the ray …

a) b) c) d) e) f)

there is no mistake; this is a proper drawing passes by the center of the aperture stop originates from an off-axis object point passes by an off-axis image point crosses the center of the entrance pupil passes by the edge of the exit pupil

60) Back to the drawing in Q 59. The entrance pupil and the exit pupil are (two correct answers) … a)

entrance pupil: real image of aperture stop

b)

entrance pupil: virtual image of aperture stop

c)

exit pupil: real image of aperture stop

d)

exit pupil: virtual image of aperture stop

61) The principal ray from the outermost off-axis object point passes through the center of the aperture stop and the (two correct) … a) b) c) d) e) f)

edge of the field stop, if the field stop is a real image edge of the field stop, always center of the exit port, always edge of the exit port, if the exit port is a real image edge of the entrance port, if the entrance port is a virtual image center of the entrance port, always

62) The principal/chief ray is a ray from …

a) b) c) d) e)

passes by the center of the aperture stop originates from an on-axis object point passes by the edge of the entrance pupil passes by the edge of the exit pupil does not cross the on-axis image point

64) The marginal ray crosses the optical axis at (two correct answers) … a) b) c) d)

the center of the aperture stop the object on-axis point the image on-axis point (if the image is real) the center of the entrance pupil

65) The marginal ray is a ray from … a) b) c) d) e)

an off-axis object point passing through the center of the aperture stop an off-axis object point passing through the center of the lens the on-axis object point passing via the edge of the entrance pupil the on-axis object point passing through the secondary focal point an off-axis object point passing through (pointing to) the center of the entrance pupil

66) The top drawing shows an imaging arrangement with a known lens, aperture stop, entrance pupil,

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GEOMETRICAL OPTICS

and exit pupil. We draw the marginal ray. Which one of drawings A, B, C, or D is correct?

a)

A

b)

B

c)

C

d) D

67) The top drawing shows an imaging arrangement with a known lens, aperture stop, entrance pupil, and exit pupil. We draw the principal ray. Which one of drawings A, B, C, or D is correct?

e) none a)

A

b) B

c) C

d) D

e) none

c)

the same number of cattle with increased magnification the same number of cattle with decreased magnification

Field of View / Field Stop and Ports 68) You are looking into a cardboard tube with a 3 cm diameter opening and a 20 cm length. What is the angular aFoV? a) b) c) d) e)

0.15° 3.0 cm 8.6° 15° 0.26 m

69) You are looking into a cardboard tube with a 4.5 cm diameter opening. If the aFoV is 17°, what is the tube length? a) b) c) d) e)

0.065 cm 4 cm 7.5 cm 15 cm 25 cm

70) You are viewing a farm via an instrument and counting cattle. You may not move or change anything in the optical instrument other than the field of view. A larger field of view affords imaging of … a) b) 7-384

more (a greater number of) cattle fewer (a lesser number of) cattle

d)

71) You are in a flying plane and look through the open window by your seat. The field of view via this window (two correct answers) ... a) b) c) d) e) f)

increases if you shut half the window shade decreases if you move closer to the window increases if you move closer to the window decreases if you if you squint your eyes (reduce the pupil diameter) is constant if you squint your eyes (reduce the pupil diameter) increases if you if you squint your eyes (reduce the pupil diameter)

72) Select two correct statements. The field stop … a) b) c) d) e)

is the image of the entrance pupil via all elements (e.g., the lenses in between) can never be situated on the aperture stop affects the field of view affects the image brightness affects the image resolution

FINITE TRANSVERSE OPTICS

73) Select two correct statements. The field stop … a) b) c) d) e)

is the image of the aperture stop via the elements that follow it can be situated exactly on the image plane affects image brightness can be situated at the aperture stop is a real, physical, restrictive element

74) Which of the following illustrations correctly describes the field stop and its images?

b) c) d)

the entrance pupil size and the axial separation between the entrance port and entrance pupil the entrance port size and the axial separation between the entrance port and exit pupil the entrance port size and the axial separation between the entrance port and exit port

79) Name the images of the field stop: a) b) c) d) e)

entrance pupil, exit pupil entrance port, exit port entrance pupil, entrance port entrance pupil, exit port entrance port, exit port

80) Which two of the following are physical elements? a) b) c) d) e) a)

A

b) B

c) C

d) D

75) You reduce the diameter of the field stop by half. What happens to the image surface area? a) b) c) d)

it is reduced to ¼ it is reduced to ½ it remains the same it doubles

76) You reduce the diameter of the field stop by half. What happens to light through the system? a) b) c) d)

it is reduced by ¼ it is reduced by ½ it remains the same it doubles

77) Select two true statements. In an imaging system the … a) b) c) d) e)

aperture stop can also be the entrance pupil field stop can also be the exit port aperture stop can also be the field stop entrance pupil can also be the entrance port exit pupil can also be the exit port

78) The object-space field of view (aFoV), when expressed angularly, is dependent on two parameters. These are (note: size = half diameter): a)

the entrance port size and the axial separation between the entrance port and entrance pupil

aperture stop field stop exit port entrance port Easter bunny

81) An astronomical telescope is constructed by an objective (F1 = +5.0 D, diameter 10 cm) and an eyepiece (F2 = +50.0 D, diameter 3 cm). The field stop is the … a) b) c) d)

objective lens image of the objective via the eyepiece eyepiece lens image of the eyepiece via the objective

82) Back to Q 81. The eyepiece is (two correct answers): a) b) c) d) e) f)

the field stop the aperture stop the entrance pupil the exit pupil the entrance port the exit port

83) Back to Q 81. The diameter of the entrance port is … a) b) c) d) e)

1 cm 3 cm 10 cm 20 cm 30 cm

84) Back to Q 81. The diameter of the exit port is … a) b) c)

1 cm 3 cm 10 cm 7-385

GEOMETRICAL OPTICS

d) e)

20 cm 30 cm

d)

there is no field stop

85) Back to Q 81. The entrance port is a … a) b) c) d)

real image of the eyepiece via the objective real image of the objective via the eyepiece virtual image of the eyepiece via the objective virtual image of the objective via the eyepiece

86) Back to Q 81. The entrance port is located … a) b) c) d)

2.2 cm to the left of the eyepiece 2.2 cm to the right of the eyepiece 2.2 m to the left of the objective 2.2 m to the right of the objective

87) In a system comprising a lens and an aperture stop as shown, what is the field stop?

a) b) c) d)

the lens the image of the AS via the lens in image space the image of the AS via the lens in object space there is no field stop

88) Back to Q 87. The lens is also (two answers) … a) b) c) d)

the exit pupil the exit port the entrance pupil the entrance port

89) In an array of four collinear apertures illuminated with collimated light, the front aperture has the smallest opening. The field stop is the … a) b) c) d)

front aperture second aperture last aperture aperture forming the smallest angle from the center of the front aperture

90) In a system comprising a lens and an aperture stop as shown, what is the field stop? a) b) c)

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the lens the image of the AS via the lens in image space the image of the AS via the lens in object space

91) Back to Q 90. The lens is also the (two correct) … a) b) c) d)

exit pupil exit port entrance pupil entrance port

92) In an imaging system, which of the following are optical conjugates (three correct)? a) b) c) d) e) f)

the entrance port and exit port the field stop and entrance pupil the field stop and exit port the entrance port and field stop the object and exit port the image and entrance port

93) In a system comprising lenses, an aperture stop, and a field stop, identify the proper conjugate pairs among the following (select two): a) b) c) d)

on-axis object point and on-axis image point on-axis object point and center of field stop center of exit port and on-axis image point center of field stop and center of entrance port

94) The exit port is (FS = field stop; AS = aperture stop) … a) b) c) d) e)

always a real image of the FS in image space always a virtual image of the FS in image space possibly a real image of the AS in object space either a real or a virtual image of the FS always a virtual image of the FS in object space

95) The entrance port is (FS = field stop; AS = aperture stop) … a) b) c) d) e)

always a real image of the FS in image space possibly a virtual image of the FS in image space either a real or a virtual image of the FS always a real image of the FS in object space possibly a virtual image of the AS in object space

96) In an imaging system, the entrance port diameter is 4 mm, and the entrance pupil diameter is 2 mm, axially separated from the entrance port by 20 mm. What is the angular aFoV?

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a) b) c) d)

5.7° 11.4° 22.6° 43.6°

97) Back to Q 96. We change the entrance pupil diameter to 4 mm. What is the aFoV? a) b) c) d)

5.7° 11.4° 22.6° 43.6°

98) Back to Q 96. We now change the entrance port diameter to 8 mm. What is the aFoV? a) b) c) d)

5.7° 11.4° 22.6° 43.6°

99) In an imaging system, the entrance port diameter is 6 mm, and the entrance pupil diameter is 2 mm, axially separated from the entrance port by 5 cm. What is the aFoV? a) b) c) d)

4.6° 6.9° 9.1° 43.6°

100) Back to Q 99. What is the field of uniform illumination (FUI)? a) b) c) d)

4.6° 6.9° 9.1° 43.6°

101) Back to Q 99. What is the total field of view (tFoV)? a) 4.6° b) 6.9° c) 9.1° d) 43.6° 102) Back to Q 99. What would you do to make the values of the aFoV, FUI, and tFoV nearly identical? a) b) c) d)

increase the entrance pupil from 2 to 4 mm increase the entrance pupil from 2 to 6 mm reduce the entrance pupil from 2 to 0.1 mm reduce the entrance pupil from 2 to 1.5 mm

103) Which of the following increases if the port size (semi-diameter) increases (four possible answers)? a) b) c) d)

the angular field of view (aFoV) the total field of view (tFoV) the field of uniform illumination (FUI) the field of (at least one) half illumination

104) Which of the following increases if the pupil size (semi-diameter) increases (one correct)? a) b) c) d)

the angular field of view (aFoV) the total field of view (tFoV) the field of uniform illumination (FUI) the field of (at least one) half illumination

105) An increase in the axial separation between the port and the pupil results in a decrease in which of the following (four possible answers)? a) b) c) d)

the angular field of view (aFoV) the total field of view (tFoV) the field of uniform illumination (FUI) the field of (at least one) half illumination

