Introduction to Visual Optics: A Light Approach [1 ed.] 0323875343, 9780323875349, 9780323875370, 9780323875387

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Introduction to Visual Optics

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Introduction to Visual Optics A Light Approach

Samantha Strong, PhD, BSc (Hons) Lecturer, Optometry Vision Sciences Aston University Birmingham, UK

Copyright © 2024 by Elsevier Ltd. All rights reserved. No part of this publication may be reproduced or transmitted in any form or by any means, electronic or mechanical, including photocopying, recording, or any information storage and retrieval system, without permission in writing from the publisher. Details on how to seek permission, further information about the Publisher’s permissions policies and our arrangements with organizations such as the Copyright Clearance Center and the Copyright icensing gency, can be found at our website www.elsevier.compermissions his book and the individual contributions contained in it are protected under copyright by the Publisher other than as may be noted herein.

Notice Practitioners and researchers must always rely on their own eperience and knowledge in evaluating and using any information, methods, compounds or eperiments described herein. ecause of rapid advances in the medical sciences, in particular, independent verication of diagnoses and drug dosages should be made. o the fullest etent of the law, no responsibility is assumed by lsevier, authors, editors or contributors for any inury andor damage to persons or property as a matter of products liability, negligence or otherwise, or from any use or operation of any methods, products, instructions or ideas contained in the material herein. ISBN: 90249

Senior Content Development Manager: omodatta oy Choudhury Senior Content Strategist: ayla olfe Senior Content Development Specialist: Priyadarshini Pandey Publishing Services Manager: hereen ameel Project Manager: ishnu . ii Design Direction: ridget oette Printed in nited ingdom ast digit is the print number         

To Howard, for everything; To Decaf and Alex, for believing in me; To my students, for supporting me with enthusiasm and a willingness to learn; To my cats Mun and ooie, for their love and company every step of the way; And to tea, because this boo would not exist without tea

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Acknowledgements Writing a textbook involves the support of several people, so I’d like to take a moment to thank everyone who helped me with this endeavour. Firstly I’d like to thank Professor Leon avies ston niversity for handing me the optis module that’s fullled my love of teahing ever sine I oined ston, and I’d like to thank r. ob ubbidge  ollege and r. ebekka eitmar niversity of udderseld for helping me get to grips with the pratial side of optis when I was learning. I’d also like to thank r. oward ollins ston ni versity and r. laine allam ston niversity for their suggestions on the ontent of the book and for their es sential help with my understanding of the linial side of optis.

I should also thank ayla Wolfe at lsevier for helping me put forward this proposal and for advising and support ing me along the way. I’d also like to thank my family and friends for putting up with me talking about this for suh a long time, but speial thanks go to my sister atherine eaf  trong for her motivating messages, endless stream of loveheart emo is, and suh a passionate investment in me nishing the book that I felt it would have let her down not to do it Finally, I’d like to express an extra set of humongous thanks to r. oward ollins for being my kind and pa tient proofreader, for believing in me even when I struggled to believe in myself and, importantly, for helping me to reognise when I went a bit ‘too orkshire for a textbook’.

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Preface is book has been an idea I’ve been nurturing since I rst started trying to learn about optics myself. I remember sitting at my desk, surrounded by black-and-white textbooks that I considered too dicult to even attempt to read and deciding in that moment that probably I ust really hated science, or possibly that I ust didn’t have the right type of brain or capacity or intelligence for understanding physics this probably sounds quite dramatic but if you’re a student you may empathise a little with me here. owever, through necessity, I had to persevere, and once I managed to break through the initial o-putting nature of the content, I began to nd the content fascinating, inspiring and all-together very appealing. is obviously reuired a re-think of my capacity for science, and I realised that actually, uite contrary to hating science, I absolutely loved it It turns out that I ust was very easily de-motivated as a student, and with not much of a science background I studied art, graphic design, philosophy and psychology at -level, I found it dicult to understand content that was written with the assumption that I had prior knowledge that I really didn’t possess – which, when you think about it, is fair enough really. hen I rst started teaching optics, it seemed sad to think that some students like myself  might be put o by a subect because it feels too dicult or because it feels too boring, and so the rst thing I did was scour the web for resources that made optics sound fun or used lots of colourful

diagrams to help explain the concepts. It didn’t take me long to realise that nothing available uite seemed to live up to what I wanted, and so I started to design -sied handouts for my students I dubbed them ‘mini-guides’, and the feedback was far better than I could ever have hoped for tudents told me they stuck them up in their accommodation to help them revise, and said they made it look ‘less frightening’, which eventually inspired me to write this little book. e beginning of the book covers fundamental principles of light and geometrical optics chapters –, before moving on to discuss physical optics chapter , clinical applications chapters – and then attempting to inspire a love of science by providing you with an opportunity to do some experiments at home chapters –. I hope that this book will serve as a way of showing all students that optics doesn’t have to be extremely complicated and can instead be interesting possibly even fun. o that end, I’ve attempted to make this book a light-hearted, simplied introduction to the topic in the hopes that it will inspire you to do some further reading of the more advanced content once you realise how lovely optics can be. o, without further ado, I wish you a lovely time reading my book, and if you have any feedback for me, please do be encouraged to reach out and share your thoughts. Samantha Strong, PhD, BSc (Hons)

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Contents Section 1: Geometric and Basic Optics, 1  1.  2.  3.  4.  .  .  7.  .  9.

Basics of Light and Colour, 3 Vergence and Refraction, 11 Thin Lenses, 27 Thick Lenses, 39 The Reduced e and herical and Clindrical Lenses, 47 Reection, 7 Ra Tracing,  isersion and Chroatic erration, 77 riss, 3

Section 2: Physical Optics, 95 1. uerosition, nterference and iraction, 97

Section 3: Clinical Applications, 10 11. ocietr, 19 12. hotoetr, 113 13. tical nstruents and Lo Vision ids, 123

14. olarisation, 131 1. aging the e and easuring Refractie rror, 139 1. aefront errations and datie tics, 149 17. tical Coherence Toograh, 1

Section : periments to o at ome, 11 1. 19. 2. 21. 22. 23.

Create our n Caera scura, 13 Create a Blue k at oe, 17 Create a ris, 171 easure the eed of Light, 17 Create a ‘Cornea’, 179 itchen Thin il nterference, 13

Section 5: estion Ansers, 1 nsers to ractice uestions, 19 nsers to Test our noledge uestions, 21

nde, 209

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Video Table of Contents Video 1.1: Light is comprised of dierent wavelengths, and these wavelengths can combine to produce dierent colours of light. This video shows an example of additive colours, demonstrating that red, green and blue light mae white light, whereas dierent combinations of two of the lights e.g. red and blue or red and green produce dierent colours e.g. magenta or ellow. This same principle also explains how shadows can be colourful, which is demonstrated towards the end of the video. Video .1: This video teaches ou about refraction and images, as seen through curved refractive surfaces in this case, a round glass of water. eal and virtual images are considered with reference to the obect’s distance relative to the surface, and we also consider how the refractive index and curvature of the glass can mae obects appear to loo dierent when placed in water. Video 1.1: iraction is a process b which light ‘bends’ round the corners of apertures or obstacles, given the right circumstances, and the amount of bending is determined b what’s called the diraction angle. ruciall, if diracting through an aperture or slit, the diraction angle will depend on how large the wavelength is relative to the sie of the slit. n general, large wavelengths e.g. red will have a larger diraction angle than shorter e.g. blue wavelengths when passing through the same slit. This video utilises a diraction grating comprising man tin slits to demonstrate this eect. Video 1.1: This video shows ou an example of a completed ‘blue s’ experiment, as described in chapter 1. nitiall the torchlight appears

white when shone through the water, but when a small amount of mil is added to the water, the particles in the mil ‘scatter’ the short blue wavelengths out of the torchlight, maing it appear ellow-orange. f ou complete this experiment at home and obtain dierent results, please see the troubleshooting table in chapter 1 for help. Video .1: This video shows successful completion of the ‘create a prism’ experiment, using water, a mirror and a torch to demonstrate dispersion at home. f ou complete this experiment at home and obtain dierent results, please see the troubleshooting table in chapter  for help. Video 1.1: This video shows clearl how to complete the ‘speed of light’ experiment b melting chocolate in a microwave and measuring the distance between the melting ‘hotspots’ that form. The video contains several advice to help ou complete it at home if ou’d lie to. f ou complete this experiment at home and obtain dierent results, please see the troubleshooting table in chapter 1 for help. Video .1: This video shows how to prepare materials and complete the ‘create a cornea’ experiment from the textboo. This experiment should demonstrate how refractive index changes and curvature can impact the appearance of the nal image. creenshots from this video can be seen in the textboo. f ou complete this experiment at home and obtain dierent results, please see the troubleshooting table in chapter  for help.

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Introduction to Visual Optics

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SECTION

1

Geometric and Basic Optics 1. Basics of Light and Colour, 3

. Reection, 

2. Vergence and Refraction, 11

. Ra Tracing, 

3. Thin Lenses, 2

. ispersion and Chromatic erration, 

. Thic Lenses, 3 . risms, 3 . The Reduced e and pherical and Clindrical Lenses, 

1

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1 Basics of Light and Colour C H A P T ER O U TL INE Introduction

Colours

What Is Light?

Absence of Light

Ho oes Light Interact With Obects?

Test our noledge

O  E C TI E After working through this chapter, you should be able to: plain hat light is in relation to the aeparticle dualit plain hat the aelength of light means in terms of appearance and energ

plain ho light interacts ith oects plain ho e perceie colours

Introduction

whether light could be classed as ‘particles’ or as ‘waves’ went on for centuries before scientists nally agreed that light must possess something called wave-particle duality, mean ing it ehibits properties of both particles and waves o ., and this is particularly relevant when we start to think about the energy associated with the electromagnetic spectrum It can be slightly biarre to think of ‘light’ as a type of energy, so let’s discuss that in more detail. e energy pack ets uanta associated with the electromagnetic spectrum are called photons, and these are considered the basic unit of all light. e amount of energy emitted per photon is dictated by its wavelength. Wavelength is a term to describe the distance between two eual points on a wave. In ig. ., you can see that on the shorter wavelength, if I take the distance between two peaks the very top of a wave, then it should be identical to taking the distance between two troughs the very bottom of a wave. ypically we measure distances from peak to peak because it is easier to identify the peak of a wave in a diagram than somewhere halfway up. Also, because it is a measure of distance, we need to use distancerelated units, for eample, nanometres, millimetres, centimetres, or metres. Importantly, the wavelength of a light source is also re lated to its frequency. reuency is dened as the number of complete cycles of the wave that pass any given point in  second. o, for eample, if we shine a laser light at a wall and measure how many full cycles pass a point halfway along the beam within  second, we can calculate the wavelength

If you’re anything like me, then at some point in your life you will have had one of those moments when you have a deep, philosophical thought about what light is, or what a shadow is, or how colours are produced, or why things look distorted in water. Alternatively, if you’ve never thought in too much detail about this – have a think about it now. What even is ‘light’ anyway is chapter will seek to answer these uestions by re viewing some key physical principles and dening some very important terms. o this end, this chapter will lay the foun dation for the rest of the book, so please make sure to obtain a solid understanding of this content before moving on.

What Is Light? When we talk about ‘light’ as something that allows us to see obects that eist in the real world, we are actually talk ing specically about something referred to as ‘visible light’. e special feature of visible light is that it’s detectable by the sensory receptors in the human eye, which is how we use light to help us to see. It can originate from natural e.g. from the sun, or articial e.g. from a lamp sources, but in all cases, visible light will illuminate light up obects, al lowing us to perceive their eistence. owever, when we start to think more deeply about what light is, or what it’s composed of, things start to be come a little more complicated. or eample, research into

3

4 S EC TI ON 1

•BOX 1.1

eoetric and asic Otics

What Is the Dierence Between a Particle and a Wave?

Let’s start by using a football as an example particle. If you have a football sitting stationary on the ground in front of you, then the particle is ‘at rest’, but if you approach the football at speed and kick it like Harry Kane, then it ill hopefully go ying off into the distance. his happens because you have transferred energy to the football – in this case, kinetic movementrelated energy. his energy allos the football to travel the distance reuired to land safely beteen to goalposts or a neighbour’s garden. Hoever, aves are different. or this example, let’s think of dropping a stone into a still pond. e ould expect the kinetic

energy from the stone to move the ater and produce a ripple, emanating outards and taking the energy ith it. Hoever, even in a single ripple, the energy is contained ithin the full ‘ave’, meaning that ith aves the energy is much more spread out ig. .. he key difference, then, is that particles ill collide ith each other and change direction, hereas aves ill pass through one another imagine to footballs colliding relative to to ripples in a pond. his idea ill be covered in more detail in later chapters of the book.

Pa rti c

Wave l

• Fig. B1.1

iagram shoing the difference in ho energy transfers in a particle relative to a ave. lue arros indicate the direction of the energy.

λ1

Shorter wavelength

complete cycles should pass through a particular point within a second, specifying a higher freuency. c Equation 1.1 Equation 1.1 (explained)

λ1

λ2

• Fig. 1.1

Longer wavelength

wavelength 

speed of light frequency of light

et’s consider the relationship between wavelength and energy in more detail. If you look at ig. ., you can eam ine the entirety of the electromagnetic spectrum, with visi ble light sitting roughly in the middle. Importantly, in the electromagnetic spectrum, as wave length increases to the right of ig. ., energy will decrease. is is because wavelength and energy are inversely propor tional. In uation . below, energy  is eual to lanck’s constant h multiplied by the speed of light c, all divided by the wavelength of the light source . is shows that as the value for wavelength increases, the value for energy gets smaller, and vice versa.

o example aves, shoing identication of avelength.

Equation 1.

 lambda by using uation . to divide speed of light c by freuency n nu o .. is euation also high lights that as wavelength decreases, freuency should increase, which makes sense because if the distance between two corresponding points of the wave is small, more

Equation 1. (explained)

E energy

hc

planck's constant speed of light wavellength of light

If we apply this principle to ig. ., we can see that cosmic rays have more energy than radio waves. As a top tip,

CHAPTER 1

•BOX 1.2

asics of Light and Colour

The Speed of ight

he speed of light, denoted in euations by the letter c hich stands for ‘constant’, is a knon speed, recorded to be ,, ms. Hoever, it is important to note that this is the speed that light is recorded to travel in a vacuum, such as that found in outer space. hen light is on arth in the atmosphere, it travels almost as fast as it ould in a vacuum, but if the light comes into contact ith any obectmaterial, then it can be sloed don if it is made to change direction slightly. ne easy ay to think about this is that hen light is in a vacuum, there are no

Light travelling through a vacuum

electrons or particles to get in the ay – it’s ust smooth sailing. Hoever, in the arth’s atmosphere, or in ater, the light ill need to take very minuscule detours every time it comes into contact ith an atom – hich inevitably slos it don ig. .. or example, if light passes through a diamond a highdensity obect, it ill be sloed don to ,, ms, hich is still a great deal faster than you or I could ever hope to run, but it’s been reduced to less than half of the speed of light in a vacuum

Light travelling through the atmosphere

• Fig. B1.

iagram shoing difference in ho light blue arro travels unimpeded in a vacuum left, relative to small detours in the arth’s atmosphere right.

Shorter wavelength higher frequency higher energy

Cosmic rays

• Fig. 1.

X-rays

Longer wavelength lower frequency lower energy

Visible UV

Infrared

Microwaves

adiowaves

chematic of electromagnetic spectrum. s avelength increases, energy decreases.

one way to remember this is to apply it to eisting knowledge. or eample, we know that rays can damage tissue and that  ultraviolet light can burn the skin over reasonably short eposure times. is suggests that they possess higher energy levels than visible light or radio waves for eample because they do not cause damage to tissue in the same way.

Ho oes Light Interact With Obects? We have already discussed that light allows us to see obects, but in order to understand how that happens, we need to rst think about the way that light can interact with obects. In simple terms, when light falls on an obect, one of three

things can happen  the light is transmitted and passed through the material,  the light is asored and stopped by the material, or  the light is reected backwards by the material ig. .. et’s consider an eample obect, like window glass. Window glass is transparent, meaning it’s seethrough, so what do you think happens to the light in this instance Is it transmitted, absorbed, or reected I’ll bet that you’ve gone for ‘transmitted’, suggesting that light travels through the glass in order for you to see through it, and in a way this is correct. owever, in most circumstanc es and with most obects, light eperiences a combination of all three possible interactions.

5

6 S EC TI ON 1

eoetric and asic Otics

Some visible light is reflected

Light rays Transmit

Some visible light is transmitted

Visible light

Absorb UV light

Reflect

Material

• Fig. 1.

imple diagram shoing ho light blue arro interacts ith

materials.

or eample, when we look out of a car window, we can see through it, clearly indicating that some of the light is be ing transmitted through however, we can usually also see a reection of the inside of the car, meaning that some light is being reected back towards us as well. imilarly, the reason we can’t get sunburnt through a window is because the harm ful  rays are absorbed by the glass and can’t reach us is means the correct answer to what happens to light when it reaches window glass would be that some light is transmitted, some is reected and some is absorbed. A nice, easytoremember eample of this is a sunglasses lens ig. ., as  rays are absorbed to prevent them from reaching our eyes, some light is reected to contrib ute to reducing the brightness, and some light is transmit ted so we can see through them.

Colours If I ask you to think about a particular colour, let’s say blue, you can probably think of lots of eamples of bluecoloured Additive colours (as with light mixing)

• Fig. 1.

UV light is absorbed

• Fig. 1. imple diagram shoing ho light blue arro interacts ith materials.

things, for eample, the sky, forgetmenot owers, sapphires. . . . owever, if I ask you to tell me what ‘blue’ is, that’s where things start to get a little tricky. o start with, we need to understand that natural, white light e.g. from the sun contains every wavelength within the visible light spectrum, from  to  nm see ig. .. ese wavelengths mi together as waves can, so when they reach the receptor cells in our eyes, they are processed at the same time. is means that colours, as we eperience them, are simply combinations of light waves from the visible light spectrum, and the perceived hue is determined by the wavelength. o, for eample, a short  nm wavelength will appear blue, whereas a long  nm wavelength of light will appear red. When light waves combine together, ‘additive’ colours are created, which can be uite counterintuitive if you have any art eperience. In ig. ., I have sketched out the dif ferences between subtractive colour miing like with paint and additive colour miing like with light. ou can see that with subtractive colour miing e.g. with paint, the primary colours are red, yellow and blue, and they mi to create predictable secondary colours for eample, red and blue make purple, blue and yellow make green and yellow and red make orange. When you mi all of them together, it produces black. Subtractive colours (as with paint mixing)

xamples of colour mixing, shoing the difference beteen additive mixing left and subtractive mixing right.

CHAPTER 1

asics of Light and Colour

Absence of Light

• Fig. 1.

Illustration depicting a yello sunoer.

owever, with additive colour miing, the primary colours are red, green and blue, which correspond to the long red, middle green and short blue wavelengths of light that make up the visible light spectrum. When these colours mi together, they produce di¤erent colours from what might be epected for eample, red and blue make magenta pink, blue and green make cyan light blue and green and red make yellow. Importantly, when all of these colours mi together, they produce white. e reason this type of miing is called ‘additive’ colour is that when the wavelengths of light combine together to produce the new colour, all wavelengths are still detectable by our eyes – therefore suggesting that the perceived colour is the result of the combination of the individual wavelengths. When we view an obect, such as that shown in ig. ., providing we have typical colour vision, we are able to easily detect the colour, and we can describe the obect in the gure as a yellow sunower. ut how do we perceive it as being yellow if the light is white When we look at any obect, the colour we perceive is a result of the reecting wavelengths. As depicted in ig. ., the white light containing all possible wavelengths reaches the obect, and depending on the obect’s physical structure, some of the wavelengths are absorbed and some are reect ed. e reected wavelengths ‘add’ together to produce the resultant wavelength that the receptor cells in our eyes detect. is is what gives rise to colour perception

White light from the sun reaches the object

White light is made up of lots of different wavelengths (colours)

• Fig. 1.

efore we end the chapter on the basics of light, we need to talk a little bit about what happens when light is stopped in its path. enerally speaking, light travels in straight lines in a homogeneous medium – this means that if light stays in one constant medium, it won’t deviate this is called the rectilinear propagation of light. owever, we should note that if there are density changes within that medium, then it is possible for light to change direction within a single medium see chapter , ‘e Weird and Wonderful World of efraction rough a ingle aterial’, for more informa tion, but for now let’s assume the simple eplanation that if light stays within a single medium, it won’t deviate. or eample, if light travels from a lamp in your house to the physical or digital pages of this book, it will travel in a straight line from the lamp to the book. owever, if you place the book in a bucket of water, the light will deviate slightly as it enters the water, as illustrated in ig. . is is important because it means that if light is not transmitted through a material, then it will be either absorbed or reected, meaning that the light will be stopped in some way by that material. is is how shadows are produced. e type of shadow produced depends on the light source and the distance of the obect from the source. If, for e ample, we have a point light source, where light originates from a small point, then the straight light rays will be stopped by the obect and produce a welldened shadow called an umbra from the atin for shade, as shown in ig. .A. In clear conditions, the sun acts as a point light source because it’s so far away from us, which is why we get such crisp shadows of ourselves on sunny days. owever, in most articially lit environments e.g. a home or work place, there are etended light sources ig. ., which produce two types of shadow – an umbra and a penumbra from the atin for almost shade. With an etended light source, the light is emanating from a source that covers a wider area than the point light source. is means that light from one side of the source will fall on the obect from one direction and light from the other side of the source will fall on the obect from a slightly di¤erent direction. is means that there will be an

Some of the And some wavelengths The reflected wavelengths get are reflected by the wavelengths are what absorbed by the object object we perceive as colour

Illustration explaining ho obects appear to have a ‘colour’.

7

8 S EC TI ON 1

eoetric and asic Otics

Homogenous

Heterogenous Air Water

Air Air

• Fig. 1.

Illustration shoing that light blue arro ill travel in a straight line in one medium homoge neous, left but ill bend if moving from one medium to another heterogeneous, right.

A

Point light source

B

Extended light source

Side view

View on screen

Umbra

Floating tennis ball

Screen

Side view

View on screen

Umbra

Floating tennis ball

Penumbra

Screen

•

Fig. 1. Illustration shoing ho point light sources  produce crisp, tidy shados, hilst extended light sources  produce to types of shados – an umbra and a penumbra.

area where none of the light can reach, because the obect obstructs the light from all parts of the source, and there will be an area where only some of the available light is stopped by the obect. is produces a dark area where no light from the source can reach umra and an area with a lighter, less dened shadow penumra. ypically, the more etended the light source is, the less dened the shadow will be, be cause more light will have the opportunity to travel past the obect. Another thing to think about is that shadows are almost always black or grey, because they represent the blockage of white or whiteish light, meaning that the shadow repre sents an absence of white light. owever, it is also absolutely possible to have coloured shadows in the right lighting conditions. or eample, we can replace our single point light source with three light sources, and instead of using white light, we can make sure each one is a di¤erent colour representing each wavelength – red, green and blue ig. .. In typical conditions, these three light sources ‘add’ together to produce white light see ig. . for

Side view (dark room)

View on screen (3)

(1)

Coloured light sources

• Fig. 1.1

(2)

Floating marble

No green No blue No red

Screen

Illustration shoing ho coloured light sources can produce coloured shados if one or more is blocked by an obect. hen light from all three sources combines, it produces hite light , hereas hen light from all three sources is blocked, it produces a black shado . n the screen, there are sections here light from all three sources is reaching the screen hite and areas here one of the sources is blocked, leaving a shado that exhibits the additive colour of the other to sources .

CHAPTER 1

a reminder of this, but in our scenario, there is a small marble in the way is means that there will be parts of the screen where one of the lights is blocked, which in turn means that the resultant light falling in the shadow of that blocked light will be the sum of the two remaining

asics of Light and Colour

lights. o, for eample, if the green light is blocked see ig. ., no green, then the red and the blue will com bine to produce a magenta shadow. ou can see an eample of this in the video content related to this book through the lsevier website.

Test Your Knowledge ry the following uestions to determine whether you need to review any sections again. All answers are available in the back of the book. .1.1 ow do we measure ‘wavelength’ .1. oes ultraviolet light have higher or lower energy than visible light .1. oes ultraviolet light have a larger or smaller wavelength than visible light

.1. ink about a regular table. o you think light approaching the table is absorbed, reected, transmitted or a combination of a few of these .1. What colour would be produced if we combined green and red wavelengths .1. What is a shadow

9

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2 Vergence and efraction C H A  T R   TL I Introduction Key Terms Relating to Light Vergence Vergence Calculations Refraction Calculating Angles Lateral Displacement Spherical Curved Surfaces Power

Focal Length and Focal Points Limiting Spherical Aberration ect and Image Vergence The eird and onderful orld of Refraction Through a Single aterial A Load of Hot Air The Atmosphere References

   C TI VS After working through this chapter, you should be able to: Explain what ergence and refraction are Explain how sign conention wors and be able to determine whether a distance should be positie or negatie nderstand how refractie index and curature of a surface can aect the emergent ergence

e able to calculate use ergence euations to determine where an image will form after refraction

Introduction

distance from the light source increases, the wavefronts will start to get relatively ‘atter’. In terms of describing the path of the light, this can be depicted as a wave (a wiggly line) or a ray (an arrow), as shown in Fig. . is boo uses wavefronts, waves and rays interchangeably in diagrams to illustrate ey points, but please remember that they are all connected.

In chapter  we began to learn about light – what it is, how it interacts with materials and how light allows us to see colour and obects. e current chapter will see to eplore these elements a little further by going over some ey terms and introducing the principles of how light changes direction when it meets certain materials (vergence and refraction).

Key Terms Relating to Light et’s start with some discussion of ‘light’. s discussed in chapter , light possesses wave-particle duality, meaning it ehibits characteristics of both waves and particles. If we tae a point light source, we can imagine that light will be emit ted in all directions, and we can lien this to ripples on a pond after throwing a pebble into the water. owever, with light, these ‘ripples’ are referred to as wavefronts. uch lie our water ripples, these wavefronts remain consistently spaced (unless interfered with – see chapter ), and as the

Vergence s discussed in chapter , when light travels unimpeded in a homogenous medium, it will travel in a straight line (rectilinear propagation of light) however, this is a slight oversimplication. s you can see from Fig. ., any individual light ray will be travelling in a straight line, but if we draw a number of light rays to show how the light is truly leaving the light source, we can see that the light would be travelling in straight lines in all directions (Fig. .). o, if we select a few rays close together, nown as a pencil, the individual rays travel in their respective straight lines, but this means that they are actually travelling further away from each other (Fig. .). 11

12 S EC TI ON 1

eometric and asic ptics

It is important to note that the vergence of the rays is associated with the wavefronts and how they are sampled (Fig. .) – all wavefronts are curved (lie ripples from throwing a stone in a pond), but the closer they’re sampled to the source, the more curved they will appear to be. In Fig. . you can see that the closer the observer is to the light source, the more curved the wavefronts will be when they reach the observer’s pupil. ltimately, then, if the sam pled wavefronts appear to be curved, then divergence or convergence is indicated, whereas wavefronts that appear to be at indicate parallel vergence. is also shows that the further away the wavefronts are from the source, the atter they appear to become, so parallel vergence is most often associated with sources that are a reasonable distance away, referred to in optics as innity. erefore, for eample, light from the sun, which is pretty far away, would be considered to have parallel vergence when it nally reaches the arth. owever, if the obect is closer to the observer than optical innity (, metres), then wavefronts will appear to be di verging, as shown in Figs . and . (and this also nicely highlights how distance is related to vergence, which we will discuss in the net section). ow, ust to mae things slightly more complicated, in optics, as well as thining of light as being emitted from a light source (e.g. a lamp), we can also consider the light to be coming from an object, as we now that in order for us to be able to see an obect, we need light to be reecting o

Wa ve f s nt ro

Point light source

Light ray Light wave

• Fig. 2.1

Diagram showing a light source producing wavefronts in all directions. The direction of the light can be represented as waves or rays. Wavefronts are typically considered to correspond to the peaks of the waves.

e dierence in direction of each ray within a pencil represents the vergence, which can be classied as one of three types (Fig. .) • Divergence (diverging rays) – individual light rays travel away from one another • Parallel (parallel rays) – individual rays remain in line with one another (this reuires a light source at innity) • Convergence (converging rays) – individual light rays travel towards one another

A

B

•Fig. 2.2

Diagram showing a point light source producing light rays (blue) that travel in straight lines in all directions (A) and showing that individual rays are travelling away from each other (B).

Diverging pencil

•Fig. 2.3

Parallel pencil

Converging pencil

The three types of pencil diverging (rays are travelling away from each other) parallel (rays are staying in line with each other) and converging (rays are travelling towards each other).

Vergence and Refraction

CHAPTER 2

13

TABLE Common Refractive Materials and Their 2.1 Respective Refractive Indices (n1

Material

Point light source

Refractive Index (n)

Air

.

Water

.

lastic ()

.

rown glass

.

Point light source

n1.00

n 1.333

n 1.523

Air

Water

Crown glass

• Fig. 2.4

Diagram showing position of observer (eye) relative to light source will alter the relative vergence sampled by the pupil of the eye. n the top image the observer is very close and so the wavefronts reaching the eye are very curved (diverging) whereas in the bottom image the observer is further away and so the wavefronts from the same light source are relatively much atter.

it. elpfully though, ust lie with a light source, the light reected from obects has parallel vergence when the obect is at innity and diverging rays when it’s in proimity, so the principle remains the same.

Vergence Calculations p to this point, we have only considered vergence in de scriptive terms, but it’s also important to be able to calculate values of vergence. ow, we already now that vergence is associated with the distance of wavefront from the light sourceobect, as a greater distance will lead to atter wave fronts. o you’ll be pleased to now that the calculation for obect vergence (the vergence of light reected from an obect) mostly revolves around the distance of the obect from the point of measurement. athematically, vergence is given the symbol , is measured in dioptres (D), and is calculated using uation .. ow ever, in order to calculate vergence we need to now two things the refractive index of the primary medium (in which the obect eists), indicated by n and, as mentioned above, the distance of the object from the point of measurement (usually a surface of some ind, e.g. a lens), represented by l. n l

uation 

L

uation  (explained)

refractive index of medium object vergence  object distance from surface

efractive index is a term that relates to the density of the material that the light is moving through, so the value of the refractive inde indicates how fast (or slow) light will travel through that particular material, relative to how fast it would travel in a vacuum. e refractive inde of air is ., meaning that light will travel (approimately) as fast in air

•Fig. 2. llustration highlighting that the primary refractive inde (refrac tive inde that the obect eists in – here the obect shown is a mug) will change depending on what material it starts in for eample air water and glass.

as it would in a vacuum, whereas the refractive inde of crown glass is ., meaning light would travel . times slower in crown glass compared to a vacuum. Impor tantly, this means that refractive inde of a material will never be below .. ome of the most common eamples of materials and their respective refractive indices are listed for your information in able .  ow, when we say that the letter n in uation . indi cates the refractive inde of the primary medium, we mean the refractive inde of the medium in which the object eists. o, for eample, in Fig. . the refractive inde changes when the mug is placed in water or encased in glass. Finally, as a top tip, if an optics uestion doesn’t state otherwise, it is safe to assume that the obect eists in air (so, if you are not provided with a material or a refractive inde, assume a value of n 5 .). hen considering obect distance (l) we need to remem ber that, in optics, distances are always measured from the surfacelens to the obect and are always measured in metres. is is because dioptres are I (International ystem) nits and so the distance also needs to be the correct I unit in order for the numbers to come out correctly. istance mea sures will also need to be assigned either a positive or nega tive value depending on the location of the obect from the surface. In typical optics convention, we assume that light will always travel from left to right across the page, as shown in Fig. ., and this is called linear sign convention. e reason we do this is because it helps to mae diagrams easy to read, and it helps to mae the mathematical euations wor. ow, assuming a direction of light where light begins on the left and travels to the right, and understanding that distances are always measured from the surfacelens to the

14 S EC TI ON 1

eometric and asic ptics

Direction of light

•Fig. 2.

DEMO QUESTION 2.1—cont’d

n optics light is always assumed to travel from left to right. Surface

ct

Object

om t fr

je ob

h Lig

Distance from surface to light source or object (I)

•Fig. 2.

chematic showing how to measure obect distance from a surface in this eample the distance would be assigned a negative value as the distance is measured against the direction of light accord ing to our linear sign convention. ote that the light is diverging as it leaves the obect.

obect, if a distance is measured with the direction of light (e.g. from left to right) then the distance is assigned a positive value or ‘sign’ (e.g. 1. m). owever, if the distance is measured in the opposite direction of that of light, then the distance is assigned a negative value or ‘sign’ (e.g. 2. m). e very careful not to forget the minus sign when recording these measurements in your calculations, as it’s one of the most liely areas for errors to occur In the eample image (Fig. .) the obect is left of the surface, which means that we measure from the surface on the right to the obect on the left, a direction that is opposite to the direction of light (Fig .). us the distance (l) would be recorded as negative. ssuming a refractive inde of . then, uation . tells us that any obect distance which is negative should produce a negative vergence value, and any positive distance should pro duce a positive vergence value. Importantly, negative values for vergence (e.g. 2. ) describe diverging rays, and positive values (e.g. 1. ) describe converging rays. DEMO QUESTION 2.1 f an obect is placed  cm in front of a surface what is the obect’s vergence at the point where light from the obect meets the surface tep  Determine what we need to calculate obect vergence L tep  Dene variables l 5 2. m (the question stated the object was 10 cm ‘in front of’, meaning it’s to the left of, the surface. Therefore, it will have a negative distance and we need to convert to metres n 5 . (nothing is secicall mentioned so we assume the object is in air

tep  Determine necessary euation  5 nl (quation .1 tep  alculate  5 nl  5 . .  5 2.D (don’t forget the 6 sign and the units)

Practice Questions 2.1: 2.1.1 f an obect is placed  cm in front of a surface what is the obect’s vergence at the point where light from the obect meets the surface 2.1.2 f an obect is placed  cm in front of a surface what is the obect’s vergence at the point where light from the obect meets the surface

owever, it is important to note that as we always as sume light is travelling from left to right, and that light will always originate from a light source or an obect, this means that our sources and obects will always be pictured on the left in any diagrams (which will be important to remember in future chapters). is means that the source or obect should always have a negative (or innite) distance and will therefore always have diverging or parallel light rays – so how do we produce convergence e answer is that when a light ray travels through a surface (e.g. a lens or a glass bloc or water), it’s path may be altered by either a change in re fractive inde or a change in power (see ‘pherical urved urfaces’) e view of the obect after the rays change direc tion is called the image, and an image can have converging, diverging or parallel rays. e distance of the image from the surface is denoted as l9 (pronounced little el prime or little el dash). In Fig. ., the image is on the right of the surface, which gives it a positive distance and indicates the light rays are converging to produce it. ow, although light usually travels in straight lines, it can change direction if it travels into a dierent medium. is change in direction is called refraction and is one of the core underlying principles of visual optics.