106) You have a circular frame with no lens, which you may fit with a high-power minus lens, a lowpower minus lens, or a low-power plus lens, while viewing the frame from the same distance. The FoV is greatest when the frame is fitted with … a) b) c) d)

no lens a low-power minus lens a high-power minus lens a low-power plus lens

107) A circular frame of 3 cm diameter is fitted 18 mm in front of the wearer’s pupil. What is the (monocular) aFoV when there is no lens and a –8.0 D lens is fitted on the frame? a) b) c) d)

frame only: 79.6°; minus lens: 71.0° frame only: 71.0°; minus lens: 71.0° frame only: 71.0°; minus lens: 79.6° frame only: 79.6°; minus lens: 87.3°

108) You are viewing a virtual-reality cardboard, whose circular opening has a 24 mm diameter; your pupil is 5 cm from the center of that opening. What is the (monocular) aFoV when there is no lens and a +10.0 D lens is fitted on the opening? a) b) c) d)

frame only: 27.0°; plus lens: 13.7° frame only: 13.7°; plus lens: 27.0° frame only: 27.0°; plus lens: 39.6° frame only: 39.6°; plus lens: 27.0°

109) Back to Q 108. You want to maintain the same aFoV achieved with the +10.0 D plus lens, but this time you want an inverted field. You will use a … a) b) c) d) e)

–30.0 D lens –10.0 D lens +10.0 lens, rotated along its axis +20.0 D lens +30.0 D lens

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GEOMETRICAL OPTICS

Depth of Field / Resolution / Blur 110) A clear image is formed if an object is placed 20 cm in front (to the left) of an imaging lens. If the dioptric depth of field (DoF) is ±1.0 D, where are the proximal and distal points of the DoF? a) b) c) d) e)

(proximal) 40 cm and (distal) 66.6 cm 4 cm and 6 cm 33 cm and 50 cm 16.6 cm and 25 cm 40 cm and 16.6 cm

111) Back to Q 110. What is the vergence of the proximal point and the distal point of the DoF? a) b) c) d)

(proximal) –5.0 D and (distal) –7.0 D –7.0 D and –4.0 D –5.0 D and –6.0 D –6.0 D and –4.0 D

112) A clear image is formed if an object is placed 50 cm in front (to the left) of a lens. If the linear DoF extends from 40 cm to 66.6 cm, what is the dioptric DoF? a) b) c) d)

±0.25 D ±0.50 D ±1.00 D ±2.00 D

113) In an optical system, you halve the diameter of the aperture stop (decrease it by a factor of two). The following changes occur (DoF = depth of field): a) b) c) d)

DoF doubles; the Rayleigh-resolution limit (minimum resolvable angle) halves DoF halves; the Rayleigh-resolution limit (minimum resolvable angle) halves DoF doubles; the Rayleigh-resolution limit (minimum resolvable angle) doubles DoF halves; the Rayleigh-resolution limit (minimum resolvable angle) doubles

114) In an optical system, you increase the diameter of the aperture stop by a factor of two. The following changes occur: a) b)

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DoF doubles; the Rayleigh-resolution limit (minimum resolvable angle) halves DoF halves; the Rayleigh-resolution limit (minimum resolvable angle) halves

c) d)

DoF doubles; the Rayleigh-resolution limit (minimum resolvable angle) doubles DoF halves; the Rayleigh-resolution limit (minimum resolvable angle) doubles

115) In the human eye, the depth of focus represents … a) b) c) d)

the image-space interval in which the retinal image blur has no noticeable difference the object-space interval over which objects can be placed while seen at the same clarity the closest distance at which an object can be clearly seen the distance between the nodal and the posterior focal length of the eye

116) A +60.0 D lens forms an image from an object placed 10 cm to its left. The distal limiting position of the DoF is 12.5 cm to the left of the lens. What is the proximal limiting position? a) b) c) d)

0.8 cm to the left of the lens 8.33 cm to the left of the lens 2.0 cm to the right of the lens 12.5 cm to the right of the lens

117) Back to Q 116. What is the dioptric depth of field? a) b) c) d)

±0.25 D ±0.50 D ±1.0 D ±2.0 D

118) Back to Q 116. What is the linear depth of field? a) b) c) d) e)

4.16 cm 8.33 cm 12.5 cm 41.6 cm 50.0 cm

119) Back to Q 116. What is the dioptric depth of focus? a) b) c) d)

±0.25 D ±0.50 D ±1.0 D ±2.0 D

120) Back to Q 116. Where is the conjugate image point in image space? a)

8.0 cm to the left of the lens

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b) c) d)

8.33 cm to the left of the lens 2.0 cm to the right of the lens 12.5 cm to the right of the lens

127) Which of the following is an effective way to increase the resolving power of a telescope?

121) Back to Q 116. What is the linear depth of focus? a) b) c) d) e)

4.16 cm 2.00 cm 1.923 cm 0.213 cm 0.16 cm

±0.25 D ±0.50 D ±1.0 D ±2.0 D

a) b)

2.0 cm to the right of the lens 2.66 cm to the right of the lens 5.0 cm to the right of the lens 6.66 cm to the right of the lens

f/2.5 for pupil diameter 8 mm f/2.5 for pupil diameter 2 mm f/8.6 for pupil diameter 8 mm f/8.6 for pupil diameter 2 mm

c) d)

NA = 0.23 refractive index 1.333, half-angle 3 mm, focal length 18 mm NA = 0.41 refractive index 1.333, half-angle 6 mm, focal length 18 mm

130) The greatest clear aperture opening corresponds to a lens with… (f/# = F-number)

124) Back to Q 122. What is the linear depth of focus (n΄ = 1.333)? a) b) c) d) e)

a) b) c) d)

129) Which optical system has a greater capacity to gather light (NA = numerical aperture)?

123) Back to Q 120. Where is the conjugate image point in image space (n΄ = 1.333)? a) b) c) d)

increase the diameter of the eyepiece lens increase the power of the eyepiece lens increase the diameter of the objective lens increase the power of the objective lens

128) Two values for the f/# of the human eye are f/2.5 and f/8.6. Find the two correct matches between the f/# and the eye pupil diameter:

122) A twist to Q 116. We now fill the image space with a liquid whose refractive index is n΄ = 1.333. What is the dioptric depth of focus? a) b) c) d)

a) b) c) d)

4.16 cm 2.00 cm 1.923 cm 0.213 cm 0.16 cm

a) b) c) d)

f/1.4 f/22 200 mm focal length, diameter 72 mm 50 mm focal length, diameter 45 mm

131) Geometrical blur is more prominent in a lens with (two correct answers) …

125) Which picture best represents low resolution but high contrast?

a) b) c) d)

a small aperture (small NA, large f/#) a large aperture (large NA, small f/#) a small depth of field a large depth of field

132) Guess which lens Dr A. will NOT use to take a portrait photograph of his loved one …

a)

A

b)

B

c)

C

d)

D

a) b) c) d)

a lens set at f/22 he shoots portraits with a 100 mm lens a lens with f/# as close to f/2.8 as possible are you still looking for the answer?

126) Back to Q 125. Which picture best represents low contrast and low resolution? a)

A

b)

B

c)

C

d)

D 7-389

GEOMETRICAL OPTICS

7.9 TRANSVERSE OPTICS SUMMARY Rays Principal / Chief ray This ray originates from an object point with a non-zero height and passes through the center of the aperture stop. The field-height principal ray passes by the edge of the field stop. Either the ray (if the pupil is a real image of the AS) or its extrapolation (if the pupil is a virtual image of the AS) passes through the centers of the entrance and exit pupils.

Figure 7-121: A principal ray path through an optical system. Here both pupils are real images of the AS.

Marginal rays These rays originate from object points with zero height (on-axis) and pass by the periphery (edge) of the aperture stop. Either the marginal ray (if the pupil is a real image of the AS) or its extrapolation (if the pupil is a virtual image of the AS) passes by the edge of the entrance pupil and the edge of the exit pupil.

Figure 7-122: The paths of pair of marginal rays through an optical system. Here both pupils are real images of the AS. 7-390

FINITE TRANSVERSE OPTICS

Apertures Aperture stop (AS) This is the physical entity that limits the marginal rays. It can be an aperture or a lens edge. The aperture stop restricts the light passage (brightness) through the optical system and determines the image sharpness. Α larger aperture stop results in • • •

increased image brightness, in proportion to the stop area / the square of stop diameter, a sharper image being formed (better resolution) at the image-forming plane, and development of a larger geometrical blur (more defocus) away from the image-forming plane.

Field stop (FS) This is the natural aperture that restricts the principal/chief ray. It may be an aperture, the lens edge, or the image frame itself. The field stop limits the field of view. A larger field stop • •

increases the field of view, in proportion to the stop diameter, increases the image surface area, in proportion to the stop area / the square of the stop diameter.

Figure 7-123: Aperture stop and field stop comparison.

Aperture images Entrance pupil and Exit pupil The aperture stop is the real object for the formation of either pupil. The object-space image of the aperture stop (formed by any lenses before the stop) is the entrance pupil. The image-space image of the aperture stop (formed by any lenses after the stop) is the exit pupil. Their sizes are not equal. The pupil magnification expresses the ratio of the exit pupil size to the entrance pupil size. The pupils determine the maximum cross-section that a pencil of rays can have in order to propagate unobstructed through the optical system. 7-391

GEOMETRICAL OPTICS

Figure 7-124: Exit and entrance pupils.

Entrance window/port and Exit window/port The field stop is the real object for the formation of either the entrance or the exit window/port. The object-space image of the field stop (formed by any lenses before the stop) is the entrance window/port. The image-space image of the field stop (formed by any lenses after the stop) is the exit window/port. These images may be real or virtual. The ports determine the maximum angular subtense that a pencil of rays can deviate from the axis (field height) in order to propagate unobstructed though the optical system.

Angular breadth of the rays entering the system

Aperture Stop Limits the cone of rays accepted into the system

Entrance Port / Window

Field Stop

Limits the cone of rays from the entrance pupil

Limits the extent of the object that can be imaged

Food for thought

Images in Image Space

Entrance Pupil

Physical Components

Images in Object Space

Figure 7-125: Exit and entrance ports. Exit Pupil Angular breadth of the rays leaving the system Exit Port / Window Limits the cone of rays from the exit pupil

:

In an optical instrument we reduce the diameter of the aperture stop by half. What happens to the image?

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FINITE TRANSVERSE OPTICS

The brightness drops by ¼, and the resolution drops by about ½. The image surface area is unaffected. Now, we reduce the diameter of the field stop by half. What happens to the image? The brightness and resolution are unaffected; the aFoV is reduced by ½, and the image surface area drops by ¼.

Field of View The spatial relationships between the pupils and the ports determine the extent of the field that can be imaged and the illumination that reaches the image. The angular field of view (aFoV) is the angular subtense of the entrance port/window from the entrance pupil and can be visualized as the subtense of the principal ray, which grazes the entrance port and passes through the center of the entrance pupil. The measure of the aFoV depends on the size of the entrance port (its semi-diameter a) and on its separation dp from the entrance pupil: Field of View (object space):

a  dp   

aFoV = 2  tan −1 

(7.13)

The same relationship applies to the image-space aFoV, the difference being that in this case we use the exit port semi-diameter a and its separation dp from the exit pupil.

Figure 7-126: Pupils and ports in an optical system.

The total field of view (tFoV) and the field of uniform illumination (FUI) assume, in addition to the above (a and dp), a semi-diameter b of the entrance pupil. Then, Total Field of View (tFoV):

a+b tFoV = 2  tan −1   dp   

(7.14)

Field of Uniform Illumination (FUI):

FUI = 2  tan −1 

a−b  dp   

(7.15)

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The total field of view is the extended field that includes object points that form partially illuminated images; it is defined by rays crossing the opposite-side edges of the entrance point and the entrance pupil. The field of uniform illumination is the most restrictive field of all. The FUI requires that all light from an object point that is admitted by the entrance pupil reaches the entrance port; it is defined by rays crossing the same-side edges of the entrance point and the entrance pupil. The field of (at least one) half illumination is a soft compromise between the tFoV and the FUI. It is commonly accepted as the ‘working’ aFoV. The following statements are true for all fields: •

As the separation dp between the pupil and port increases, all fields of illumination decrease; as the pupil and port move closer together (dp decreases), all fields of illumination increase.

•

As the size of the entrance port (semi-diameter a) increases, all fields of illumination increase; as the size of the entrance port decreases, all fields of illumination decrease.

•

As the size of the entrance pupil (semi-diameter b) increases, tFoV increases, aFoV is unaffected, and FUI decreases.

Numerical Aperture (NA) and F-number The numerical aperture of an optical system is determined by the angle formed by the marginal rays and the optical axis. In very good optical systems, the NA approaches the value of 1.0. The F-number (f/#) is reciprocal to the NA. It expresses how many times the lens diameter (pupil) fits in its focal length. Both the NA and f/# relate to the image quality and depth of field: NA = n · sin(φ/2)

f/# (F-number) = 1 / (2 · NA)

The larger the NA or the smaller the f/#, the better the resulting image quality (small blur, good resolution) on the image-forming plane. At the same time, a large bokeh defocus brings about a strong geometrical blur resulting from a shallow depth of field.

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LECTURES IN OPTICS, VOL 2

8 OPTICAL ABERRATIONS

8.1 IMAGING TO AN IDEALIZED POINT If all of the rays from an object meet at a single common point, the formed image is called stigmatic. Stigmatism refers to an assumed image-formation property of an optical system to focus a point source in object space to a single point in image space. The concept that such a proposition is realizable belongs to the realm of geometrical approximation. We have indeed followed this approximation up to now, employing terms such as the image or the focal ‘point.’ A point is a mathematical concept with no dimension, volume, surface area, or length. It is, however, absolutely essential in geometry! As it applies to optics, the concept of the point is used for denoting a very small surface area. Although the focal point is a conceptional point, it has a physical spatial extent, called the blur circle. The ‘point’ at which all rays from an object beam converge may be considered an image point if the corresponding area is very small.

Figure 8-1: As we observe the ‘point’ of convergence with an increasingly larger magnification, we note that it is not a point but has a small, specific surface area. The image ‘point’ is not a point! 8-395

GEOMETRICAL OPTICS

Figure 8-2: Focusing from a lens does not lead to an ideal point. Hey, nobodie’s prefect!

In every real optical system, there exist deviations from the ideal that result in a ‘not perfectly’ blurred image. These imperfections are called optical aberrations. The name aberration derives from the Latin aberrare, to wander away. In the area where we would ideally expect the formation of a focal ‘point,’ we instead have a spot, or a defocus blur disk, whose smallest size is called the circle of least confusion.

Figure 8-3: The circle of least confusion situated at the minimum of the transverse/axial extent of the aberration.

Circle of Least Confusion • Its extent relates to the severity of the aberrations in an optical system. • It corresponds to the smallest transverse cross-section at the image plane. • According to geometrial optics, it should be an ideal point. • Even in the total absence of aberrations, the wave nature of light sets a limit to how small this 'point' can be.

The location of the circle of least confusion has practical significance, as it is likely the location of the image plane. Its extent relates to the severity of the aberrations of the optical system. Attempts to minimize (or compensate for) aberrations are evaluated by their effects on the size of the circle of least confusion. The design of an optical system is not a trivial task; even perfectly designed systems contain residual optical aberrations because it is impossible to completely eliminate all 8-396

OPTICAL ABERRATIONS

aberrations. For this reason, the aim of optical design is a so-called ‘corrected’ optical system with the best-possible reduction of optical aberrations. Even in highly corrected systems, we accept tolerances on lens parameters. In a system with typical aberrations, the differences between an idealized and an actual corrected system must not exceed certain limits for a given range of object location, size, and orientation. Typically, any optical system is a combination of optical elements (single refractive surfaces, lenses, mirrors, prisms), each of which carries, to a greater or lesser extent, aberrations. The task of optimization involves tuning parameters as well as introducing certain apertures with the goal to achieve the least possible number of aberrations over a large field of view and/or a large range of object locations. Then this system is considered corrected. The correction of optical aberrations is not an easy undertaking. Corrected systems can use a large number of optical elements (and, therefore, can use many surfaces). Other corrected systems use elements whose aberrations mutually cancel each other; an example of such mutual compensation is the achromatic lens. Aspects such as the size and location of the aperture and field stop are important. All of these parameters relate to the scope and mission of the optical instrument. For example, in a telescope with a small-field, collimated object beam, the correction of spherical error is more important than the correction of coma or oblique astigmatism. Lens designers take into account the following: •

the radius of curvature of the various refractive / reflective surfaces,

•

the surface separation, including lens thickness and locations,

•

the material properties such as refractive index, which varies by type of material and wavelength; the latter is a property called dispersion,47 and

•

8.1.1

the size and location of the aperture stop along the optical axis.

The Origin of Optical Aberrations

Many factors can be responsible for the presence of aberrations. Light usually consists of different wavelengths. We know that the value of the refractive index depends (among other parameters) on the color, or the wavelength. A rainbow is indisputable proof of this.48 The optical power of a refracting surface depends on the refractive index (§ 2.4); for this reason, the optical power and the focal length of a lens are different for the red compared to the blue, etc.

47

Wave Optics § 3.3.2 Dispersion in Optical Glass.

48

Introduction to Optics § 3.5.2.1 Rainbow.

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GEOMETRICAL OPTICS

This dependency is the chromatic aberration associated with lenses (in general, refracting elements). Chromatic aberration does not affect the reflecting surfaces since the law of reflection does not involve the refractive index, nor is there any other dependence on the wavelength.

Aberrations

Refractive index dependence on λ Chromatic aberration

Surface geometry

Spherical aberration and other monochromatic aberrations

Wave nature of light

Airy disk instead of focal point (diffraction)

In order to circumvent chromatic aberration, we can use monochromatic light. Now we consider aberrations that do not relate to this ‘color’ dependence of the refractive index. These aberrations apply to both refracting and reflecting optical elements. The two remaining factors are (1) the geometry of the refracting/reflecting surfaces and their placement/orientation with respect to the optical axis, which may not be ideal, and (2) the wave nature of light. The geometrical factor is responsible for a certain class of such aberrations called monochromatic aberrations. The wave nature of light, the fact that light is a wave, means that, in every passage of light from any aperture such as a lens or a mirror, the effect of diffraction is present. This is significant enough when the aperture / obstacle dimensions are comparable to the wavelength and affect image quality. Therefore, instead of a point, we have the formation of a small blurry spot, the Airy disk, which, in the best-case scenario, is diffraction limited (as discussed in § 7.5).

8.1.2 The Paraxial Approximation Spherical surfaces are commonly employed in many optical applications. However, the basic condition for stigmatic image formation is the principle of minimum optical path. One of the imperatives of this principle is that the reflecting or refracting surface must have a parabolic shape (as shown in § 2.2.3.) What is this parabolic surface? Mathematically, a parabola is the curve produced by the function ψ = α x2, where α is a number. The parabolic function is quite well known, as many physical phenomena follow such a path. For example, the trajectory in an oblique shot is

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OPTICAL ABERRATIONS

parabolic. Mathematically, a parabolic surface is produced if we completely rotate (360°) this curve around its axis of symmetry. Because it originates from a curved rotation, the parabolic surface (the parabola, specifically) is said to be rotationally symmetrical and belongs to a larger family of surfaces called conic surfaces. These are spherical, elliptic, and even the flat surfaces whose crosssections correspond to a sphere, ellipse, or line, respectively.

Figure 8-4: Reflection properties of spherical (left) and parabolic (right) surfaces drawn by Leonardo da Vinci. Note how the reflection from the parabolic surface results in all rays crossing the same focal point, as opposed to the reflection from the spherical surface, in which they do not. (Images from da Vinci’s The Codex Arundel, made freely available through a collaboration between The British Library and Microsoft, Turning the Pages 2.0.)