Refraction If a light ray changes direction, this is called refraction. is can occur if there is a change in velocity of the light as it travels from one material (primary medium, e.g. air) to another (secondary medium, e.g. glass) where the refractive indices of the materials are dierent.  reallife eample of this is when you try to loo at your arms or legs under water when you’re in a swimming pool – water has a dier ent refractive inde to the air and so the image of your limbs loos distorted. is is refraction in action ou can try it at home by placing a pencil in a glass of water (or by watching the demo video on the associated lsevier web content) – does it loo distorted in any way lease also see chapter 

Vergence and Refraction

CHAPTER 2

15

Surface Object

om t fr

ct

Ligh

je

ob

t pro

h Lig

duc

ing

ima

ge

Image

Distance from surface to light source or object (l)

Distance from surface to image (l’)

• Fig. 2.

chematic showing a surface can change the direction of light to alter the vergence. n this eample it has produced converging rays and a minied upsidedown image (magnication will be dis cussed in more detail in chapter ).

Angle of deviation

h Can e ee Ice in ater

y t ra

n

e rg

e Em

Angle of incidence

gh t li

Normal

nt lig h

tr ay

Angle of refraction

Glass

ide

As a general rule when light travels within a single medium it trav els in a straight line providing the refractive inde doesn’t change (see section on ‘The Weird and Wonderful World of efraction Through a ingle aterial’ for more discussion on this). imilarly if light travelled from one medium to another of an identical refractive inde no refraction (bending of the light) would occur. ow in the case of ice cubes in a glass of water ice is literally made of water and yet it refracts light differently – this suggests it probably has a different refractive inde somehow (which indeed it does). This is because refractive inde is determined by both density and tem perature. ce is less dense than water (we know this because ice oats in water – a clear indication of lower density) and usually a lot colder which means its refractive inde is lower sitting at approimately . (relative to water’s .). This means light can technically travel slightly faster through ice than it can through water – which will affect the re fraction that we see. et time you have an ice cube in a drink see if you can link it back to what you’ve learned here

Inc

•BOX 2.1

• Fig. 2. Diagram showing how light rays can refract (bend) when entering a material with a different refractive inde.

to complete your own demonstration of how refractive in de can inuence an image, and see o . to investigate why we can see ice in water. In Fig. ., light rays are seen approaching a glass bloc. nother way of phrasing this is to say the light rays are incident upon the surface of the glass bloc, and so these are called the incident light rays. ou will hope fully notice that the incident light rays are approaching the glass bloc at a particular angle relative to the normal, which is a hypothetical line drawn perpendicularly to the surface itself (o .). is particular angle between the incident ray and the normal is nown as the angle of incidence (i). ow, because the glass bloc (n9 .) has a dierent refractive inde to air (n .), the light rays will change direction (refract) as they enter it and thus alter the angle of the light ray relative to the normal. is new angle is called the angle of refraction (i9), but please be careful here, as students often confuse the angle of refrac tion with the angle of deviation, so it is important to re member that both the angle of incidence and the angle of

refraction are always measured relative to the normal. In this eample, the light rays have refracted (or bent) towards the normal, meaning that the angle of refraction (i9) (o .) is smaller than the angle of incidence. In contrast to this, the angle of deviation describes the angu lar dierence between the emergent (refracted) light ray and the original path of the incident light ray, but don’t worry too much about that right now.

Calculating Angles In order to mathematically calculate the angles of incidence or refraction, we can use something called nell’s law (ua tion .) to wor out one from the other if we now the refractive indices of the materials (see able .). lso, ust as a word of caution, in this tet I have dened the angles of incidence and refraction as i and i9 respectively, but you may encounter other resources that refer to them as u and u9 instead – they represent the same thing and simply depend upon whichever convention the author prefers to use

16 S EC TI ON 1

•BOX 2.2

eometric and asic ptics

hat Is the ‘ormal’

The ‘normal’ is a hypothetical line that we imagine eists relative to the surface of refractive or reective material. rucially the normal alwas eists at a ° angle to the surface and intersects the point where the light ray is meeting the surface. n this tetbook it is always represented as an orange dashed line. t is important to remember how to draw the normal so that we can remember how to measure our angles of incidence (i) and refraction (i ) correctly. emember that this is true even if the surface is curved. ee ig. B. for a demonstration. Incorrect

Incorrect

nn′ ii′

nn′ ii′ i

i

n n′

i′ i

n n′

•Fig. 2.1

Diagram demonstrating the relationship between angles of incidence and refraction relative to the refractive inde of the medium.

Correct

nell’s law helpfully shows us that the refractive indices of the primary (without the primes) and secondary (with the primes) mediums will aect the sie of the angles. ssentially, without getting into too much detail, if the light ray moves from a low refractive inde to a high one, the light ray will bend towards the normal (producing a smaller angle of re fraction), whereas if the light ray moves from a high refractive inde to a low one, the light ray will bend away from the normal (producing a larger angle of refraction). is is shown in Fig. .

Correct

DEMO QUESTION 2.2

• Fig. 2.1

Diagram highlighting that the normal must be ° to the surface at the point where the light ray intersects the surface.

•BOX 2.3

h the ittle postrophe

n optics the apostrophe is a prime (or dash) meaning that if we were talking about the angle of refraction for eample repre sented as (i ) we would pronounce this as ‘i prime’ (though  like to call it ‘little i prime’ to save myself the confusion when we some times use capital letters). As a handy tip for you try to remember that the apostrophe usually means that the variable is representing something after a refractive or reective process has taken place and so it is related to the secondary (after refraction) variables. As an eample of this the refractive inde of air might be written as n 5 . but if the light moves from air (the primary medium) to a glass block (the secondary medium) we would write the refractive inde of the glass block as n9 5 ..

uation  uation  (explained)

n(sin i )

n (sin i )

primary refractive index (sin angle of incidence ) secondary refractive index (sin angle of refraction)

f a light ray is incident on a glass block (refractive inde .) at an angle of ° what is the angle of refraction tep  Determine what we need to calculate angle of refraction i’ tep  Dene variables i 5 ° (angle of incidence n 5 . (nothing is secicall mentioned so we assume the rimar medium is air n9 5 . (refractive inde of secondar medium – glass bloc tep  Determine necessary euation n (sin i) 5 n (sin i ) (quation . tep  alculate n (sin i) 5 n (sin i ) . (sin ) 5 . (sin i ) ... 5 . (sin i ) (solve the left side ... . 5 (sin i ) (rearrange n  ... 5 (sin i ) (solve the left side sin (...) 5 i (rearrange sin – this maes it an inverse sin i9 5 .° (don’t forget the units) Also remember to double check your answer. f n is larger than n (as in this case) then i should be smaller than i.

Practice Questions: 2.2.1 f a light ray is incident on a glass block (refractive inde .) at an angle of ° what is the angle of refraction 2.2.2 f a light ray is incident on a piece of plastic (refractive inde .) at an angle of ° what is the angle of refraction 2.2.3 f a light ray refracts out of a glass block (refractive inde .) at an angle of ° what was the angle of incidence

Vergence and Refraction

CHAPTER 2

1

Lateral Displacement

Em er ge

nt

lig

ht

ra y

If a light ray enters a glass bloc, chances are it will leave the glass bloc on the other side. e term lateral displacement describes a phenomenon when a ray of light enters a material and, upon leaving the material, has ehibited a parallel shift to the side. If you’ve ever played a video game involving the use of portals, it can be liened to entering a portal at the rst surface of a glass bloc and then leaving a portal shifted slightly to the side on the second surface of a glass bloc, as shown in Fig. .. It also means that the angle of incidence at the rst face will be identical to the angle of emergence at the second face. Importantly, when an emergent ray is parallel to the incident ray, the emergent ray forms two hypothetical rightangled triangles within the material that we can use to calculate how much displacement (s) has taen place (Fig. .). In Fig. ., we can see that the displacement (s) is actually the same distance as the side  of our triangle , so in order to calculate the displacement, we’re going to need to investigate our triangles, and as we’ll see below, the ey component is the path of the light through the bloc (side ). Firstly, though, we need to do a little bit of revision of vertically opposite angles (o .), because we can use this principle to deter mine the angle ‘’ (). his principle tells us that i and the mystery angle () should add up to eual i on the opposite side of the surface (Fig. .). his means that subtracting i from i should reveal the sie of the mystery angle. ∠BAD  i1  i1

i2′ i2 i1′

Inc ide nt

lig

ht

ra y

i1

• Fig. 2.11

Diagram showing that even though refraction occurs here at two surfaces (the front and back surface of the glass block) the incident and emergent light rays are parallel ust displaced slightly to the side from the original path of the light ray.

C

s D i2′ i2 A

i1′

B

N

i1

d • Fig. 2.12 Diagram showing that when the emergent light ray eiting a material is parallel to the incident light ray it forms two hypothetical rightangled triangles (highlighted in colour) within the glass block.

nce we now the angle , we can then use trigonometry to solve for triangle  – all we have to do is substitute in our calculation for nding the mystery angle ∠ D/AB (remember, we can substitute BAD for i1 – i1 ) sin (i1 i1 ’) BD/AB which can be rearranged to show that  5  sin (i – i ) agic e’ve managed to nd one way of epressing side . et’s now use the same principles of trigonometry to solve for triangle , but for this triangle we now the angle cos (angle) 5 adacent  hypotenuse cos (i ) 5    which can be rearranged to show that  5   cos (i ) erfect hat we’ve managed to achieve here is two dierent ways of calculating the length of the side , which represents the path the light ray taes within the glass bloc. is means that these euations are euivalent and can be written without needing to now  at all. (  cos (i )) 5  sin (i – i ) is rearranges to  5 ( (sin (i – i )))  (cos (i ))

1 S EC TI ON 1

•BOX 2.4

eometric and asic ptics

Revision of erticall pposite ngles

When thinking about angles along a straight line several as sumptions can be made. or eample we know that angles on each side of the straight line have to add up to ° so that the full circle euals ° (ig. B.A). This means that if we bisect that original line with one running perpendicular to it it will cre ate four angles each eualling ° (ig. B.B). This helps us

A

see that if we rotate that second line by ° two of the angles will increase to eual ° 1 ° whilst two of the angles will decrease to eual ° ° but crucially the total angles all still add up to ° (ig. B.). This means that vertically opposite angles (angles opposite each other at a verte) will be identical (ig. B.D).

B

C 180°

90°

90°

90°–25°

D 90°25°

x°

180°–x°

180° 180° 90° 180°

90° 90°25°

90°–25°

x° 180°–x°

•Fig. 2.2

llustration showing angles along a straight line eual ° (A) so if a second line is placed on top at a perpendicular angle all angles will now eual ° (B). f we then rotate the bisecting line by ° two angles will increase by ° and two will shrink by ° () showing that vertically opposite angles will be identical (D).

DEMO QUESTION 2.3

i2 ? A

i1′

i1

• Fig. 2.13

llustration applying the principle of vertically opposite an gles to angle BAD from ig. ..

s a nal step, we ust need to substitute in the variables from Fig. . to obtain our calculation for determining lateral displacement (uation .) uation 

uation  (explained)

s

lat. disp.

d sin ( 1 cos i1

1

)

width of block sin ( 1 cos i1

1

)

alculate the lateral displacement of a light ray that enters a  cm wide glass block (refractive inde .) at an angle of °. tep  Determine what we need to calculate lateral displacement s tep  Dene variables i 5 ° (angle of incidence n 5 . (nothing is secicall mentioned so we assume the rimar medium is air n9 5 . (refractive inde of secondar medium – glass bloc d 5 . (deth of glass bloc in metres tep  Determine necessary euation n (sin i) 5 n (sin i ) (quation . adated for lateral dislacement q ( ut 1s in net to the i variables to show that it’s the refraction at the rst surface see ig. .1 s 5 (d (sin (i – i )))  (cos (i )) (quation . tep  alculate n (sin i) 5 n (sin i ) . (sin ) 5 . (sin i ) ... 5 . (sin i ) (solve the left side ... . 5 (sin i ) (rearrange n  ... 5 (sin i ) (solve the left side sin (...) 5 i (rearrange sin – this maes it an inverse sin i9 5 .° s 5 (d (sin (i – i )))  (cos (i )) s 5 (. (sin(–.)))  (cos (.)) s 5 ...  . s 5 . m s 5 . cm (don’t forget the units)

CHAPTER 2

Vergence and Refraction 1

DEMO QUESTION 2.3—cont’d

Practice Questions: 2.3.1 alculate the lateral displacement of a light ray that enters a  cm wide glass block (refractive inde .) at an angle of °. 2.3.2 alculate the lateral displacement of a light ray that enters a  cm wide glass block (refractive inde .) at an angle of °.

Spherical Curved Surfaces p to this point we have only considered at, ‘plane’ sur faces, but we also need to understand how light interacts with curved surfaces. In optics, a spherical curved surface is any surface or boundary between two refractive materials that forms part of a sphere in its shape (Fig. .). is sec tion will go through some important elements of vergence and refraction at spherical curved surfaces.

oer pherical curved surfaces can alter the vergence of incoming light through refraction, and the degree of change they are capable of introducing is indicated through their power, measured in dioptres (). For eample, if a spherical surface increases convergence, this suggests it is maing the ver gence more positive (see ‘ergence’ section), thereby indi cating they have a positive power (e.g. 1. ). hereas if a spherical surface increases divergence, this suggests it is maing the vergence more negative (see ‘ergence’ sec tion), thereby indicating they have a negative power (e.g. 2. ). e power of a surface determines the degree to which the light will converge or diverge. For eample, a surface with a power of 1.  will converge light to a greater etent than a surface of power 1. . Importantly, spherical curved surfaces will also have a centre of curvature, which we can lien to the very central point of a sphere. et’s consider an eample of a football which is spherical and therefore easy to identify the

•Fig. 2.14

Diagram highlighting that spherically curved surfaces will t along a circle (A) whereas aspherical (not spherical) curved surfaces will not (B).

• Fig. 2.1 ample showing the centre of curvature (pink star) re mains the same distance away for the whole football (left) and for a small section of the football (right).

centralmost point, termed the centre of curvature (Fig. .). It is reasonable then to assume that any distance measured from the outer edge of the football to the centre of curvature would euate to the football’s radius. If we then cut out a small section of the football, as shown in Fig. ., although it no longer maes the completely spherical, foot ball shape, you can see that the curvature of the outer sur face still falls along the ‘sphere’ of the original shape. is means that it will maintain the same centre of curvature as the original football. ith spherical refractive surfaces the idea is the same – although we only see part of the sphere in the surface, it will still possess a centre of curvature () whose distance from the surface euates to the radius of curvature (Fig. .). ou will also notice that Fig. . has a line called the optical axis – this is a straight line drawn in optical diagrams that passes through the centre of curvature of the system. s discussed previously, because the radius of curvature will be a measure of distance, the radius will need to be as signed a positive or negative value depending on which side of the surface it is on and how it is measured relative to the direction of light (see Fig. .), and it will also need to be reported in metres. emember, because we always assume that light travels from left to right across the page and all distances are measured from the surface, if radius of curva ture is measured from left to right with the direction of light, then the distance is positive (e.g. 1. m). owever, if the radius of curvature is measured from right to left against the direction of light, then the distance will be nega tive (e.g. 2. m). ypically this means that if a surface is convex and the centre of curvature is on the right, then the radius will be positive, whereas if the surface is concave with a centre of curvature on the left, then the radius will be negative. is assignment of a positive or negative sign is crucial for calculations determining the power of the sur face, so mae sure you have a good understanding of this before proceeding (o .). ogically, a small radius of curvature would predict a higher degree of curvature in the surface than a larger radius of curvature, meaning that the radius of curvature must pre dict the curve of the surface which is intrinsically lined to the power of the surface. It is therefore no surprise to see that the radius of curvature (r) can be used to calculate the

2 S EC TI ON 1

eometric and asic ptics

Curved surface

om t fr

t

jec

ob

Ligh

igh

L

t pro

duc

ing

ge Image

Centre of curvature (C) Light source/ object

ima

Optical axis

Radius of curvature (r)

•Fig. 2.1 Diagram showing vergence of light from an obect and its constituent image following refrac tion through a spherically curved surface. The radius of curvature (distance from surface to centre of curvature ‘’) is on the right because this surface is conve.

•BOX 2.5

Is the Centre of Crvatre on the Right or the eft

n terms of how to determine on which side to put the centre of curvature we can pretend the spherical surface is part of a full sphere and then ust draw a dot in the centre like in ig. B. n the eample on the left in ig. B. the surface is conve (it bows out towards the direction we assume light to be originating from – the left) which means the centre of curvature is on the

right and will therefore have a positive value. The opposite would be true if the surface was concave as that would produce a centre of curvature on the lefthand side with a negative distance. ogically then using uation . we can assume that • onve spherical surfaces are positively powered • oncave spherical surfaces are negatively powered

Convex surface

Concave surface

Centre of curvature (C)

Centre of curvature (C)

Positive radius of curvature (r)

Negative radius of curvature (r)

Direction of light

• Fig. 2.3

Diagram showing relationship between conve and concave surfaces and their centre of

curvature.

power using uation .. emember that distances always need to be measured in metres so that the power is calcu lated correctly. uation 

uation  (explained)

F

r

power secondary refr.index primary refr.indeex radius of curvature

DEMO QUESTION 2.4 Determine the power of a conve spherical glass surface (refractive inde .) with a radius of curvature of  cm. tep  Determine what we need to calculate power  tep  Dene variables r 5 1. m (because the surface is conve the radius is ositive and in metres (see o . n 5 . (nothing is secicall mentioned so we assume the rimar medium is air

CHAPTER 2

Vergence and Refraction

21

Convex surface

DEMO QUESTION 2.4—cont’d n9 5 . (refractive inde of secondar medium – glass surface tep  Determine necessary euation  5 (n – n)  r (quation . tep  alculate  5 (n – n)  r  5 (. – .)  .  5 1. D (don’t forget the 6 sign and the units) Also remember to double check your answer – if the surface is conve it should have a positive power and if the surface is concave it should have a negative power.

Secondary focal point Positive focal length (f’)

Concave surface

Practice Questions: 2.4.1 Determine the power of a conve spherical glass surface (refractive inde .) with a radius of curvature of  cm. 2.4.2 Determine the power of a concave spherical glass surface (refractive inde .) with a radius of curvature of  cm. 2.4.3 Determine the power of a conve spherical glass surface (refractive inde .) with a radius of curvature of  cm.

Secondary focal point Negative focal length (f’)

• Fig.

2.1 Diagram showing how parallel light can help decide whether the secondary focal point is on the right of the surface (conve top) or the left (concave bottom).

ocal Length and ocal oints s discussed previously, the power of a surface indicates the degree of vergence it will add to or remove from incoming light rays, but the value assigned for power is based on what happens to incoming parallel light (from innity). arallel light possesses ero vergence as the rays are neither diverg ing away nor converging towards one another. hen these rays refract through a spherical surface, they will form an image at a point referred to as the secondary focal point (F9) whose distance away from the ape (tip) of the surface is referred to as the secondary focal length (f9). e will cover how to calculate focal length in chapter , so for now ust mae sure you understand what it is (Fig. .). Importantly, as focal length is measured as a distance, it needs to be recorded in metres, and it needs to be measured from the refracting surface. erefore it will have a positive value if the secondary focal point is on the right of the sur face, and it will have a negative value if the secondary focal point is on the left of the surface. o decide which side the secondary focal point will be on, you can thin about what happens to light when it meets a conve (positively pow ered) or concave (negatively powered) surface. hen parallel light enters a positively powered (conve) surface, the surface converges the light towards the optical ais. is means that light will bend towards the optical ais on the righthand side of the surface, indicating the second ary focal point will be on the right (see Fig. ., top). ow ever when parallel light enters a negatively powered (con cave) surface, the surface will diverge the light away from the optical ais, meaning that the light rays need to be proected bacwards (left) to ever meet the optical ais. is is referred to as the point where the light rays appear to originate from,

and it means the secondary focal point will always be on the left of a concave surface (see Fig. ., bottom).

Limiting Spherical Aerration ith a curved surface, the normal needs to remain at ° to the surface at every position on the surface. ypically, we’d epect that parallel light (from innity) should form a lovely, focused image at the secondary focal point of the system. owever, light rays within a wide beam of light from innity will all refract slightly dierently depending on where they intersect the surface (as shown in Fig. .) as their angles of incidence (i), relative to the normal, will all be slightly dif ferent. In this case, light will be focused along the optical ais instead of producing at a single focus, and the light rays will follow what’s called a caustic curve. Importantly, the tip of the caustic curve will coincide with the focal point of the surface. is is called spherical aberration which means the uality of the image will be poor because it will be blurred along this section of the optical ais (we will discuss aberrations in more detail in chapter ). e good news, however, is that paraxial rays (rays very close to the optical ais) are negligibly aected by these in creasingly large dierences in vergence, and so we will always use paraial rays in our euations to mae the maths easier, hooray

ect and Image Vergence ining bac to ‘ergence’, we now that obect vergence () denes the path of the light rays leaving the obect, and

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eometric and asic ptics

Convex surface

C

au

r

s ti

c cu

rve

Optical axis

Secondary focal point (F’)

•Fig. 2.1

Diagram showing a wide pencil of parallel rays will fail to focus at a single point instead pro ducing a caustic curve and spherical aberration (blurring).

we have recently learnt that the power of a spherically curved surface can alter the vergence at the point of refraction. is means that we also need to be able to calculate the vergence of the image rays (image vergence 9), which represents the path of the light rays (following refraction) that form the image (this is also true for reection at curved surfaces, which will be discussed in chapter ). reviously we learnt only how to calculate obect vergence (see uation .), because we hadn’t yet learnt about power, but lucily for us we can now learn the full set of euations Image vergence (9) is calculated from the sum of the obect vergence () and the power (F) of the surface (uation .), which then allows us to determine where the image will form (the image distance, l9 uation .). F

uation  uation  (explained) uation  uation  (explained)

power L

n l l

image distance 

n l

secondary refr.index image vergence

If you are thining that uation . loos very similar to uation . then you’re right, because it relies on the same principles – that vergence depends on distance and the refractive inde of the material. owever, notice all the prime symbols (which loo lie little apostrophes) indicat ing that these are all secondary variables this time, so for eample, n9 would be the refractive inde of the material into which the light travels. Importantly, the surface is curved so we need to ensure we always measure distances (e.g. for obect, image, radius) from the ape (the pea) of the surface because we are using paraial rays which lie close to the optical ais.

DEMO QUESTION 2.5 An obect is placed  cm in front of a conve spherical glass surface (refractive inde .) with a radius of curvature of  cm. Where does the image form tep  Determine what we need to calculate image distance l tep  Dene variables l 5 2. m (the question stated the object was 1 cm ‘in front of’, meaning it’s to the left of the surface. This means it will have a negative distance (see ig. ., and we need to convert to metres r 5 1. m (because the surface is conve the radius is ositive and in metres (see o . n 5 . (nothing is secicall mentioned so we assume the rimar medium is air n9 5 . (refractive inde of secondar medium – glass bloc tep  Determine necessary euation(s)  5 n  l (quation .1  5 (n – n)  r (quation . 9 5  1  (quation . 9 5 n  l (quation . tep  alculate  5 (n – n)  r  5 (. – .)  .  5 1. D 5nl  5 .  .  5 2. 9 5  1  9 5 . 1 . 9 5 1. 9 5 n  l l9 5 n   (rearrange to get l  l9 5 .  . l9 5 1. m The image forms . cm right of the surface. (don’t forget the 6 sign and the units) We know the image forms to the right of the surface because the distance is positive.

CHAPTER 2

DEMO QUESTION 2.5—cont’d

Practice Questions: 2.5.1 An obect is placed  cm in front of a conve spherical glass surface (refractive inde .) with a radius of curvature of  cm. Where does the image form 2.5.2 An obect is placed  cm in front of a concave spherical glass surface (refractive inde .) with a radius of curvature of  cm. Where does the image form

In Fig. . we can see that the surface is conve, meaning it is positively powered, which also means that the surface will add convergence to the rays after they pass through it. ou can see that there is an obect drawn on the left, illustrated as an arrow pointing upwards. In optics, we usually draw obects to loo lie this to help mae diagrams easier to understand (see chapter  for more information on optical diagrams). imilarly the image is drawn on the right hand side and is also depicted as an arrow. ne crucial thing to note is that in this eample illustration, the image formed is upside down, or ‘inverted’ (o .). hen describing image characteristics in optics, we need to tae special care to determine which way up the image is, it’s position relative to the surface, and whether it is bigger (magnied) or smaller (minied) than the obect.

The eird and onderful orld of Refraction Through a Single aterial In this nal section of the chapter, we will discuss specic reuirements that cause light to change direction and re fract even when within a homogenous medium (i.e. staying within one material).

A Load of Hot Air ave you ever noticed that on a really hot day, there seems to be ‘invisible’ wiggly lines emanating o warm surfaces

Vergence and Refraction 23

(e.g. the road or cars, or above a re in a replace) (If not then it might be worth looing for this eect net time it’s really warm out). is distortion occurs because the light travelling through the ‘heat’ to get to your eyes is travelling through dierent temperatures of air, which have dierent densities. For those of you who aren’t natural eperts in ther mal physics, it’s important to now that warm air is less dense than cold air, and the dierence between the densities of warm and cold air results in small refractive inde dierences. s we learned in sections on ‘ergence’ and ‘efraction’, if light passes through materials with dierent refractive indi ces, then it may refract (change direction). In a way, then, we can assume that if there is a pocet of hot air within a larger pocet of cold air, then the hot pocet of air will ind of behave as if it’s a very wea lens. us if an obect is on the opposite side of the hot air pocet to you, even though it is only travelling through one material (air), it will produce a refracted image as it passes through the hot pocet. nother ey factor here is that the heat coming o the warm surface in these eamples varies rapidly over time, hence the density also varies over time, which ultimately varies the refractive inde dierence over time. is results in the moving ‘wiggly’ image that you can see on a hot day or above a hot item, for eample, a candle ame heating the air around it. e cool thing about this is that, because the dierences in air density aect the light that travels through it, we can pho tograph the density changes associated with heat using spe cialised chlieren imaging (also see o .). chlieren im aging is the name ascribed to imaging systems that can detect density changes within a material (and thereby can detect pat terns of heat). e setup reuired to get a chlieren image is uite comple because we need to somehow get rid of the light that isn’t passing through the density changes – as otherwise the density changes will be obscured by the rest of the light. ne way to do this is outlined in Fig. ., which shows a laser light source in front of a positively powered lens, a can dle, another positively powered lens, a pin and a screen (to view the image). Importantly, the laser light source (produc ing divergent light) is placed at the focal point of the rst positively powered lens (for content on lenses, please see

Curved surface n’ Apex (A)

n h

Image (I)

Object (O)

h’ Distance from apex of the surface to the object (l)

•Fig. 2.1 surface.

Optical axis

Distance from apex of the surface to image (l’)

implied diagram showing formation of image on the right of a positively powered spherical

24 S EC TI ON 1

•BOX 2.6

eometric and asic ptics

Image Characteristics

mages produced following refraction (or reection) will either be real or virtual (ig. B.). The following is always true about real images • inverted (upside down) • can be proected onto a screen • drawn with solid line The following is always true about virtual images • erect (upright) • cannot be proected onto a screen • drawn with dashed line

Virtual image (I) Real image (I)

•Fig. 2.4

•BOX 2.

Diagram showing real vs virtual images.

hadograph

o far this chapter has taught us that density differences in air can cause light to refract differently and that this can be viewed as ‘wiggly’ images above a hot surface or through a chlieren image (in a lab setting). owever there is another way to view these changes in air density and that’s through a method called shadowgraph. ow if we eamine the etymology of the word ‘shadowgraph’ we can see it breaks down as ‘shadow’ and ‘graph’ (which means a way of producing images) so we can safely assume that this method will produce an image of a shadow – which is called a shadowgra. ore specically though this techniue can be used to identify the density changes within a single material for eample density changes in air associated with differences in temperature. or this to work light must be shone onto a heat source and onto a screen (see ig. B.). As the light travels through the

Dark room

Light source

chapters , ,  and ). t this point in the boo, we now that parallel light entering a surface (or lens) will focus at the focal point (see section on ‘Focal ength and Focal oints’), but for lenses, the opposite is also true. If an obect (or light source) is placed at the focal point of a lens, it will produce parallel vergence leaving the lens. erefore, as the laser light travels through the positively powered lens, it leaves the rst lens with ero vergence (parallel rays). e light then travels past the candle, and still has ero vergence as it approaches the second positively powered lens, meaning the light will focus at the focal point of that lens. owever, in our setup we’ve sneaily placed a pin at the focal point of the second lens in order to bloc the light that travels unimpeded through the system. owever, we now the heat near the candle will low er the density of the air around it, which will alter the path of any laser light that travels through the pocet of hot air. is light will have altered vergence as it approaches the second lens (see Fig. .), which means it doesn’t focus at the pin and instead carries on to produce an image on the screen. is image will be a wiggly image, showing that the density prole of the air changes with time, but a rough approimation of the image is shown in Fig. ..

different densities of air it will change direction and leave areas where less (or no) light is able to reach the screen which produces shadows associated with the density changes. ig. B. shows an eample shadowgram that  produced at home with a candle. rom this you can see that the hot air from the candle is rising relatively straight for a while before then becoming very wiggly and interesting. This shows us that there are density changes in

Screen

air Hot

Candle

• Fig. 2.

Diagram showing setup of a simple shadowgraph tech niue. n a dark room a light source is shone onto a lit candle to produce a shadow on a screen. n principle the hot air from the candle should produce a shadowgram showing the density changes.

•Fig. 2.

A photograph of a shadowgram produced by the air above a lit candle.

CHAPTER 2

•BOX 2.

Vergence and Refraction 25

hadograph—cont’d

the air above the candle (and all  used was a torch – how fun is that). or another (slightly less dangerous) eample see ig. B. – this time  produced a shadowgram of a wine glass (can you see

•Fig. 2.

A photograph of a shadowgram produced by glass.

Positive lens 1 Laser light source

Candle

•Fig. 2.2

Screen

Positive lens 2

ir ta Ho

Focal point

the different patterns of concentric lines appearing in the shadow on the desk). This shows that the density of the glass is variable from the bottom to the top – you can try this one at home and see if you can nd your ‘best’ optical glass

Example image seen

Pin

Focal point

A diagram showing one possible setup reuired to produce a chlieren image. ere you can see that laser light source is placed at the focal point of the rst positive lens which (if no heat was pres ent) would produce parallel light that would focus at the focal point of the second positive lens (lighter red light). owever in this case a pin is placed in eactly the same point that the light focuses meaning only light that has been refracted (changed direction) through air temperature changes (caused by the candle) will be visible (darker red light). A rough approimation of the type of image you might see is shown on the right.

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Real sun and image of sun Image of sun

Z

zz

Z

zz

Real sun

• Fig. 2.21

Diagram showing eaggerated effects of atmospheric refraction using the sun as the light source. The atmosphere is made up of gradually increasing density as it gets closer to the arth produc ing a gradient of refractive inde changes (shown in purple). When the sun is directly above the observer (A) it will travel in a straight line through the atmosphere but when it is in any other location the light will bend as it travels through the refractive indices within the atmosphere. This is particularly evident at sun set and sunrise as the sun will be visible above the horion even when it is below the horion (B).

The Atmosphere nother eample of how light can change direction within a single medium is when light from the sun (or moon or a star) travels through the arth’s atmosphere. For this to mae sense, we need to picture the arth, surrounded by an atmosphere, as shown in Fig. . where we are using the sun as our eample, but the same principle applies to light from the moon or stars as well. Fig. . demonstrates that there is a density change within the atmosphere – the atmo sphere closer to the arth is denser than the atmosphere further away, and this change in density can change the direction of light that travels through it if the circumstances are appropriate. Interestingly, this refraction is dierent to that which occurs when light travels directly from one ma terial into another (e.g. light moving from air into water), because the refractive inde change within the atmosphere is gradual, lie a gradient. is can produce a similarly gradual change in the direction of light, which is called atmospheric refraction

For eample, if light from the sun travels in a straight line (along the normal) through the atmosphere, lie it does at noon on a summer’s day (when in a enith position directly above the observer), then the light will travel in a straight line through the dierent densities within the atmosphere (see Fig. .). owever, when the sun is in any other posi tion relative to the observer, the light will gradually bend (and curve gradually towards the arth) as it travels through the refractive indices within the atmosphere. is is particularly evident when the sun is below the horion, at which point the light from the sun will curve so much that it will produce an image of the sun which appears much higher up than its ac tual location. is means that, at sunset, the sun is still visible after it has disappeared below the horion (see Fig. .). s a little bonus thought to end the chapter – we also now that temperature of the air can aect the way light travels through air (and the atmosphere), which means that on a warm day (with lower overall air density) the sun may appear to set faster than on an euivalent day that is slightly cooler in temperature.

Test Your Knowledge ry the uestions below to see if you need to go over any sections again. ll answers are available in the bac of the boo.  hat is a collection of light rays called  hat does parallel vergence tell us about the ori gin of the light rays  ow do we decide whether an obect distance will be negative or positive

References . unnaclie , irst . Optics. nd ed. ondon,  ssocia tion of ritish ispensing pticians .

 If a light ray moved from a medium with a refrac tive inde of . into a medium with a refractive inde of ., would it bend towards or away from the normal  If you wor out that following refraction an image will have a vergence of 1. , are the rays converging or diverging  ould a concave spherical surface possess a nega tive or positive power hy

. avidhay . Introduction to hadowgraph and chlieren Imag ing. I cholar ors . vailable at httpscholarwors. rit.eduarticle. ccessed th anuary .

3 Thin Lenses    P T E R   TL I E Introduction What Is a ‘Thin Lens’? Power of a Thin Lens Power of a Single Thin Lens Focal Length Vergence Relating to a Thin Lens Linear Magnication Image Distance

Multiple Thin Lenses Separated by a Distance Vertex Power Step-Along Method Equialent Lenses Principal Planes ewton’s Formulae Magnication sing ewton’s Formulae Test our nowledge

Multiple Thin Lenses Multiple Thin Lenses in Contact

  E  TI VE After working through this chapter, you should be able to: Dene ‘thin lens’ and apply the ergence euations from chapter  to thin lenses Determine linear magnication of an image nderstand the dierence between bac erte power front erte power and euialent power

Condently apply stepalong ergence euations and ewton’s Formulae to determine image formation with multiple lens systems

Introduction In chapter 2 we covered the basics of vergence and refraction at plane and spherical curved surfaces, but within clinical optics we also need to be able to understand how light interacts with lenses. is chapter will seek to cover the fundamentals of thin lenses for calculating image location, power, vergence and focal lengths of the system.

it can be completely ignored. ese types of lenses, in which we ignore the refractive inde, are called thin lenses In optical diagrams, thin lenses are drawn as vertical lines with arrowheads, where positive converging conve lenses are simplied to be outward pointing arrows, and negative diverging concave lenses are simplied to be inward pointing arrows ig. .2. e will cover this in more detail in chapter , but it is good to be aware of this now as lenses will commonly be depicted as arrows throughout this chapter.

What Is a ‘Thin Lens’?