For small angles with respect to the optical axis, we can assume that all conic surfaces coincide. This is the paraxial approximation (παρα-, near & -αξονική, axial), which relies on the assumption that sinϑ ≈ ϑ [rad]. Expanding the sine function in a Taylor / McLaurin series, sin  =  −

3 5 7 + − + ... 3! 5! 7!

(8.1)

we note that only the first term has numerical significance for paraxial rays. In other words, the paraxial approximation is valid for rays striking near the center of an optical system. In practice, the angles of incidence may be large, resulting in the third term being significant. In this case, simple approximations are not applicable, and significant imperfections in the image are noted, the most notable example being spherical aberration. At the limit of a small angle of incidence, we might be able to construct a lens from spherical surfaces and restrict rays to small angles around the optical axis, in other words, to only use rays forming a small angle with the normal to the surface. The well-polished refracting and reflecting surfaces used in optical devices are, for practical construction reasons, indeed, primarily spherical and not parabolic surfaces.

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GEOMETRICAL OPTICS

Figure 8-5: Cross-sections of conic surfaces. For small angles, a circular surface and a parabolic surface almost coincide. This is the paraxial approximation.

8.1.3 Classification of Optical Aberrations There are two ‘schools’ that describe the classification of aberrations. The traditional ‘old school’ classifications are named after the German ophthalmologist Erich Seidel and the German physicist and astronomer Karl Schwarzschild. The ‘new school’ classifications are named after the optical physicist Frits Zernike, the winner of the 1953 Nobel Prize in Physics and inventor of phase-contrast microscopy. The Seidel classification investigates deviations from the paraxial approximation when the term ϑ3 is included, while the Schwarzschild classification incorporates an additional term ϑ5 in Eq. (8.1). There is a limited number of aberrations in each classification. The Zernike classification treats optical aberrations from a wave perspective. For this reason, Zernike aberrations are also known as wave aberrations. In this classification, there is no limit to how many aberrations can exist because each successive aberration order has an increasing number of aberrations. We can, of course, limit our Zernike aberration analysis to an order of our choosing, for example, up to the fifth order. Some terms are shared between these two classifications, such as defocus, spherical aberration, coma, and astigmatism. However, as the complexity increases, there is no true oneto-one association; for example, there are two Zernike functions that describe astigmatism. Additionally, there are terms such as field curvature and field distortion that are only found in the Seidel/Schwarzschild classification and terms such as trefoil, second-order spherical, and others that are only found in the Zernike classification.

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Optical Aberrations Monochromatic (both lenses and mirrors)

Zernike Classification

Seidel Classification Point Aberrations

Chromatic Aberration (only lenses)

Field Aberrations

Low Order

High Order

Spherical aberration

Curvature

Defocus

Spherical aberration

Coma aberration

Distortion

Astigmatism

Coma aberration

Astigmatism

Higher-order astigmatism

Seidel aberrations are distinguished as point aberrations, which affect image sharpness (how fine the image is), and as field aberrations, which affect the geometry of the formed image (for example, if it is distorted). In Zernike classification we distinguish between low-order and high-order aberrations. Low-order Zernike ‘aberrations’ have closely matched Seidel aberrations and are known as spherocylindrical. High-order Zernike functions are particular applicability in visual optics, as they can express pupil-matched fitting functions to describe ocular aberrations, while the Seidel classification, being more traditional, is more often employed in optical engineering and instrument analysis. Note

: Zernike ‘aberrations’ are not by themselves optical aberrations. They are polynomial fitting

functions that help describe optical aberrations in an orthogonal manner over the unit circle of the exit pupil of an optical system. The usefulness of Zernike functions is exactly this: They can reduce a complex optical system into a set of numbers, each function contributing the weight, or significance, if you will, of each component to the ‘soup’ of optical aberrations. They can be scaled to fit a specific pupil size so can be very useful in describing a potential improvement or deterioration of a planned change in the refractive system, such as the eye.

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8.2 CHROMATIC ABERRATION As we already know from the lens-maker’s formula (§ 2.4.1), the focal length depends on the refractive index of the medium, which in turn depends on the wavelength/color. Therefore, for a different wavelength (color), there is a different focal length; usually, longer for the red. This is because in normal dispersion, which is the usual case in most materials such as glass in the visible,47 longer wavelengths correspond to lower values of the refractive index. Typically, the differences (for example, n = 1.685 for the indigo-violet up to n = 1.645 for the red) may not appear to be very significant, but they are. Even such small differences result in different focal lengths (and therefore different optical powers) for the various components of the spectrum. The blue, in other words, focuses at a different axial location than the red. The dimensionless Abbe number (also known as constringence), named after the German physicist Ernst Karl Abbe, is an expression of the medium’s dispersion properties: Abbe Number:

V =

nY/d − 1 nB/F − nR/C

(8.2)

where nR/C corresponds to the red (R) hydrogen spectral line (C at 656.3 nm), nY/d corresponds to the yellow (Y) sodium line (d at 587.6 nm), and nB/F corresponds to the blue (B) hydrogen line (F at 486.1 nm). The Abbe number for the human eye has a value of 50.23.

Figure 8-6: Abbe numbers and refractive indices for various optical glasses.

The dispersive power Δ is the reciprocal of the Abbe number: Dispersive Power:

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 =

nB/F − nR/C nY/d − 1

(8.3)

OPTICAL ABERRATIONS

If the indices nB/F and nR/C are quite different, the Abbe number (also called the Vnumber) has a relatively small value (for example, V < 55), indicating that the glass is strongly dispersive (flint glass is a typical example). If the indices nB/F and nR/C differ slightly, the Vnumber has a relatively large value: If V > 55, the material is characterized by low dispersion (such as crown glass). When a broad spectrum of wavelengths (such as from a multicolor or white light source) participates in image formation via any dispersive refractive element, and a sharp blue ‘image’ is formed, the red components are not yet in focus; when the red ‘image’ is focused, the blue components are at defocus. This is chromatic aberration.

Figure 8-7: Chromatic aberration: The red sharp focus and the blue sharp focus do not coincide.

The separation of the different focal points along the axis constitutes longitudinal (or axial) chromatic aberration (LCA). Typically, shorter wavelengths correspond to shorter focal lengths (for example, the blue and the green focus first), while longer wavelengths (red) correspond to longer focal lengths. In this case, the aberration is due to normal dispersion, and LCA is positive. If the opposite occurs, the aberration is due to anomalous dispersion,49 and LCA is negative. Thus, in LCA, a succession of sharp images can be presented along the optical axis. This difference in focal length leads to colored fringes surrounding the images that arise because not all colors can be displayed in focus. When the blue components are focused, the red are not focused, and the red part casts a red fringe due the larger geometrical image blur. Conversely, when the red components are focused, the blue components are not, and a blue fringe is cast at the borders of the main image. This is perceived as transverse (or lateral) chromatic aberration, which is a manifestation of a prismatic effect associated with chromatic aberration. The similarity between the chromatic aberration in a lens and the effect of a prism is Anomalous dispersion occurs near zones of increased absorbance. Shorter wavelengths (blue) may then have a lower refractive index value. More on the topic in Wave Optics § 3.3.1 Dispersion in Thin Media. 49

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not a coincidence. Chromatic aberration is the principle of operation in prisms, as they act as color component separators; it is also the principle of operation of the rainbow.

Figure 8-8: Images with (left) low and (right) significant transverse chromatic aberration. Note the color fringes around the flag–sky interface in the right image.

Figure 8-9: The chromatic aberration characteristic curve for a specific lens or lens system presents a plot of the wavelength as a function of the corresponding focal length.

The red-green, or duochrome, test is an application of chromatic aberration. The patient is asked to compare the clarity of letters on two adjacent green and red backgrounds. With optimal spherical correction, the green part is slightly in front of the retina, while the red part is slightly behind the retina. Both parts appear to be equally clear. Because this test is based on chromatic aberration and not on color discrimination, it is used even in individuals with color vision deficiency.

Figure 8-10: Longitudinal chromatic aberration in the human eye.

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OPTICAL ABERRATIONS

Longitudinal chromatic aberration is directly proportional to the lens power F and inversely proportional to the lens material Abbe number V: Longitudinal Chromatic Aberration:

LCA =

F =F·Δ V

(8.4)

Example ☞: What is the LCA in a +8.00 D lens made of very low-dispersion crown glass (V = 65)? We simply use Eq (8.4): LCA = F/V = +8.00 D/65 = +0.125 D.

Lateral chromatic aberration is proportional to the prism power P [Δ] of the lens50 times the dispersive power Δ of the material: Transverse/Lateral Chromatic Aberration:

TCA = P[Δ] · Δ

(8.5)

which can be re-arranged as follows, considering that Δ = 1/V, P[Δ] = d · F, and LCA = F · Δ: Transverse/Lateral Chromatic Aberration:

TCA = d · LCA

(8.6)

where d is the image distance from the optical center of the lens. Thus, a material with a smaller

V-number has more chromatic aberration, both longitudinal and lateral. Recall that if the Vnumber has a relatively small value (e.g., V < 55), the glass is strongly dispersive (flint glass). If the V-number has a relatively large value (e.g., V > 55), the material is characterized by low dispersion (crown glass).

8.2.1

Management of Chromatic Aberration

The first adjustment might be to use a material with the smallest possible dispersion (large V). However, there are two problems with this approach. First, this would not correct the aberration. Second, materials with large V-numbers also have lower refractive indices (Figure 8-6), which would require a lens with larger radii of curvature (and perhaps, thickness); this lens would, in turn, induce other aberrations. Complete elimination of chromatic aberration for all values of wavelengths across the visible spectrum is not possible. To reduce the overall presence of chromatic aberration, a combination of two lenses of different dispersion curves is used. In this combination, a converging lens of crown glass (low refractive index, lower dispersion) is cemented to a diverging lens of flint glass (high refractive index, higher dispersion). The compound lens can be specifically designed to have zero chromatic aberration for two wavelengths, such as the red 50

Introduction to Optics § 3.3.5 Prism Power & Visual Optics § 10.4.1 Prentice’s Rule.

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and the blue. In such a compound lens, the aberration of one lens compensates for the aberration of the other. This lens is called an achromatic doublet lens. In this lens, chromatic aberration is relatively low for the wavelengths between the two wavelengths for which total aberration elimination is calculated. Lenses that correct chromatic aberration for three wavelengths are called apochromatic triplet lenses.

Figure 8-11: Compound achromatic lens consisting of a plus crown glass lens and a minus flint glass lens.

The design of an achromatic lens is based on the principle that the chromatic aberration induced by lens 1 (LCA1) is of equal magnitude and opposite sign compared to the chromatic aberration induced by lens 2 (LCA2), calculated for at least two different wavelengths. In addition, the two powers add up to the desired combined lens power (for thin lens approximation, Fset = F1 + F2). Based on Eq. (8.4), the power F1 of lens 1 can be found using the known values of the Abbe number value V1 for lens 1 and V2 for lens 2: LCA1 = –LCA2 ⇒ F1/V1 = –F2/V2 ⇒

F1 = Fset·(V1)/(V1 – V2) and/or F2 = Fset·(V2)/(V2 – V1) (8.7)

Example ☞: To create an achromatic doublet of Fset =+7.00 D power, using crown glass (n1 = 1.523, V1 = 58) and Schott High-Lite glass (n2 = 1.701, V2 = 31), the first component has power F1 = (+7.00)·(58)/(58 – 31) = +15 D. Assuming thin lenses, the second component has power F2 = Fset – F1 = (+7.00) – (+15.00) = –8.00 D. Therefore, the achromat doublet consists of a converging crown glass lens of +15.00 D power and a diverging flint-type meniscus lens of –8.00 D.

Practice achromatic doublet lens examples

:

Use F1 crown glass (n1 = 1.523, V1 = 58) and F2 Schott High-Lite glass (n2 = 1.701, V2 = 31). For Fset =+5.00 D, F1 = +10.75 D and F2 = –5.75 D; for Fset = –5.00 D, F1 = –10.75 D and F2 = +5.75 D. For Fset =+3.50 D, F1 = +7.50 D and F2 = –4.00 D; for Fset = –3.50 D, F1 = –7.50 D and F2 = +4.00 D.

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OPTICAL ABERRATIONS

Another technique for correcting chromatic aberration (applied in Huygens and Ramsden eyepieces) is the use of two plus lenses with opposite chromatic aberration at a distance equal to half the sum of their focal lengths: d = (f1 + f2)/2. We apply the relationship of two lenses in air (n = 1.0) separated by d (using formulas presented in § 6.5.1):

d 1 1 Fset = F1 + F2 − F1  F2 = + − n f1 f2

f1 + f2 1

2 1 1 =

f1 f2

f  1 1 1  f1 1 1 1 1 = + −  + 2  =   +  =  ( F1 + F2 ) f1 f2 2  f1  f2 f1  f2  2  f1 f2  2

(8.8)

8.3 MONOCHROMATIC ABERRATIONS 8.3.1

Spherical Aberration

The world became aware of this optical aberration due to the early problems relating to the images obtained by the Hubble Space Telescope. The analysis, employing Zernike polynomials, showed that the problem was a slight, but apparently significant enough, spherical aberration. It was caused by the periphery of the main mirror having been polished very slightly too flat by a few microns! 51 Today, what had been perhaps the most ‘superstar’ application of wavefront aberrometry is extensively employed for the measurement and management of ocular aberrations in ophthalmic imaging.52

Figure 8-12: (left) Hubble’s main mirror being polished before installation. (right) Astronaut Kathryn C. Thornton lifts the Corrective Optics Space Telescope Axial Replacement (COSTAR) into Hubble during the First Servicing Mission. Astronaut Thomas D. Akers, who assisted in the COSTAR installation, is at the lower left. (Image credit: NASA JSC Digital Image Collection.)

51

Ocular Imaging § 5.2.1 Spherical Aberration – insert.

52

Ocular Imaging § 6.2 Types of Aberrometry.

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Figure 8-13: Photograph of galaxy M100 taken with (left) the Wide-Field/Planetary Camera 1 and (right) the Wide-Field Planetary Camera 2 shows the greatly improved image quality after the telescope’s spherical aberration was corrected. (Image credit: NASA.)

The principle of least optical path requires that, in order for all rays (which will be considered monochromatic from now on in this text)—either those on the optical axis (paraxial rays) or those away from the optical axis (for instance, the tallest object)—to converge to a single image point, the shape of the reflecting or refracting surface must be parabolic. In reality, the majority of optical surfaces are spherical. To form a satisfactory image, we restrict imaging to a small solid cone, in which the rays form very small angles ( n2 (second, external medium). Prisms A prism is an optical medium that can be defined by two refracting surfaces that form an angle, called the apical angle A. Typically, a prism is represented by a triangle. The side opposite the apex is called the base. A ray incident on one side exits through the other side of the prism at the angle defined by the apical angle. The ray deviates toward the base with a deviation angle ϑE: Prism Deviation Angle ϑE:

ϑE = ϑi1 + ϑt2 – A

where ϑi1 is the angle of incidence at the first surface, and ϑt2 is the angle of refraction at the second surface. In certain configurations, the angle of deviation can be minimized. Then, the minimum angle of deviation ϑE MIN is calculated using this relationship: Calculation of Minimum Angle of Deviation:

438

n  sin

A 2

= sin

A + E MIN 2

APPENDIX

In thin prisms (apical angle less than 30°), the deviation angle ϑE is nearly constant: ϑE = (n – 1) · A

Thin Prism Deviation: Prentice Rule:

prism power P [Δ] =

ray deviation [cm] distance measured [m]

=

100  ray deviation [cm] distance measured [cm]

Vergence, Optical Power and Focal Lengths, and Imaging Sign Conventions A positive algebraic sign is assigned to the direction of light travel. The origin, or point zero (0, 0), is the intersection of the optical axis with the optical element (e.g., the lens, a refracting surface, or a mirror).

Cartesian sign conventions for directional distances along the optical axis.

Vergence The physical entity that expresses the degree of convergence is vergence L. Vergence is reported in diopters [D], the reciprocal of a meter [m–1]. In air, vergence is the reciprocal of the distance to convergence. The distance needed for a wavefront to converge (focus) is denoted by either l or x, and is expressed in meters. Vergence and Distance to Convergence (air):

L [D] =

1

x

=

1 distance to convergence [m]

439

If the medium has a refractive index n, the generalized relationship is

L [D] =

Reduced Vergence:

n refractive index = x distance to convergence [m]

Downstream and Upstream Vergence Downstream vergence is the value of the vergence along the propagation of light. When vergence L is known at a certain point, then at a point that is located d [m] downstream of the initial (known) vergence, the (unknown) vergence L΄ is Downstream Vergence (in air):

L 1− d  L

L΄ =

The distance d is positive, since downstream means along the direction of light propagation. The same formula can be used for the upstream vergence; the only difference is that the distance d is negative in upstream vergence. The more generalized expression that includes the fact that the space between is filled with a medium with refractive index n is Downstream Vergence (medium with n):

L΄ =

L d 1−  L n

Optical Power Optical (or refractive) power is the fundamental property of an optical element that can converge (or diverge) a collimated ray beam. The unit of the power is the diopter D, which is the reciprocal of the meter (m–1). Single Spherical Refracting Interface Power A single spherical refracting interface (SSRI) is a surface that has a radius of curvature r and separates two media with two different refractive indices. To the left of the interface the refractive index is n, and to the right of the interface the refractive index n΄. The optical power is described by n΄

SSRI Optical Power:

F optical power, expressed in D

=

−

refractive index after the interface

n refractive index before the interface

r radius of curvature, expressed in m

440

APPENDIX

For an SSRI that separates air (n = 1.0) from glass (n΄ is the SSRI glass material refractive index), SSRI Optical Power (in air):

F = (n΄ – 1.0) / r

The most generally applicable formula relating the optical power and the focal length states that the optical power is the reciprocal of the focal length multiplied by the refractive index: Optical Power F [D] =

Refractive Index Focal Length f [m]

More generally, when the refractive index of image space is n΄ and the refractive index of object space is n, we call the image-space focal length secondary and the object-space focal length primary (the two focal lengths point in opposite directions!). In this case, n΄

Optical Power F and Focal Length(s) f :

F =

n

image space refractive index

f΄

= −

secondary focal length

object space refractive index

f primary focal length

Lens Power A lens comprises two SSRIs. If the lens is thin, we can simply add the two surface powers or we can use the lens-maker’s formula: Lens Optical Power in Air (lens-maker’s formula):

F =

1 1 = ( n − 1)   −  f  r1 r2 

1

where n is the lens material refractive index, r1 is the first surface radius of curvature, and r2 is the second surface radius of curvature. This formula assumes that the lens is surrounded on both sides by air. If the lens is surrounded by a medium of refractive index next, the formula takes the form Lens Optical Power (surrounded by medium next):

1 1 F = ( n − next )   −   r1 r2 

Thick Lens Power For a thick lens with refractive index n and center thickness t, in which F1 is the front surface power and F2 is the back surface power (which can be calculated as simple SSRIs powers), the equivalent power is provided by Gullstrand’s formula.

441

Optical Power (in a thick lens):

Fe =

n − 1 1− n t  n − 1   1− n  t + −    = F1 + F2 − F1  F2 r1 r2 n  r1   r2  n

The same formula applies to the total optical power of a lens system; instead of lens thickness t, we use the separation between the two lenses d and pay attention to use the refractive index of the space between the two lenses (which is often, but not always, air). The equivalent focal length (distance) fe is reciprocally related to the effective optical power. It is measured from the corresponding principal planes in either object space or image space: Equivalent Focal Length (in air):

fe =

1

Fe

Vertex Powers The back vertex power and the back focal length correspond to image space. The back focal length is measured from the image-space vertex point of a thick lens: Back Vertex Power:

FBVP =

F1 + F2 t 1 −  F1 n

The front vertex power and the front focal length correspond to object space. The front vertex power is also known as the neutralizing power, or the front neutralizing power. The front focal length is measured from the object-space vertex point of a thick lens: Front Vertex Power:

FFVP = F1 +

F2 t 1 −  F2 n

Mirror Power

Mirror Optical Power (in air):

Fmirror =

1

f

=

2

r

where f is the mirror focal distance, and r is the radius of curvature. In a concave mirror, the focal length and radius of curvature are positive. In a convex mirror, the focal length and radius of curvature are negative.