Power of a Thin Lens

Lenses, by denition, are refractive devices that possess two surfaces one on the front and one at the back ig. . and are available in several forms ig. .. In the last chapter we discussed at length that the curvature of a surface and the refractive inde of a material can have an impact on the refractive power of a material. owever, if the thickness is slight enough relative to the radius of curvature of each surface, then it’s assumed that the refractive inde of the lens material has a negligible eect on the power, so

e refractive power  of a thin lens can be determined by either knowing the power or curvature of each surface, or by knowing the focal length f9 of the lens. emember, power is measured in dioptres

Power of a ingle Thin Lens ith a single thin lens, the approimate overall power can be calculated by adding the front  and back surface 27

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A

B

ECv

BCv

PCv

PM

Back surface

Front surface

Positive lenses

Negative lenses

ECc

BCc

PCc

MM

•Fig. 3.1

Lenses possess both a front and back surface (A) relative to the conventional direction of light (blue arrow) and come in a variety of forms (B). Positive lenses can be equiconve (v) biconve (Bv) planoconve (Pv) or plus meniscus (P) whereas negative lenses can be equiconcave (c) bicon cave (Bc) planoconcave (Pc) or minus meniscus ().

Positive

Negative

DEMO QUESTION 3.1—cont’d tep  alculate  5  1   5 1. 1 1.  5 1. (don’t forget the 6 sign and the units)

Practice Questions:

• Fig. 3.2

Positive and negative lenses are commonly depicted as straight lines with outward (positive) or inward (negative) facing arrows on either end.

powers 2 together uation ., for eample, in ig. ., the front and back surface powers add together to suggest the lens depicted has an approimate power of 2.. Equation 3.1 Equation 3.1 (explained)

F

1

2

power

DEMO QUESTION 3.1 A biconve thin lens has a front surface power of 1. and a back surface power of 1.. hat is the overall power of this lens tep  etermine what we need to calculate power  tep  ene variables  5 1.  5 1. tep  etermine necessary equation  5  1  (Equation 3.1)

3.1.1 A biconcave thin lens has a front surface power of . and a back surface power of .. hat is the overall power of this lens 3.1.2 A plus meniscus thin lens has a front surface power of 1. and a back surface power of .. hat is the overall power of this lens

ocal Length In chapter 2 we learned the basics of a surface’s focal length by discussing that the secondary focal length (f9) describes the distance between the surface and the secondary focal point (F9). e also learned that conve spherical surfaces will have a positive focal length whilst concave surfaces will have a negative focal length. or thin lenses, this principle is the same, with positively powered conve lenses converging light towards the optical ais, resulting in a secondary focal point on the right-hand side of the lens, and therefore a positive secondary focal length. In contrast, negatively

F1 –2.00D

F2  –1.50D

F  (–2.00) + (–1.50)  –3.50D

•Fig. 3.3 iagram showing how to calculate approimate power () of a single thin lens by adding the front surface power ( ) and the back surface power () together.

CHAPTER 3

powered concave lenses diverge light away from the optical ais and therefore possess a secondary focal point on the left of the lens, leading to a negative secondary focal length. owever, we now also need to start to consider what happens if light travels backwards through the system, from right to left. ith a thin lens, we assume that the lens will do the same to light travelling from either direction, so the focal length will be identical on either side. e only dierence lies in the nomenclature, as when light travels backwards through the system it produces a primary focal point (F) at a distance corresponding to the primary focal length (f ). is distinction will be very important for chapter , but for now, please see ig. . for an illustration of this in action. e can use the secondary focal length f9 and the secondary refractive index of the surrounding material n9 to calculate the power of the lens using uation .2. ssuming that refractive indices of materials will always be $, then if you’re a fan of maths, this euation should demonstrate that positively powered lenses will have a positive focal length and vice versa, and negatively powered lenses will have a negative focal length and vice versa. n f

Equation 3.

F

Equation 3. (explained)

power 

surrounding refr index secondary focal length

A

Thin Lenses

B

F’

F

f

f’

F’

F

f’

f

•Fig. 3.4 iagram showing light from the left (blue) and light from the right (green) interacting with lenses. he positive lens (A) converges light towards the optical ais and so the secondary focal point ( ) is on the right of the lens and the primary focal point () is on the left. he negative lens (B) diverges light away from the optical ais and so the secondary focal point ( ) is on the left of the lens and the primary focal point () is on the right.

Vergence Relating to a Thin Lens or thin lenses in a homogenous material – e.g. the entire lens is in air – you can use the same vergence euations we learned in chapter 2 to calculate where the image will form for revision, see uations 2., 2., and 2., but we can also use euations to determine how big or small the image will be relative to the obect. is is called linear magnication m.

Linear Magnication DEMO QUESTION 3.2 A biconve thin lens has a focal length of  cm. hat is its power tep  etermine what we need to calculate power  tep  ene variables n9 5 . (nothing is specically mentioned so we assume the surrounding medium is air) f9 5 1. (the question stated the lens is biconvex, maing it positively powered – this means the ocal length will be to the right o the lens, maing it positive. e also need to convert to metres) tep  etermine necessary equation  5 n  f (Equation 3.) tep  alculate 5n f  5 .  1.  5 1. (don’t forget the 6 sign and the units)

Practice Questions: 3.2.1 A biconve thin lens has a focal length of  cm. hat is its power 3.2.2 A biconcave thin lens has a focal length of  cm. hat is its power 3.2.3 A biconve thin lens has a focal length of  cm. hat is its power in water (refractive inde .) 3.2.4 A biconcave thin lens has a power of .. hat is its focal length

29

e term ‘linear magnication’ is a measure of how big or small the image is relative to the original obect. It’s important to note here that because we are using paraxial rays which are, by denition, etremely close to the optical ais, the angle of incidence i and the angle of refraction i9 are etremely small and hardly dierent from one another. ltimately this helps us to use substitute tan into nell’s Law and use trigonometry to determine the linear magnication of the image, but we don’t need to worry about how that works providing we can remember how to apply the euation to calculate linear magnication m uation .. It is also important here to understand that although magnication is related to power, it is not the same. or eample, an obect cm in front of a 1. lens will produce an image with dierent magnication to an obect 2cm in front of the same lens. In that way, magnication is related to both the power of the lens and the distance of the obect.  top tip here is to make a note to yourself to ensure you understand the dierence between power of a lens and magnication of an image. h l L m Equation 3.3 h l L Equation 3.3 (explained) mag 

image height image dist object verg   object height object dist image verg

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will usually be identical on each side nn9, for eample if the lens is “in air” then the refractive inde would be . on both sides ig. ..

e best feature of a magnication calculation is that the answer can tell you an awful lot about the characteristics of the image. o be specic, the numerical value indicates the sie of the image relative to the obect, and the sign 12 will tell you if the image is upright or inverted. or eample, if the magnication is below ero negative, the image is inverted classied as a real image, whereas if it is above ero positive, it is upright classied as a virtual image. imilarly, if the magnication is between 2 and 1 ecluding ero, then it is minied smaller than the obect, and if it is below 2 or above 1, then it is magnied bigger than the obect. or eample, a magnication of 2. would be inverted and minied, whilst a magnication of 12. would be upright and magnied. Importantly, if the magnication euates to a value of  or -, then it is the same sie as the obect, and the sign will tell you if it is inverted or upright. ee ig. . for a handy way to remember this. or a real world eample of magnication in action, if you know someone who wears glasses then you may have noticed that their eyes appear to be larger magnied or smaller minied when seen through their lenses. is is an eample of the light from the obect in this eample, the person’s eyes being refracted to produce an image a dierent sie the image is the view of the person’s eyes that you see when you look at them. opefully though their eyes will always still be upright, indicating a magnication value above ...

DEMO QUESTION 3.3 An obect is placed  cm in front of a biconve thin lens of power 1.. (a) here does the image form (b) hat is the magnication of the image tep  etermine what we need to calculate image distance l magnication m tep  ene variables l 5 2.m (negative distance and we need to convert to metres)  5 1. n 5 . (nothing is specically mentioned, so we assume the primary medium is air) n9 5 . (nothing is specically mentioned, so we assume the secondary medium is air) tep  etermine necessary equation L 5 n  l (Equation .1) L9 5 L 1  (Equation .) L9 5 n  l (Equation .) m 5 (h  h) 5 (l  l) 5 (L  L ) (Equation 3.3) tep  alculate L5nl L 5 .  . L 5 2. L9 5 L 1  L9 5 2. 1 1. L9 5 2. l9 5 n  L (rearranged) l9 5 .  . l9 5 .m (a) he image forms . cm left of the lens. m 5 (h  h) 5 (l  l) 5 (L  L )

Iage istance tilising euations from chapter 2 and from the previous linear magnication section, we can calculate where an image forms after the light from the obect refracts through a thin lens. emember, with thin lenses we don’t consider the refractive inde of the lens at all, and so the refractive inde Inverted

Upright –1

Magnified

•Fig. 3.5

0

Minified

1 Minified

Magnified

chematic for helping you remember what the magnication value tells you about the image.

Lens

ce

en

rg t ve

c

je Ob

Ima

ge v erge

nce

Object

Image

Distance from lens to object (l)

•Fig. 3.6

Optical axis

Distance from lens to image (l’)

iagram showing relationship between vergence and obectimage distance with a thin lens.

CHAPTER 3

Thin Lenses

DEMO QUESTION 3.3—cont’d

DEMO QUESTION 3.4

m 5 L  L (choose which part we want to use) m 5 2  . (b) m 5 1. (don’t forget the 6 sign)

wo thin lenses of powers 1. and 1. are in contact with each other. f an obect is placed  cm in front of the rst lens where does the image form tep  etermine what we need to calculate image distance l tep  ene variables l 5 2.m (negative distance and we need to convert to metres)  5 1.  5 1. n 5 . (nothing is specically mentioned, so we assume the primary medium is air) n9 5 . (nothing is specically mentioned, so we assume the secondary medium is air) tep  etermine necessary equation e 5  1  (Equation 3.) L 5 n  l (Equation .1) L9 5 L 1  (Equation .) L9 5 n  l (Equation .) tep  alculate e 5  1  e 5 1. 1 1. e 5 1. L5nl L 5 .  . L 5 2. L9 5 L 1  L9 5 2. 1 1. L9 5 1. l9 5 n  L (rearranged) l9 5 .  1. l9 5 1.m he image forms . cm right of the second lens. (don’t forget the 6 sign (or the direction) and the units)

Practice Questions: 3.3.1 An obect is placed  cm in front of a biconve thin lens with a power of 1.. here does the image form 3.3.2 An obect is placed  cm in front of a biconcave thin lens with a power of .. hat is the magnication of the image 3.3.3 An obect is placed  cm in front of a biconcave thin lens with a focal length of  cm. here does the image form

Multiple Thin Lenses p to this point we have considered thin lenses in isolation, but sometimes optical systems have more than one lens present, for eample, in telescopes. is means we also need to think about systems with more than one thin lens, and how to calculate power, vergence and magnication for these systems.

Multiple Thin Lenses in ontact e simplest form of system containing more than one thin lens is when all the lenses are in contact with one another. hen positioned like this to have negligible distance between each lens, ust like how we can ignore the refractive inde, we can pretend the distance doesn’t eist o that end, we can simply add up the powers of each of the lenses in contact in order to work out the overall power of the system ig. ., and then we can calculate vergence using the same euations as in the section titled “ergence elating to a in Lens”. is time, however, we will refer to the calculated power as the equivalent power e as we are working out the euivalent single power of the multiple lenses, as shown in uation .. Equation 3.

Fe

1

2

Practice Questions: 3.4.1 wo thin lenses of powers 1. and 1. are in contact with each other. f an obect is placed  cm in front of the rst lens where does the image form 3.4.2 wo thin lenses of powers 1. and . are in contact with each other. f an obect is placed  cm in front of the rst lens what is the linear magnication 3.4.3 hree thin lenses of powers 1. . and 1. are in contact with each other. f an obect is placed  cm in front of the rst lens where does the image form

Equation 3. (explained) equivalent power

1

2

F 6.25D F4.25D

F1.50D

Fe5.75D

F 3.75D

F 1.00D

Fe 11.00D

F2.25D

F 1.75D

Fe0.50D

• Fig. 3.7 iagram showing that the equivalent power of thin lenses in contact with each other can be calculated through adding up the individual powers.

31

32 S EC TI ON 1

eoetric and asic ptics

Multiple Thin Lenses eparated  a istance If the thin lenses are separated by a distance, then we need to take that distance into consideration when performing our calculations. is is because vergence depends on the distance from the surface, so increasing the distance d of the second lens will have an impact on the sampled vergence of the light as it reaches the second lens ig. .. Lens systems of this nature still have secondary and primary focal points, but they will no longer be euidistant on either side, as the powers of the individual lenses will likely be dierent, so light will travel dierently in each direction. Instead, we need to consider what’s called the back and front verte power.

Vertex Power erte power refers to the vergence of light as it leaves the system, assuming that the incident light is parallel ero vergence, leading to light focusing at one of the focal points. If the light travels forwards through the system from left to right, the vergence of light leaving the second lens will be called the ac vertex power v9, sometimes referred to as emergent vergence at 2. owever, if the light travels backwards, from right to left, then the vergence of light leaving the rst lens will be called the front vertex power v, or sometimes referred to as emergent vergence at . ow, because the incident approaching light is parallel, the light will form a focus at the secondary 9 or primary focal point , depending on the direction. e distance between the second lens 2 and the secondary focal point 9 is called the ac vertex focal length fv9, and the distance between the rst lens  and the primary focal point  is called the front vertex focal length fv. ig. . depicts these distances in a multiple lens system. o calculate verte power, we need to take the distance between the lenses d into consideration. ee uations . and . for back and front verte calculations, but be careful because these euations are etremely similar and easy to confuse for one another. y top tip for successful application of

Lens 1 (F1) Incident vergence

Distance between lenses (d)

•Fig. 3.8

Ver afte gence r le ns 2

Lens 2 (F2)

F

F’

fv

fv

•Fig. 3.9 iagram showing light from the left (blue) and light from the right (green) interacting with a multiple lens system. he distance be tween lens  and the secondary focal point ( ) equates to the back verte focal length (fv ) whereas the distance between lens  and the primary focal point () equates to the front verte focal length (f v). otice that these distances are not identical.

these euations is to try to understand what the calculations are doing. or eample, if we’re determining the emergent vergence at 2 after light has entered the system through  back verte power v9 , then we need to know how the vergence from  will have changed across the distance d between the lenses. s the incident light is assumed to be parallel, the power of the lens at  will be identical to the vergence leaving it, and so the euation for back verte power takes into account the power of  in the denominator and the opposite is true for front verte power v. Equation 3.

dF1 F2 1 dF1

1

Fv

2

Equation 3. (explained) back vertex power Equation 3.

1

Fv

2 (distance 1 1 (distance lens1) 1

2)

dF1 F2 1 dF2 2

Equation 3. (explained) front vertex 1 2 (distance 1 power 1 (distance lens2)

Lens 2 (F2)

Verge n after le ce ns 1

Lens 1 (F1)

2)

DEMO QUESTION 3.5 F’

Back vertex focal length (fv’)

iagram showing light (blue) passing through two thin lenses separated by a distance (d). As the incident light is parallel the nal image will form at the secondary focal point ( ) but the distance be tween the back lens and the secondary focal point is called the back verte focal length (fv ).

wo thin lenses of powers 1. and 1. are separated by a distance of  cm. hat is the back verte power of the system tep  etermine what we need to calculate back verte power v tep  ene variables  5 1.  5 1. d 5 .m (we need to convert to metres) tep  etermine necessary equation v9 5 ( 1  d)  ( d) (Equation 3.)

CHAPTER 3

Thin Lenses

DEMO QUESTION 3.5—cont’d

DEMO QUESTION 3.6—cont’d

tep  alculate v9 5 ( 1  d)  ( d) v9 5 (1. 1 1. – (.)(1.)(1.))  ( (.)(1.)) v9 5 1.  . v9 5 1. (don’t forget the 6 sign and the units)

tep  alculate v 5 ( 1  d)  ( – d) v 5 (1. 1 1. – (.)(1.)(1.))  ( (1.)) v 5 1.  . v9 5 1.… fv 5 2 (n  v) fv 5 2 (  1.…) fv 5 2.m he front verte focal length is . cm. (don’t forget the 6 sign and the units)

Practice Questions: 3.5.1 wo thin lenses of powers 1. and . are separated by a distance of  cm. hat is the back verte power of the system 3.5.2 wo thin lenses of powers 1. and . are separated by a distance of  cm. hat is the front verte power of the system

If we know the back or front verte power of a lens system, then we can use that information to help us calculate the ac vertex focal length (fv9) and the front vertex focal length (fv). uations . and . eplain the dierence for each. e euations refer to the surrounding refractive inde n, but usually this will be air, so you can substitute n for the number  in this euation. It is also important to notice the minus sign in uation . front verte focal length, students often miss this, but it is crucial as we’re assuming light is travelling backwards when we solve for front verte focal length. n fv Equation 3. Fv Equation 3. (explained) back vertex surrounding refractive index focal length  back vertex power Equation 3.

fv

n Fv

Equation 3. (explained) front vertex focal length

surrounding refractive index front verttex power

Practice Questions: 3.6.1 wo thin lenses of powers 1. and . are separated by a distance of  cm. hat is the back verte focal length of the system 3.6.2 wo thin lenses of powers . and . are separated by a distance of  cm. hat is the front verte focal length of the system

ese euations help us a lot when there’s ero vergence in the incident light, but if the incident rays don’t originate from innity, then we need to utilise something called the step-along method.

Step-Along Method e step-along method allows us to calculate the image vergence emergent vergence, L9 of light leaving a multilens system if the incident vergence L is not eual to ero. or this method, we need to utilise but slightly modify the vergence euations we learned in chapter 2 uations 2., 2. and 2., and we need to learn a brand-new euation uation . to help us take into account that the lenses are separated by a distance d. e idea is that we work out the incident and emergent vergence at each lens individually, as if we’re ‘stepping along’ the lens system ig. .. ou’ll notice that, this time, we’ve used subscript numbers to indicate whether the vergence is related to lens   or lens 2 2, but remember, if there’s a prime 9 then the

DEMO QUESTION 3.6 wo thin lenses of powers 1. and 1. are separated by a distance of  cm. hat is the front verte focal length of the system tep  etermine what we need to calculate front verte focal length fv tep  ene variables  5 1.  5 1. d 5 .m (we need to convert to metres) n 5 . (nothing is specically mentioned so we assume the primary medium is air) tep  etermine necessary equation v 5 ( 1  d)  ( – d) (Equation 3.) fv 5 2 (n  v) (Equation 3.)

(.)

Lens 1 (F1)

Lens 2 (F2)

L1

L2

L1 Object

F

Object distance (l)

L2 F’

Distance between lenses (d)

Image

Image distance (l’)

• Fig. 3.10 iagram highlighting the incident and emergent vergence steps at each lens when the obect is closer than innity. his eample uses two positively powered lenses.

33

34 S EC TI ON 1

eoetric and asic ptics

variable represents emergent vergence, otherwise it represents incident vergence. Equation 3.

L2

L1 1 dL1

Equation 3. (explained) verg at lens 2

verg leaving lens1 1 (distance verg leaving lens1)

Equialent Lenses In this section we are still considering an optical system where two thin lenses are separated by a distance, but now we are going to see if we can determine a way of hypothetically replacing the two-lens system with a single lens of equivalent power e. is single lens would be referred to as an equiv alent lens and, importantly, although it will have a dierent power to the rst and second lens, it will need to focus the light at the focal points  and 9 in order to truly be ‘euivalent’ to the two-lens system. e appropriate euivalent power of the system can be calculated using uation ..

DEMO QUESTION 3.7

Equation 3.1

wo thin lenses of powers 1. and 1. are separated by a distance of  cm. f an obect is placed  cm in front of the rst lens where will the image form tep  etermine what we need to calculate image distance after second lens l tep  ene variables  5 1.  5 1. d 5 .m (we need to convert to metres) l 5 2.m (the question stated the obect was  cm ‘in ront o’ the rst lens, meaning it’s to the let o the surace this means it will have a negative distance and we need to convert to metres) n 5 . (nothing is specically mentioned, so we assume the primary medium is air) n9 5 . (nothing is specically mentioned, so we assume the secondary medium is air) tep  etermine necessary equation(s) L 5 n  l (odied Equation .1) L9 5 L 1  (odied Equation .) L 5 L  ( – dL ) (Equation 3.) L9 5 L 1  (odied Equation .) L9 5 n  l (odied Equation .) tep  alculate L 5 n  l L 5 .  . L 5 2. L9 5 L 1  L9 5 2. 1 1. L9 5 2. L 5 L  ( – dL ) L 5 2.  ( – (. 32.)) L 5 2.… L9 5 L 1  L9 5 2.… 1 1. L9 5 1.… l9 5 n  L l9 5 .  1.… l9 5 1.m he image will form . cm right of the second lens. (don’t forget the 6 sign (or the direction) and the units)

Equation 3.1 (explained)

Practice Questions: 3.7.1 wo thin lenses of powers 1. and 1. are separated by a distance of  cm. f an obect is placed  cm in front of the rst lens where will the image form 3.7.2 wo thin lenses of powers . and 1. are separated by a distance of  cm. f an obect is placed  cm in front of the rst lens where will the image form

e

equiv. power

1

1

2

dF1 F2

2 (distance

1

2)

DEMO QUESTION 3.8 wo thin lenses of powers 1. and 1. are separated by a distance of  cm. hat is the equivalent power of the system tep  etermine what we need to calculate equivalent power e tep  ene variables  5 1.  5 1. d 5 .m (we need to convert to metres) tep  etermine necessary equation(s) e 5  1  – (d) (Equation 3.1) tep  alculate e 5  1  – (d) e 5 (1) 1 (1) – (. 3 (1) 3 (1)) e 5 1. he equivalent power of the system is 1. (don’t forget the 6 sign and the units)

Practice Questions: 3.8.1 wo thin lenses of powers 1. and 1. are sepa rated by a distance of  cm. hat is the equivalent power 3.8.2 wo thin lenses of powers . and 1. are separated by a distance of . cm. hat is the equiva lent power

owever, we know from the section titled “erte ower” that the back verte power v9 describes the vergence of light leaving the system from left to right, and we know this light would form a focus at the secondary focal point 9. e also know that if the back verte power of the system diered to the power of the euivalent lens e, then they must also have dierent focal lengths. is means that, in most cases, the euivalent lens will need to be placed somewhere dierent to that of one of the eisting lenses. o how do we determine where it should go

Principal Planes e answer to the uestion in the previous section is that the euivalent lens will need to have two uniue positions along

CHAPTER 3

the optical ais one position for when light is travelling left to right in order to focus light at the secondary focal point 9, and a separate position for when light travels backwards through the system right to left to focus light at the primary focal point . ese uniue positions are called principal planes. e plane at which the euivalent lens would need to be located to focus light at the secondary focal point is called the secondary principal plane 99, and the location of the euivalent lens to focus light at the primary euivalent focal point is called the primary principal plane  ig. .. nce you know the euivalent power, the distance between the principal planes and the focal points can be calculated. e distance between the secondary principal plane and the secondary focal point is called the secondary equiva lent focal length fe9, and the distance between the primary principal plane and the primary focal point is called the pri mary equivalent focal length fe. ese can be calculated using uations ., .2 and .. n Equation 3.11 f e Fe Equation 3.11 (explained) secondary equiv. focal length  Equation 3.1

fe 

surround refr. index equivalent power

n Fe

primary equiv. focal length 

e

F1

surround refr index equivalent power

e

HP

Equation 3.13 (explained)

DEMO QUESTION 3.9 wo thin lenses of powers 1. and 1. are separated by a distance of  cm. hat is the secondary equivalent focal length of the system tep  etermine what we need to calculate secondary equivalent focal length fe tep  ene variables  5 1.  5 1. d 5 .m (we need to convert to metres) n 5 . (nothing is specically mentioned, so we assume the primary medium is air) n9 5 . (nothing is specically mentioned, so we assume the secondary medium is air) tep  etermine necessary equation(s) e 5  1  – (d) (Equation 3.1) fe9 5 n  e (Equation 3.11) tep  alculate e 5  1  – (d) e 5 (1) 1 (1.) – (. 3 (1) 3 (1.)) e 5 1. fe9 5 n  e fe9 5   1. fe9 5 1.m he secondary equivalent focal length is 1. cm. (don’t forget the 6 sign and the units)

Practice Questions:

Equation 3.1 (explained)

Equation 3.13

Thin Lenses

F2

3.9.1 wo thin lenses of powers . and 1. are separated by a distance of  cm. hat is the secondary equivalent focal length of the system 3.9.2 wo thin lenses of powers . and 1. are separated by a distance of  cm. hat is the primary equivalent focal length of the system 3.9.3 A multiple lens system has a secondary equivalent focal length of 1 cm. hat is the primary equivalent focal length of the system

H’P’

ewton’s orulae F

F’ fe

fv

•Fig. 3.11

fe

fv

iagram highlighting the secondary ( P ) and primary (P) principal planes for an optical system with two positively powered lenses (grey). he primary (fe) and secondary equivalent focal lengths (fe ) are identical albeit inverted to one another which is always true for equivalent focal lengths (see quation .).

If you can determine the euivalent focal lengths of a multiple lens system, then you can utilise a method called ewton’s formulae in order to calculate obect and image distances. owever, for ewton’s formulae to be useful, we need to be given the oect distance from the primary fo cal point , which is very dierent to the obect distance relative to the rst lens l, and we will also need to utilise the euivalent focal lengths fe and fe9. is will allow us to calculate the image distance relative to the secondary focal point 9. ig. .2 utilises the same lens system as demonstrated in ig. ., but this time it shows the relationship between obect distance  and image distance 9 relative to the focal points. ewton’s formulae can be used to calculate image distance using uation . or . see o . for advice

35

36 S EC TI ON 1

eoetric and asic ptics

Lens 1 (F1) HP

on how to know when to use step-along vergence or ewton’s formulae.

Lens 2 (F2) H’P’

Equation 3.1 Object

F

F’ fe

Image

fe’

Object distance (x)

Image distance (x’)

• Fig.

3.12 iagram highlighting the important points required for ewton’s formulae. his eample uses two positively powered lenses.

•O 3.1

Step-Along or Newton’s Forle

f presented with an optical question that provides you with an obect distance in front of a multiple lens system then you might be wondering how to decide whether it’s more appropriate to use a stepalong method or ewton’s formulae. he answer to this riddle lies in the description of the obect distance . f the obect distance is measured relative to the rst lens then you have been given the variable l (little l) and you need to use stepalong methods. n this instance there is no need to calculate equivalent power as you will consider each lens individually. . f the obect distance is given relative to the primary focal point then you have been given the variable  and you need to use ewton’s formulae. n this instance it is essential to calculate equivalent power to help you determine the equiva lent focal lengths. ig. B. shows the difference between the two distances.

A

Lens 1 (F1)

Object

Lens 2 (F2)

F

Object distance (l)

fe

2

xx

Equation 3.1 (explained) ( secondary equiv. focal length )2 Equation 3.1

fe fe

xx

Equation 3.1 (explained) rimary equiv. focal length

secondary equiv. focal length

DEMO QUESTION 3.10 wo thin lenses of powers 1. and 1. are separated by a distance of  cm. f an obect is placed  cm in front of the primary focal point of the system where will the image form tep  etermine what we need to calculate image distance (relative to secondary focal point)  tep  ene variables  5 1.  5 1. d 5 .m (we need to convert to metres)  5 2.m (negative distance and we need to convert to metres) n 5 . (nothing is specically mentioned, so we assume the primary medium is air) n9 5 . (nothing is specically mentioned, so we assume the secondary medium is air) tep  etermine necessary equation(s) e 5  1  – (d) (Equation 3.1) fe9 5 n  e (Equation 3.11) (fe ) 5 2 (Equation 3.1) tep  alculate e 5  1  – (d) e 5 (1) 1 (1) – (. 3 (1) 3 (1)) e 5 1. fe9 5 n  e fe9 5   1. fe9 5 1.m (fe ) 5 2 (.) 5 2 ( .)  ... 5 2 ( .)  ...  . 5 2 . 5 2 9 5 1.m he image will form . cm right of the secondary focal point. (don’t forget the 6 sign and the units)

Practice Questions:

B Object

F

Object distance (x)

• Fig. B3.1

llustration showing the difference between l (A) and  (B). t might help to do a little diagram like this when solving equations to decide which method to use.

ist.

3.10.1 wo thin lenses of powers . and 1. are separated by a distance of  cm. f an obect is placed  cm in front of the primary focal point of the system where will the image form 3.10.2 wo thin lenses of powers 1. and 1. are separated by a distance of  cm. f an obect is placed  cm in front of the primary focal point of the system where will the image form 3.10.3 A multiple lens system has a secondary equivalent fo cal length of 1. cm. f an obect is placed  cm in front of the primary focal point of the system where will the image form

CHAPTER 3

Thin Lenses

Magnication sing ewton’s orulae

DEMO QUESTION 3.11—cont’d

ow that we know how to use ewton’s formulae to calculate image distance, it’s also possible to determine the relative image sie h9 and orientation of the image by calculating the linear magnication (m) with uation ..

tep  etermine necessary equation(s) e 5  1  – (d) (Equation 3.1) fe9 5 n  e (Equation 3.11) (fe ) 5 2 (Equation 3.1) m 5 2 e (Equation 3.1) tep  alculate e 5  1  – (d) e 5 (1.) 1 (1) – (. 3 (1.) 3 (1)) e 5 1. fe9 5 n  e fe9 5   1. fe9 5 1.…m (fe ) 5 2 (.…) 5 2 ( .)  ... 5 2 ( .)  ...  . 5 2 ... 5 2 9 5 1... (don’t forget to keep all the numbers long in the calculator) m 5 2 e m 5 2 ( ...) 3 (1.) m5 2. he image is inverted (real) and .3 smaller than the obect. (don’t forget the 6 sign and the units)

Equation 3.1

m

fe x

x fe

x Fe

Equation 3.1 (explained) mag

primary equiv. focal length obj dist img dist.

secondary equiv. focal length

img dist.

equiv. power

DEMO QUESTION 3.11 wo thin lenses of powers 1. and 1. are separated by a distance of  cm. f an obect is placed  cm in front of the primary focal point of the system what is the linear magni cation of the image tep  etermine what we need to calculate linear magnication m tep  ene variables  5 1.  5 1. d 5 .m (we need to convert to metres)  5 2.m (negative distance and we need to convert to metres) n 5 . (nothing is specically mentioned, so we assume the primary medium is air) n9 5 . (nothing is specically mentioned, so we assume the secondary medium is air)

Practice Questions: 3.11.1 wo thin lenses of powers . and 1. are separated by a distance of  cm. f an obect is placed  cm in front of the primary focal point of the system what is the linear magnication of the image 3.11.2 wo thin lenses of powers 1. and 1. are separated by a distance of  cm. f an obect is placed  cm in front of the primary focal point of the system what is the linear magnication of the image

Test Your Knowledge ry the uestions below to see if you need to go over any sections again. ll answers are available in the back of the book. .3.1 hat is the denition of a thin lens .3. hat would a magnication of 2. tell us about the nature of the image .3.3 ow would you calculate euivalent power of two thin lenses in contact with one another

Reference . unnaclie , irst . Optics. 2nd ed.  ssociation of ritish ispensing pticians .

.3. hat is a principal plane .3. hat is the dierence between back verte focal length and secondary euivalent focal length .3. hat determines whether we use step-along vergence or ewton’s formulae to nd image distance with a multiple lens system

37

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4 Thick Lenses C H  P T   O  TL I  Introduction

Step-Along Method

Lens Thickness

Fresnel Lenses

Thick Lens Power Thick Lens Focal Length

Test Your Knowledge

Image Vergence and Virtual Objects Virtual Object Method

O    C T IV  After working through this chapter, you should be able to: Dene what a ‘thick lens’ is and calculate how to easure thickness nderstand the relationship between thickness curature o the suraces reractie inde and how these contribute to the power o the lens

alculate iage position using the irtual object euations or the step-along ergence euations Dene what a ‘Fresnel lens’ is

Introduction

because this, in combination with the curvature and refractive index, helps to understand the power of the lens. o start calculating the thickness, we need to calculate the sag s of a spherical lens, but in order to do this, we need to know the radius of curvature r of the surface and the value that euates to half the lens diameter . ig. . shows these variables labelled on an example lens. ag can be calculated using uation ..

Up to this point, we have considered lenses with such negligible thickness that it is possible to ignore the refractive index of the material and still reach reasonable estimates of image location and magnication, etc.; these are called thin lenses. However, it is more accurate to take both the thickness and the refractive index into account when performing these calculations. If the thickness is considered, the lens is referred to as a thick lens, and in these circumstances the individual lens surfaces need to be considered. is chapter will cover some of the fundamentals of thick lens theor and calculations.

(

s

Equation 4.1

2

y2 )

Equation 4.1 (explained)

(

sag

2

half lens diameter 2 )

Lens Thickness o measure the thickness of a lens, usuall corresponding to the central point of the lens centre thickness, it is rst important to be able to determine the sagitta (sag) and the edge thickness. e sag is dened as the height of a segment of a circle or sphere from arc to base, whereas the edge thickness is the phsical thickness at the edges of the lens. ig. . shows a diagram to explain these concepts visuall. It is important to know the thickness of the lens

tc

s te

• Fig. 4.1

Illustration showing a planoconvex lens with centre thickness (tc), edge thickness (te) and sag (s) labelled.

39

40 S EC TI ON 1

eometric and asic Otics

te

PCv

s

BCv

te

PM

te

y r

A

s1

s

s2

s1 s 2

• Fig. 4.2

Illustration showing a planoconvex lens with sag (s), radius of curvature (r) and half the diameter of the lens (y) labelled.

tc tcs  te

DEMO QUESTION 4.1 A -cm wide planoconvex lens has a radius of curvature of  cm. hat is the sag of the lens tep  etermine what we need to calculate sag, s tep  ene variables y 5 . m (good practice to convert to metres) r 5 . (refractive index of lens) tep  etermine necessary euation s 5 r – √(r – y) (Equation 4.1) tep  alculate s 5 r – √(r – y) s 5 . – √(. – .) s 5 . m s 5 . cm (don’t forget the units)

Practice Questions: 4.1.1 A -cm wide planoconvex lens has a radius of curvature of  cm. hat is the sag of the lens 4.1.2 A -cm wide planoconvex lens has a radius of curvature of  cm. hat is the sag of the lens 4.1.3 A -cm wide planoconvex lens has a radius of curvature of  cm. hat is the sag of the lens

e centre thickness tc can onl be calculated if ou know the sag s and the edge thickness te, and the euation for this will change depending on the tpe of lens. s shown in ig. ., the relationship between the front and back surface of the lens, and whether those surfaces are convex or concave, will aect the euation for centre thickness. nother important point to note here is that the sag is determined b the curvature of the surface, which helps to contribute to the power of the lens. at means that for high powered lenses, the thickness can increase so much that the lenses become heav. In an optical imaging sstem this would be ok, but if ou’re a person who needs glasses, heav lenses are not ideal uckil, the thickness can be reduced whilst maintaining the same power b changing the material of the lens to something of a higher refractive index. emember that higher refractive indices indicate that light is slowing more, leading to higher amounts of refraction. is ultimatel means that higher index lenses won’t need to be as curved leading to a smaller sag and smaller thickness to produce the same power.