442

APPENDIX

In mirrors, the object location (and vergence) follows the same sign conventions as in lenses or refracting interfaces, as shown in the figure.

Since light propagation flips in mirrors following reflection, the image location (and vergence), radius of curvature, and focal distance are positive if their directional distances are pointing to the left (which is the new direction of light propagation). Mind the sign!

Imaging Relationships Magnification is defined as the ratio of any image linear dimension to an object linear dimension perpendicular to the optical axis (for example, the length of an arm between the image and the object), called height h. It is called linear, lateral, or transverse magnification, or simply magnification. Often, we calculate the magnification using the ratio of the image distance (x′) to the object distance (x) or of the object vergence (L) to the image vergence (L′).

443

Magnification:

m

h΄ x΄ L = = h x L΄

The expression involving the vergence ratio (instead of the distance ratio) is generally more applicable, particularly if the object and image space have different indices of refraction (the distance ratio does not apply here). In vision, it is more sensible to use the angular magnification:

M =

Angular Magnification:

perceptual angular subtence object angular subtence

=

i o

Imaging relationships for an object at location x (object vergence L = 1/x), an image at location

x΄ (image vergence L΄ = 1/x΄ ), and a lens (or mirror) with focal length f΄ (optical power F = 1/f΄ ) state that the object vergence is added to the power of the imaging element to produce the image vergence: Vergence and Optical Power:

+

L object vergence

1 x

Object and Image Locations:

=

F optical power

1 f΄

+

object location

L΄ image vergence

1 x΄

=

focal length

image location

The above relationships assume that the optical medium surrounding the lens (imaging element, in general) is air. If the refractive index in object space is n and in image space is n΄, then the more generally applicable form of the imaging relationship is: n x object location

+

n΄ f΄ focal length

=

n΄ x΄ image location

A real object is placed in front of the lens (or mirror) and has a negative vergence. A virtual object (will be) formed after the lens and has a positive vergence. A real image is formed after the lens (or mirror) and has a positive vergence. A virtual image is (perceived as being) formed before the lens and has a negative vergence.

444

APPENDIX

Optics of the Human Eye Optical Power and Focal Length:

F =

Visual Angle:



Retinal Image Size:

nvitreous f vitreous

 60 D

object size h distance d

h΄ [mm] ≈ 17 · ϑ [rad]

Rule to convert angles to distance at the retina, expressed in mm (angles from the nodal point): 1 mm 0.06 rad = 3.345° Parameters of Gullstrand’s exact schematic eye model.

L L΄

Thickness

Refractive index

Radii of curvature

Cornea

0.5 mm

1.376

7.7 mm (anterior) 6.8 mm (posterior)

Aqueous humor

3.1 mm

1.336

–

Crystalline lens

3.6 mm

1.386 (cortex) 1.406 (nucleus)

10.0 mm (anterior) −6.0 mm (posterior)

Vitreous body

17.2 mm

1.336

–

Axial length

24.4 mm

–

–

Near Distance for Acceptable Sharpness: Angular Magnification (magnifying lens):

Δ = 25 cm M ANG =

Δ

f

−

Δ

x΄

When the image is formed at the near point (x΄ = −Δ = −25 cm), the magnification has its largest value and is called the maximum lens magnifying power: Lens Magnifying Power:

M ANG =

25 cm +1 f

445

APPENDIX

ANSWERS TO QUIZ QUESTIONS (Answers are provided only to odd-numbered questions.) Chapter 1 Refraction Quiz 1) 2) 3) 4) 5) 6) 7) 8) 9) 10) 11) 12) 13) 14) 15)

d b b b a d a b d c b c b and c a d

16) 17) 18) 19) 20) 21) 22) 23) 24) 25) 26) 27) 28) 29) 30)

c a c a d e c a c d a c (62.7°) b f c

31) 32) 33) 34) 35) 36) 37) 38) 39) 40) 41) 42) 43) 44) 45)

a d b c a d c e e c b b a and b a b

46) 47) 48) 49) 50) 51) 52) 53) 54) 55) 56) 57) 58) 59) 60)

c a b and d c and d d d b a d c d a b d f

61) 62) 63) 64) 65) 66) 67) 68) 69) 70) 71) 72) 73) 74) 75)

d d c (=+2.0 D) f (=+3.5 m) b c b c a b a and d d a c a

15) 16) 17) 18) 19) 20) 21)

e a c e a and b d a and c

22) 23) 24) 25) 26) 27) 28)

a e b and f b b d c

29) b 30) d

Chapter 2 Lens Power Quiz 1) 2) 3) 4) 5) 6) 7)

a c a b a and f a b

8) 9) 10) 11) 12) 13) 14)

b d a c e c a

Chapter 3 Vergence and Imaging Concepts Quiz 1) 2) 3) 4) 5) 6)

d c b d a a

7) 8) 9) 10) 11) 12)

d a b d a b

13) 14) 15) 16) 17) 18)

c e b c f e

19) 20) 21) 22) 23) 24)

d c d a a b

25) 26) 27) 28) 29) 30)

d c b c and d a a

19) 20) 21) 22) 23) 24) 25) 26) 27)

d c a d a c a d e

28) 29) 30) 31) 32) 33) 34) 35) 36)

b e c c c d d a c

37) 38) 39) 40) 41) 42) 43) 44) 45)

b a d (F = +20/3) a b e d a b

5) 6)

d b

7) 8)

b c

9) c 10) c

Chapter 4 Lens-Imaging Quiz 1) 2) 3) 4) 5) 6) 7) 8) 9)

a b a a and d a and b c and d a f e

10) 11) 12) 13) 14) 15) 16) 17) 18)

e d d d d c c b d

Chapter 5 Mirror Imaging Quiz 1) 2)

c e

3) 4)

c b

447

11) 12) 13) 14) 15) 16) 17)

b e c and d b a a a

18) 19) 20) 21) 22) 23) 24)

d b c a and d b and d b and c a

25) 26) 27) 28) 29) 30) 31)

a c e a a c b

32) 33) 34) 35) 36) 37) 38)

d a d d a b b and c

39) 40) 41) 42) 43) 44)

b and d a d c a b

39) 40) 41) 42) 43) 44) 45) 46) 47) 48) 49) 50) 51) 52) 53) 54) 55) 56) 57)

d d b b c c a b c b b e d c a c c d a

58) 59) 60) 61) 62) 63) 64) 65) 66) 67) 68) 69) 70) 71) 72) 73) 74) 75) 76)

d b a d b a a a and d d b a b c c a and d c d c a and c

77) 78) 79) 80) 81) 82) 83) 84) 85) 86) 87) 88) 89) 90) 91) 92) 93) 94) 95)

a and b b a a and c b a c b b a b f f b a b c e b

55) 56) 57) 58) 59) 60) 61) 62) 63) 64) 65) 66) 67) 68) 69) 70) 71) 72) 73) 74) 75) 76) 77) 78) 79) 80) 81)

d a a and f d and e f a and c b and d e a and e b and c c a d c d a c and e b and c b and e b a c a and b a b a and b c

82) a and f 83) e 84) b 85) a 86) c 87) a 88) b and d 89) d 90) a 91) b and d 92) a, c, and d 93) a and d 94) d 95) c 96) b 97) b 98) c 99) b 100) a 101) c 102) c 103) a, b, c, and d 104) b 105) a, b, c, and d 106) c 107) d 108) a

109) e 110) d 111) d 112) b 113) c 114) b 115) a 116) b 117) d 118) a 119) d 120) c 121) e 122) d 123) b 124) d 125) c 126) d 127) c 128) a and d 129) d 130) d 131) b and c 132) a

15) 16) 17) 18) 19) 20) 21)

a a d c a a, b, and d a

22) 23) 24) 25) 26) 27) 28)

29) 30) 31) 32)

Chapter 6 Thick Lens and Lens Systems Quiz 1) 2) 3) 4) 5) 6) 7) 8) 9) 10) 11) 12) 13) 14) 15) 16) 17) 18) 19)

b b c c d a c and e a, b, and e a and e b and d a d c a and c d c b, c, and f a, d, and f c

20) 21) 22) 23) 24) 25) 26) 27) 28) 29) 30) 31) 32) 33) 34) 35) 36) 37) 38)

a d a, d, and f b b and c g a b c a c d c d d a c, d, and e b c

Chapter 7 Pupils and Stops Quiz 1) 2) 3) 4) 5) 6) 7) 8) 9) 10) 11) 12) 13) 14) 15) 16) 17) 18) 19) 20) 21) 22) 23) 24) 25) 26) 27)

a d e d d a c d b b a c c d a e a c c b a d c f a b a

28) 29) 30) 31) 32) 33) 34) 35) 36) 37) 38) 39) 40) 41) 42) 43) 44) 45) 46) 47) 48) 49) 50) 51) 52) 53) 54)

c a b d a, b, and f a and c c c a and b c a a b b a a a and e c a b b a, c and f c c a and e c a

Chapter 8 Aberrations Quiz 1) 2) 3) 4) 5) 6) 7)

448

a a and b c d b d a

8) 9) 10) 11) 12) 13) 14)

a c b a b and c a and d b

c b c b c b a

b b and c a c

APPENDIX

INDEX on-axis ...............................................................................8-419

A

radial...................................................................................8-418

Abbe number....................................................................... 8-400