Thick Lens Power In general terms, an lens in which the thickness exceeds an amount that is acceptable for thin-lens assumptions is

PCc te

B

s

tc tcte – s

tc

tc

tcs1  s2  te BCc

te

s2

s1

tc tcte – (s1  s2)

tc(s1  te) – s2 MM

te

s1

s2

tc tc(te – s2)  s1

•Fig. 4.3

Illustration showing sag (s), centre thickness (tc) and edge thickness (te) for positively powered (A) planoconvex (v), biconvex (v) and plus meniscus () lenses, and negatively powered () planoconcave (c), biconcave (c) and minus meniscus () lenses. ach lens type utilises a slightly different euation for calculating centre thickness.

classed as a thick lens. However, in some cases, lenses that are phsicall ver thin are classed as thick lenses if their front vertex and back vertex powers are substantiall dierent, for example, contact lenses lenses that sit on the front surface of the ee. e ke thing to note for thick lenses is that ou need to take into account the refractive index of the material, so it is a little like having two thin lenses separated b a distance, with a refractive index change in the middle. Importantl, with thick lenses, we can’t assume that incident light ras will onl refract once as the pass through; instead, we think of them as refracting at each individual surface of the lens. ig. . depicts an example thick lens to show that the light will refract twice as it passes through the lens; once at the front surface, and again at the back surface. o determine the overall power of a thick lens, we need to calculate the power of the front  and back  surfaces individuall, and we need to know how thick t the lens is. o start with then, to calculate the power of each surface, we can use a familiar calculation from chapter  which has been updated in uation .. is will involve using what we know about the radius of curvature of each surface r and r and the refractive index of the lens and surrounding material n in order to calculate the respective powers. However, we need to be careful to remember that we

Thick Lenses

CHAPTER 4

It’s ke here to make sure we know whether the radius of curvature will be positive or negative – remember that measurements alwas start from the relevant surface to the point of measurement in this case the centre of curvature, . s we learned in chapter , if the radius distance is measured from left to right in the same direction as light then it will be positive; otherwise, it will be negative. e trick is to remember that if the front surface is convex, it will have a positive radius of curvature, whereas if it’s concave, it will have a negative radius of curvature. imilarl, if the back surface is convex, it will have a negative radius of curvature, and if the back surface is concave, it will have a positive radius of curvature ig. .. In order to determine how the thickness of the lens and the refractive index work together to refract the light, we need to calculate the reduced thickness t¯; pronounced teereduced or tee-bar. e do this b dividing the measured thickness t b the refractive index of the material ng, as shown in uation .. t Equation 4.3 t  ng

F’

f’

•Fig. 4.4

Illustration showing parallel light entering a convex thick lens. otice that the light refracts at both the rst and second surface before coming to a focus at the secondary focal point.

n1

n1’/n2

n2 r1

c2

c1

r2

Equation 4.3 (explained) reduced thickness  •Fig. 4.5

Illustration showing relationship between radius of curvature for the front (r) and back (r) surface of the lens, along with the refractive index variables. he diagram also shows the centre of curvature of the front surface () and the back surface ().

nce we have all the information needed, we can utilise and modif a familiar euation from chapter  to calculate the euivalent power of the lens uation .. Equation 4.4

alwas assume light travels from left to right, meaning that at the rst surface , the primar refractive index n will be air, and the secondar refractive index n9 will be the lens; however at the back surface  this will be swapped as the light will travel from the lens n to the air n9. ig. . explains the relationship between the radius of curvature values and the refractive index values. n1 n1 n2 n2 F1 F2 Equation 4.2 r1 r2

2

t F1 F2 1

1

+ve r2

c1 c1

2)

A -cm thick biconvex lens (refractive index .) has a front surface with a radius of curvature of  cm and a back surface with a radius of curvature of  cm. hat is the power of the lens tep  etermine what we need to calculate euivalent power, e

B +ve r1

2

DEMO QUESTION 4.2

A

–ve r2

1

(reduced thickness

secondary refr.index primary refr.indeex radius of curvature

c2

e

Equation 4.4 (explained)

Equation 4.2 (explained) power

thickness refractive index of lens

–ve r1

c2

•Fig. 4.6 Illustration showing relationship between convex (A) and concave () surfaces and whether they possess a positive (1ve) or negative ( ve) radius of curvature.

41

42 S EC TI ON 1

eometric and asic Otics

DEMO QUESTION 4.2—cont’d tep  ene variables t 5 . m (we need to convert to metres) ng 5 . (refractive index of lens) r 5 1. m (because the front surface is convex the radius is positive and in metres) r 5 2. m (because the back surface is convex the radius is negative and in metres) n 5 . (nothing is specicall mentioned so we assume the primar medium is air) tep  etermine necessary euation  5 (n – n)  r (Equation 4.)  5 (n – n)  r (Equation 4.) t 5 t  ng (Equation 4.) e 5  1  – t  (Equation 4.4) tep  alculate  5 (n – n)  r  5 (. – .)  .  5 1.… (remember to keep the number long in the calculator)  5 (n – n)  r  5 (. – .)  .  5 1. t 5 t  ng t 5 .  . t 5 .… (remember to keep the number long in the calculator) e 5  1  – t  e 5 .… 1 . – (.… 3 . 3 .) e 5 1. (don’t forget the 6 sign and the units)

ow, in order to determine the location of the secondar 9 and primar focal points  relative to the front and back surface of the lens, we need to calculate the front and back vertex focal lengths, which means we rst need to calculate front and back vertext power. emember that vertex power describes the vergence of light leaving a sstem in a particular direction, providing the incident approaching light is parallel ero vergence. athematicall, we can calculate front (v) and ack (v9) vertex poer b modifing the thin lens euation for a multiple lens sstem see chapter  to consider the reduced thickness instead of the distance between the lenses uations . and .. nce we have those values we can calculate front (fv) and ack (fv9) vertex focal length uations . and .. Equation 4.

F1

Fv

F2 t F1F2 1 t F1

Equation 4. (explained) lens1 lens 2 (thickness lens1 lens 2) back vertex power 1 (thickness lens1) Equation 4.

Fv

F1

F2 t F1 F2 1 t F2

Equation 4. (explained) front vertex power

lens

lens (thickness lens1 1 (thiickness lens2)

Practice Questions: 4.2.1 A thick lens (refractive index .) has a convex front surface with a radius of curvature of  cm. hat is the power of the front surface 4.2.2 A thick lens (refractive index .) has a concave back surface with a radius of curvature of  cm. hat is the power of the back surface 4.2.3 A -cm thick lens (refractive index .) has a front surface power of . and a back surface power of 1.. hat is the power of the lens

Equation 4.

Equation 4. (explained) surrounding refractive index back vertex focal length  back vertex power Equation 4.1

fv 

Equation 4.1 (explained)

If it’s possible to calculate the euivalent power of a thick lens, then it is also possible to calculate the secondary (fe9) and primary (fe) equivalent focal lengths uations . and .. emember that the euation for the primar euivalent focal length reuires ou to include a minus sign. n fe Equation 4. Fe

front vertex focal length 

secondary equiv. focal length  Equation 4.

fe

Equation 4. (explained) primary equiv. focal length

primary refractive index equivalent poweer n Fe primary refractive index equivalent poweer

n Fv

fv

Thick Lens Focal Length

Equation 4. (explained)

lens2 )

n Fv

surrounding refractive index front vertex power

ow, thinking about this logicall, we know both that vertex focal lengths describe the distance between the surfaces of the lens and their respective focal points and that the euivalent focal lengths describe the distance between the principal planes and the same focal points. is means that the location of the principal planes can be determined mathematicall b working out the dierence between the vertex focal length and the euivalent focal length for each focal point. or example, if I knew m secondar focal point was 1 cm right of the back surface which would correspond to the back vertex focal length and m secondar euivalent focal length was 1 cm, then I can deduce that the secondar principal plane must exist 2 cm left of the back surface 1 cm 2 1 cm 5 2 cm. is is highlighted in uations . and ., where we can calculate the distance between the primar

CHAPTER 4

principal plane and the rst surface of the lens (e), and the distance between the secondar principal plane and the back surface of the lens (e9) Equation 4.11

1

v

e

Equation 4.11 (explained) 1

Equation 4.12

Thick Lenses

DEMO QUESTION 4.3—cont’d here is the secondary principal plane, relative to the back surface of the lens 4.3.2 Imagine a -cm thick biconvex lens (surface powers 1. and 1. refractive index .). here is the primary principal plane, relative to the back surface of the lens

Image Vergence and Virtual Objects 2

v

e

Equation 4.12 (explained) 2

DEMO QUESTION 4.3 Imagine a .-cm thick biconvex lens (surface powers 1. and 1. refractive index .). here is the secondary principal plane, relative to the back surface of the lens tep  etermine what we need to calculate secondary principal plane location, A tep  ene variables  5 1.  5 1. t 5 . m (we need to convert to metres) n 5 . (nothing is specicall mentioned so we assume the primar medium is air) n9 5 . (refractive index of lens) tep  etermine necessary euation(s) t 5 t  ng (Equation 4.) v9 5 ( 1 - t )  (- t ) (Equation 4.) fv9 5 n  v (Equation 4.) e 5  1 - t  (Equation 4.4) fe9 5 n  e (Equation 4.) A9 5 e9 5 fv – fe (Equation 4.1) tep  alculate t 5 t  ng t 5 .  . t 5 .… v9 5 (1- t )  (- t ) v9 5 ( 1 - ...3  3 )  (- (...3 )) v9 5 1.... fv9 5 n  v fv9 5 .  1... fv9 5 1... e 5 1- t  e 5 1- ...3  3  e 5 1... fe9 5 n  e fe9 5 .  1... fe9 5 1... A9 5 e9 5 fv – fe A9 5 e9 5 1... 2 1... A9 5 e9 5 2. he secondary principal plane for this lens exists . cm left of the back surface of the lens. (don’t forget the 6 sign and the units)

Practice Questions: 4.3.1 Imagine an .-cm thick biconvex lens (surface powers 1. and 1. refractive index .).

nce we know the power of each surface, it’s possible to calculate image location l9 b determining the vergence approaching and leaving each surface of the lens. ere are two possible methods for this, and so in order to be thorough and for extra maths practice, I’ll go through them both.

Virtual Object ethod e rst method we’ll go over is called the ‘virtual obect’ method, because it assumes that the image formed at the rst surface will become the new virtual obect for the second surface. e euations used for is method will be extremel familiar as we’ve alread seen them in chapters  and , but I have included them here with updated descriptions. s we have two surfaces to consider, we will also be utilising our hand subscript numbers to help us work out which surface we’re dealing with. n L Equation 4.13 l Equation 4.13 (explained) primary refractive index object vergence  object distance from surface F

Equation 4.14 Equation 4.14 (explained)

power Equation 4.1

L

n l

l

n L

Equation 4.1 (explained) image distance 

secondary refr.index image vergence

ook at the example diagram shown in ig. ., where a biconvex thick lens refractive index . is located in air. emember that the primar refractive index will represent air when light is entering the lens at the front surface, but it will also represent the lens when light is leaving the lens again at the back surface, so be careful to check our numbers here. nother thing to think about is the presence of the virtual obect. is describes the location of the image that would be formed b the rst surface  if the second surface didn’t exist. It’s referred to as a virtual obect because this hpothetical image then acts as the obect for the second surface. e onl problem is that we’ll have initiall

43

44 S EC TI ON 1

eometric and asic Otics

F1

t

F2

Object

Image l2

Virtual object l2 = l1 –t

l1 l1

• Fig. 4.7

iagram using example biconvex lens to highlight the relationship between surface powers ( and ), lens thickness (t), obect distance from the apex of the rst surface (l ), virtual obect distance from the apex of the second surface (l  5 l -t) and nal image distance relative to the second surface (l  ).

calculated its distance relative to the rst surface , and so we’ll need to account for the thickness of the lens t when we perform our calculations for the back surface . lease note that when using the virtual obect method, we use the genuine thickness of the lens and not the reduced thickness. DEMO QUESTION 4.4 An obect is placed  cm in front of a -cm thick biconvex lens (refractive index .) with a front surface power of 1. and a back surface power of 1.. here does the image form relative to the back surface of the lens tep  etermine what we need to calculate image distance relative to the back surface, l tep  ene variables l 5 2. m (negative distance and we need to convert to metres)  5 1.  5 1. t 5 . m (we need to convert to metres) n 5 . (nothing is specicall mentioned so we assume the primar medium is air) n9 5 . (refractive index of lens) tep  etermine necessary euation(s)  5 n  l (Equation 4.1) 9 5  1  (Equation 4.14) 9 5 n’  l’ (Equation 4.1) tep  alculate  5 n  l  5 .  .  5 2. 9 5  1  9 5 2. 1 . 9 5 1. 9 5 n  l l9 5 n   (rearrange to get l1 ) l9 5 .  . l9 5 1.… (this is the distance of the virtual obect from the front surface) l 5 l’ – t (need the distance of the virtual obect from the back surface) l 5 1.… – . l 5 1.…  5 n  l  5 .  1.… (the primar refractive index is now the lens)

DEMO QUESTION 4.4—cont’d  5 1.… 9 5  1  9 5 .… 1 . 9 5 1.… 9 5 n  l l9 5 n   (rearrange to get l ) l9 5 .  .… l9 5 1. m he image forms . cm right of the back surface of the lens. (don’t forget the 6 sign and the units)

Practice Questions: 4.4.1 An obect is placed  cm in front of a -cm thick biconvex lens (refractive index .) with a front surface power of 1. and a back surface power of 1.. here does the image form relative to the back surface of the lens 4.4.2 An obect is placed  cm in front of a .-cm thick biconcave lens (refractive index .) with a front surface power of . and a back surface power of .. here does the image form relative to the back surface of the lens

telong ethod e euations used for the ‘step-along method’ will also be extremel familiar, as we’ve alread seen them in chapter  in relation to thin lenses. e onl dierence is that instead of two thin lenses separated b a distance, we have two surfaces of a lens of a certain thickness. ll we need to do is replace the distance d variable with one for reduced thickness t¯ which takes into account the distance and the refractive index, as shown in uation .. L1 L2 Equation 4.1 1 t L1 Equation 4.1 (explained) verg. at F2

verg. leaving F1 1(thickness verrg. leaving F1 )

o show that this method is euivalent to the virtual obect method, I have used exactl the same uestion in the demo – ou can check it ourself at home

CHAPTER 4

Side view

DEMO QUESTION 4.5 An obect is placed  cm in front of a -cm thick biconvex lens (refractive index .) with a front surface power of 1. and a back surface power of 1.. here does the image form relative to the back surface of the lens tep  etermine what we need to calculate image distance relative to the back surface, l tep  ene variables l 5 2. m (negative distance and we need to convert to metres)  5 1.  5 1. t 5 . m (we need to convert to metres) n 5 . (nothing is specicall mentioned so we assume the primar medium is air) n9 5 . (refractive index of lens) tep  etermine necessary euation(s)  5 n  l (Equation 4.1) 9 5  1  (Equation 4.14) t 5 t  ng (Equation 4.)  5   ( – t ) (Equation 4.1) ’ 5 n  l (Equation 4.1) tep  alculate  5 n  l  5 .  .  5 2. 9 5  1  9 5 2. 1 . 9 5 1. t 5 t  ng t 5 .  . t 5 .… (remember to keep the number long in the calculator)  5   ( – t )  5 .  ( – (.…3 .))  5 1.… (remember to keep the number long in the calculator) 9 5  1  9 5 .… 1 . 9 5 1.… 9 5 n  l l9 5 n   (rearrange to get l ) l9 5 .  .… l9 5 1. m he image forms . cm right of the back surface of the lens. (don’t forget the 6 sign and the units)

Practice Questions: 4.5.1 An obect is placed  cm in front of a -cm thick biconvex lens (refractive index .) with a front surface power of 1. and a back surface power of 1..

Thick Lenses

Front view

A

B

•Fig. 4.8 iagram showing side and front views of a biconvex lens (A) and the resnel euivalent (). DEMO QUESTION 4.5—cont’d here does the image form relative to the back surface of the lens 4.5.2 An obect is placed  cm in front of a -cm thick biconcave lens (refractive index .) with a front surface power of . and a back surface power of .. here does the image form relative to the back surface of the lens

Fresnel Lenses s we discussed brie¢ earlier in the chapter, one thing that is true about lenses is that as the power increases, usuall the thickness and the curvature also increases. is naturall increases the weight and sie of the lens, which can sometimes be a problem when applied to real-world optical problems. or example, I wear a 2. lens in m glasses, which can be so thick and heav in relative terms that I usuall have to purchase high index lenses, but I actuall prefer to wear contact lenses for ease. n alternative solution to this lens problem, however, is to utilise resnel pronounced Freh-nel lenses, which are lenses divided up into concentric rings, as illustrated in ig. .. Here ou can see that each small section of the resnel lens  possesses the euivalent curvature of the planoconvex lens , but with a dramaticall reduced thickness. is leads to thinner, lighter lenses, but one of the limitations of this tpe of lens is that image ualit will be slightl reduced due to diraction see chapter  occurring at the ridges between each curved ring.

Test Your Knowledge r the uestions below to see if ou need to review an sections again. ll answers are available in the back of the book. .4.1 hat is the denition of a thick lens relative to a thin lens .4.2 hat is the dierence between the sag and the edge thickness

eerence . avis , ühnlen . ptical design using resnel lenses basic principles and some practical examples. Optik & Photonik. ;-.

.4.3 In an euiconcave lens, does the back surface have a positive or negative radius of curvature .4.4 hat is a virtual obect .4. hat causes the resolution of a resnel lens to be reduced relative to a ‘regular’ lens

45

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5 The Reduced Eye and Spherical and Cylindrical Lenses C H A  T E R   T I E Introduction The Human Eye

Cylindrical Error Cross-cylinder Technique

A Reduced Eye

Focimetry

Spherical Refractive Error Myopia Hyperopia

Test Your Knowlede

   E C T I ES After working through this chapter, you should be able to: Outline all the important features of a reduced eye Outline the dierence beteen a spherical refractie error and a cylindrical refractie error Eplain hat the far point of the eye is and be able to calculate it

nderstand ho poer of a lens ill need to be altered as its distance from the eye increases nterpret a poer cross

Introduction

mportantly, we can imagine that if light travelled from one section of air into another section of air that was curved but had the same refractive index, it probably wouldn’t do very much to the path of the light we can even conrm this mathematically using uation . from chapter   5 n9 – n  r – here we can see that if the primary n and secondary refractive indices n9 are identical then no matter the curvature, the power will always be ero. is means that a key feature of the human eye focusing light is that the cornea has a dierent refractive index to the air, and this has been measured to be roughly .. is means that the human eye can focus light because  it is curved, and  it is a dierent refractive index to air please see chapter  for a do-it-yourself demonstration of this. is also explains why your vision is blurry if you try to open your eyes underwater – the refractive index dierence between water . and the cornea . is much smaller than that of air . and the cornea ., so the light doesn’t focus properly. s one nal fact about the human eye before we get into the depths of chapter  light can enter the eye through the convex cornea because the cornea is completely transparent please see biology textbooks for a cool explanation of how this is possible, which means the light will then travel

Now that we know the basics of thin and thick lenses, we need to spend a little bit of time focusing on clinical examples of corrective lenses. However, in order to do that eectively, we need to rst consider the power of the human eye.

The Human Eye s you will no doubt be aware, the human eye is a particularly complex anatomical structure, and one of its many functions is to focus light onto the back of the eye the retina. nterestingly for us, the human eye is able to focus light by adding convergence to the incoming light. is means it must be positively powered, which if we think back to chapter  suggests that its front surface must be curved and, in particular, convex. Now, given that it’s likely that you have at least one eye, this probably comes as no surprise to you because we know that the front surface the cornea is domed and pokes outwards in a convex way from the front surface of the eye ig. ..

47

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Retina Fovea

Cornea

Cornea

•Fig. 5.1

An illustration of the side view (prole) of a person’s face. The box outlined with a blue dashed line indicates that the image on the right is a zoomed-in cross section of the side view of the ee. n both images the cornea is labelled as the front convex dome on the front of the ee. n the image on the right the retina (ellow) and fovea are identied as well.

through the pupil to be focused on the retina. e retina is the physiological structure at the back of the inside of the eye that processes the light signal and transduces it into a signal the brain can interpret. e central part of the retina is called the fovea, and this is important because this part of the eye provides us with our central vision.

A Reduced Eye e previous section has briey covered the complexities of human vision to explain that the human eye has a positive power and that the goal is to focus the incoming light onto the retina at the back of the eye. Now, as we’ll remember from chapter  vergence, the amount of power of a surface will greatly aect where the image forms. n the human eye, we need the image to form on the back of the eye, which is approximately . mm behind the front surface of the eye, and this means there’s little room for error. However, this distance from the front of the eye to the back will vary depending on individual factors such as height and age, etc. roviding the incoming light is focused with the right power for the distance of the retina from the cornea, then the eye is appropriately proportioned and the image will be in focus emmetropic eye. f, however, the light focuses too far in front of the retina myopic or too far behind the retina Emmetropic eye

Myopic eye

Fovea

•Fig. 5.2

hyperopic, then this is called a refractive error, meaning the eye has refracted the light inappropriately. ig. . shows examples of this in a simple diagram of an eye. e important thing to remember here is that because the emmetropic eye in general terms is powered to focus light from innity ero vergence at the retina, this means the distance between the cornea and the retina is the secondary focal length of the eye, and the fovea corresponds to the secondary focal point (F9) However, if we wanted to calculate how light travels through the eye mathematically, there would be a lot to consider – the power of the cornea, the refractive index changes cornea, aueous, lens and vitreous, the power of the lens and the relative distances of all the front and back surfaces of the structures. uckily for us, we can use a simplied model of an eye in order to make this easier this simplied model is referred to as a reduced eye e reduced eye assumes a dioptric corneal power of 1., meaning the convex corneal radius of curvature is 1. mm. is also then assumes an axial length length of the eye from cornea to retina of 1. mm. mportantly, the eye also possesses a dierent refractive index n9 5 . relative to the air outside. ig. . shows how the reduced eye can be drawn and ox . shows how this can be calculated. Hyperopic eye

Fovea

imple diagram of refractive errors. The fovea is the central part of the retina. mmetropic ees focus light (shown as blue lines) at the fovea on the retina whilst mopic ees focus light in front of the fovea and hperopic ees focus light behind the fovea.

Fovea

CHAPTER 5

F = +60.00D n = 1.00

Reduced eye n’ = 1.333

C e (cornea)

F

–16.67 mm (f)

F’ (Fovea)

+5.55 mm (r)

+22.22 mm (f’)

• Fig. 5.3

educed model ee with power distances and refractive indices labelled appropriatel.

Now, we know that not all obects in the world will be an ‘innite’ distance away from us, but we are able to account for changes in distance by using our lens accommodation. is book will not discuss accommodation, but it is good to be aware that we are only discussing distant obects as examples.

Spherical Refractive Error s discussed previously, in nature not all eyes are appropriately powered for their length, which can lead to myopia or hyperopia, both of which are examples of a spherical refractive error measured in dioptres DS or D.  spherical error is dened by a refractive error that is the same in all meridians, so for example, if an eye had a spherical error of 2., the error would be 2. along all possible axes, as shown in ig. .

yopia yopia shortsightedness is characterised by the eye either being too powerful or the axial length of the eye being too long – both of which result in an image that forms in front of the retina ig. .. e result is that people with myopia will see blurry images of obects at far-away •BOX 5.1

The Reduced Eye and Spherical and Cylindrical enses

49

distances. However, they will still be able to see obects closer to them although, how close will depend on how large the refractive error is. or example,  am myopic, and  can only see clearly without my lenses to about . cm away from my eye before things start to get a little blurry, meaning that . cm corresponds to the far point of my eye. However, as  can still see obects clearly at . cm, the vergence of an obect at that location will be focused on my retina, and we can use that information to calculate my refractive error. Ln / l L  1.00 /  0.1481 L 6.75 uation . from chapter  helps us to determine that an obect . cm in front of my eye will have a far point vergence of 2. at my cornea, which is then focused neatly onto the retina of my eye. is means that my eye needs 2. of vergence removed from the visual signal in order to focus correctly, so we can therefore refer to the refractive error of my eye as 2.. f we assume that my cornea is appropriately powered at 1., we can calculate the image vergence and then the image distance which will euate to the axial length of the model eye. L L  F L6.75  60.00 L53.25 l n / L l 1.333 /  53.25 l0.02503m l25.03mm (converted to mm) ese euations show that, assuming a ‘typical’ corneal power and the refractive index of the reduced eye, my eye would be estimated to be nearly  mm longer than the model, reduced eye . mm. aken together, this means that in order to allow me to see farther than a measly . cm,  need corrective, spherical lenses. ese lenses could be contact lenses, resting on the front surface of my eye, or lenses in a pair of glasses frames, sitting slightly in front of my eye.

How Can We Calculate the Reduced Eye?

As discussed in the main text the reduced ee assumes a refractive index of . and a corneal curvature of . mm. As our model cornea would be a convex spherical surface relative to the light entering the ee we now the centre of curvature needs to be positioned to the right of the cornea (within the ee) and that it will therefore have a positive distance. e can then use uation . from chapter  to determine the power of the cornea whilst being careful to remember to convert to metres 5n -nr  5 . – .  1.  5 1.

e can then use the power of the cornea to calculate the emmetropicall appropriate distance of the retina whilst being careful to remember that the refractive index of the ee is not the same as air f 5 n9   f 5 .  1. f 5 1. m f 5 1. mm (converted to mm)

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–2.00D –2.00D

–2.00D –2.00D

–2.00D –2.00D –2.00D

–2.00D

•Fig. 5.4

A diagram to show that spherical error is the same along all axesmeridians of the ee.

Now, in simple terms, if my eye has a refractive error of 2., as we’ve determined, then there is 2. of vergence that needs correcting at my cornea, so  could prescribe myself a contact lens of the matching power in order to correct my vision. is is because with contact lenses, we need the secondary focal point 9 of the lens to correspond to the far point of the eye, and there aren’t any complicated gaps between the lens and my cornea to think about. However,  usually like to be over-corrected, so ’ll always ask for a 2. lens, but that’s because ’m an awkward patient n contrast to this, when prescribing glasses where the lens sits in a frame there is a small amount of air between the back surface of the lens and the front of the eye that will aect the vergence of light as it approaches the eye. is means this distance needs to be taken into account when we prescribe lenses, and this distance between the lens and the cornea is referred to as the back vertex distance ig. . for an example.

e trick to determining the appropriate power of the lens reuired is by measuring the back vertex distance on the patient’s chosen frames and then taking that distance away from the far point distance. n the example in ig. ., the back vertex distance  is 2. cm relative to the far point, which is 2. cm, meaning the lens placed within the frames needs to have a secondary focal length that corresponds to the distance between the back surface of the lens and the far point. e can calculate this by subtracting the back vertex distance from the far point, for example 2. cm – 2. cm 5 2. cm. e can then use this distance as our focal length f  which will allow us to calculate the power  of the lens. is can be achieved using the power calculation as shown below Fn / f F  1.00 /  0.1331 (converted to metres) F 7.51D is suggests  would need a 2. lens if it were to sit . cm in front of my eye and also shows that the further away from my eye it is, the stronger the lens will need to be, because decreasing the focal length will increase the power respectively.

Hyperopia Now, hyperopia also known as long-sightedness is characterised by the refractive power of the eye being too weak or the axial length of the eye being too short – both of which result in an image that forms behind the retina see ig. .. e result of this is that people with hyperopia will see

Sam’s myopic eye Object

Fovea

–14.81 cm (far point

Object

Fovea

Any distance further than far point

• Fig. 5.5

A reduced ee showing m brain’s interpretation (shown as thought bubbles) of an obect if it was at the far point (top) relative to being farther awa (bottom).

CHAPTER 5

The Reduced Eye and Spherical and Cylindrical enses

to slightly complicate things, the light needs to be converging as it approaches his cornea and will therefore be producing a virtual obect behind the front surface of the eye with a positive distance for the corresponding far point. athematically, we can determine the distance of this virtual obect by using the vergence

Frames

ln / L l  1.00 /  1.00 l 1.00 m l 100.00 cm

Back vertex distance

•Fig. 5.6

A diagram showing the bac vertex distance.

Here we can see that, to focus on the retina, the virtual obect will need to be  cm behind the front surface of the eye, as shown in ig. . f we assume, then, that my partner’s cornea is appropriately powered at 1., we can calculate the image distance which will euate to the axial length of the model eye. L L  F L1.00  60.00 L61.00 l n / L l  1.333 /  61.00 l0.02185 m l21.85 mm (coonverted to mm)

–14.81 cm

Fovea

Far point Lens BVD –1.5 cm

• Fig. 5.7

A diagram showing the bac vertex distance () relative to the far point of m example mopic ee.

blurry images of obects placed at close-up near distances. However, they will still be able to see obects far away although how well they see them will depend on how large the refractive error is. or example, my partner is hyperopic by approximately 1.. is means that if left uncorrected and ignoring the contribution of the accommodative system, light will only focus correctly on his retina if it has 1. of vergence at the cornea of his eye. However,

ese euations show that, assuming a ‘typical’ corneal power, my partner’s eye would be estimated to be approximately . mm shorter than the model, reduced eye. aken together, this means that in order to allow my partner to be able to see completely clearly, he would need corrective, spherical lenses. gain, these lenses could be contact lenses, resting on the front surface of his eyes, or lenses in a pair of glasses frames, sitting slightly in front of his eye. ust as with myopia, a refractive error of 1. means there is 1. of vergence that needs correcting at his cornea, so my partner could wear a contact lens of the matching power. is is because with contact lenses, we still need the secondary focal point 9 to correspond to the far point of the eye without troubling ourselves with any pesky back vertex distances, which in this example is 1. cm behind the front surface of the eye.

Sam's OH's hyperopic eye Object

Fovea

+100.00 cm (far point)

• Fig. 5.

A reduced ee showing m partner’s or other half’s () brain’s interpretation (shown as a thought bubble) if light is producing a virtual obect at the far point behind the front surface of the ee.

51

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+100.00 cm

Fovea Far point Lens BVD –1.5 cm

•Fig. 5.

A diagram showing the bac vertex distance () relative to the far point of m partner’s example hperopic ee.

However, when prescribing glasses, we again need to consider the back vertex distance. e dierence is that with myopia the increasing back vertex distance reduces the reuired focal length and therefore strengthens the reuired lens the further away from the eye it gets. ith hyperopia, the increasing back vertex distance will be increasing the focal length, which will, in turn, reuire a lower-strength lens relative to the contact lens. gain, to calculate this, we must measure the back vertex distance on the patient’s chosen frames and then take that distance away from the far point distance. n the example in ig. . the  is 2. cm relative to the far point, which is 1. cm, meaning the lens placed within the frames needs to have a secondary focal length that corresponds to the distance between the back surface of the lens and the far point. e can calculate this by subtracting the back vertex distance from the far point, for example 1. cm – 2.cm 5 1. cm. e can then use this distance as our focal length f  which will allow us to calculate the power  of the lens we would need to use. is can be achieved using the power calculation as shown below F n / f F  1.00 / 1.015 (converted to metres) F 0.99D is suggests that my partner would benet from a 1. lens if it were to sit . cm in front of his eye, and it also conrms that the further away from his eye it is, the weaker the lens will need to be, because increasing the focal length will decrease the power.

Cylindrical Error o far, we have discussed spherical errors, dened as errors in which the eye possesses the same refractive error along all meridians. However, it is possible, and common, to have a type of refractive error that has a dierent power along a particular meridian, called a cylindrical error measured in dioptres DC or D. linically, people with this type of error are described as having astigmatism. ig. . shows a simplied demonstration of this.

–0.50D –0.50D

–2.00D

–0.50D

–0.50D –2.00D

–0.50D –0.50D

• Fig. 5.1

A diagram to show that clindrical error would involve a difference in power along one axis of the ee.

o correct for astigmatism, it’s necessary to utilise a type of cylindrical lens. e simplest cylindrical lens is a planocylindrical lens, shown in ig. .. is type of lens only possesses power along one axis, called the power meridian. e perpendicular axis axis meridian or cylinder axis will not possess any power .. ith a planocylindrical lens, the optical axis would exist perpendicular to both the front and back surface of the lens. f appropriately powered and angled along the astigmatic axis of the eye, it can correct the incoming light and help it remain in focus at the retina. t is possible, however, to have a lens that corrects for both spherical and cylindrical refractive error, called a spherocylindrical lens ig. .. n this type of lens, the power meridian still carries the power of the cylindrical element of the lens, but this time the axis meridian will carry the power of the spherical element. ne way clinicians can diagrammatically depict the power of a spherocylindrical lens is by using a power cross. is is where a very simplied lens is drawn as two perpendicular lines ig. .. f both the vertical and horiontal lines have the same power then it is depicting a spherical lens, whereas if one line is plano pl or a dierent power, then it is a cylindrical lens. o determine the combined power of a spherocylindrical lens, you can add together the two lenses, as shown in ig. .. lease note that the example in ig. . shows minus-cylinder form where the cylindrical power is depicted as negative, but some people

CHAPTER 5

0.00D

AM

5.00D

OA

5.00D

PM

0.00D

• Fig. 5.11

A diagram showing a planoclindrical lens. ere ou can see that the power of the lens exists along the power meridian ( the curved surface) whilst there is zero power along the axis meridian (A the plane surface). The optical axis (A) is shown perpendicular to both front and bac surfaces of the lens.

Spherical

Cylindrical

The Reduced Eye and Spherical and Cylindrical enses

use positive-cylinder form where the cylindrical power is positive. Now, as we’ve mentioned, the cylindrical power will likely sit along a particular axis of the eye, and so when writing the prescription of a patient with astigmatism, clinicians need to report the spherical power, the cylindrical power and then the axis of the cylinder. mportantly, the axis of the cylinder will correspond to the orientation where the cylinder lens has no power the axis meridian, which is perpendicular to the power meridian – and this is something that often causes errors because it feels counterintuitive to describe the cylinder by its nonpowered axis  nd it helps to think of it as a rotating cylinder – for example, if asked to describe the image in ig. ., although we know the power exists along the power meridian , we’d probably describe the lens as being oriented vertically, which corresponds to the nonpowered meridian . However, when describing the orientation of the axis meridian in a cylindrical lens, we need to understand the standard notation for orientation. s a general rule, we measure the angle by using the  o’clock position right, middle as °, and then measure the angle in an anticlockwise direction from that point ig. .. is is the same no matter which eye is being refracted. Now, we also only need to consider angles between ° and °, as anything between ° and ° has a corresponding angle between ° and °. mportantly, the horiontal line is considered °, not °. is means in our example in ig. ., the lens would be described as having an axis of °. o describe the power of a lens in a way that’s easy for people to understand, the typical lens notation is written as Spherical power

•Fig. 5.12

A diagram showing a cross section of an example spheroclindrical lens comprising planoclindrical (clindrical) and positive meniscus (spherical) components.