B

Aberration ............................................................................. 8-394 axial .................................................................................... 8-401 chromatic ......................................................................... 8-396 classification ................................................................... 8-398 longitudinal..................................................................... 8-394 monochromatic............................................................. 8-405 Petzvál ............................................................................... 8-422 Schwarzschild................................................................. 8-398 Seidel ................................................................................. 8-398 transverse ........................................................................ 8-394 Zernike .............................................................................. 8-398 Achromatic lens .................................................................. 8-404 Afocal system .......................................... 6-253, 6-269, 7-371

Ball lens .................................................................... 6-219, 6-263 Base curve power................................................................6-214 Blur and aberration................................................................8-393 and diffraction, example ............................................7-348 circle ...................................................................................8-393 circle diameter................................................................7-358 geometrical....................................................... 7-356, 7-360 Bokeh .......................................................................................7-358 examples ...........................................................................7-360 Brightness ............................................................... 7-302, 7-351

aFoV ............................................... See Field of View (angular)

C

Airy disk ................................................................................. 7-352 Angle critical ..........................................................................1-7, 1-34

Cardinal points .....................................................................6-227

of incidence ................................................................ 1-2, 1-4

in a thick lens system ..................................................6-255

of incidence, normal ........................................................ 1-6

in a thin lens system ....................................................6-250

of reflection ..............................................................1-2, 3-75

Cartesian sign convention................................... 3-77, 3-106

of refraction ......................................................................... 1-4

Caustic curve.........................................................................8-407

Angle of View ...................................................................... 7-323

Chief ray................................................................... 7-315, 7-388

Anish Kapoor ....................................................................... 5-193

Chromatic aberration

Anisoeikonia ......................................................................... 6-219

in the human eye ..........................................................8-402

Aperture stop ........................................................ 7-295, 7-389

management ...................................................................8-403

identification and location ......................... 7-297, 7-361

Circle of confusion .............................................................7-346

of the human eye ......................................................... 7-312

circle of least confusion............................... 8-394, 8-412

Aplanatic ................................................................................ 8-417

Coddington position factor ............................................8-409

Apochromatic lens ............................................................ 8-404

Coddington shape factor ................................................8-408

Apparent depth ..................................................................... 3-95

Coma ........................................................................................8-413

Aspect ..................................................................................... 7-296

and aperture stop size ................................................8-416

Aspheric element ............................................................... 8-410

dependence of lens shape ........................................8-417

Astigmatism

management ...................................................................8-416

management .................................................................. 8-419

negative ............................................................................8-414

oblique .............................................................................. 8-418

positive ..............................................................................8-414

449

sagittal .............................................................................. 8-415 tangential......................................................................... 8-415 Concave surface concave interface example.......................................... 1-11 Confocal aperture .............................................................. 7-362 Conic surface ....................................................................... 8-397 Contrast.................................................................................. 7-351 examples .......................................................................... 7-355 Convex surface convex Interface example ............................................ 1-11 Cornea .........................................................................2-59, 6-219 optical power ................................................................. 6-220 Crystalline lens optical shape ..................................................................... 2-62 Curvature center of ........................................ 1-9, 1-34, 5-168, 5-169 in reflecting surfaces ................................................... 5-168 in refracting surfaces ............................................1-9, 1-34 in wavefronts ..................................................................... 3-82 Cylinder axis of ................................................................................... 2-55

D Depth of Field ...................................................................... 7-345 aperture stop dependence....................................... 7-348 dioptric ............................................................................. 7-346 examples .......................................................................... 7-359 in the human eye ......................................................... 7-349 linear .................................................................................. 7-345 Depth of Focus.................................................................... 7-348 in the human eye ......................................................... 7-350 Deviation angle ........................................................................ 1-5 Diffraction ............................................................................. 7-352 and image blur .............................................................. 7-360 Diffraction-limited optics................................................ 7-352 Diopter definition................................................................... 3-82, 437 in refracting surfaces ..................................................... 1-10 Dioptric power ........................................... See Optical power Dispersion ............................................................................. 8-400 Distortion barrel ................................................................................. 8-423 pincushion ....................................................................... 8-423

450

E Ellipse .......................................................................................5-168 Enantiomorphism ...............................................................5-164 Entrance port ......................................................... 7-329, 7-390 real image ........................................................................7-330 Entrance pupil ....................................................... 7-304, 7-389 of the human eye ........................................... 7-312, 7-371 real image ........................................................................7-310 virtual image .................................................... 7-310, 7-318 Entrance window........................................ See Entrance port Exit port.................................................................... 7-329, 7-390 real image ........................................................................7-329 Exit pupil .................................................................. 7-304, 7-389 in microscope .................................................................7-370 of the human eye ..........................................................7-313 real image ........................................................................7-308 virtual image .................................................... 7-310, 7-316 Exit window .................................................. See Entrance port Eye relief .................................................................................7-306 Eyepiece lens ......................................................... 6-254, 7-327

F Faceform tilt ..........................................................................8-418 Field curvature .....................................................................8-420 Field distortion .....................................................................8-420 Field of at least one-half illumination, .......................7-341 Field of uniform illumination .........................................7-341 Field of View .......................................................... 7-323, 7-391 angular ..............................................................................7-323 diagonal ............................................................................7-325 horizontal .........................................................................7-325 human eye .......................................................................7-325 image-space....................................................................7-374 linear .................................................................... 7-323, 7-336 object space ....................................................................7-372 object space vs image space..................... 7-335, 7-391 size ......................................................................................7-333 total.....................................................................................7-340 vertical ...............................................................................7-325 Field stop................................................................. 7-325, 7-389 at the lens plane ............................................................7-331 of the eye .........................................................................7-326

APPENDIX

Focal length

erect ...............3-80, 4-115, 4-121, 5-182, 5-191, 5-196

and geometry ................................................................... 2-61

height ..................................................................... 3-78, 5-185

and lens orientation ....................................................... 2-58

intermediate ..................................................... 6-257, 6-292

and surrounding medium............................................ 2-59

inversion ...........................................................................4-116

back focal length (BFL)............................................... 6-231

inverted ................................................................... 2-49, 3-80

definition................................................................. 1-13, 1-35

location .............................................................................4-110

effective ................................................................................. 440

magnified .......................................................... 4-121, 4-149

equivalent .......................................................... 6-214, 6-221

parity changes ................................................................5-164

front focal length (FFL)............................................... 6-231

real ..... 3-74, 3-90, 3-107, 4-120, 5-188, 5-189, 5-202

in a ball lens .................................................................... 6-263

reflection............................................................................. 3-76

in a lens ............................................................................... 2-56

retinal .................................................................................4-119

in a mirror ........................................................................ 5-174

reversal ..............................................................................4-116

in a thick lens ..................................... 6-221, 6-243, 6-291

secondary .......................................................... 6-257, 6-292

in an SSRI ...................................................... 1-17, 1-35, 439

vergence ...........................................................................4-111

Focal line................................................................................... 2-55

virtual ........................................... 3-74, 3-90, 3-107, 5-182

Focal plane............................................................................... 2-51

Image space .............................................................. 3-72, 3-107

curved .................................................................................. 2-53 Focal point

Imaging definition ............................................................................ 3-72

in a lens ............................................................................... 2-49

propagation to the left ...............................................4-114

in a mirror ........................................................................ 5-172

with concave mirror ...................................... 5-188, 5-189

in an SSRI ................................................................ 1-13, 1-35

with convex mirror ........................................ 5-180, 5-185

primary .............................................1-13, 1-35, 2-49, 2-68

with flat mirror ................................................... 3-75, 5-163

secondary ........................................1-13, 1-35, 2-49, 2-68

with positive lens ..........................................................4-118

Focus .................................................................... See Focal point

with two or more lenses ............................................6-256 with virtual object ............................ 4-144, 4-145, 4-148

G Gaussian thin-lens-imaging formula ......................... 4-110 Geometrical image blur .....................................................7-356 Glare ........................................................................................ 7-344 Glare stop .............................................................................. 7-344 Gravitational lensing ............................................................ 2-45 Gullstrand’s relationship ................................... 6-213, 6-287

Imaging relationship in a spherical mirror .....................................................5-178 in a thin lens ..........................................................4-110, 442 with object and image locations ............................4-109 with vergence and optical power ...........................4-109 Incidence normal ....................................................................................1-6 Index of refraction .. See Refractive index, See Refractive index Intraocular Lenses................................................................. 2-60

H

Isoplanatic ..............................................................................8-417

Höegh meniscus ......................................................2-63, 6-269

L

Hyperfocal distance .......................................................... 7-350

Lagrange invariant ...............................See Optical invariant

I Image definition............................................................................. 3-71

Lens and wavefront transformation .................................. 2-44 ball lens .............................................................................6-263 biconcave ........................................................................... 2-53

451

biconvex ................................................................2-53, 6-214 converging ......................................................................... 2-38 cylinder ................................................................................ 2-55

Microwaves reflectors ...........................................................................5-170 Mirror

definition............................................................................. 2-37

flat ........................................................................... 3-79, 5-163

diverging ................................................................. See minus

multiple imaging ...........................................................5-166

Fresnel .................................................................................. 2-41 Graded Refractive Index (GRIN) ................................ 2-44 gravitational....................................................................... 2-45

spherical ............................................................................5-170 Mirrors right-angle .......................................................................5-167

meniscus ...............................................................2-53, 6-220 minus .................................................................................... 2-39

N

planoconcave .................................................................... 2-53 planoconvex ........................................................2-53, 6-272 plus ........................................................................................ 2-39 positive .........................................................................See plus principle of operation.................................................... 2-40 size lens ............................................................................ 6-269 spherical .............................................................................. 2-53 thick ................................................................................... 6-213 thickness .......................................................................... 6-213 thin ........................................................................................ 2-48 Lens system .......................................................................... 6-247 lenses in contact ............................................. 6-247, 6-248

Neutralization .......................................................................4-131 Newtonian imaging formula ..........................................4-117 Nodal point in a lens .............................................................................6-227 in a mirror ......................................................... 5-174, 6-230 in an SSRI............................................................................ 1-20 location, in relation to the principal ....... 6-227, 6-290 Nominal lens formula .......................................................6-287 Numerical Aperture ...........................................................7-313 in the human eye ..........................................................7-314

Lens-maker’s formula.................................... 2-56, 2-69, 439 light-forming cone ............................................................ 7-294

O

Listing method ........................................ 4-140, 4-141, 6-242 Object

M Magnification angular ..................................................................... 3-80, 3-96 axial .........................................................................3-80, 8-421 in lens imaging .............................................................. 4-115 in mirror Imaging ......................................................... 5-165 in mirror imaging, using vergence ........................ 5-165 lateral......................................................................3-79, 3-108 longitudinal........................................................................ 3-80 negative............................................................................... 3-80 overall.................................................................. 6-257, 6-292 positive ................................................................................ 3-80 transverse .............................................................3-79, 3-108 Magnifying lens .................................................................. 7-337 magnifying power ............................................................. 443 Marginal ray ........................................................... 7-316, 7-388 Microscope ........................................................................... 7-370

452

definition ............................................................................ 3-71 extended ................................................................. 3-72, 3-76 height ................................................................................... 3-78 intermediate ..................................................... 6-257, 6-292 location .............................................................................4-110 point ..................................................................................... 3-72 real ................................. 3-73, 3-89, 3-106, 4-133, 5-184 vergence ...........................................................................4-111 virtual .............. 3-73, 3-89, 3-100, 3-106, 4-131, 4-143 Object space ............................................................. 3-72, 3-106 Objective lens ........................................................ 6-254, 7-328 Oblique incidence method .............................................4-140 Optical axis .............................................................................. 3-77 in lenses .............................................................................. 2-40 in reflecting surfaces....................................................5-168 in refracting surfaces ........................................................1-9 secondary ....................................................1-9, 2-52, 5-168 Optical conjugates

APPENDIX

definition............................................................................. 3-71

in a biconvex lens ............................ 6-225, 6-244, 6-273

in a lens ............................................................................ 4-110

in a meniscus lens .........................................................6-236

principal planes ............................................................. 6-224

in a thick lens ..................................................................6-225

pupils and aperture stop........................................... 7-305

location (formula) .......................................... 6-224, 6-235

Optical density

optical conjugates ........................... 6-224, 6-240, 6-258

in relation to refractive index ....................................... 1-3

ray tracing ......................................................... 6-239, 6-291

Optical infinity ........................................... 1-13, 4-109, 4-122

ray tracing ........................................................................6-239

image with concave mirror ...................................... 5-194

ray tracing ........................................................................6-291

image with lens ............................................................. 4-141

Principal point

Optical invariant ................................................................. 4-138

in a mirror ........................................................................6-230

Optical path length

in a thick lens ..................................................................6-221

definition............................................................................... 1-3 Optical power

in an SSRI..........................................................................6-229 Principal ray............................................................ 7-315, 7-388

air to the left...................................................................... 1-14 and focal length ............................................................... 1-16 and lens geometry .......................................................... 2-61 and lens orientation ....................................................... 2-58 and surrounding medium............................................ 2-59

Principle of Least Optical Path ...... See Principle of Least Time Principle of Least Time and Lenses ......................................................................... 2-42 Pupil

formula .......................................................... 1-14, 1-34, 438

identification and location ........................................7-364

in a mirror ........................................................................ 5-174

Pupil matching .....................................................................7-306

in a thick Lens ........................................ 6-213, 6-287, 439 in an SSRI ................................................................ 1-14, 1-34

R

in lenses in contact ...................................................... 6-247 of the human eye .............................................................. 442 summation ...................................................................... 6-215 units ............................................................................ 1-14, 438 Optical thickness ..................................................................... 1-3

P

Radius of curvature in lenses ........................................ 2-48, 2-54, 2-69, 6-214 in reflecting surfaces....................................................5-169 in refracting surfaces ............................................ 1-9, 1-34 infinite .................................................................................. 2-54 negative ........................................ 1-10, 2-54, 2-56, 5-169 positive ............................................. 1-9, 2-54, 2-56, 5-169

Pantoscopic tilt ................................................................... 8-419 Parabola ................................................................... 5-168, 8-396 Parabolic surface ................................................................... 2-43

sign conventions ................................................................1-9 units ...................................................................................... 1-10 Ray

Paraxial approximation .................................................... 8-397

focal .......................... 4-124, 4-135, 4-136, 5-181, 6-240

Pinhole .................................................................................... 7-362

in a thick lens ..................................................................6-240

Plane of incidence

marginal ............................................... 7-316, 7-370, 7-388

in reflection .......................................................................... 1-2

meridional ........................................................................8-418

in refraction ......................................................................... 1-5

nodal ........................ 4-124, 4-135, 4-136, 5-181, 6-240

Port

principal ............................................... 7-315, 7-370, 7-388

identification and location ....................................... 7-366

random ..............................................................................4-140

Principal planes ................................................................... 6-221

sagittal ...............................................................................8-418

ball lens ............................................................................ 6-263

tangential .........................................................................8-418

crossed................................... 6-226, 6-250, 6-265, 6-268

wild......................................................................................4-140

curved ................................................................. 6-223, 6-264

wild, in a thick lens .......................................................6-242

453

Ray height ............................................................................. 7-296 Ray spot diagrams

in other media .....................................................................1-3 in vacuum ..............................................................................1-2

in spherical aberration ............................................... 8-408 Ray tracing

Spherical aberration ..........................................................8-405 and lens shape ...............................................................8-407

construction rays ............................................ 4-124, 4-135

longitudinal ...................................................... 8-406, 8-410

principal planes ............................................................. 6-239

management ...................................................................8-410

rules, convex mirror..................................................... 5-181

negative ............................................................................8-406

rules, minus lens ............................... 4-131, 4-161, 5-210

positive ..............................................................................8-406

rules, plus lens ................................... 4-124, 4-160, 5-209

transverse .......................................................... 8-407, 8-410

rules, plus lens, virtual object .................................. 4-144

Spherical reflecting surface ............................................5-168

rules, SSRI .............................................................1-36, 4-135

SSRI ....................... See Single Spherical Refracting Surface

rules, thick lens ................................................ 6-238, 6-240

Stigmatic imaging .................................................. 2-42, 8-393

Rayleigh criterion ............................................................... 7-353

Stray light ................................................................ 7-344, 7-362

Reduced system ................................................................. 6-255 Reflecting power ............... See Optical power: in a mirror

T

Reflecting surface ................................................................... 1-2 Reflection law ........................................................................................... 1-2 Law .......................................................................................... 436 Refracting surface ................................................................... 1-4 flat .......................................................................................... 3-94 Refraction definition............................................................................... 1-4 external .................................................................................. 1-6 in water .................................................................................. 1-4 internal ................................................................................... 1-6 Law ..................................................................... 1-5, 1-34, 436

Telecentric system ..............................................................7-307 Telescope astronomical (Kepler) type ........................................6-254 Galilei-type ....................................................... 6-254, 7-371 Hubble Space Telescope................................ 2-47, 8-405 principle of operation .................................................6-254 Thickness of lens ................................................................................6-213 of lens, critical.................................................................6-215 Total Internal Reflection........................................................1-7

Refractive index definition..................................................................... 1-3, 435

U

water ....................................................................................... 1-3 Refractive power ....................................... See Optical power Resolution ............................................................................. 7-353

Uniaxial system ....................................................................6-247

and f/# .............................................................................. 7-354

V

limit .................................................................................... 7-353 minimum angle of ....................................................... 7-353 Resolving power ................................................................. 7-354

Veiling glare ..........................................................................7-344 Vergence

S Semi-diameter..................................................................... 7-296 Single Spherical Refracting Interface definition............................................................................... 1-9 Snell’s Law .................................................. See Refraction: Law Speed of light

454

and lenses .......................................................................... 3-88 and mirrors ......................................................................5-176 and propagation ............................................................. 3-85 and reflection.................................................................... 3-87 definition .................................................... 3-82, 3-107, 437 downstream .......................................................... 3-85, 3-93 formula ...................................................................... 3-82, 437 of image ............................................... 4-111, 4-112, 5-179

APPENDIX

of image, real .................................................................... 3-90

Vertex power

of image, virtual ............................................................... 3-90

back ....................................................................................6-231

of object ............................................... 4-111, 4-112, 5-179

front ....................................................................................6-231

of object, virtual ............................................................... 3-73

Vignetting ..............................................................................7-343

reduced .......................................................3-83, 3-108, 438 units ............................................................................ 3-82, 437

W

upstream ................................................................. 3-85, 3-92 Vertex in a thick lens ................................................................. 6-225 in mirrors ......................................................................... 5-168 in reflecting surfaces ................................................... 5-169

Warp.........................................................................................8-418 Wavelength in a medium ........................................................................ 435

in refracting surfaces ....................................................... 1-9

455

George Asimellis, PhD, serves as Associate Professor of Optics and Research Director at the Kentucky College of Optometry, Pikeville, Kentucky, which he joined in 2015 as Founding Faculty. He oversees development and coordination of the Geometric Optics and Vision Science courses and development of the Laser Surgical Procedures course. In the past, he served as head of Research at LaserVision.gr Institute, Athens, Greece, and as faculty in: the Physics Department, Aristotle University, Greece; Medical School, Democritus University, Greece; and the Electrical Engineering Department, George Mason University, Virginia. His doctorate research involved advanced optical signal processing and pattern recognition techniques (PhD, Tufts University, Massachusetts), and optical coherence tomography (Fellowship, Harvard University, Massachusetts). He then worked on research and development of optoelectronic devices in a number in research centers in the USA. He has authored more than 75 peer-reviewed research publications, 8 scholarly books on optics and optical imaging, and a large number of presentations at international conferences and meetings. He is on the Editorial Board of eight peer-reviewed journals, including the Journal of Refractive Surgery, for which he serves as Associate Editor. He received the 2017 Emerging Vision Scientist Award by the National Alliance for Eye and Vision Research (NAEVR). His research interests include optoelectronic devices, anterior-segment (corneal and epithelial) imaging, keratoconus screening, ocular optics, and ophthalmological lasers. His recent contributions involve publications in clinical in vivo epithelial imaging and corneal cross-linking interventions.