Spherical

Cylindrical

2.50D

0.00D (pl)

B

2.50D  Spherical

• Fig. 5.13

7.50

15.00

–1.50D

2.50D

2.50D –1.50D 

Cylindrical

x-axis

ake sure to include the sign of the powers 1  2. n ig. . we have an example spherocylindrical lens comprising 1. spherical power and 2. cylindrical power at an angle of °, leading to a lens notation of

0.00D (pl)

2.50D

A

Cylindrical power

1.00D

Sphero-cylindrical

A diagram showing power crosses for spherical and clindrical lenses (A). The combined powers along each axis can produce a power cross for a spheroclindrical lens (). A zero-power plano (pl) orientation is depicted as pl.

x 

53

54 S EC TI ON 1

eometric and asic ptics

A

s some people work in minus-cylinder forms and some work in positive-cylinder forms, it may be necessary to convert one notation to the other. o convert the minuscylinder form to positive-cylinder form, we need to transpose the values. is is achieved by adding the spherical power to the cylinder e.g. 2. 1 . 5 2., then changing the sign but not the power of the cylinder 1.. f the axis is between ° and °, we add ° to the axis, whereas if the axis is between ° and °, then we subtract °. e transposed example from before is shown here.

B

AM PM

2.50

• Fig. 5.14

lanoclindrical lens with axis meridian (A) and power meridian () labelled (A) showing that it is sensible to describe its orientation along the A line (highlighted blac arrow) even though it is not the one with the power ().

45

45

135

180

f the cylinder is at an obliue noncardinal angle, then the power cross also needs to be drawn at an obliue angle, as shown in ig. .

CrossCylinder Techniue

10.25

90

•Fig. 5.15

iagram showing how to label the orientation of the clindrical lens. The blac arrow indicates the direction of increasing angles.

ou can also see in ig. . that using minus-cylinder form, we can work out the nal lens notation even ust by using the combined power cross on the right, as the spherical power will correspond to the most positive power 1. and the cylindrical power will correspond to the dierence between the two nal powers which is 2..

5.00D +

0.50

x 

is suggests that the spherical power is 1., and the cylindrical power is 2. at ° remember the axis tells us the axis meridian, not the power meridian. is means that along the ° axis there is only the spherical power 1., so to get down to the eual opposite sign at °, the lens needs to take 2. o 1. to achieve 2. in that meridian. e neat feature about these lenses is that, apart from two cylindrical axes, they also possess a ‘ip’ axis ig. .. is ‘ip’ axis shows that if the clinician turns the handle of the lens so that the lens rotates °, then the cylindrical axes would swap places. is allows a direct

0.00D (pl)

5.00D

5.00D –7.50D =

–2.50D

5.00 – 7.50  90

•Fig. 5.16

x 

o help uantify the power and direction of astigmatism, clinicians can use a techniue called the cross-cylinder techniue, also called ackson cross-cylinder (CC). is utilises a specialised lens that has two eual but oppositely powered cylinders at ° to one another. f the lens was made out of two planocylindrical lenses, the power meridians of each of the lenses would be perpendicular as shown in ig. .. However, for ease of manufacture, most  lenses are actually made to be spherocylindrical, which means that the cylindrical power of the lens is twice that of the sphere, and of the opposite sign, for example, if we had a  lens with the following notation

90 135

17.50

An example spheroclindrical lens power cross with its corresponding lens notation.

CHAPTER 5

–4.00D

–4.00D

pl

pl

The Reduced Eye and Spherical and Cylindrical enses

Flip axis

ve

55

Flip axis –ve

45

45

–ve

ve

•Fig. 5.17

–0.25D

e dl an H

e dl an H

An example power cross shown with clindrical axis along an obliue angle.

• Fig. 5.1 0.25D 0.25D –0.25D

•Fig. 5.1

A simplied version of a acson cross-clinder () lens if made of two planoclindrical lenses. n order to get eual opposite powers perpendicular to one another the curved (power) axes need to be perpendicular to one another.

comparison to be made between the positive and negative cylinders, which can inform the nal prescription. imilarly, if the clinician changes the orientation of the handle, this will euivalently alter the axes of the , which allows netuning of the axis and power. n clinical practice, clinicians

iagram showing a acson cross-clinder () lens with positive (1ve) and negative ( ve) axes perpendicular to one another. The ‘ip’ axis is a hpothetical axis that exists in line with the handle. This allows the lens to swap power axes when spun round (rotated ° as indicated b the arrow).

will most often use  towards the end of an examination to conrm and rene their refraction.

Focimetry n clinical practice, it’s useful to be able to determine a patient’s prescription before you begin an eye test, but if it’s a new patient then there won’t be any records, and the patient is very unlikely to know their refractive error. n these instances, if the patient is wearing a pair of glasses, clinicians can use a focimeter to determine the spherical and cylindrical power of the lens before the eye test begins. e will discuss focimetry in more detail in chapter .

Test Your Knowledge ry the uestions below to see if you need to review any sections again. ll answers are available in the back of the book.  hat is the presumed power of the reduced eye  hat is the dierence between spherical and cylindrical refractive error  f the far point of a patient’s eye is 2.cm, what is their refractive error

Reference . abbetts . Bennett and Rabbetts’ Clinical Visual Optics. th ed. xford,  utterworth-Heinemann, lsevier .

 hich meridian in a cylindrical lens is written as the ‘axis’  f you work out that following refraction an image will have a vergence of 1., are the rays converging or diverging

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6 eection C   P   R   L I  Introduction

Linear Magnication

What Is Reection?

Reection at spherical Curved Surfaces

Laws of Reection

est our nowledge

Image Formation – Plane Surfaces Multiple Plane Mirrors Reection at Spherical Curved Surfaces Focal Length and Focal Points Image Formation

    C  I S After working through this chapter, you should be able to: Dene the laws of reection Calculate the angle of reection Understand image formation at a plane mirror

Understand how the equations for spherical mirrors dier from those of refractie surfaces and lenses Determine linear magnication of an image formed  reection

Introduction

Laws of Reection

By this point we’ve covered the basics of refraction, but this is only half the story. Indeed, light can also be reected by surfaces, and this is an important principle of optical systems. is chapter will focus on reection at plane (at and curved surfaces.

In simple terms, there are two laws of reection to remember, and they apply to both plane and curved surfaces First law of reection: e incident light ray and the reected light ray lie in one plane. is means that the reected ray eists along the same line (and on the same side of the reecting surface as the incident (approaching ray. Second law of reection: e angle of incidence (i) is equal to the angle of reection (i9). ngles of incidence and reection are always measured relative to the ‘normal’ of the surface, with the normal being the hypothetical line that eists perpendicular to the surface at the point where light meets the surface. In ig. . you can see that, relative to the normal, the angle of incidence (i is identical to the angle of reection (i9. is second law of reection also leads on to the reversibility principle, which means light would reect bac along the same path in ig. . this would mean if the reected ray became the incident ray, the incident ray would become the reected ray.

What Is Reection? hen we say something is ‘reected’, we mean that it is returned in the plane it originally came from. e principle itself can be liened to bouncing a ball at a wall (providing we pretend gravity isn’t a factor – the ball cannot travel through the wall, so instead it will bounce bac as a specied angle depending on the angle it approached the wall (ig. .. is is the same as the principle of reection in optics. In some instances, light encounters a material that will reect the rays bac, which, at its core, is the ey principle of how we see obects in the real world, as all of our vision depends on light reecting o obects to reach our eyes (see chapter .

57

58 S EC TI ON 1

eometric and asic ptics

• Fig. 6.1

Diagram showing tennis ball behaving like a reected light ray when bounced at a wall.

Re fle c

ted

lig ht ra y

Normal

i’ i

ay tr gh i l t en cid n I

Plane mirror

of the mirror, so it travels bac along eactly the same line. It is also interesting to note that the image of you in the mirror will be positioned as far within the mirror as you are standing in front of it, so if you’re  cm in front of the mirror, your alternate dimension mirror self will also be  cm within the mirror. nother way of saying this might be that the obect distance (l is eual to the image distance (l9 with a plane mirror (ig. .. If we now consider a slightly more comple eample, let’s say we have an obect in front of our mirror, lie shown in ig. .. ow you can see that the incident light rays travelling along the normal (labelled  produce reected rays that travel bac along the same path. owever if we stood in a dierent position without moving the obect, then the image of the obect would stay in the same place, but the incident and reected rays from the obect that we utilise to ‘see’ it will now be travelling along dierent paths. In ig. . this is shown as the labelled rays  and . rucially, what we can also notice about the image of ourselves (or an obect in a plane mirror is that it will be laterally inversed (ipped from left to right, so if we raise our right hand, our mirror self appears to raise their left hand. imilarly the image will be reversed (ipped front to bac, so we see ourself facing the front side of ourselves, indicating our image is facing the opposite way to us. ig. . shows an eample of this. o summarise . e image is as far within the mirror as the obect is in front. . e imaginary line oining the obect and the image is perpendicular to the mirror surface (irrespective of the position it is viewed from. . e image is the same sie as the obect. . e image is virtual, reversed and laterally inverted. In summary, if we now the angle of incidence (i of a ray of light approaching a ed, plane mirror, then we can also

Position of observer

• Fig. 6.2

Diagram showing an incident light ray reecting off a plane (at) mirror. You can see that i 5 i (relative to the normal).

Plane mirror

i’

Image Formation – Plane Surfaces e easiest way to understand a plane mirror (at reective surface is to nd a mirror in your house that does not possess magnication or produce distortion. sually a standard hallwaybathroomfull-length mirror will be a plane mirror, unless it’s designed to be curved. If you stand directly in front of a mirror lie this, you will see an image of yourself which is the same sie as you are in the real, physical world. Importantly, the image of yourself will appear to be standing directly in front of you – this is because the light leaving your body to meet the mirror is travelling along the normal

i 2

Object

1 I

• Fig. 6.3

Image I’

Diagram showing that the image of an object remains stationary even if the observer is standing away from the object. he image also shows that the object distance (l) is eual to the image distance (l9).

CHAPTER 6

•Fig. 6.4

Diagram showing that the image of an object in a plane mirror is laterally inversed (left to right) and reversed (front to back).

easily determine the angle of reection (i9. o what if we complicate things a little by introducing a second plane mirror into the mi

ultiple Plane irrors p to this point, we have discussed the angle of reection (i9 but not the angle of deviation (d) – dened as the angle between the original path of the ray and the actual reected ray. ost of the time this angle isn’t necessary with reection at plane surfaces, but when we have two mirrors it becomes incredibly relevant, so let’s start by doing some revision of angles with a single plane mirror. In ig. ., you can see that a ray of light is incident upon a plane mirror and is reected at an angle identical to the angle of incidence (relative to the normal, shown as the orange dashed line. owever, in this image, I’ve also drawn a hypothetical continuation of the incident ray (blue dashed line, into the mirror, to show what the ray

A

would have done had the mirror not been obstructing its path. e angle of deviation (d, then, will be measured as the angle from the continuation line and the reected ray. irst though, let’s determine what we now about the angles – as you can see in ig. .B, the total angle of deviation (d, shown in green can be split into two smaller angles which are separated by the mirror. ese angles will be eual to one another, and we now that because we can determine that the angle between the incident (blue solid ray and the mirror will be eual to the angle between the reected (purple solid ray and the mirror. en, we can use the law of opposite angles to determine that the angle between the continuation (blue dashed ray and the mirror will be the same as the angle between the incident (blue solid ray and the mirror (for revision on opposite angles, see chapter , Bo .. reat, but what has this got to do with two plane mirrors, you as ell, in optics, if we have two plane mirrors inclined at an angle towards one another, we can use these principles to determine the total angle of deviation (d, even though it has reected o two mirrors (ig. .. In this scenario, two plane mirrors are inclined towards one another at an angle (a – importantly, this angle must be smaller than ° because otherwise the light wouldn’t be able to reect o both mirrors. If a ray of light is incident at one of the mirrors and reects in such a way as to reect o the second mirror, then the light will reect bac away from the mirrors, at an angle of deviation (d – relative to the original, incident ray. Importantly, both reected rays (from mirror  and mirror  obey the laws of reection (angle of incidence is eual to angle of reection, and so we also now (thans to ig. .B that the individual angles of deviation can be split into two identical angles. t the rst mirror these angles are referred to as a (alpha, and at the second mirror, these angles are referred to as g (gamma. If you’re good at visualising, you might also be able to see that the two a angles and the two g angles would add together to produce the total angle of deviation. is

B d

i’

i’

i

i

Plane mirror

Reection

a1 a1 a1

Plane mirror

•Fig. 6.5 Diagram showing how light might have travelled () if the mirror had not been in the way (blue dashed line) relative to how it is reected with the mirror in the way (purple solid line). he diagram also eplains how angle of deviation (d) can be determined whether you know the angle between the mirror and the angle of incidence as this will be eual to the angle between the mirror and the angle of reection as well as the angle between the mirror and the original undeviated (blue dashed) path of light ().

59

60 S EC TI ON 1

eometric and asic ptics

hich we can rearrange to be a

Plane mirror 1

 d

Equation 6.2

 Plane mirror 2



f two plane mirrors are inclined towards one another at an angle (a) and light is incident upon one of the mirrors in a way that causes it to reect off both mirrors (blue solid line) then the total angle of deviation (d) will be euivalent to the deviation at each mirror (a and g).

means that we can calculate the angle of deviation using uation .. d

d

(a

)

Equation 6.1 (explained) deviation

2(180

) or

360

2

Equation 6.2 (explained)

• Fig. 6.6

Equation 6.1

a

is now informs us that the sum of the angles a and g will be eual to  minus the angle between the mirrors (a. If we thin bac to uation ., we now that the total angle of deviation can be calculated by adding twice these values together, which means that we can use this information to prove uation ..



a

180°

2(

1

2)

owever, we probably don’t actually now the values of a and g, which could hinder our ability to utilise this euation somewhat. Instead, let’s mae use of the natural triangle formed by the angle between the mirrors (a and the rst reected ray of light (shown in yellow in ig. .. In this triangle, we now that all internal angles need to add up to °, and we can also apply our nowledge of opposite angles (shown in ig. . to determine that the previously unnown angle within the triangle must be eual to g – but what does this tell us ell, it tells us that

360 (2

)

DEMO QUESTION 6.1 wo plane mirrors are inclined at an angle of ° towards one another. hat is the angle of deviation tep  Determine what we need to calculate deviation d tep  Dene variables a 5  tep  Determine necessary euation d 5  – a (Equation 6.2) tep  alculate d 5  – a d 5  – () d 5  –  d 5 ° (don’t forget the units)

Practice Questions: 6.1.1 wo plane mirrors are inclined at an angle of ° towards one another. hat is the angle of deviation 6.1.2 wo plane mirrors are inclined at an angle of .° towards one another. hat is the angle of deviation

 now let’s mae it interesting – what happens if the reective surface is curved instead of at

Reection at Spherical Curved Surfaces

Plane mirror 1   d

a   Plane mirror 2 

•Fig. 6.7

his is the same diagram as that in ig. . but this time a yellow colour is highlighting the triangle that eists between the two mirrors and the rst reected ray of light. sing knowledge of opposite angles we can determine that the three internal angles of the triangle are a a and g

ust as lenses can be conve (positively powered or concave (negatively powered, mirrored (reective surfaces can also be curved in a conve or concave way. owever, the ey difference is that, with mirrors, a conve surface will be negatively powered, whilst a concave surface will be positively powered. o start with, we will consider spherical reective surfaces, meaning that the mirror forms a small part of a sphere and therefore will have a centre of curvature (ig. .. nd, ust lie with refractive surfaces, spherical reective surfaces can produce spherical aberrations if a wide aperture of light approaches the surface. s with lenses, light will focus along the optical ais, instead of in a single location, and the light rays will follow what’s called a ‘caustic curve’ (see ig. . and chapter , section titled ‘imiting pherical berration’. is means the uality of the image will be poor

CHAPTER 6

Convex surface

Reection

Concave surface

Normal

Normal Centre of curvature

Centre of curvature

Positive radius of curvature (r)

Negative radius of curvature (r)

Direction of light

• Fig. 6.8

Diagram showing relationship between conve and concave surfaces and their centre of

curvature. Concave surface

aberration. If you tae the image of the caustic curve produced by the concave mirror in ig. . (top and imagine the light incident on both sides of the optical ais, you can see that the aberration would form a pinched shape, curving towards the focal point, ust lie the refection in the ring eample in ig. .. ive it a go

Focal Length and Focal Points

Convex surface

s discussed previously in chapters  and , the poer ( of a surface indicates the degree of vergence it will add to or remove from incoming light rays. or lenses, we can calculate the focal length (f  to give us the lens’s primary ( and secondary (9 focal points, indicating where light will focus if originating from innity. owever, with spherical mirrors, there will only be one focal point () and one associated focal lengt (f ), which maes things slightly easier. Importantly for spherical mirrors, the focal length is eual to half the radius of curvature (r (see uation .. Incidentally, this means that the radius of curvature is also eual to twice the focal length, so if you now one, you can easily calculate the other. Equation 6.

f 

r 2

Equation 6. (explained) • Fig. 6.9

Diagram showing a wide pencil of parallel rays will fail to focus at a single point instead producing a caustic curve and spherical aberration (blurring).

because it will be blurred along this section of the optical ais (we will discuss aberrations in more detail in chapter . s before, paraial rays are hardly aected by these aberrations, and so we will assume paraial rays in our euations. or a real-world eample of reective spherical aberration, if you have a ring handy, you can place it on the table and shine a torch at it. ou’ll see the light produces a strange, curved pattern within the ring (ig. . – this is spherical

focal length 

radius of curvature 2

emember, all distances relating to power need to be measured in metres, and they will have a positive or negative sign, depending on their location relative to the surface (because we always measure from the surface. In the eample in ig. ., the radius of curvature is left of the reective surface because it is concave (see ig. . for an eplanation of this. s our optics convention dictates that we always measure from the surface to the point of interest (in this case centre of

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Equation 6. (explained) power 

refractive index focal length

DEMO QUESTION 6.2

• Fig. 6.10

llustration of spherical aberration demo you can do at

home.

Concave surface

Centre of curvature

Focal point

Negative radius of curvature (r)

 concave spherical mirror has a radius of curvature of  cm. hat is its power tep  Determine what we need to calculate power  tep  Dene variables r 5 2. m (because the surface is concave the radius is negative and in metres) n 5 . (nothing is specically mentioned so we assume the primary medium is air) tep  Determine necessary euation f 5 r   (Equation 6.3)  5 2 (n  f) (Equation 6.) tep  alculate f5r f 5 2.   f 5 2. m  5 2 (n  f)  5 2 (.  2 .)  5 1.D (don’t forget the 6 sign)

Practice Questions: 6.2.1  conve spherical mirror has a radius of curvature of  cm. hat is its power 6.2.2  concave spherical mirror has a focal length of  cm. hat is its power

Image Formation Negative focal length (f)

•Fig. 6.11

Diagram showing relationship between radius of curvature and focal length for a concave spherical mirror.

curvature and the focal point, we measure from right to left, which is the opposite direction to that of light (which we always assume goes left to right, so in this eample, both the radius of curvature (r and the focal length (f are negative. If the surface was conve, both distances would be positive. ust as we have previously seen for refractive surfaces, the focal point of a system is the point at which light from innity will focus, so once we have determined the focal length, we can use euation . to calculate power (. otice that it’s a similar euation to the one we use to calculate power of a refracting surface, but it has a negative sign in order to account for the fact that with reective surfaces the rays eist within one plane (rst law of reection instead of passing through the surface. Equation 6.

F

n f

ow that we now spherical mirrors possess power (, and we now how to calculate it, we can start to thin about how and where they form images. e already now that an obect at innity (relative to a curved mirror will produce an image at the focal point, but what if the obect is positioned anywhere else or this situation, we can use modied vergence euations (taen from what we learned about refractive surfaces to calculate image distance. o do this, we eep the obect vergence (uation . and image vergence (uation . euations the same as for refractive surfaces, but we need to remember that the light will reverse direction when reected, and so we need to adust our nal image distance euation (uation . in order to ensure we calculate the correct direction of light. n l Equation 6. (explained) Equation 6.

L

object vergence  Equation 6.6

refractive index of medium object distance from surface F

CHAPTER 6

Equation 6.6 (explained) power Equation 6.

L

n l

l

n L

Equation 6. (explained) image distance

secondary refr.index image veergence

Reection

nother (slightly uicer but more challenging way to calculate image distance is to use uation ., which utilises the nown relationship between the obect and image distance, relative to the focal length and power of the surface. emember, if an euation has multiple euals signs (5 in it, then you can choose the two elements you want to use. o, for eample, if I am given the focal length (f  and the obect distance (l, then I can ignore the part of the euation that uses the radius of curvature (r. Equation 6.

1

1

2 r

1 f

Equation 6. (explained) DEMO QUESTION 6.3 n object is placed  cm in front of a concave spherical mirror with a radius of curvature of  cm. here does the image form tep  Determine what we need to calculate image distance l tep  Dene variables l 5 2. m (the obect is  cm ‘in front of’ meaning it’s to the left of the surface which gives it a negative distance – and we need to convert to metres) r 5 2. m (because the surface is concave the radius is negative and in metres) n 5 . (nothing is specically mentioned so we assume the primary medium is air) tep  Determine necessary euation(s) f 5 r   (Equation 6.3)  5 2 (n  f) (Equation 6.)  5 n  l (Equation 6.) ’ 5  1  (Equation 6.6) ’ 5 2 (n  l ) (Equation 6.) tep  alculate f5r f 5 2.   f 5 2. m  5 2 (n  f)  5 2 (.  .)  5 1.D 5nl  5 .  .  5 2. (remember to eep the number long in the calculator) 9 5  1  9 5 2. 1  9 5 1. 9 5 2 (n  l ) l9 5 2 (n   ) (rearrange to get l ) l9 5 2 (.  .) l9 5 2. m he image forms . cm left of the surface. (don’t forget the 6 sign and the units) e know the image forms to the left of the surface because the distance is negative.

Practice Questions: 6.3.1 n object is placed  cm in front of a conve spherical mirror with a radius of curvature of  cm. here does the image form 6.3.2 n object is placed  cm in front of a concave spherical mirror with a power of 1.D. here does the image form

1 image dist

1 object dist.

2 radius

1 focal length

DEMO QUESTION 6.4 e’ll use the same question as in demo 6.2 to show that it’s an equivalent process. n object is placed  cm in front of a concave spherical mirror with a radius of curvature of  cm. here does the image form tep  Determine what we need to calculate image distance l tep  Dene variables l 5 2. m (the obect is  cm ‘in front of’ meaning it’s to the left of the surface which gives it a negative distance – and we need to convert to metres) r 5 2. m (because the surface is concave the radius is negative and in metres) n 5 . (nothing is specically mentioned so we assume the primary medium is air) tep  Determine necessary euation(s) (  l ) 1 (  l) 5 (  r) (Equation 6.) tep  alculate (  l ) 1 (  l) 5 (  r) (  l ) 1 (  .) 5 (  .) (ubstitute in our values) (  l ) 5 (  .) – (  .) (earrange) (  l’) 5 2. (olve the right side) l9 5   . l9 5 2. m he image forms . cm left of the surface. (don’t forget the 6 sign and the units)

Practice Questions: 6.4.1 n object is placed  cm in front of a conve spherical mirror with a radius of curvature of  cm. here does the image form 6.4.2 n object is placed  cm in front of a concave spherical mirror with a power of 1.D. here does the image form

Linear agnication s you’ll recall from our chapter about thin lenses (see chapter , ‘linear magnication’ refers to the sie of an image relative to the original obect. e can use uation . to calculate magnication of an image produced by a spherical mirror, providing we now the obect (l and image distances (l9. owever, please be very careful when

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using this euation as, again, you’ll notice that it is dierent from the one we use for refractive surfaces as it includes a minus sign. lso I would always recommend using the obect (l and image (l9 distances to calculate magnication, rather than the vergence values. is is because it’s easy to get confused about the image vergence as everything is ipped relative to a lens. or eample, a converging image (positive vergence will have a negative distance. h l L m Equation 6. h l L image height object height

Paraboloidal

Ellipsoidal

• Fig. 6.12

Equation 6. (explained) mag.

Spherical

image dist. object dist.

object verg. image verg.

owever, one thing that remains the same is how useful the magnication value is – it can tell us whether an image is virtual or real, and magnied or minied. ee chapter , ig. ., for revision on this if necessary.

llustration showing how light focuses with three types of mirror spherical paraboloidal (aspherical) and ellipsoidal (aspherical).

DEMO QUESTION 6.5—cont’d m 5 2( .  .) m 5 2. (minied inverted real) (don’t forget the 6 sign and the  represents the units for magnication)

Practice Questions: DEMO QUESTION 6.5 n object is placed  cm in front of a concave spherical mirror with a radius of curvature of  cm. hat is the magnication of the image tep  Determine what we need to calculate magnication m tep  Dene variables l 5 2. m (the obect is 2 cm ‘in front of’ meaning it’s to the left of the surface which gives it a negative distance – and we need to convert to metres) r 5 2. m (because the surface is concave the radius is negative and in metres) n 5 . (nothing is specically mentioned so we assume the primary medium is air) tep  Determine necessary euation ou can choose which equation to use to determine image distance here – ’ll use 6. (  l ) 1 (  l) 5 (  r) (Equation 6.) m 5 (h  h) 5 2 (l  l) 5 (   ) (Equation 6.) tep  alculate (  l ) 1 (  l) 5 (  r) (  l ) 1 (  .) 5 (  .) (ubstitute in our values) (  l ) 5 (  .) (  .) (earrange) (  l ) 5 2. (olve the right side) l9 5   . l9 5 2. m (remember to eep the number long in the calculator) m 5 (h  h) 5 2(l  l) 5 (   ) m 5 2(l  l) (choose which part we can use)

6.5.1 n object is placed  cm in front of a concave spherical mirror with a radius of curvature of  cm. hat is the magnication of the image 6.5.2 n object is placed  cm in front of a concave spherical mirror of power 1.D. hat is the magnication of the image

Reection at spherical Curved Surfaces In this brief, nal section we will consider curved reective surfaces that fail to fall along a sphere. ese types of surfaces are called asperical irrors. ese types of mirrors are capable of producing high-uality images because they do not suer from spherical aberration, which means that they can produce clear images even with a wide pencil of light. or eample, we can use paraboloidal mirrors to focus light from innity, and we can use ellipsoidal mirrors to focus light from a distance closer than innity, as shown in ig. . ese mirrors can be utilised as wing mirrors on cars to dramatically increase the eld of view available to the driver however, in order to increase the eld of view (how much space you can see behind you, these mirrors also minify the image, meaning that other cars, pedestrians and obects may seem farther away than they actually are.

Test Your Knowledge ry the uestions below to see if you need to review any sections again. ll answers are available in the bac of the boo. .6.1 hat are the two laws of reection .6.2 If an obect is  cm in front of a plane mirror, where will the image form

.6. escribe the nature of an image formed in a plane mirror. .6. plain why the formula for calculating the focal length (f  of a spherical mirror has a ‘minus sign’ in it. .6. If a spherical mirror has a radius of curvature of 1 cm, what is the focal length (f 

7 Ray racing     T E R   TL I E Introduction What Is a Ray Diagram? Ray Tracing – Single Thin Lens Drawing Lenses Drawing Objects and Images Rules for a Positively Powered Lens Rules for a Negatively Powered Lens

Ray Tracing – Spherical Mirrors Drawing Mirrors Rules for a Positively Powered erical Mirror Rules for a Negatively Powered erical Mirror o to Scale our Diagrams? Test our noledge

Ray Tracing – Equivalent Lenses Rules for Equivalent Lenses

  E  TI ES After working through this chapter, you should be able to: Elain wat a ray diagram is Draw an accurate ray diagram for a conve lens Draw an accurate diagram for a concave lens

Draw an accurate ray diagram for a conve mirror Draw an accurate ray diagram for a concave mirror

Introduction

Ray Tracing – Single Thin Lens

By this point in the textbook you might have become a bona de maths wizard, but there’s also a chance that you might be sick of looking at equations, so let’s take a short break to learn how to understand image formation using line drawings. es, you heard me correctly, line drawings n the optics business, we refer to this as ray tracing because we can use scale line drawings to trace the path of the rays from the obect to the image, and my best advice is to read this chapter with a pad of paper, a pencil and a ruler so that you can learn to apply the principles as you go.

emember that with a thin lens, the eect of the refractive index of the material is considered so negligible that it is ig nored completely. is means that with a thin lens in air, we can assume that the primary refractive index n and the secondary refractive index n9 are identical both 5 .. f we think back to the equations from chapter , we know that the focal length of a thin lens is determined by the power  of the lens and the refractive index in which it resides. o that end, given that the refractive index is identical on either side of the lens, we can assume that the focal length would be the same distance away from the lens in either direction. ow, for our equations we always assume that light is travelling from left to right, meaning that for a positively powered lens that converges light, the secondary focal length f9 will exist to the right of the lens to form the secondary focal point 9. owever, for ray tracing, we can start to consider what happens if the light were to change direction and travel from right to left. eoretically, because we’re assuming a thin lens, the light should behave identically in either direction, so if made to travel backwards through the system, it will produce a focus at the primary focal point . e distance between

What Is a Ray Diagram?  ray diagram is a carefully measured line drawing that traces the path light takes from an obect to its image formed by an optical system. ese diagrams can be produced suc cessfully for both thin lenses and spherical mirrors, but in all cases, we are assuming that we are using paraxial rays rays that lie close to the optical axis.

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of 2. m to the left of the lens and a primary focal length of 1. m to the right of the lens. n these diagrams, you’ll also probably notice that we’ve added points referred to as f and f9. ese are important points for ray diagrams and correspond to twice the secondary focal length f9 and twice the primary focal length f. ow ever, as a word of warning, when were learning this content it can be very easy to mix up f with f9 as the description twice the focal length for f sounds like a similar idea to the sec ondary focal length f9 as they both relate to the number . y best advice here is to make sure you understand what they each relate to, which may involve going back over chapter  for some revision on secondary focal lengths if necessary.

the lens and the primary focal point is called the primary focal length f  and should be equivalent to the negative of the secondary focal length quation . and ig. .. or example, a 1.  lens will have a secondary focal length of 1. m to the right of the lens and a primary focal length of 2. m to the left of the lens. Equation 7.1 Equation 7.1 (explained) h or a negatively powered lens that diverges light, light origi nating from innity will diverge after refracting through the lens, which means we have to draw the rays back to see where they appear to originate from. s ig. . shows, this produces a negative secondary focal length f9 to the left of the lens to form the secondary focal point 9, which is the exact opposite to that of a positive lens. f the light then travelled backwards through the system ig. .B, it would diverge on the opposite side of the lens, meaning the rays would appear to originate from the primary focal point  on the right of the lens. e dis tance between the lens and the primary focal point is called the primary focal length f  and, again, should be equiva lent to the negative of the secondary focal length. or example, a 2.  lens will have a secondary focal length

Draing Lenses ne of the key aspects of a ray diagram is that the person who views it should know whether it’s a positively pow ered or negatively powered lens. is can be achieved through looking at the behaviour of the rays, but the easi est way to do this is to draw the lens in line with the set specication. e guidelines state that all lenses are drawn as straight, vertical lines, but positively powered lenses are drawn with outward facing arrows as in ig. . whilst negatively powered lenses are drawn with inward facing arrows as in ig. ..

Positive lens

A

Secondary focal point (F’)

2F’

2f’ Secondary focal length (f’) Positive lens

B Primary focal point (F)

2F

•Fig. 7.1 2f Primary focal length (f)

Basic principles of ray tracing with a positive lens. When light originates from innity and travels from left to right (A), it produces a focus at the secondary focal point (F9) with a positive focal length. f the direction of light is reversed (B) then light will focus at the primary focal point (F) and will have a negative (ut euivalent) focal length.

CHAPTER 7

67

which allows us to assume the distance of the tip of the image relative to the optical axis which gives us the image height. hen we come to draw our image, this will also be de picted as an arrow with the base starting at the optical axis, but the image will be upright or inverted, depending on the type of lens and where the obect is positioned.

Negative lens

A

Ray Tracing

Secondary focal point 2F’ (F’)

Rules or a ositively oered Lens 2f’ Secondary focal length (f’) Negative lens

B

Primary focal point (F) 2F

2f Primary focal lenght (f)

• Fig. 7.2

Basic principles of ray tracing with a negative lens. When light originates from innity and travels from left to right (A), it diverges as if originating from the secondary focal point (F ) with a negative focal length. f the direction of light is reversed (B) then light will diverge as if originating from the primary focal point (F) and will have a positive (ut euivalent) focal length.

Draing ects and Images n ray diagrams, we will learn to draw our obects as upward pointing arrows, the base of which starts at the optical axis – the idea is that it helps to make the diagram easy to inter pret, but  do agree that it’s not very imaginative. en, once we have our arrowshaped obect, we need to think about how we’ll draw the light rays. echnically, an obect is an extended source with light rays emanating from every part of it ig. ., but it is simpler and quicker to limit our drawings to the light emanating from the tip of the oect ig. .B. is helps us to use ray tracing to determine the location of the tip of the image after refraction or re¢ection,

•Fig. 7.3

Object

B

Object

A

ects drawn as upward-pointing arrows. Although oects are ‘etended’ sources (A), we only use the tip of the oect for our rays (B).

ow we’re ready to learn how to draw the rays in our ray diagrams. or positive and negativepowered lenses there are three rays that we need to learn, but they’re drawn slightly dierently for each lens type, so ’ll start with posi tively powered lenses in this section and then move onto negatively powered lenses in the next section. e rays are outlined below and demonstrated in ig. . P-ray – drawn from the tip of the obect, parallel to the optical axis refracts through 9 C-ray – drawn from the tip of the obect, undeviated through the optical centre intersection of lens and opti cal axis F-ray – drawn from the tip of the obect, through  re fracts parallel to the optical axis n the example in ig. . you can see that the rays all come to a neat little intersection on the right side of the lens, which is the point corresponding to the tip of the im age mportantly, light traveling along the normal optical axis through the optical centre will ust travel in a straight line, so we know the base of the obect and the base of the image both form along the optical axis. e can draw our image by connecting the intersection of the rays to the opti cal axis with a very straight, vertical line. n this example, the diagram shows that with this lens and obect distance, the image forms slightly to the right of 9, and it’s inverted upsidedown and slightly magnied. ow, if drawn to scale with the correct focal length and obect distance, and even obect height when specied, this diagram can tell you the exact image distance and exact im age height for the height of the obect. ou can try this for yourself at home with a ruler and a pencil. e can also do a quick mathematical condence check to prove that our ray tracing makes sense. or this, we can ust use sensible estimates for example, if we assume the lens in ig. . is 1. a power that for this example has been chosen at random, then our secondary focal length f9 would be 1. m and our primary focal length f  would be 2. m see chapter  for revision on this if needed. is means our  point on the left corre sponds to 2. m twice the primary focal length. e obect itself looks to be slightly closer to the lens than , so let’s estimate its distance l as 2. m. sing our vergence equations from chapter , we can then determine that the obect vergence  for our estimated obect dis tance would be 2., which would make our image ver gence 9 1.. is means mathematically, the image l9 should form 1. m right of the lens, which would

68 S EC TI ON 1

eometric and asic ptics

P-ray C-ray F-ray

Parallel to the optical axis

h rou gh

Und

evia ted t

hrou

2F

gh c entr e

F

Object

F’ F’

2F’

h

ro u

gh F Parallel to the optical axis

Image

• Fig. 7.4

ample ray diagram for a positively powered lens. All three rays are drawn and laelled. he tip of the image is identied as the point where the rays intersect one another.

correspond to slightly right of 9. . . hich looks pretty spoton in ig. . to me or positive lenses, the position of the image relative to the obect can be determined by learning the contents of able ., but the highlights are that • n obect at innity will form an image at 9 • n obect at  will produce an image at innity • n obect at  will produce an inverted image the same size at 9 • n obect between  and the lens will produce an up right image left of the lens mportantly, you can also see that as the obect moves from innity towards , the image moves from  towards innity, so when you become more experienced with ray

diagrams, you will be able to estimate the position of the image based on knowing roughly where the obect is, relative to the focal points. s you may be able to tell from the table, when the obect is between  and the lens, this is a special case where the obect produces an upright image even though it’s a posi tive lens. n this instance, the three rays will diverge away from each other on the righthand side of the lens  always think diverging rays resemble spiders’ legs, which means they won’t form a neat intersection to identify our image position. nstead, when the rays diverge like this, we need to draw the refracted rays backwards to see where they appear to originate from as seen in ig. ., which is also a key principle of ray tracing for negative lenses.

TABLE Table Showing Relationship Between Object Position an ssociate age haacteistics 7.1 with Positie enses

Object Distance (l)

Image Location

Image Distance (l9)

Inverted/Upright

Image Size

.F

ight of lens

.F9  ,F9



,

F

ight of lens

F9



5

.F  ,F

ight of lens

.F9



.

innity

ight of lens

F9



F



nnity



,F

eft of lens

.l



2F

F

F’

.

2F’

Image Object

P-ray C-ray F-ray

• Fig. 7.5 ay diagram showing how to draw the rays (and identify the location of the image) when an oect is etween F and the lens in front of a positively powered lens. ou can see it forms a virtual, upright, magnied image.

CHAPTER 7

e rays for negatively powered lenses are outlined below and demonstrated in ig. . P-ray – drawn from the tip of the obect, parallel to the optical axis refracts as if originated from 9 and drawn backwards on the left of the lens C-ray – drawn from the tip of the obect, undeviated through the optical centre intersection of lens and optical axis F-ray – drawn from the tip of the obect, aiming for  when it meets the lens, refracts parallel to the optical axis and draws backwards on the left of the lens nlike with a positively powered lens, negatively pow ered lenses diverge the rays away from one another on the right of the lens, meaning they will never form an intersec tion think spiders’ legs again. nstead, we draw the refract ed rays backwards and look for an intersection on the left to identify where the rays appear to originate from. is pro duces an upright image, as shown in ig. . or negative lenses, the position of the image relative to the obect can be determined by learning the contents of able ., but the highlights are that • n obect at innity will produce an image at 9 • n obect anywhere other than innity will produce a minied, upright image on the left, between 9 and the lens and closer to the lens than the obect

Ray Tracing – Equivalent Lenses , but what do we do if there is more than one lens in a system n chapter , we learned about multiple lens sys tems and equialent lenses, and in this section we will F’

P-ra -ra F-ra

Parallel to the optical axis

Towar ds F Parallel to the optical axis 2F’

F’

Object

F Image

Und

evia te

d th

2F

roug

h ce

ntre

• Fig. 7.

ample ray diagram for a negatively powered lens. All three rays are drawn and laelled. he tip of the image is identied as the point where the rays appear to originate from.

TABLE Table Showing Relationship Between Object Position an ssociate age haacteistics ith 7.2 egatie enses

Object Distance (l)

Image Location

nnity 

69

learn how to draw a ray diagram for a lens system with two lenses. t may be useful to refer back to some of the equa tions in chapter  to help you understand some of the steps. mportantly, for these equivalent lens ray diagrams to be drawn successfully, you need to know the positions of six cardinal points. ese points include the two focal points (F F9), the two principal planes (P 9P9) and two points called the nodal points ( 9). e rst step would be to draw the lenses separated by their distance d – this can be scaled down as long as all other distances are scaled down by the same amount. en you need to draw in the focal points, which you can determine by calculating the back vertex focal length fv9 distance between second lens and secondary focal point and the front vertex focal length fv distance between the rst lens and the primary focal point. emember to make a note of whether the distances are positive or negative, as this will tell you whether the focal point is to the right of a lens if positive or to the left of a lens if negative. ext, we can determine the location of the principal planes by calculating the secondary equivalent fo cal length fe9 distance between secondary focal point and secondary principal plane and the primary equivalent focal length fe distance between primary focal point and pri mary principal plane – see ig. . inally, then, we can calculate the location of our nodal points. odal points are present in all optical systems and exist in a location that means an incident approaching ray of light heading towards the rst nodal point  would leave the system along exactly the same angle, but shifted to emerge from the second nodal point 9. is means that the incident and emergent rays are parallel to one another. f you’ve ever played any of the portalcreatingbased game series then this will make complete sense, but if not then

Rules or a egatively oered Lens

m Fro

Ray Tracing

Image Distance (l9)

Inverted/Upright

Image Size

eft of lens

F9



,

eft of lens

,l



,

70 S EC TI ON 1

eometric and asic ptics

HP

H’P’

fv

Rules or Equivalent Lenses e rays for equivalent lenses are outlined below and dem onstrated in ig. .

fv

F

N

F’

N’

fe

fe

n easy way to remember whether the rays need to travel to 99 or  is to think about whether the ray is going to intersect 9 or  – remember that primes 9 always link together

• Fig. 7.7

llustration to remind us of the difference etween the ac verte focal length (fv ) and front verte focal length (fv) relative to the secondary euivalent focal length (fe ) and the primary euivalent focal length (fe).

HP

F

N

H’P’

N’

P-ray – drawn from the tip of the obect, parallel to the optical axis, runs all the way to 99 refracts through 9 -ray – drawn from the tip of the obect, towards , leaves system from 9 at parallel angle to incident ray F-ray – drawn from the tip of the obect, through  when it meets , refracts parallel to the optical axis

F’

O

Ray Tracing – Spherical Mirrors or the nal section of this chapter, we need to consider ray diagrams for positively powered concave and negatively powered convex spherical mirrors. mportantly, in order to draw an accurately scaled ray diagram for a spherical mir ror, we would need to know the focal length f  and the radius of curvature r however, the good news is that if you know one then you can calculate the other using the equa tions we learned in chapter  quations . and .. or example, if we have a concave mirror with a power of 1. , we know the focal length is 2. m and so the radius of curvature will be 2. m. hen practising drawing the rays, you can draw the fo cal length and radius of curvature any length you like to t on the page providing you make sure that the radius is twice the focal length and remember to scale your answers cor rectly – see section titled ow to cale our iagrams for help with this.

Draing Mirrors • Fig. 7.

llustration showing how nodal points wor. ncident light (shown in lue) enters the system aiming at the primary nodal point () and then leaves the system at the secondary nodal point (9) at the same angle it entered . he point at which the light ray would have crossed the optical ais is identied as the optical centre () of the system.

please see ig. .. mportantly, the point at which the path of light crosses or appears to cross the optical axis is de ned as the optical centre of the system . e good news for us is that when the obect and image exist within a medium comprising the same refractive index as each other for example, if the obect and image are both in air then the primary nodal point  will correspond to the point where the primary principal plane  intersects the optical axis, and the secondary nodal point 9 will correspond to the point where the secondary principal plane 99 intersects the optical axis – so now we’re all set

s with lenses, the rst thing we need to know is how to draw the mirror for a ray diagram so that anyone who sees the diagram knows whether the mirror is positively pow ered or negatively powered. imilarly to lenses, mirrors are drawn with straight, vertical lines, but this time in stead of using directional arrows, we put a bowlshaped hat on the top of the line as shown in ig. .. is helps to dierentiate it as a mirror as opposed to a lens, and the direction of the bowl tells the viewer whether it is concave or convex.

Rules or a ositively oered oncave Spherical Mirror or positive and negativepowered mirrors there are four rays that we need to learn, but they’re drawn slightly dier ently, so ust like before ’ll start with positively powered mirrors in this section and then move onto negatively pow ered mirrors in the next section.

CHAPTER 7

Parallel to the optical axis to H’P’

Th

ro to ugh HP F

HP

H’P’

h

F

N

F’

N’

F’ Em

Object

71

P-ray N-ray F-ray

Th ro ug

To N

Ray Tracing

erge

Image

und

eia ted

Parallel to the optical axis

•Fig. 7. ample ray diagram for an euivalent lens system compsising two positively powered lenses (greyed out). All three rays are drawn and laelled. he tip of the image is identied as the point where the rays intersect. Convex

Concave

P-ra C-ra F-ra -ra

Parallel to the optical axis Tow ards inte rsec tion Th rou gh F F

C

Object Parallel to the optical axis

iagram showing the ray diagram convention for conve (left) and concave (right) spherical mirrors.

e rays are outlined below and demonstrated in ig. . P-ray – drawn from the tip of the obect, parallel to the optical axis re¢ects through  C-ray – drawn from the tip of the obect, undeviated through the centre of curvature re¢ects back along the same line F-ray – drawn from the tip of the obect, through  re¢ects parallel to the optical axis -ray – drawn from the tip of the obect to the intersection of the mirror and optical axis then re¢ects at an identical angle underneath the optical axis n the example in ig. ., you can see that the rays all come to a neat little intersection on the left side of the mir ror, which is the point corresponding to the tip of the image gain, light traveling along the optical axis will re¢ect back along a straight line, so we know the base of the obect and the base of the image both form along the optical axis. is means we can draw our image by connecting the intersection of the rays to the optical axis with a very straight, vertical line. n this example, the diagram shows that with this mirror and obect distance, the image forms between  and , left of the mirror, and it’s inverted real and minied. lease note, as a little word of warning here, that unless you

F gh ro u Th

• Fig. 7.1

eath ern und ngle s t a lec Ref t same a

Un

de via

te d

th ro ug h

C

• Fig. 7.11 ample ray diagram for a positively powered concave mirror. All four rays are drawn and laelled. he tip of the image is identied as the point where the rays intersect one another.

are a wizard with a protractor or drawing the diagram digi tally,  would never recommend drawing the ray. t will almost certainly lead your diagram astray, as even a very tiny error in the angle of the second line underneath the optical axis will lead to increasingly large errors as the line gets lon ger. erefore if it was me, d only ever use the other three rays as they are much more reliable. ust like with lenses, if drawn to scale with the correct focal length and obect distance, this diagram can tell you the exact image distance and exact image height for the height of the obect. ou can try this for yourself at home with a ruler and a pencil. or positive spherical mirrors, the position of the image relative to the obect can be determined by learning the con tents of able ., but the highlights are that • n obect at innity will produce an image at  • n obect at  will produce an image at innity • n obect at  will produce an inverted image the same size at  • n obect between  and the mirror will produce an upright image right of the mirror

72 S EC TI ON 1

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TABLE Table Showing Relationship Between Object Position an ssociate age haacteistics ith Positie 7.3 Spheical ios

Object Distance (l)

Image Location

Image Distance (l9)

Inverted/Upright

Image Size

.

eft of mirror

.F  ,



,



eft of mirror





5

.F  ,

eft of mirror

.



.

nnity

eft of mirror

F



F



nnity



,F

ight of mirror

.l



gain, ust like with positive lenses, when the obect is between  and the mirror, this is a special case where the obect produces an upright image. n this instance, the rays will diverge away from each other after re¢ecting again, think spiders’ legs, which means they won’t form a neat in tersection to identify our image position. nstead, this means we need to draw the re¢ected rays backwards to see where they appear to originate from as seen in ig. ., which is also a key principle of ray tracing for negative mirrors.

Rules or a egatively oered Spherical Mirror emember that for a negatively powered, convex mirror, the radius of curvature and therefore focal length will be positive, which will put them on the opposite side of the mirror to the obect – this makes ray diagrams slightly more complicated. e rays for negatively powered spherical mir rors are outlined below and demonstrated in ig. . P-ray – drawn from the tip of the obect, parallel to the optical axis re¢ects as if originating from  and drawn backwards on the right of the mirror P-ray C-ray F-ray V-ray

F C

• Fig. 7.12

Object

Image

ay diagram showing how to draw the rays (and identify the location of the image) when an oect is etween F and the mirror in front of a positively powered mirror. ou can see it forms a virtual, upright, magnied image ehind the mirror itself.

.

C-ray – drawn from the tip of the obect towards the centre of curvature drawn as if to continue along the same line on the right of the mirror F-ray – drawn from the tip of the obect towards  re¢ects parallel to the optical axis and drawn backwards on the right of the mirror -ray – drawn from the tip of the obect to the intersection of the mirror and optical axis then re¢ects at an identical angle underneath the optical axis drawn backwards on the right of the mirror s we know, negatively powered mirrors add divergence to light rays, and so the rays we draw will diverge away from one another after re¢ection, meaning they will never form an intersection. is means to nd our image we need to draw the re¢ected rays backwards and look for an intersec tion on the right to identify where the rays appear to origi nate from. is will always produce an upright image, as shown in ig. ..  will also oer the same advice here as  did previously for positively powered mirrors do not use the ray unless absolutely necessary as it is notoriously un reliable. or negative spherical mirrors, the position of the image relative to the obect can be determined by learning the con tents of able ., but the highlights are that • n obect at innity will produce an image at  • n obect anywhere other than innity will produce a minied, upright image between  and the mirror on the righthand side and closer to the mirror than the obect

o to Scale our Diagrams ow that we’ve learned how to draw these ray diagrams, it’s important to learn another key skill – how to scale them down to t onto a piece of paper. et’s say that we want to produce a ray diagram to show how an image forms when an obect is placed  cm in front of a 1.  lens. • bect distance 2 cm • ens power 1.   distance of  cm is already too large to t on an  piece of paper, and that’s only considering the left side of the

CHAPTER 7

Ray Tracing

ay Aw m fro

Unde

Pray Cray

F

Parallel to the optical axis

iated to wards C

C Object

F

Fray ray Tow ards

inte rsec

Towa rds

F

tion

C

Parallel to the optical axis Object

F

th r nea nde gle u s an lect Ref t same a

•Fig. 7.13

ample ray diagram for a negatively powered mirror. All four rays are drawn and laelled, ut they have een split across two diagrams for clarity. he -ray and -ray are illustrated in the top image, whereas the F-ray and -ray are illustrated in the ottom image. he tip of the image is identied as the point where the rays intersect one another.

TABLE Table Showing Relationship Between Object Position an ssociate age haacteistics ith 7.4 egatie Spheical ios

Object Distance (l)

Image Location

Image Distance (l9)

Inverted/Upright

Image Size

nnity

ight of mirror

F



,



ight of mirror

,l



,

lens f we calculate the focal length f 5 n 5  5 1. m, we can see that in order to draw our ray which goes through the secondary focal point 9, we’d also need  cm right of the lens too. is is already totalling a fairly massive  cm and is getting out of hand. nstead of sticking lots of pieces of paper together and digging out our metre ruler, we can scale the drawing down to make it smaller. hen done correctly, scaling the dia grams down will be like resizing a digital photograph – as long as all the proportions are sensibly maintained, then it will look identical ust smaller. ere is also the added benet that smaller diagrams are less prone to errors, which is a nice bonus. is scaling down can be done in one of two possible ways. • ption  ivide all the numbers by a set number

or this method, we calculate all our distances using the same example numbers from before 2 cm for obect distance 1 cm for focal length f  1 cm for f en we divide them all by any number we like. or ex ample, if we divide all our values here by  a nice, easy number, then the distances become 2  cm for obect distance 1. cm for focal length f  1 cm for f is easily ts on the page. mportantly, this method works beautifully providing you remember to apply it to all

73

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P-ray C-ray F-ray

Object

F’

2F

2F’ Image

F

• Fig. 7.14

orrect ray diagram for the eample discussed in the section titled ow to cale our iagrams. o matter how the diagram is scaled, it should produce an inverted image further than f on the right-hand side of the lens.

distance measurements. e only downside of this method is that sometimes it can leave you with a decimal point or two or three, which might impact the overall accuracy of the diagram. • ption  alculate proportional dierences or this method, we need to calculate the proportional relationship between the obect distance and the focal length. roportions tell us numerically how big or small something is relative to something else. n our case, we would want to know how big or small our obect distance is relative to our focal length, so all we need to do is divide one by the other proportion 5 obect distance  focal length. is will give us a number, so a proportion , suggests the obect distance is smaller than the focal length, a proportion of  suggests they are identical and a proportion . suggests the obect distance is larger than the focal length. n our example, we know the obect distance  cm is larger than the focal

length  cm, so we’re expecting a number larger than , and mathematically it comes out as .. is means that our obect distance is . times larger than our focal length. ith this information, we can now draw our ray dia gram literally any size we want and then use our new focal length to determine the appropriately scaled proportional obect distance. or example, let’s say that we want to draw our ray diagram with a focal length of . cm chosen be cause it’s small enough to t on a page. e know our obect distance needs to be . times larger so we can multiply the new focal length . cm by our proportion . to discern that our new, scaled obect distance needs to be  cm. f you’re keen, you’ll have noticed that these are the same numbers  used in the previous example from ption , which were chosen deliberately to prove that it’s a valid method. ry it yourself at home and see how it works out e nal ray diagram should look something like ig. .

Test Your Knowledge ry the questions below to see if you need to review any sec tions again. ll answers are available in the back of the book. .7.1 sing your knowledge of optical systems, can you explain why a ray will pass through 9

2F’

F’ Object

.7. xplain how you could use a ray diagram to de cide whether an image distance would be positive or negative. .7. n the diagram below, which image is drawn cor rectly xplain your answer.

F A

B

2F

CHAPTER 7

.7. raw a ray diagram to show where an image would form if an obect was placed at  in front of a positively powered lens. .7. raw a ray diagram to show where an image would form if an obect was placed between 9 and the lens in front of a negatively powered lens. .7. raw a ray diagram to show where an image would form if an obect was placed left of  in front of a positively powered mirror.

Ray Tracing

.7.7 raw a ray diagram to show where an image would form if an obect was placed at any location in front of a negatively powered mirror. .7. o get an inverted, minied image, what kind of lens would be needed positive or negative and where would the obect need to be placed

75

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8 Dispersion and Chromatic Aberration C H A P T E R   T I E Introduction

Chromatic Aberration

Dispersion Rainbows Prisms

Chromatic Aberration in the Human Eye Test Your Knowledge

  E C TI E After working through this chapter, you should be able to: Explain what dispersion is Understand how dispersion occurs Explain how rainbows are produced

Explain what chromatic aberration is Discuss the relevance of chromatic aberration in the human eye

Introduction

Rainbows

In this chapter we will start to think about the constituent wavelengths that make up ‘light’ and how white light (e.g. from the sun) can be separated into these wavelengths through a process called dispersion

hen we see a rainbow in the sk (ig. .1), we will perceive what appears to be a continuous spectrum from longwave length light (red) all the wa to shortwavelength light (blue violet). ow, it is no coincidence that the distribution of colours in a rainbow falls in eactl the same pattern as we would epect if we ordered them b wavelength, as the rea son that dispersion occurs is because the amount that the light undergoes refraction is partl dependent on the mate rial and partl dependent on the wavelength of the light it self. ticking with the rainbow eample, we probabl know that rainbows onl occur when there is water in the air (e.g. when it rains), which indicates that it is the refractive prop erties of the water droplets (the rain) that produces the ef fect. ost of the time, white light from the sun (all visible wavelengths) can refract through the water droplets in the sk without an trouble, but as we can see in able .1, the refractive inde of materials changes slightl depending on the wavelength of the incident light. or eample, shorter wavelengths of light will eperience higher refractive indices than longer wavelengths of light. Importantl, this indicates that there can be a dierence in the speed each wavelength travels through the material. ore specicall, shorter wave lengths are slowed down to a greater etent than the longer wavelengths when the enter these materials, which means that the refract to a greater etent (think back to nell’s law from chapter ) and therefore will have a greater angle of deviation relative to longer wavelength light. ometimes we teach this as ‘blue bends best’ to help students learn that

Dispersion As discussed in chapter 1, white light can be produced when the light source emits all the wavelengths of visible light (e.g. a continuous spectrum), or when the light source emits a large enough range of wavelengths (e.g. a discrete spectrum) that it appears whiteish due to the additive nature of light. or eample, sunlight looks white and comprises a continu ous spectrum, but if I had a red bulb, a green bulb and a blue bulb, the resultant light would look white but would be classed as a discrete spectrum due to the limited range of wavelengths. owever, in some circumstances it is possible for white light to be separated into the individual wave lengths well enough that we can actuall see the dierent colours ascribed to each wavelength (longred shortblue). is happens through a process called dispersion. ome of ou reading this ma have heard about dispersion happening as part of the refractive process when light passes through a prism, but I epect all of ou will be familiar with a form of dispersion that happens when it rains on a sunn da to produce a rainbow in the sk o that end, let’s eplain dis persion using the eample of a rainbow to start with, and then we can move on to discuss prisms and lenses.

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Rain dr

Long wavelength

p

t nt ligh Incide Short wavelength

• Fig. 8.1 An illustration showing the colours that can be found in a rainbow. Red wavelengths are longer than violet wavelengths. TABLE Relationship Between Wavelength (Dened 8.1 as Visible Colou and Reative nde o

the ateial Water (n 5 1.333)

Crown Glass (n 5 1.523)

Red

1.331

1.512

Orange

1.332

1.514

Yellow

1.333

1.51

reen

1.335

1.51

lue

1.33

1.524

iolet

1.342

1.53

Wavelength

short wavelengths (blueend of the spectrum) will refract at a greater angle than long wavelengths (redend of the spec trum), however please remember that this onl applies to dispersion and refraction – the opposite is true for dirac tion which we will learn about in chapter 1. owever, as we’ve alread discussed, most of the time the white light remains white after refracting through the water droplets, so how does the dispersion produce a rainbow e answer is that rainbows occur when the sunlight refracts into the droplet, and then reects back o the farthest sur face of the water droplet to make it refract back out of the side it originated from. ig. . illustrates what this looks like within a raindrop and how the raindrop can disperse the white light into all the colours of the rainbow. In short, when white light is incident upon a raindrop, each constitu ent wavelength is refracted b slightl dierent amounts. e light begins to disperse, but when it reaches the other side of the raindrop, some of the light is reected back again, and so it is then refracted a second time as it leaves the rain drop. It is the combination of the two refraction opportuni ties that disperse the light to produce the rainbow – the re ection that occurs on the back surface does not contribute to the dispersion process as reection does not alter for dif ferent wavelengths, but as the wavelengths will alread be slightl dispersed, the will each be reected slightl dier entl as the will reect o dierent locations of the water droplet. owever, now ou ma be wondering – wh doesn’t

• Fig. 8.2 An illustration showing how a raindrop can split white light into constituent wavelengths through refraction and reection.

the whole sk look multicoloured in that case h is it restricted to a neat little rainbow shape, ell, we then per ceive this process in the sk as a rainbow as we are onl able to perceive the light when it reaches our ees at a particular angle (which makes the rainbow the characteristic arc shape). ow, if I asked ou what shape a rainbow was, what would ou sa I suspect ou’d sa ‘arch’, ‘bridge’, ‘semi circle’ or something similar to that, and ou’d be half right. hat if I told ou that actuall all rainbows are complete circles, but we can onl see the top half In fact, we (as observers close to the ground) are onl able to see light re ected above the horion, and the position of the rainbow will be dictated b our own position on the ground. is means that if I see a rainbow in the sk outside m oce and I tet m colleague across campus to demand that the look outside, the will see an entirel dierent rainbow. l timatel, the ke factor aecting the appearance of the rain bow is the antisolar point, dened as the point 1 degrees awa from the sun relative to the observer (see ig. . for a visual eplanation of this). echnicall, the central point of the rainbow (raincircle?) coincides with the antisolar point, and so logicall, in order to see the rainbow ou’d have to face awa from the sun (which makes sense now that we know rainbows are formed from reections within rain drops). is also indicates that the height of the sun will directl inuence the apparent height of the rainbow,

Sun

Antisolar point

• Fig. 8.

Observer

llustration showing that the antisolar point will eist 1 degrees awa fro the sun relative to the observer. On a bright sunn da it will correspond to the shadow of our head as this highlights the line fro the sun to our ees.

CHAPTER 8

Dispersion and Chromatic Aberration

a li White

Sun Visible part of rainbow

ght

42

Observer

•Fig. 8. Antisolar point

Hidden, mysterious part of rainbow

• Fig.

8. llustration showing how the antisolar point dictates the centre of the rainbow. he radius of the rainbow will be 42 degrees fro this point relative to ou as the observer.

meaning rainbows produced when the sun is low in the sk will appear to be much larger in appearance than those produced when the sun is up high. Interestingl, rainbows form at a specied degree angle between the observer and the antisolar point (ig. .), which roughl determines the radius of the rainbow itself. is means if the antisolar point falls below the horion, the rainbow will be set low down relative to the horion, whereas if the antisolar point is more level with the horion, then the rainbow will form  degrees above that. If the light undergoes two reections within a raindrop (caused b interacting with higher raindrops in the sk), then a second (double) rainbow will be produced with the order of the colours reversed, and this sight is occasionall portraed as ver eciting in viral videos on the internet. Importantl this second rainbow should have a radius that roughl corresponds to  degrees so it should be larger than the main rainbow. In summar, rainbows are produced as a feature of refrac tion that relies on the wavelength properties of light, and the are actuall circular – this means it is unlikel (arguabl impossible) to nd the end of one, and therefore unlikel that there will be an accompaning pot of gold (though never sa never).

he apical angle a of a triangularshaped pris is the angle between the surfaces where the light enters and leaves the pris. hite light will be dispersed.

refractive inde is higher. In the prism eample, dispersion occurs when white light refracts at the rst surface, and it occurs again when the white light eits the prism at the sec ond surface. e angle between the surfaces where the light enters and eits is called the apical angle (a) (see ig. .). Importantl, remember that dispersion is dened as the splitting or separating of wavelengths that make up a light source. is means that dispersion can onl occur if the light source contains more than one wavelength, so if we used a light source made of onl one wavelength (e.g. a red laser), then the light will all refract euall and no disper sion will occur (ig. .). If ou’re interested in the principles of dispersion and feel like it would be benecial to see it in action, ou can follow instructions to make our own prism at home in chapter .

Chromatic Aberration ispersion itself is a neat eample of the relationship be tween refractive inde and wavelength of light, and it pro duces some beautiful patterns of light in the sk and in the optics lab, however it can cause problems in optical sstems. If we think back to chapters 1 and , we know that the rectilinear propagation of light states that light must travel in a straight line when in a homogenous medium, but we also know that following dispersion, the wavelengths of light will be leaving the material at dierent angles. is means that the farther the light travels awa from the material, the more ‘spread out’ the wavelengths will be. imilarl, large

Prisms In optical phsics, we are most likel to consider dispersion occurring as light travels through a prism, as shown in ig. .. risms will be discussed in more detail in chapter , but for now all we need to know is that the word ‘prism’ denes an transparent obect that comprises two surfaces that are at an angle towards one another. ost commonl, prisms are thought of in the classic ‘triangular’ shape such as that in ig. ., but a glass cube would also be classed as a prism (a suare one). owever, for the sake of simplicit in this chapter we will onl consider triangular prisms. ow, in prisms this process of dispersion is identical to that described with the raindrops before, but this time the prism will be made of plastic or glass (as opposed to water) and so the

a ght Red li

•Fig. 8.

Red lig ht

he apical angle a of a triangularshaped pris is the angle between the surfaces where the light enters and leaves the pris. onochroatic single wavelength light e.g. red will be refracted but cannot disperse.

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A

B

• Fig. 8. A beautiful iage of a bird taen through a telephoto lens attached to a obile phone caera A. he edges of the lens have introduced a sall aount of chroatic aberration into the iage highlighted b the two arrows in the ooedin section . age taen b ohn ollins.

amounts of dispersion can occur over shorter distances through the use of high powered lenses, which cause the light to bend a lot. is can cause problems when viewing images, as dispersion can produce varing degrees of distor tion in the nal image called aberrations. A particularl important eample of this in optics is chromatic aberration (colourrelated distortion), which can occur in magni fing optical sstems (e.g. telescopes, binoculars, low vision aids) if not carefull considered. is is a term used to de scribe imperfections in an image related to the splitting of colours, as shown in ig. .. Interestingl, chromatic aber rations can be categorised across two tpes transverse chro matic aberrations (often abbreviated as A) and longitu dinal chromatic aberrations (often abbreviated as A). e ke dierence is that transverse chromatic aberration is identied when the sie of the image changes due to the wavelength of the incident light (see ig. .A), whereas longitudinal chromatic aberration is identied when indi vidual wavelengths in the incident light focus at dierent points along the optical ais (see ig. .). is chapter will primaril focus on longitudinal chromatic aberration, but please be encouraged to read about the subect if ou want to know a little bit more. e amount of dispersion that a material will produce can be calculated as the Abbe number or constringence number (two names for the same thing) of the material, denoted b the letter  in uation .1. e n variables in uation .1 are showing the dierent refractive indices for ellow (nd), blue (nf ) and red (nc) light. Equation 8.1

V

nd

1

f

c

Equation 8.1 (explained) ref . ind . yellow 1 constringence ref . ind . blue ref . ind . red In simple optical sstems, lenses are often made of either crown glass or int glass. ow, even within the same ‘tpe’ of glass, the can possess dierent constringence numbers depend ing on their densit, but for simplicit in this chapter we will consider common tpes of crown glass with constringence

A

Lens TCA White light

B

Lens LCA White light

•Fig. 8.8

iagra illustrating the difference between transverse chro atic aberration A A where sie of iage is affected b wave length relative to longitudinal chroatic aberration A  where point of focus on the optical ais is affected b wavelength.

values of , and common tpes of int glass with constrin gence values of . It is important to note here that high constringenc indicates low levels of dispersion and low con stringenc indicates higher levels of dispersion, so if we had two identical lenses, one made of crown glass and one of int glass, the crown glass lens ( 5 ) should disperse light less than the int glass lens ( 5 ). In optical sstems, if this constringenc is not accounted for, it can produce the chromatic aberration demonstrated in ig. .. is chromatic aberration (A) increases as the power () of the lens increases, as indicated in uation .. Equation 8.2

CA 

F V

Equation 8.2 (explained) chromatic abb 

power constringence

CHAPTER 8

Dispersion and Chromatic Aberration

DEMO QUESTION 8.1

DEMO QUESTION 8.2

hat is the chroatic aberration of a crown glass lens with a power of 15.  tep 1 eterine what we need to calculate chroatic aberration A tep 2 ene variables  5  the nown constringenc value for crown glass  5 15. tep 3 eterine necessar euations A 5    (Equation 8.2) tep 4 alculate A 5    A 5 5   A 5 . a A value of  would be an ‘aberrationfree’ iage

hat power of int glass would be reuired to reove chroatic aberration fro a lens ade partl of crown glass 14.  tep 1 eterine what we need to calculate power  f tep 2 ene variables c 5  the nown constringenc value for crown glass f 5 3 the nown constringenc value for int glass c 5 14. A 5  the uestion hasn’t stated it but we’re aiing to reove all chroatic aberration that would be a value of ero tep 3 eterine necessar euations A 5 c  c 1 f  f (Equation 8.3) tep 4 alculate A 5 c  c 1 f  f  5 4   1 f  3  5 ... 1 f  3  ... 5 f  3 ... 5 f  3 ... 3 3 5 f 2.  5 f he int glass in this achroatic doublet would need to be 2.  to reove chroatic aberrations.

Practice Questions: 8.1.1 hat is the chroatic aberration of a crown glass lens with a power of 12.  8.1.2 hat is the chroatic aberration of a int glass lens with a power of 11. 

owever, it is possible to design lens sstems in such a wa to undo the potential for some of the chromatic aberration b utilising a doublematerial lens called an achromatic doublet. e know that dierent materials possess dierent constringenc values, so it is reasonabl logical then to un derstand that we might be able to use two dierent materials to bring the chromatic aberration down to ero. A good e ample would be to use a combination of crown glass and int glass to produce the power of image that is reuired whilst also removing all the chromatic aberration (ig. .). e can calculate the power of one tpe of glass lens re uired to undo the chromatic aberration produced b the lens made of the other tpe of glass b using a modied ver sion of uation . shown as uation .. Fc F f CA uation . Vc V f uation . (eplained) power (cr ) constring (cr )

chromatic abb Crown glass

power ( fl ) constring. (fl )

Flint glass

White light

Achromatic image Achromatic doublet

•Fig. 8.

hen a lens is ade of two aterials and designed in such a wa as to produce a chroatic aberration value of ero it is called an achroatic doublet.

81

Practice Questions: 8.2.1 hat power of int glass would be reuired to reove chroatic aberration fro a lens ade partl of crown glass 1.  8.2.2 hat power of crown glass would be reuired to re ove chroatic aberration fro a lens ade partl of int glass 12. 

In conclusion, dispersion occurs when light comprising man wavelengths is refracted through a material. is dis persion can be managed b using a light source comprising onl a single wavelength, or b utilising a special lens (ach romatic doublet) that removes an impact of dispersion on the nal image.

Chromatic Aberration in the Human Eye p to this point in the chapter, we’ve discussed chromatic aberration in optical sstems and lenses, but it might inter est ou to know that longitudinal chromatic aberration is also present in the human ee. is occurs because the shorter wavelengths refract more than the longer wave lengths when transmitting through the optics of the ee (ust like in lenses). is means that it is usuall assumed that shorter (‘blue’) wavelengths will focus slightl in front of the retina and longer (‘red’) wavelengths will focus slightl behind the retina, as illustrated in ig. .1 (re member, ‘blue bends best’). owever, interestingl, we (as observers) don’t usuall notice this because (1) it’s uite a small eect and () the brain is ver good at overcoming predictable (freuentl eperienced), small errors like this. In optometric practice, optometrists attempt to measure refractive error of the ee (see chapter ) and then test dif ferent corrective spherical or clindrical lenses to see if the

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Emmetropic eye

A

B

Too much positive power

Too much negative power

• Fig. 8.1

llustration of longitudinal chroatic aberration in the hu an ee. onger wavelength light focuses behind the retina red whilst shorter wavelength light focuses in front of the retina blue.

C

Correct

•Fig. 8.11

aple iage of a duochroe test. n clinical practice the wavelengths would need to be carefull calibrated and so this printed or digital eaple will not wor as effectivel if ou’re tring it for ourself at hoe.

can improve visual acuit. is assignment of corrective lenses is referred to as a patient’s spectacle prescription. Importantl, optometrists can utilise the naturall occur ring chromatic aberration in the ee b using a test called a duochrome (twocolour test) to check the prescription is accurate. In this test, two sets of targets are presented to the patient, one set on a green background ( nm) and one set on a red background ( nm)1 (see ig. .11 for an eample). If the prescription is accurate, then the targets should look euall clear on both the green and red backgrounds, indicating that the ee is fo cused between the red and green (as is appropriate). ow ever, if the patient has been given too much positive power

• Fig. 8.12

A diagra showing the patient’s perspective of a duo chroe and the associated focusing of the light on the retina when overplussed A overinussed  and appropriatel corrected .

in the lens (‘overplussed’) or too little negative power (‘underminussed’), the target on the red will look clearer – indicating that the combination of the ee and lenses are bending the light too much and more negative power is need ed (see ig. .1A). n the other hand, if the patient has been given too much negative power in the lens (‘overminussed’), or too little positive power (‘underplussed’), the target on the green will look clearer – indicating that the combination of the ee and lenses aren’t uite bending the light enough and more positive power is needed (see ig. .1). e clinician can then adust the prescription as reuired.

Test Your Knowledge r the uestions below to see if ou need to review an sec tions again. All answers are available in the back of the book. TY.8.1 hat wavelength refracts the most – red or blue TY.8.2 plain how raindrops produce a rainbow. TY.8. ould dispersion occur if we shone a red laser through a prism plain our answer.

Reerence 1. abbetts . Bennett and Rabbetts’ Clinical Visual Optics. th ed.  utterwortheinemann, lsevier .

TY.8. hat is chromatic aberration TY8. If a person is slightl underminussed in their glasses prescription, will green or red wavelengths be more likel to focus on the retina

9 Prisms    P  E   L INE Introduction

Images

What Is a prism?

Prismatic Power

Deviation of Light Minimum Angle of Deviation Total Internal Reection and Critical Angle

Prismatic Lenses and Base Notation Prismatic Eects in Spherical Lenses est our nowledge

 B E  I ES After working through this chapter, you should be able to: Describe what a prism is Eplain what the minimum angle of deviation is Dene a ‘critical angle’

Discuss how light deviates through a prism relative to the base and ape Calculate prismatic power

Introduction

wavelength beam of light as this prevents dispersion, see chapter  for revision on this if necessary. o start with, let’s consider the deviation of the light path. If you look at the solid lines in ig. ., these correspond to the path that the light takes as it travels through the prism. n important feature of prismatic refraction refraction through a prism is that light will always deviate towards the base, shown by the emergent light ray being angled downwards in this eample. or clarity, the ‘base’ of the prism is the side opposite to the apical angle, so in ig. . it is the horiontal line along the bottom of the prism. ow, if you look instead at the dashed lines in ig. ., you can see that they correspond to the original path the light would have taken if it hadn’t been deviated by this pesky prism being in the way. o this end, the angle between the original path of light and the true, deviated path of light is called the angle of deviation (d). It is also crucial to note that this angle of deviation d is the sum of the angles of deviation at each surface of the prism d and d – see uation . for an eample of this.

In the previous chapter, we introduced the topic of prisms and discussed that they are capable of dispersing white light. In this chapter, we will discuss prisms in more detail and learn all about their power, minimum angle of deviation, total internal reection and critical angle.

What Is a Prism? When I read the word ‘prism’, I think of a glass, triangularshaped obect with ve sides like the one in ig. .. echnically my instinct is correct, and this is a prism, but in optics the true denition of the word ‘prism’ encompasses any obect that possesses two at refracting surfaces inclined at an angle towards each other. is means that all of the glass obects illustrated in ig. . are also technically prisms, but this chapter is going to focus on the triangularly shaped prisms.

Deviation of Light ow, if we take a cross section of a triangular-shaped prism, we can imagine that it would be made of a material other than air – suggesting it will have a dierent refractive inde np relative to the surrounding air ns ig. .. Importantly, prisms also possess an apical angle a, which is dened as the angle that eists between the rst and second refracting surface where the light enters and leaves the prism. is is demonstrated in ig. . with a monochromatic single

Equation 9.1

1

2

Equation 9.1 (explained) 1

2

e

ut, how do we calculate these angles of deviation in the rst place e answer is maths as always and cyclic quadrilaterals. ow, if you haven’t heard of cyclic uadrilaterals 83

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before, don’t worry, you can utilise the information in o . to help you learn what they are. e information provided in o . highlights that we can edit uation . to look like uation . depending on the information we have at the time

A

d

Equation 9.2 B

(1

2

) a

Equation 9.2 (explained) dev

.

. 2

.

) apical ang.

is shows us that the angle of deviation d is directly related to the angle that the light enters the prism i, the angle light leaves the prism i9, and the apical angle a of the prism itself – neat.

• Fig. 9.1

Drawing of ‘classic’ triangular prisms (A) and examples of other types of prism (B).

ns

1

in O r ig

a

al p

ath ns

d

d1

d2

np

Dev

iated of lig path ht

•Fig. 9.2

Illustration showing refractive index outside the prism (ns) will differ to the refractive index inside the prism (np). If a monochromatic (single wavelength) eam of light is incident upon the surface the light will refract at oth the incident rst surface and the emergent second surface producing two angles of deviation (d and d). he overall angle of deviation (d) is the difference etween the original path of the light relative to the deviated path of light.

•BOX 9.1

Opposite Angles, Cyclic Quadrilaterals and Prisms

If we thin ac to chapter  we rememer that angles of inci dence (the angle at which rays of light approach a refractive sur face) are measured relative to the ‘normal’ of the surface – a hy pothetical line that exists at a perpendicular angle to the surface itself (at the point where light intersects the surface). e also learnt all aout opposite angles (see Box . for revision on this) and with understanding deviation in prisms opposite angles hold the ey. or example in ig. B. we can see that we can apply the nowledge that vertically opposite angles are eual in order to calculate the deviation at the rst surface (d) showing that d 5 i i he same is true for the second angle of deviation (d) at the second face of the prism (ig. B.) showing that d 5 i92i And as uation . taught us we now that total deviation is eual to d 1 d so that means we can also say that d 5 (i i ) 1 (i92i) ow let’s tae a moment to thin aout cyclic quadrilaterals. he denition of a cyclic uadrilateral is that it must have four sides and all four vertices (corners) must connect to a hypotheti cal circle (see ig. B.B). yclic uadrilaterals also reuire all in ternal angles to add up to ° and opposite angles within the uadrilateral must add up to °. ow you’re liely thining ‘Why is Sam telling us all this?’ and the answer is ecause the ‘normal’ lines of each surface (at the point where light enters and leaves the prism) form a cyclic uadrilateral (ig. B.A). his means that us ing our newfound nowledge of cyclic uadrilaterals we now that

d1 i1

i1’

• Fig. B9.1 Illustration showing how principle of opposite angles ap plies to the rst surface of the prism. Image highlights that i  5 d 1 i9 and therefore d 5 i – i9.

CHAPTER 9

•BOX 9.1

Prisms

85

Opposite Angles, Cyclic Quadrilaterals and Prisms – cont’d the apical angle (a) and the angle formed y the two normal lines () must add up to °. owever we also now that angles along a straight line eual ° which means that this hypothetical angle outside of the cyclic uadrilateral in ig. B.B would also e eual to that of the apical angle. Another important mathematical fact aout triangles is that external angles (along a straight line) will e eual to the sum of the opposite two interior angles (see ig. B.). his is ecause the internal angles of a triangle all add up to ° and angles along a straight line add up to ° so if we consider ig. B. for a moment we now that ° 5  1 i9 1 i and ° 5  1 a herey proving that Equation B9.1:

i2

a 5 i 1 i

i2

hus highlighting that we can rearrange our earlier euation to turn d 5 (i 2 i ) 1 (i9 2 i) into d 5 (i 1 i ) – a (see uation .)

d2

• Fig. B9.2

Illustration showing how principle of opposite angles ap plies to the second surface of the prism. Image highlights that i 9 5 d 1 i and therefore d 5 i – i

A

B a

a

a

b

b a

a + b = 180

C

i1’

i2 b a

• Fig. B9.

i1’ + i2 + b = 180 a + b = 180 therefore a = i1’ + i2

(A) Diagram showing how the two ‘normal’ lines (orange dashed lines) of the prism create a cyclic uadrilateral (lue). (B) Diagram showing properties of cyclic uadrilaterals to prove that the external angle (right) will e eual to the apical angle (a) of the prism. () Diagram showing that the apical angle (a) will e eual to the sum of the two internal angles formed y the light ray (i 9 and i).

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eometric and Basic ptics

second face i9 decreases. owever, it also shows us that the total angle of deviation will reach a point where it can’t go any lower. is point is called the iniu angle of deviation (din) of the prism, and as you might epect by now, it’s related to the apical angle a and the refractive indices of the prism np and the surroundings ns. We can calculate the minimum angle of deviation using uation .

DEMO QUESTION 9.1 A ray of light is incident upon a prism at an angle of ° and leaves the prism at an angle of °. If the prism has an apical angle of ° what is the angle of deviation of the light tep  Determine what we need to calculate angle of deviation d tep  Dene variales i 5 ° (angle of incidence at rst surface) i9 5 ° (angle of refraction at second surface) a 5 ° (apical angle) tep  Determine necessary euation d 5 (i 1 i ) a (Equation 9.2) tep  alculate d 5 (i 1 i ) – a d 5 ( 1 ) –  d 5 ° (don’t forget the units)

Equation 9.

np

0.5 ( min ) sin0 .5 (a )

ns

Equation 9. (explained) RI prism RI surround

0.5 (

. . ang . dev .) sin0 .5(apical ang.)

Practice Questions: DEMO QUESTION 9.2

9.1.1 A ray of light is incident upon a prism at an angle of ° and leaves the prism at an angle of °. If the prism has an apical angle of ° what is the angle of deviation of the light 9.1.2 A ray of light is incident upon a prism at an angle of .° and leaves the prism at an angle of .°. If the prism has an apical angle of ° what is the angle of de viation of the light

inimum ngle of Deviation y this point in the chapter, we’ve learned that prisms deviate light towards the base, and we know that the deviation of light partly relates to the angle of incidence i of the incoming light. is means that as the angle of incidence changes, the angle of deviation d will also change – but how are the two variables related Well, thanks to uation ., we can determine that if we know the apical angle of a specic prism, we can calculate the angle of deviation for lots of dierent angles of incidence. on’t worry, though I’ve done the maths for you, so all you need to do is enoy looking at the data in able . is table shows us that as angle of incidence at the rst face of a prism i increases, angle of refraction at the

alculate the minimum angle of deviation of a prism con structed of refractive index . of apical angle °. tep  Determine what we need to calculate minimum angle of deviation dmin tep  Dene variales np 5 . ns 5 . (not specied so we assume air) a 5  (apical angle) tep  Determine necessary euation np  ns 5 (sin.(a 1 dmin))  (sin.(a)) (Equation 9.) tep  alculate np  ns 5 (sin.(a 1 dmin))  (sin.(a)) (ut we need to rearrange the equation so that we can calculate dmin) ns (sin.(a 1 dmin)) 5 np (sin.(a)) . (sin.(1dmin)) 5 . (sin(. 3 )) sin.(1dmin) 5 . (sin(.)) sin.(1dmin) 5 ... .(1dmin) 5 sin(...) .(1dmin) 5 ... . 1 .(dmin) 5 ... .(dmin) 5 ... – . .(dmin) 5 ... dmin 5 ...  . dmin 5 .° (don’t forget the units)

TABLE ata oing at as Angle o ncidence at te First Face o a Prism i1 9.1 ncreases, Angle o eraction at te econd Face i29 ecreases, ut otal

Angle o eiation d Can’t o oer tan .9° ee old tet

i1 (deg)

i29 (deg)

d (deg)



.

.



.

.



.

38.9



.

.



.

.



.

.

CHAPTER 9

DEMO QUESTION 9.2 – con’

Practice Questions: 9.2.1 alculate the minimum angle of deviation of a prism constructed of refractive index . of apical angle °. 9.2.2 alculate the minimum angle of deviation of a prism constructed of refractive index . of apical angle °.

otal Internal eection and ritical ngle or this section of the chapter, let’s do some uick revision of how light refracts through surfaces. o you remember in chapter , we discussed that when light is incident upon a surface, it can be transmitted through, absorbed, reected or a combination of all of these things If not please see chapter  for revision on this. Well, along that same line of thought, when we discuss light refracting through surfaces, we are oversimplifying the process by assuming that  of the light is transmitted through the surface. In reality, some of the light is reected back, and the amount of light reected and the angle of reection i9 will depend on the angle of incidence i. ow, this common, small amount of reection occurs when light travels from a low-refractive inde medium to a higher refractive inde medium, and vice versa, but now we are going to discuss a phenomenon called total internal reection which can only occur when light travels from a higher refractive inde into a lower one. In chapter , nell’s law uation . taught us that if light travels from a high refractive inde medium to a lower refractive inde medium, then the angle of refraction i9 will be larger than the angle of incidence. is is important because, as you can see in ig. ., if the angle of incidence i is large enough, eventually the refracted light will travel along the surface of the material it’s leaving ig. ., meaning that its angle of refraction i9 is eual to ° relative to the normal. e specic angle of incidence reuired for this to Light source

A ic

Prisms

happen is called the critical angle ic. Importantly, if the angle of incidence increases beyond the critical angle, then all the light will be reected back into the material it originated in, which is called total internal reection ig. .. e critical angle itself is entirely reliant upon the refractive inde of the material, meaning that, for eample, the critical angle of a prism made of diamonds n 5 . will be dierent to that of a prism made of perspe n 5 .. With prisms, we can use uation . to calculate the critical angle of each specied prism sin(ic ) 

Equation 9.

1 np

Equation 9. (explained) sin(critical angle ) 

1 ref. index prism

rucially, in normal circumstances in which a prism is likely sitting in a room surronded by air, the critical angle of the prism can only eist at the second surface, as light is trying to leave the prism. is is because at the rst surface, the light isn’t moving from a high- to a low-refractive inde as air is the lowest refractive inde available to us, whereas as it leaves the prism, it is moving from a high- to a low-refractive inde, so total internal reection can occur. is means that if light leaving the prism leaves at ° to the normal, then the angle of incidence at the second face i will be eual to the critical angle ic of the prism ig. .. ow, using a combination of skills we’ve learnt throughout this tetbook so far, if we know the apical angle and the refractive inde of a prism, we can calculate the smallest angle of incidence possible before total internal reection occurs. r, to phrase it dierently, we can calculate the angle of incidence needed to produce the critical angle at the second face of the prism. o do this, we rst need to calculate the critical angle ic of the prism in uestion uation ., which tells us the angle of incidence at the second surface i. We can then calculate the angle of refraction at the rst surface i9 using uation ., which leads us to be able to use nell’s law see chapter , uation . to calculate the angle of incidence reuired at the rst face to make it all happen i.

B

if i2’=90 then i2=ic C

i1 Glass block

Air

i1’

i2

i2 ’

•Fig. 9.

Diagram showing that when light travels from a high refrac tive index glass loc (.) to air (.) the angle of refraction will e larger than the angle of incidence (A) and some light will e re ected ac within the glass loc. At a specic angle of incidence nown as the critical angle (ic) the light will refract at ° to the normal and travel along the surface of the glass loc (B). If the angle of inci dence exceeds the critical angle then all light will e reected ac – this is called total internal reection ().

• Fig. 9.

Diagram showing that as light emerges at the second face at ° (i9) then the angle of incidence at the second face (i ) must e eual to the critical angle (ic).

87

88 S EC TI ON 1

eometric and Basic ptics

DEMO QUESTION 9.3

Images

Determine the smallest angle of incidence at the rst face of an euilateral triangular prism of refractive index . for light to ust pass through the second face. he apical angle of the prism is °. tep  Determine what we need to calculate angle of incidence i tep  Dene variales np 5 . ns 5 . (not specied so we assume air) a 5  (apical angle) tep  Determine necessary euation sin(ic) 5   np (Equation 9.) a 5 i9 1 i (Equation 9.) n (sin i) 5 n’ (sin i ) (Equation 2.2) tep  alculate sin(ic) 5   np sin(ic) 5   . ic 5 sin (  .) ic 5 ...° i 5 ic a 5 i’ 1 i i’ 5 a – i i’ 5  – ... i’ 5 ...° n (sin i) 5 n (sin i ) ns (sin i) 5 np (sin i’) . (sin i) 5 . (sin ...) sin i 5 . (sin ...)   i 5 sin (. (sin ...)) i 5 .° (don’t forget the units)

opefully, by this point in the chapter, we’re happy to accept that prisms will deviate the path of light, but before we move on, I think it’s also important to think about what that means in the real world. et’s use an eample of an obect – perhaps our favourite mug ig. .. We already know that in order to be able to see the obect, we need it to be illuminated lit up so that light can reect o the surface of the obect to reach the backs of our eyes – this allows us to perceive an image of the obect see ig. .. owever what we don’t often think about is that this means the light will travel along a specied path to your eye, and because light travels in straight lines in a homogenous medium see chapter  for revision on this, providing the light doesn’t undergo any refraction change in direction then the image of the obect will correspond to the point where the obect is. owever, what happens if the light ray changes direction before it reaches you I think the easiest way to illustrate this is to think about plane mirrors see chapter  for revision on this. We can hopefully understand that when we look at an obect reected in a mirror, the image of the obect will appear to eist ‘within’ the mirror itself ig. .. is occurs because the light ray is reected o the mirror at an angle relative to us. is means the light ray reaching our eyes will appear to be coming from within the mirror, which then proects the image of the obect to eist ‘behind’ the mirror owever, thankfully we understand how mirrors work, so we ust interpret it as a reection rather than a magical portal into a parallel, backwards dimension although it always surprises me when my cats successfully utilise mirrors to keep an eye on me from across the room – how do they understand how to do that?. ou can demo this principle for yourself at home by looking into a regular mirror and trying to think about where it looks like the image is instead of what we know is happening – can you see that it looks like the images are ‘within’ the mirror k, that all makes sense, but you’re probably thinking – how does this apply to prisms Well, we learned in an earlier section that light travelling through a prism will always be

Practice Questions: 9.3.1 Determine the smallest angle of incidence at the rst face of an euilateral triangular prism of refractive index . for light to ust pass through the second face. he apical angle of the prism is °. 9.3.2 Determine the smallest angle of incidence at the rst face of an euilateral triangular prism of refractive index . for light to ust pass through the second face. he apical angle of the prism is °. 9.3.3 Determine the smallest angle of incidence at the rst face of an euilateral triangular prism of refractive index . for light to ust pass through the second face. he apical angle of the prism is °.

Light from object reaches eye

Person perceives image of object

Object

•Fig. 9.

Illustration showing how light travels from an oect to our eyes – this allows us to perceive the image of the oect.

CHAPTER 9

Mirror world

Prisms

Real world Person perceives image of object

t jec e ob s ey m fro ard ht tow g i L cts le ref

Image of object will appear to exist “within” the mirror as the reflected ray appears to originate from here

Plane mirror Image

Object

•Fig. 9.

Illustration showing how light travels from an oect to our eyes if reecting off a plane mirror – this allows us to perceive the image of the oect ut the ray that reaches our eye appears to e coming from within the mirror and so we perceive the image as eing within the mirror.

deviated towards the base of the prism, but we now also know that we perceive images based on where the light appears to be coming from. is means that in order to understand where the image is in a prism, we have to imagine a straight line is proected backwards from the emergent light ray on the right side of the prism ig. .. is illustrates another important feature of prisms – the iage is always

deviated towards the apex. o now we can remember that with prisms . ight will deviate towards the base. . e image will deviate towards the ape. is understanding of how light can be deviated is a very important clinical consideration, as optometrists will sometimes need to prescribe prismatic lenses for patients if

Image of object will be projected towards the apex of the prism (the tip) as the deviated ray appears to originate from here

Person perceives image of object

Apex

Lig tow ht from ard obj s em the ba ect is erg d ent se of eviat tow light r the pr ed ard i s ey efracts sm e

Image

Base

Object

• Fig. 9.

Illustration showing how light travels from an oect to our eyes if refracting through a prism. his allows us to perceive the image of the oect ut the ray that reaches our eye appears to e coming from the apex side of the prism so we perceive the image as eing in a different location to the oect itself.

89

9 S EC TI ON 1

eometric and Basic ptics

they have, for eample, a turned eye strabismus. is will be discussed in more detail later in the chapter in the section discussing ‘rismatic enses and ase otation.’

Prismatic Power In the section titled ‘eviation of ight’ we talked a lot about angles of deviation d and how they will vary depending on the apical angle of the prism and angles of light. In this section, we will begin to consider the power of the prism , which is measured in pris dioptres (denoted as pd or ∆). In simple terms, the power of a prism dictates the light-deviating capabilities of the prism, and it can be calculated by considering the distance y and the displacement  of the image. ig. . uses a right-angled prism to help illustrate an eample of the dierences between these two terms. In order to calculate prismatic power , we can use uation ., but it’s really important to remember that with priss all the units need to be in centietres not metres ⎛ x⎞ Equation 9. P (100) ⎝ y ⎠ Equation 9. (explained) prismatic power

A

⎛ displacement ⎞ (100) ⎝ distance ⎠

DEMO QUESTION 9.4 A prism produces  cm of displacement at  cm. hat is its power tep  Determine what we need to calculate prismatic power  tep  Dene variales x 5  cm (displacement) y 5  cm (distance) tep  Determine necessary euation  5 ()(xy) (Equation 9.) tep  alculate  5 ()(xy)  5 ()(  )  5 ∆ (don’t forget the units)

Practice Questions: 9.4.1 A prism produces  m of displacement at  m. hat is its power 9.4.2 A prism produces  cm of displacement at  cm. hat is its power

ow, if you were a huge fan of maths when you did your secondary school eams, then you may have already realised that because the angle of deviation d, the distance y, and the displacement  all form part of a right-angled triangle,

Apex P = (100)(x/y) p = (100)(2/100) p = 2Δ

Distance displacement is measured at (y) d Displacement (x)

Base

B

Apex Distance displacement is measured at (y)

P = (100)(x/y) p = (100)(3/150) p = 2Δ

d Displacement (x)

• Fig. 9.

In this image a rightangled prism is deviating light (red line) towards the ase of the prism at a set angle of deviation (d). In oth examples the prism possesses the same power ( ∆) ut in (A) the displacement (x) is measured at a shorter distance (y) which means the displacement is less than that in (B). his is ecause the angle of deviation (d) is constant and so trigonometry teaches us that the distance and displacement are intrinsically related.

CHAPTER 9

we can of course use trigonometry to show the relationship between these three numbers. If you revise some trigonometry with me in ig. ., you’ll be able to see that we know the adacent distance and opposite displacement sides of the triangle, which means we can use tan to solve for the angle or side lengths. owever, what’s cool about this is that we know from uation . that the division of the displacement over distance is part of how we calculate prismatic power, which means that we can also substitute in our angle of deviation d to calculate prismatic power  using uation . Equation 9.

P (100 )(tan d )

prismatic power

angle of dev.

ow we can also determine that the power of the prism is related to the angle of deviation – which makes sense because logic tells us that if prisms deviate light, then the power of the prism should dictate how much deviation occurs. sing uation ., you can see that large angles of deviation d will produce large powers , and vice versa. owever, one nal link we need to make is between the angle of deviation and the apical angle a. s we learned in chapter , the refractive inde of a material will dictate the amount of refraction that occurs at the boundary of the surface of the material. With prisms, this means the refractive inde will have a role in determining how light is deviated. We also know from chapter  that the distance light travels will determine its vergence as it approaches a surface, and so we can also deduce that the distance between the rst and second surfaces of the prism will also have a role in how light is deviated as it leaves the prism. Importantly, this distance between the rst and second surface of the prism is intrinsically linked to the sie of the apical angle. It should come as no big surprise then to learn that we can use uation . to calculate the angle of deviation d

(

p

Equation 9. (explained) dev

refr. index prism refr. index surround

pical ang

rucially, the take-home message here is that increasing the apical angle and thereby increasing the distance between the rst and second surface, along with the sie of the base of the prism, will also increase the angle of deviation and therefore increase the overall power of the prism. o, in summary, chunky prisms will have more power and deviate light to a greater etent than their narrow counterparts.

Prismatic Lenses and Base Notation

Equation 9. (explained)

Equation 9.

Prisms

s

linically, it’s important to understand how prisms refract light, because they may be prescribed for patients who eperience diplopia (double vision). iplopia typically arises because the visual aes of the eyes are not eactly aligned, which means that the image of the world falls in two, distinct regions of the retina in each eye ig. .. In some cases, these patients may be able to have their vision corrected, but in cases where the eye is unable to be straightened eectively, one of the ways to reduce the impact of diplopia is to prescribe a prismatic lens to deviate the light to make it fall on the corresponding part of the retina, relative to the other eye ig. .. emember that in the section titled “eviation of ight” we learned that light always deviates towards the base of the prism is means that, in general, it’s relatively straightforward to determine the direction the prism would need to be oriented in order to deviate the light in the right direction once we understand the principles of course. In ig. ., A RE

LE

) a

Adjacent

A

B

θ

Hyp

oten

Opposite

tan(θ) = O/A

use

LE (turned outwards)

RE

y

B

•Fig. 9.9

d x

tan(d) = x/y

Diagram showing revision of rightangled triangles (A) so that we can understand how to apply trigonometry to the distance and displacement from the deviated light leaving the prism (B).

•Fig. 9.1

Diagram showing diplopia (doule vision) affecting a patient whose left eye is turned outwards (A). his causes the light to fall on different parts of the retina of each eye B) producing two images of the oect.

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eometric and Basic ptics

clear and in focus. owever, if my glasses slip down my nose as is their vocation in life then my eyeline will stop corresponding to the optical centre and will instead be looking through a different region entirely, which induces prisatic eects as it begins to deviate the light inappropriately ig. .. ese prismatic eects are present in both positively powered and negatively powered lenses and can be understood simply in terms of the basic shape of the lens. ig. . shows the cross-section of two lenses, and highlights that an eample positive lens and an eample negative lens could be represented simply as two prism-shaped lenses instead. e amount of prismatic power at various distances from the optical centre of a spherical lens can be calculated using rentice’s rule, which takes into account the distance in c from the optical centre (decentration 2 c) and the power of the lens () – see uation .

A RE

B

LE

LE (turned outwards)

RE

P  cF

Equation 9. Equation 9. (explained) prism power

•Fig. 9.11

Diagram showing diplopia (doule vision) can e corrected y prescriing a prismatic lens (A). his causes the light to deviate towards the ase which means the light falls on corresponding parts of the retina of each eye (B).

the left eye  is turned outwards, which means the light needs to be deviated inwards in order to match the right eye . is means we need the base of the prism to be near the inner side of the eye in order to deviate the light in that direction, so we call this a ‘base-in’ prism. ig. . shows the notation for types of base. It’s good to also remember that the amount of deviation reuired depending on the amount of ‘turn’ of the eye will dictate the power of prism reuired too.

Prismatic Eects in Spherical Lenses nother important feature to consider with prisms is that prismatic eects are present in pretty much all glasses – whether you have a prism prescription or not or eample, I wear a cool pair of roughly 2.  lenses, and when I look through the optical centre of the lens then everything is

Base

lens power

DEMO QUESTION 9.5 ow much prism power will e induced if a patient loos through their 1. D lens  mm left of the optical centre tep  Determine what we need to calculate prismatic power  tep  Dene variales  5 1. (lens power) c 5 . cm (decentration conerted to cm) tep  Determine necessary euation  5 c (Equation 9.) tep  alculate  5 c  5 . 3   5 ∆ (don’t forget the units)

Practice Questions: 9.5.1 ow much prism power will e induced if a patient loos through their . D lens  mm left of the optical centre 9.5.2 ow much prism power will e induced if a patient loos through their 1. D lens  mm right of the optical centre

Base

Base

Base Base-down (deviates light downwards)

decentration

Base-up (deviates light upwards)

•Fig. 9.12

Base-out (deviates light outwards)

Illustration of notation for prism directions.

Base-in (deviates light inwards)

CHAPTER 9

A

Positive lens

B

Prisms

Negative lens

•Fig. 9.1

ide view of a person wearing negative lenses. hen loo ing through the optical centre light focuses appropriately (A) ut if looing through part of the lens nearer the edge then prismatic effects can e induced (B).

• Fig. 9.1

enses induce prismatic effects that get greater as you move away from the optical centre of the lens.

Test Your Knowledge ry the uestions below to see if you need to go over any sections again. ll answers are available in the back of the book. .9.1 Will light deviate towards the base or the ape of the prism .9.2 What is the ‘critical angle’

.9. If a prism were submerged in water, would it deviate light dierently from in air plain your answer. .9. Would increasing the apical angle increase the power of the prism or decrease it

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SECTION

2

Physical Optics 10. Superposition, Interference and Diraction, 97

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10 Superposition, Interference and Diraction C   P T  R  U T I  Introduction

Circular Diraction

Features of a Single Wave

The Resolving Power of a Sste

Features of Multiple Waves

Thin Fil Interference The Fun Side of Interference

Utilising Interference to Measure Distances

Test our nowledge

Diraction Single Slit Diraction Multiple Slit Diraction

    C T I S After working through this chapter, you should be able to: plain the ey features of a ae aelength, phase, aplitude plain the dierence eteen phase dierence and path dierence

Dene the ters ‘superposition’, ‘interference’ and ‘diraction’ plain hat a diraction pattern loos lie plain realorld eaples of interference to your friends

Introduction

point centre line to the peak or trough of a wave, and frequency number of cycles per second. As a general rule, the energy of a wave is related to the wavelength l and the amplitude A. As we learned in chapter 1, as wavelength increases, energy decreases, and as wavelength decreases, energy increases uation 1.1,

At this point in the textbook, we have almost exclusively considered light to be travelling in straight lines, and we’ve been discussing principles of geometric optics. However, as we touched on in chapter 1, some phenomena can only be explained by thinking of light as a wave, which is why we consider light to exhibit wave-particle duality. is prin ciple highlights that it’s important for us to start thinking about the wavelike properties of light (physical optics. is chapter aims to take you through the fundamental principles of physical optics and introduce superposition, interference and diraction

Wavelngth

Peak Amplitude

Features of a Single Wave o start with, let’s review what makes up a wave and some of the key terminologies. ig. 1.1 shows the peaks/crests as the top of the wave, with the troughs being the bottom of the wave. t also shows the dierence between the wavelength distance between two corresponding points on a wave, amplitude maximum distance from the euilibrium

One cycle (frequency is number of cycles in a second)

•Fig. 10.1

Trough

Diagram showing key features of a wave.

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98 S EC TI ON 2

Phsical ptics

showing that wavelength and energy have an inverse rela tionship as one gets bigger, the other gets smaller. Equation 10.1

E

hc  •Fig. 10.3

Equation 10.1 (eplained energy

wo smaller-amplitude waves (lue) are interacting (superposition) to form a resultant wave (pink).

Plancks constant speed of light waveleength of light

n terms of amplitude, however, the rules are a little more straightforward, because as amplitude increases, energy also increases. n fact, as the amplitude is doubled, the energy of the wave will be uadrupled, as shown by uation 1.. Equation 10.

E  A2

Equation 10. (eplained energy (is proportional to ) amplitude 2 e nal feature of a wave to discuss is the phase – dened as the location of a point on the wave within a cycle.  nd this is made clearer if we imagine the wave is a point rolling on the edge of a wheel, like shown in ig. 1.. Here you can see that at the peak of this wave, its phase corresponds to °, and as half a wavelength corresponds to half a phase cycle, the bottom of the trough will be 1° further round, making it correspond to a phase of °.

Features of Multiple Waves hat happens if more than one wave exists within the same space e know that white light from the sun, for example, is made up of all the wavelengths of visible light, and we know that we could shine two laser lights at the same spot on a wall if we felt so inclined – so what happens to the waves when they meet ell, when two waves meet each other, they will overlap and interact in a process called superposition. is means that the waves will kind of mix together to form a resultant wave – and this resultant wave

will have a larger or smaller amplitude than the individual waves, depending on how they interact. ig. 1. shows an example where two smalleramplitude waves blue are in teracting to form a resultant wave pink. ow, let’s go back to think about our extremely common example of shining two lasers at the same spot on a wall. e cool feature of lasers is that they possess only a single wave length. is means that a red laser, for example, will only output light of a single wavelength in the ‘red’ end of the spectrum around  nm. f we have two identical lasers then, the waves produced will be identical in their wave length and therefore freuency, which would make them coherent. f we shine one laser at a point on the wall, de pending on the distance, it will arrive at a specied point phase of the cycle ig. 1.A. f we add another laser and shine it at the same point on the wall, and if it is the exact same distance away, it will arrive in phase at the same point in the cycle ig. 1.. However, if the distance of the second laser pointer from the wall is varied, then the second wave will arrive at a dierent point in the phase and will be out of phase with the rst wave ig. 1.. or waves to be considered to be in phase with one an other, they either need to have travelled exactly the same distance or have travelled a distance that results in a ° or ° separation. f the waves travel any other distance, re sulting in any other degree of phase separation, then they can be described as being out of phase, and the amount of separation between the waves is called the phase dierence Laser pointer

Wall

A 90

270 360 0

90 180

0 360

180

270

•Fig. 10.2 Illustration showing a wave (left) with coloured circles highlighting specic points along the waveform. Different points along the wave correspond to different angles on a circle (right) which is dened as the ‘phase’ of the wave measured in degrees. or eample for a wave like this a point corresponding to the peak of the wave (pink) will e at ° phase.

B

C

• Fig. 10.4 If a single laser is shone at a wall it will produce light (). owever if several laser pointers are shone at a wall they may arrive at different points in the phase of the wave. If they arrive at the same point in the phase they are in phase () whereas if they arrive at different points in the phase they are out of phase (). ee tet for more details.

CHAPTER 10

mportantly, if multiple waves meet and interact with one another, then they produce interference, dened as the variation in wave amplitude that occurs when multiple waves interact. et’s consider a hypothetical example in which we have two identical light sources producing two identical waves like with the laser pointers before. f the two waves are in phase, then the amplitudes will add to gether to increase the amplitude of the resultant wave, which is considered complete constructive interference  like to remember this by thinking that it ‘constructs’ a taller amplitude ig. 1.A and will produce a bright light maima. However, if the waves are 1° out of phase meaning the peaks of one wave line up with the troughs of another wave then the amplitudes will add together to pro duce a net amplitude of ero, meaning they cancel each other out. is is called complete destructive interference ig. 1. and will produce an absence of light minima. hen we talk about waves, we can discuss their phase dierence as before or their path dierence. is means that whilst phase dierence can be expressed in degrees of separation in phase, path dierence is dened as the dier ence in distance travelled between the two waves and is therefore expressed relative to the wavelength l. or con structive interference to occur, we know that we need the waves to be in phase, which means that either they need to travel the same distance path dierence l, or they need to travel a distance that euates to a whole multiple of the wavelength path dierence 1l, l, l, … nl. However, with destructive interference, we know the waves will be 1° out of phase, meaning that one will have travelled half a wavelength further path dierence .l or any multiple

Superposition Interference and Diraction

of this that isn’t a whole number path dierence 1.l, .l, .l, … n1.l. is means that if you slowly in crease the path dierence between two waves, it will cycle through constructive interference l, destructive interfer ence .l, constructive interference 1l, and so on. e exciting part of this is that it means if we have two coherent light sources identical wavelength and freuency then we can utilise the amount of interference to measure small dif ferences in distance.

Utilising Interference to Measure Distances et’s imagine we have our two hypothetical, identical light sources again, both producing identical waves of light. f we asked someone to shine them at a wall so that they interfered with one another, and then measured the resultant wave at the wall, we would be able to tell whether they were produc ing constructive or destructive interference and therefore be able to identify whether there was a path dierence. n a classic experimental setup, the ‘ichelson interferometer’, a single laser light source is shone at a beam splitter to pro duce two, identical waves. ese waves then reect back o mirrors to meet at a detector which can measure the amount of interference ig. 1.. f you move one of the mirrors by a distance that euates to . of the wavelength of light, then the wave that reects o that mirror will have to travel .l further than the other to reach the detector .l more to reach the mirror and then .l more to travel back. is would produce destructive interference at the

A = +1 A = +2

+

A A = +1

A = +1 A=0

+

B A = –1

•Fig. 10.5

Diagram showing two identical waves (left) interacting to produce a resultant wave (right). If the waves are in phase () then the amplitude of the resultant wave will increase to the sum of the contriuting waves (complete constructive interference). owever if the waves are ° out of phase () then the amplitude of the resultant wave will e cancelled out (complete destructive interference).

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Phsical ptics

Fixed mirror (or something we want to measure)

Coherent light source Moveable mirror Beam splitter

Detector

• Fig. 10.6

n eample set-up for a ichelson interferometer. he eam splitter allows some light to pass through to the moveale mirror whilst reecting the rest of the light towards the stationary mirror.

detector. ow, imagine if the stationary mirror in this setup was at a distance we didn’t know – for example, if we were wanting to measure the distance between the front of an eye to the retina axial length. y moving the second mirror back and forth, the interference pattern at the detector will change relative to the distance the light has travelled to the back of the eye, and hence we can determine the distance in this case axial length. is in very simple terms is the fundamental principle of how optical coherence tomogra phy  works, which will be discussed in chapter 1.

Diraction  so now we’re hopefully convinced that light waves can interfere with one another and thus increase or decrease the resultant amplitude. ut what if  told you that in very spe cic circumstances, light can also bend around corners is bending of light is called diraction and is o¡cially dened

as the bending of waves around corners of an obstacle or edges of an aperture into regions that should according to the rectilinear propagation of light produce shadows. n order to understand how diraction occurs, we rst need to take a step back to learn a little bit more about the theory of wavefronts. e touched on wavefronts in chapter  in relation to vergence remember the more curved the wavefront, the higher the vergence, and so at wavefronts imply a light source is an innite distance away. mpor tantly, the wavefronts can be thought of as corresponding to the peak of a wave, and we can understand how they propa gate light by understanding uygen’s principle also called Huygen’s construction. is principle asserts that at every point of a wavefront there is a source of secondary waves – a little bit like if we imagine that every single point on a wave front is a source of a new, smaller wavefront called a wavelet  promise ’m not making this up ig. 1.. e enve lope shape of all of these wavelets will form the secondary wavefront of the main light source. ow, when these wavefronts are incident upon a corner of an obstacle, it’s easy to imagine that light will be stopped by the obstacle and create a shadow, as shown in ig. 1.A. However, as Huygen’s principle asserts that each wavefront will possess several sweet little wavelets, the envelope of which will determine the next secondary wavefront, then this means there will be a small amount of overlap at the edges ig. 1., which would then potentially allow the light to bend around the corner. n terms of optometry and dispensing optics, we care about this because diraction also occurs when light passes through an aperture a hole or slit, and as the human eye possess a pupil that acts as an aperture, it’s important to know what can happen to light as it passes through these types of systems. mportantly, the sie of the aperture will play a huge part in determining what happens to the light. f the sie of the aperture termed slit width is larger than the wavelength of

Wa ve

A

B

Wavelets

fr ts

Point light source

Light wave

Secondary wavefront

•Fig. 10.7

Diagram showing how each wavefront (see ) will possess wavelets (small sources of secondary wavefronts see ). he net (secondary wavefront) will correspond to the envelope of the wavelets. lease note that although this diagram only shows three wavelets you should try to imagine they eist at every single point along the wavefront.

CHAPTER 10

Superposition Interference and Diraction

A

Torch

A Wall Predicted shadow Obstacle

B B Torch

Wall Actual shadow Obstacle

•Fig. 10.8

Diagram showing difference etween shadow predicted y rectilinear propagation of light () and actual shadow produced () (eplained y uygen’s principle).

the light, then light will pass through unimpeded however, if the slit width is smaller than the wavelength of light then it will produce diraction ig. 1..

Single Slit Diraction n the previous section, we learned that when light passes through a slit with a slit width smaller than the wavelength of the light, diraction occurs – but what does this look like ell, when light is diracted, it produces a diraction pattern – a mathematically predictable pattern of light that contains bright spots maima and dark areas minima that are produced by dierent levels of interference on the other side of the slit. n ig. 1.1, you can see that light of a particular wavelength is passing through a slit and let’s assume it’s undergoing diraction – as the light only passes through one slit, this is called single slit diraction. ingle slit diraction predicts that at the point on a wall corre sponding to the area directly in front of the slit, all the light waves will arrive in phase and therefore produce constructive interference bright maxima. However, in the area adacent to this area, the waves from each edge of the slit will travel slightly dierent distances which leads to a path dierence when this path dierence corresponds to a dif ference eual to n 1 .l, then the light will arrive at the wall out of phase and destructive interference will occur dark minima. is pattern repeats itself as you get farther away from the central maxima, but intensity and sie of the maxima

•Fig. 10.9

Diagram showing that large slit widths do not produce diffraction (). he slit width must e smaller than the wavelength to produce diffraction ().

decrease as the path dierence becomes increasingly large ig. 1.11. athematically, we can calculate the path dierence Δ of the light by using trigonometry  can tell you’re getting excited about this – see ig. 1.1. All we need to know is the angle of diraction f and the slit width a. o calculate the angle of diraction we can use uation 1., and to calculate the path dierence we can use uation 1.. Equation 10.

sin 

y 2

L

y2

Equation 10. (eplained sin(ang.diff. )

Equation 10.

dist.btw maxima slit / wall dis sin 

2

2

D a

Equation 10. (eplained sin(ang .diffr .) 

path diff slit width

DEMO QUESTION 10.1  single slit is placed  cm in front of a wall. If the rst and second maima are  cm apart from one another what is the angle of diffraction tep  Determine what we need to calculate angle of diffraction f tep  Dene variales  5 . m (converted to metres)

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A

Maxima

Slit Wall

Path difference (n+0.5)

B

Minima

Maxima

Slit

•Fig. 10.10 Wall

•Fig. 10.11

Diagram showing how single slit diffraction will produce areas of constructive interference () and areas of destructive interference ().

ample diffraction pattern produced y single slit diffraction.

Wall Single slit

 y

L

• Fig. 10.12

Diagram showing relationship etween maima distance (y) wall distance () and angle of diffraction (f).

CHAPTER 10

DEMO QUESTION 10.1 – cont’d y 5 . m (converted to metres) tep  Determine necessary euation sin f 5 y  √( 1 y) (Equation 10.3) tep  alculate sin f 5 y  √( 1 y) sin f 5 .  √(. 1 .) sin f 5 ... f 5 sin-(...) f 5 .° (don’t forget the units)

Practice Questions 10.1: 10.1.1  single slit is placed  cm in front of a wall. If the rst and second maima are  cm apart from one another what is the angle of diffraction 10.1.2  single slit is placed  cm in front of a wall. If the rst and second maima are  cm apart from one another what is the angle of diffraction

Multiple Slit Diraction ow, in the wonderful world of optics, it’s possible to dif fract one light source through multiple slits1 as opposed to ust one as outlined in the previous section. e dierence is that when we increase the number of slits, we create interference patterns at the wall that coincide with the diffraction pattern ig. 1.1.

Superposition Interference and Diraction

is is because the light coming from each slit acts as its own light source in a way which adds the interference pat tern on top of the diraction pattern, and these combine to produce the resultant pattern that will have areas of minima within the maxima ig. 1.1. t’s also possible to diract light through an item called a diraction grating – which essentially describes a small lm that comprises a number of slits all in a line next to each other. mportantly, for this to work in diraction terms, these lines need to be extremely close together and so often that it’s not possible to see the slits in a diraction grating with your naked eye. ese gratings are usually la belled as the number of lines per mm, and as a general rule of thumb, the more lines per mm, the smaller the gap slit width between them e.g.  lines per mm would have a smaller slit width than  lines per mm. iraction grat ings are used in optical systems to measure and separate wavelengths of light, as dierent wavelengths will diract dierently through the grating. is is because the slit width plays a huge part in the angle of diraction uation 1., and we learned in the section titled ‘iraction’ that the slit must be smaller than the wavelength in order to produce dif fraction – thereby indicating that the smaller the width rel ative to the wavelength, the greater amount of diraction. is suggests long wavelength light e.g. red will diract more than short wavelength light e.g. blue, which is the opposite to what we’d expect for dispersion see chapter .

The central maxima

Resultant fringe Interference

Amplitude

Diffraction

Position on wall

• Fig. 10.13

 diagrammatic eplanation of how multiple slits will add an interference pattern (lack) on top of the diffraction pattern (pink) to produce minima within the maima in the resultant fringe (lue).

•Fig. 10.14

ample diffraction pattern produced y doule slit diffraction.

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iraction of white light through a grating can be viewed in a demo in the associated video content on the lsevier website, a screenshot from which is shown in ig. 1.1

Circular Diraction e previous examples of diraction all assume that the slit or aperture is a long, thin aperture. or example, if  taped two pieces of card very close together to produce a single very narrow slit,  could produce a diraction pattern if  shone a laser pointer at the slit and in fact  have done this before, much to the amaement of a handful of students. e principle of diraction suggests that light will bend around the corners of the slit, which is why the diraction pattern spreads out at an angle perpendicular to the angle of the slit. or example, a vertical slit will produce a hori ontal diraction pattern. However, diraction can also occur for circular apertures which more easily applies the principles of dirac tion to optical systems which possess circular apertures, e.g. cameras and telescopes. n these cases, the circular aper ture would need to be very small in diameter in order to be smaller than the wavelength of light, but this will allow the light to diract in all directions around the edges of the circle is means that instead of the attish, single orienta tion diraction pattern we see with the narrow slit see ig. 1.11, we get fun circular diraction patterns as shown in ig. 1.1. e other cool thing about these types of diraction patterns is that the distance the light travels on the other side of the aperture dictates what type of dif fraction occurs. f, for example, the distance between the aperture and the screen, or the distance between the aper ture and the light source, is ‘nite’ curved wavefronts, then near-eld diraction occurs resnel diraction,

Diffraction pattern

Diffraction grating

Torch

•Fig. 10.15

hotograph showing white light passing through a diffraction grating. ong wavelength (red) light will diffract at a larger angle relative to shorter (lue) wavelength light so the diffraction pattern shows the splitting of light from lue-red as we move away from the central maima.

whereas if the screen and light sources are at ‘innite’ dis tances away from the aperture planar wavefronts, then far-eld diraction occurs raunhofer diraction. e pattern produced by raunhofer diraction is called an airy disc, and this is important in terms of diraction be cause it denes the smallest point at which a light source can be focused, and therefore is associated with the resolv able power of any optical system.

The Resolving Power of a Sste As we have ust alluded to, diraction is an important con cept because the resolution highest resolving power of all optical systems will be limited by its aperture sie, and this relationship is due to diraction. ypically, although a smaller aperture will increase the resolution by removing aberrations – see chapter 1 for more detail on aberra tions, at a certain degree of smallness, light will diract around the edges of the aperture and reduce the resolution again. e point at which resolution is highest but dirac tion doesn’t occur is called the diraction-limit of the system, and this also applies to the pupil of the human eye. nterestingly, this minimum resolvable power diraction limit of an imaging system can be determined using the ayleigh criterion. is proposes that two overlapping images airy discs will only become distinguishable from one another when separated at a distance such that the centre of the eroorder central maxima of one disc lines up with the rst minima of the other. is principle is shown in ig. 1.1

Thin Fil Interference the Fun Side of Interference efore we close this chapter, let’s take a moment to think of reallife examples of interference that we might experience in our daytoday lives. emember, interference occurs when two or more waves travel at dierent distances to each other – in some cases, this could be because they originate from two independent sources that are located at dierent points as in ig. 1., or this can happen if light from a single source is split into two or more. is can happen in a lab setting e.g. the ichelson interferometer, but it can also happen in the natural environment, and chances are we’ve all seen examples of interference before and ust not realised it or example, if light travels through a thin film of a material e.g. a very thin layer of oil on top of some wa ter on the road, or the very thin ‘film’ of a soap bubble, then interference can occur. n these circumstances, even if there is only one light source e.g. from the sun, and even if the light source contains multiple wavelengths of visible light, something called thin film interference takes place, which can make the light reflecting off the front and back surface of the film appear to be multicoloured.

CHAPTER 10

A

Screen

Superposition Interference and Diraction

View of screen

a

D

B

Screen

View of screen

a

D

• Fig. 10.16

Illustration of resnel diffraction ( curved wavefronts) and raunhofer diffraction ( planar wavefronts) with a circular aperture.

A

B

• Fig. 10.17

Diagram illustrating the ayleigh criterion. If two images (airy discs) are farther apart than the radii of their central maima then they are easily resolvale (). he smallest distance they can e separated y whilst still eing resolvale is determined y the ayleigh criterion which dictates that this smallest resolvale power occurs when the images are separated at a distance such that the centre of the ero-order (central) maima of one disc lines up with the rst minima of the other (). In oth images the lue lines indicate the intensity proles of the individual images whilst the lack lines show the ‘visile’ prole.

o understand this better, let’s stick with the soap bubble example but the same is true for the oil on the water in the road. in lm interference occurs because, as the incident white light reaches the ‘lm’ of the bubble, some of the light reects at the front surface of the lm whilst some passes through to the back surface. en at the back surface, some light transmits through, but some is reected back again, which means that even though we started with one light source, there will be in this case two waves that reect back towards us the observer. is means that we’ll have two waves reaching us, one of which has now travelled a slightly greater distance than the rst as it’s been to the back surface of the lm ig. 1.1A. f the second wave reecting o the back surface travels a whole multiple of the wavelength farther than the rst wave in phase, then constructive interference is observed light, whereas if the second wave travels a half multiple of the wavelength farther than the rst wave out of phase, then destructive interference is observed no light. emember, though, with white light from the sun, the light comprises many wavelengths of light from violet to red, and so as the white light reects at both the front and back surfaces of the soap bubble lm, sometimes the long wavelengths will be in phase, and other

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A I “w ncid hit en e” t lig ht

at cts lm e l ef fi t r of il gh face me sur So irst f

cts fle ce e r rfa ht lig d su r he on lm t sec of fi at Air outside bubble Soap bubble “film”

Soap bubble

Thickness of film and angle of incidence play a key role in this process

Inside of soap bubble

Incident “white” light

B

Patches of constructive interference for specific wavelengths of light produce visible colours

•Fig. 10.18 Diagram showing eplanation of how thin lm interference makes soap ules look remarkaly colourful. ight from a source reects at oth the front surface and the ack surface of the soap ule () which produces constructive or destructive interference depending on path difference. reas of constructive interference produce a reection that appears to e the colour associated with those wavelengths () (e.g. red).

times the shorter wavelengths will be in phase. is pro duces an interference pattern which splits the white light into its constituent wavelengths and produces a multico loured appearance on the soap bubble or the oil on the water ig. 1.1 ou can also see this for yourself in a demonstration at home if you read the instructions in chap ter  later in the book.

e circumstances for thin lm interference to occur are largely dependent on the thickness of the lm as that dic tates how far the second wave travels relative to the rst, which in turn determines the interference and the angle of incidence of the light approaching the lm as this dictates the distance between the rst reected wave and the second reected wave.

Test Your Knowledge ry the uestions below to see if you need to review any sections. All answers are available in the back of the book. .10.1 hat is the ‘phase’ of a wave .10. f two identical waves are 1° ‘out of phase’, what will happen .10. hat does Huygen’s principle predict about light when it gets blocked by an obstacle

Reference 1. oung . e akerian lecture. xperiments and calculation rela tive to physical optics. Philos Trans R Soc Lond. 111.

.10. f we increased the number of slits in a dirac tion experiment from one slit to ve slits, what do you think would happen to the diraction pattern .10. hy do soap bubbles look multicoloured some times

SECTION

3

Clinical Applications 11. Focimetry, 109

1. Imain the ye and easurin eractie rror, 139

12. Photometry, 113 13. Optical Instruments and Lo ision Aids, 123

1. aeront Aerrations and Adaptie Optics, 19 1. Optical Coherence omoraphy, 1

1. Polarisation, 131

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11 Focimetry C H A P T E    T I E Introduction

How Does a Focimeter Work?

What Is a Focimeter?

Test Your Knowledge

   E C T I E After working through this chapter, you should be able to: plain hat ocimetry is iscuss the principles underlyin ocimetry

e ale to lin ocimetry ith your understandin o spherical and cylindrical reractie errors see chapter 

Introduction

and an observation system (telescope, which focuses the image for the user) see ig. . for a schematic of this. e focusing system itself aims to produce an image of the target (illuminated lines) overlaid onto a graticule (a pattern of lines in an optical device used as a measuring scale sometimes called a reticle) and proect it to inn ity. s the focimeter is used to measure a large range of lens powers, logic tells us that the target (acting as the obect) will need to be moveable in order to position the target (acting as an obect) at the correct location so as to proect it to innity. owever, for the focimeter to work correctly, the eyepiece of the observation system also needs to be focused for your eyes (to account for any uncorrected prescription you might have) before you introduce the lens you’re trying to measure. is will mean that the target and graticule are all lovely and in focus (ig. .). ssuming that the eyepiece of the focimeter is focused appropriately for the user, the user can place a patient’s spec tacle lens onto the lens (frame) rest (ig. .). Important ly, the back surface of the lens should be sitting on the lens rest and will need to be clamped into position in order to provide an accurate reading. owever, it’s also important to note that before clamping the lens, it will need to be moved upwardsdownwards and leftwardsrightwards until the tar get is visibly aligned with the centre of the graticule. t this point, the user will have located the optical centre of the lens, and most focimeters have a process for allowing the user to mark this point on the lens (to help later on). It’s important to nd the optical centre, because this is the point where no prismatic eect will be introduced (see chapter  for review on this) and so the power will be able to be accurately determined.

In this chapter we’re going to think about lenses and how we can measure the power of lenses using an instrument called a focimeter (internationally it is also known as a lensometer – but they are the same thing). is chapter will be most useful to you if you have access to a focimeter to try out the theory and have a go, but it will still (hopefully) be interesting even if you can’t apply it practically.

What Is a Focimeter?  focimeter is a device that can be used to determine the spherical power, cylindrical power (and corresponding ais), prismatic power and the optical centre of a lens. ou may be thinking, why on arth would we need a device that does that ut in practice this is an important instrument which allows clinicians to measure the power of a patient’s glasses, even if the patient themselves doesn’t know their prescription. It also allows dispensing opticians to check the power of lenses before dispensing them to patients. In broad terms, there are three main types of focimeter . onventional (eyepiece focusing) . roection (screen focused) . utomatic electronic (automated) In this chapter we will focus on the rst type – the con ventional focimeter – but it’s good to be aware that other types eist.

How Does a Focimeter Work? e focimeter itself reuires two main parts – a focusing system (collimator, which narrows the beam of light)

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Focusing system (collimator)

Observation system (telescope)

is wheel

Target Standard Power lens wheel



Lens table

Obective raticule dustable lens eyepiece

0. 0.00 0.

• Fig. 11.1

A schematic of a focimeter, with the focusing system (including target) on the left and the observation system (including graticule) on the right. The lens you want to measure would sit on the lens rest, in between the two systems.

Rotating part of graticule i

Gr at

l

Power indicator Rotating target

• Fig. 11.2

The view of an example target (green) and example graticule (black) as seen through a focimeter eyepiece. f in focus, all the lines should be completely clear.

e net step is to determine the spherical power (though this can only be achieved after the lens’s optical centre has been identied). t this point (unless the lens has no power), the target will be out of focus (because the power of the lens is adding or removing positive or negative power to the tar get that was in focus previously). In order to determine spherical power and get the target back into focus, the user will need to turn the power wheel until the lines of the target become clear. If the lens only possesses spherical power then when the collimator part of the focimeter is correcting for the appropriate amount of power (e.g. 2. ), all the target lines will become clear, and the power can be read from the instrument at that point. In terms of the target itself (the litup green bit), the tar get in our eample is made up of nine threebythree suares, a ring of circles, and then two sets of perpendicular lines. e ring of circles will remain stationary (the ed target), whilst the suares and the lines will rotate when we turn the ais wheel (the rotating target). If we’re using the focimeter to measure a spherical lens, and the focimeter is focused, the ed target will still look like circles and the rotating part of the target will be in focus wherever it is rotated to. owever,

Patient’s lens

•Fig. 11.3

iagram showing focimeter set up with a patient’s lens on the lens rest.

CHAPTER 11

if the lens is toric (meaning it has a cylindrical power as well as spherical see chapter  for review on this), then it will only be possible to get some of the lines of the target into focus using the power wheel alone. If this is the case, the clue that cylindrical power is present will be that the ring of circles will appear stretched at a particular angle and start to resemble lines (like those shown in ig. .), even when the maimum and minimum powers are neutralised – which we’ll discuss in a moment. e lines formed for these two powers will be orientated at ° to each other, and the orientation of these lines indicate the aes for the two maor power meridians, and so the ed target allows us to locate these two meridians. e rotating target lines, which are also at ° to each other, will only be clear when they are aligned with the lines formed by the ed target – that is, when they are aligned to the two maor meridians. In ig. ., the target is clearly out of focus, but hope fully you can also see that the target seems to be stretched along an obliue angle. In this case, the rst thing to do is to adust the power wheel until the ring of circles (the ed target) comes into focus to help us identify the angle of stretch. en we can rotate the target using the axis wheel until the shorter target lines fall along the ‘stretch’ angle, as shown in ig. . en, adust the power wheel until the short lines come into focus (and are straight and unbroken – if they seem more focused but broken you may need to reassess the ais wheel) and make a note of the reading on the power wheel however, you’ll notice that even though the short lines are in focus, the longer target lines will be stretched and blurry. t this point, we need to continue to adust the power wheel (without adusting the ais wheel) until the longer lines come into focus (which should also blur the shorter lines

• Fig. 11.4

llustration of view down a focimeter. This lens has power (so the target is out of focus), and we can identify the presence of cylindrical power because the target appears to be stretched at an obliue angle.

Focimetr

111

• Fig. 11.5

The target has now been rotated so that the shorter lines of the target line up along the line of the stretch this will help us to determine cylindrical power.

again). e then make a note of this new reading from the power wheel. et’s say that the shorter lines were in focus at 2. and the longer lines were in focus at 2., with negative sphere-cylinder form we assume the most positive of the two powers is the spherical power of the lens, so in this e ample 2. is more positive than 2., which means we would say that the spherical power of this lens is 2. . e cylindrical power can now be determined as the difference between the most positive and least positive powers (e.g. 2.  and 2.  would indicate a cylindrical power of 2. ). e ais of the cylinder corresponds to the direction of the least positive power, which can be read by seeing where the most negatively powered lines in tersect the graticule in degrees. is can be achieved by ro tating the long arm of the graticule to align with these target lines to reveal the ais of the cylindrical power. ee ig. . for an eample of this. o determine whether vertical prismatic dierences are present between the two lenses, when you switch from one lens to the other you should note whether the second lens is still able to centre the target on the centre of the graticule. If so, then no prismatic dierences are present. owever, if instead, the second lens has moved the target above or below the centre of the graticule, then vertical prismatic dier ences eist between the two lenses. e amount of prismatic dierence can be determined by reading o the graticule as shown in ig. ., and if the target falls below the grati cule, then there is basedown prism, and if the target falls above the graticule centre, then there is baseup prism (be cause prisms deviate light towards the base). owever, it is still important to note that this method only identies the total prism dierence between both lenses in order to calcu late the amount of vertical prism in each lens, you would

112 S EC TI ON 3

Clinical Alications

–2.50DS / –0.50DC  145

• Fig. 11.6

A diagram showing the rst reading on a focimeter (left) with the most positive power on the power wheel corresponding to the spherical power of the lens, and the second reading (right) showing the least positive power (and the axis). or negative sphere-cylinder form, the difference between the most positive power ( .) and the least positive power ( .) is taken as the cylindrical power ( .).

need to also consider the patient’s pupil position relative to the optical centre for each lens. oriontal prism can only be determined by taking into account the patient’s interpupillary distance (I). or e ample, if a patient has an I of  mm, then once you have identied the optical centre of the rst lens, you can measure from this point across the I distance to the oth er lens and then position the lens on the focimeter using this mark as the central point. If the target is then not central but instead is slightly to one side, then horiontal prismatic dierences are present. gain, this method only identies the total prism dierence between both lenses in order to calculate the amount of horiontal prism in each individual lens, you would need to also consider the patient’s pupil position relative to the optical centre of each lens. Base-down prism 2 dioptre

• Fig. 11.7

A diagram showing a -dioptre, base-down prismatic difference between the two lenses. n this diagram, orange text has been added to help show which lines represent which prismatic power.

Test Your Knowledge ry the uestions below to see if you need to review any sections. ll answers are available in the back of the book. TYK.. hat does a focimeter do TYK.. hat is the graticule

TYK.. hy is it important to focus the eyepiece before attempting to use a focimeter TYK.. If the target falls below the centre of the grati cule, would this indicate basedown or baseup prism

12 hotometry C H A P T E R   TL I E Introduction Angles and Lights Measurements of Light Luminous Flux Luminous Intensity

Illuminance Luminance A Cool Experiment Colour Temperature Test Your Knowledge

   E C T I E After working through this chapter, you should be able to: Explain a solid angle Explain luminous ux, luminous intensity, illuminance and luminance Describe the two laws of illumination

Explain what colour temperature is and be able to interpret elins

Introduction

that we can say it’s emitting light spherically from its centre owever, this means that if we want to measure the angle of light that’s being emitted by the light source, we need to be able to determine the relevant  angle ut how do we measure  angles ell, let’s start by reviewing some prinicples of  angles n a  circle, like the one in ig , any planar (at) angle (u) measured from the centre can be represented in degrees (°) or radians (rad or r), which is typically epressed relative to pi (p) n the eample in ig , the angle is °, which able 

e word photometry can be broken down into photo(light) and -metry (measurement), so it should come as no surprise then to hear that this chapter will be focused on discussing my cats’ favourite toys… Obviously ’m oking this chapter will eplain how we can measure light in the real world and will link this to clinical considerations where appropriate

Angles and Lights n the introduction, we discussed that the term photometry means ‘measurement of light’, but more specically than that, it refers to measurement of visible light – the part of the electromagnetic spectrum that is detectable by the human eye is could include natural light (eg from the sun), but more often we use photometry to discuss articial light sources (eg from a lamp) or eample, if you’ve ever bought a lightbulb and spent a while wondering what the dierence between the watts and the lumens are (and what they mean), then this chapter is for you o start with then, let’s consider a light bulb (a bogstandard light bulb), as shown in ig  ere you can clearly see that the bulb itself is emitting light in all directions (and although this is a two-dimensional () image, please imagine it eists in three-dimensional () space), meaning

• Fig. 12.1 Diagram of a light bulb to show that it emits light in all directions. 113

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Clinical Applications

2D

ir

C

DEMO QUESTION 12.1 c

Calculate the planar angle shown as ‘’ in the image below

le 42cm

 R

?

14cm

•Fig. 12.2

Circle showing planar angle from the centre (u) which corresponds to 90°. The distance from the apex of the angle and the circumference of the circle is eual to the radius of the circle (). The distance along the circumference reuired to create the angle is called the arc (c).

TABLE Table Showing Conversion Between Turns, 12.1 Degrees and Radians in a 2D Circle

Turns (complete revolutions of a circle)

Degrees (°) Radians (rad)

0

0

0

0.

90

0.

0

p

0.

0

p

.00

0

p

p

shows us would convert to  turns around the circumference (outside edge) of the circle, or p radians owever, what if we don’t know the angle o calculate the angle we can utilise the relationship between the radius () and the length of the arc (c) through uation  (radians) and uation  (same but in degrees) emember that because these euations use  units, all measure of distance must be in metres (m) Equation 12.1.1



tep  Determine what we need to calculate planar angle u tep  Dene ariables c 5 0. m (we need to convert to metres)  5 0. m (we need to convert to metres) tep  Determine necessar euation u 5 c   (Equation 12.1.1) tep  Calculate u5c u 5 0.  0. u 5  radians (now we can divide by p to understand how this number relates to p radians (p is 3.14 so it’s probably going to be close to 1 p radians…)) u5p u 5 0.9 p rad (or we could also convert to degrees…) u 5 c    (0  p) u 5   (0  p) u 5 .9° (in either case don’t orget the units)

Practice Questions 12.1: 12.1.1 Calculate the planar angle shown as ‘’ in the image below

35cm

c R

?

Equation 12.1.1 (explained) angle (rad )  Equation 12.1.2

length of arc (m) radius of circle (m ) 

c R

180

Equation 12.1.2 (explained) angle (deg )

length of arc radius of circle

188 0

20cm

CHAPTER 12

12.1.2 Calculate the planar angle shown as ‘’ in the image below

60cm 10cm

?

O, so now (even if we don’t like them very much), hopefully we can appreciate the link between degrees and radians for  angles, but we know at some point we’re going to need to discuss  angles ree-dimensional angles are tricky o make them a little less tricky, let’s start by considering a  circle however, the  circle is so unbelievably tiny that it looks like a dot (ig ) f we put another larger circle in front of the

B

Front view

Tiny circle (point) far away

C

Side view

Join e

•Fig. 12.3

arger circle closer to s

D

Side view

Change in distance

Solid (3D) angle

d to a co gether  ne

115

tiny little circle, closer to us (ig ), we could oin the two together by their entire circumferences to make a cone (ig ) e can see from this thought eperiment that the shape of the ape (pointy bit) of the cone would be directly related to () the distance between the point and the circle (larger distance with same sie circle 5 smaller ape) and () the sie of the larger circle (larger circle at same distance 5 larger ape) (ig ) is, in essence, is how we dene  angles, which are termed solid angles o this end, the solid angle () can be considered the amount of the eld of vie (m) from a particular point (termed the apex) rucially, the solid angle of the ‘cone’ (eld of view) is measured in steradians (sr) which can be thought of as  radians n a real-world eample of this, we can imagine the ape as corresponding to an observer’s eye, and the amount of space taken up by the page of this book (or the screen of an e-book) as the eld of view (ig ) ow, if we turn our eample of a solid angle and relate it to a sphere, this can be likened to the  version of the  planar angles we discussed in ig  or planar angles within a circle, the angle is related to the distance from the centre of the circle to the outside edge (radius) and the amount of the circle that is covered by the angle (the arc) ith solid angles, this principle remains the same, but this time we need to consider the relationship between the suare of the distance from the centre of the sphere to the outside edge (radius ) and the amount of surface of the sphere that is covered by the angle (area ) (ig ) o see this in action, please review uation  (steradians)

DEMO QUESTION 12.1 – cont’d

A

Photometr

Change in size

Diagram showing concept of solid (D) angles. e start with a tin circle far awa () and we superimpose a larger circle closer to us (). f we iewed this from the side (C) we could oin the two circles from circumference to circumference to mae a cone shape. This cone shape denes our solid angle. s we can see the area of the larger circle (D pin) and the distance awa of the larger circle (D blue) will hae an impact on the solid angle.

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Observer

Clinical Applications

which can be simplied to  4π

Solid angle

Field of view (area) A

is means that when light is emitted in all directions, the solid angle of the light is considered to be p steradians (or  steradians, which is literally  multiplied by p)

Measurements of Light ow that we’re hopefully reasonably well versed with what a  solid angle is, this chapter will consider four dierent measurements of light luminous ux, luminous inten sity, illuminance, and luminance

• Fig. 12.

Diagram showing real-life example of lin between solid angle eld of iew and distance of obect from the obserer.

A

3D

sp

re he

R

• Fig. 12.

Diagram showing a solid angle () in relation to a sphere

(D shape).

and remember that because these euations use  units, all measures of distance must be in metres (m) A Equation 12.2  R2

Luminous lu is is a scientic term  love because it reminds me of an integral part of a time-travelling car ut its true denition is slightly less ‘sci-fi’ uminous ux (f) describes the measure of power (or perceived power) emitted by a light source o put this into contet, when you buy a light bulb, it will disclose the wattage and the associated lu mens (lm) e lumens are describing the power of the bulb (which is often associated with brightness), which means the higher the lumens, the higher the power and the brighter the bulb is means that when purchasing a lightbulb for your home, you can review the packaging to see how many lumens are associated with the bulb and how many watts e dierence is that the wattage tells you the amount of energy that the lightbulb consumes, whereas the lumens indicate how much power (brightness) the bulb emits uminous ¢u can be calculated using uation , which shows that the power of the light source is related to the luminous intensity (l see net section) and the solid angle () of the light source Equation 12.



l



Equation 12. (explained) e

Equation 12.2 (explained) solid angle (sr ) 

area of sphere surface (m 2 ) radius of sphere 2 (m )

ou may be thinking, how does this relate to light ell, we did mention brie¢y at the beginning of the chapter that we want to be able to account for the fact that light leaves a light source in all directions, which means that if the light bulb was at the centre of a sphere, it would emit light in such a way that the entire surface area of the sphere would be illuminated ow, the surface area of a sphere can be calculated using a formula you may remember from your secondary school days p f we substitute this into uation , we get 4 R 2 /R 2

o, what is luminous intensity

Luminous Intensit uminous intensity (l) is a term that denes the power of a light source to emit light in a specied direction t’s measured in candelas (cd) and can be calculated using uation  Equation 12. l   Equation 12. (explained) luminous intensity 

luminous flux solid angle

CHAPTER 12

o put this into perspective, let’s consider two hypothetical light sources – a spherical light bulb and a torch oth light sources emit a luminous ¢u of  lumens (f), but the spherical light bulb emits light in all directions, giving it a solid angle () of p steradians, whilst the torch emits light in a given direction which leads to a solid angle of p steradians f we substitute our values into uation , we can see that the light bulb emits a luminous intensity of  candelas, whereas the torch emits a luminous intensity of  candelas  think this serves as a nice eample, because it makes sense that if both sources emit the same power (luminous ¢u), and if one has all the power in a smaller space, it should have a higher intensity (which it does)

Illuminance oving on then, unless we’re afraid of the monsters that hide in the dark, one of the primary reasons we have light sources is to help us see obects in our environment s discussed in chapter , this reuires light to light up (illuminate) an obect and re¢ect o the obect to our eyes e can then process the light signal and interpret what it is that we’re seeing is ‘lighting up’ of obects is called illuminance () and more specically is dened as the luminous ¢u density at a point on a surface or obect – in other words, the amount of light power per unit area of the surfaceobect lluminance is measured in lux (lx) and can be calculated using uation

A

117

, which shows us that as the distance from the source increases, the illuminance will decrease, which is called the inverse square la of illumination Equation 12.

l d2

E

Equation 12. (explained) illuminance 

luminous intensity distance from source 2

 second law of illumination (the cosine square la of illumination) species that the angle of the surfaceobect relative to the source also plays a part in the illuminance (uation ) ith this law, as the angle of the source to the surface gets larger, the illuminance will decrease Equation 12.

E5

lcosA d

Equation 12. (explained) illuminance

disttance from source 2

o the take-home message here is that increasing distance, or increasing the angle of the obect away from the light source, will decrease the illuminance of an obect (ig ), which in practical terms means that if light

B

Low illuminance

Low illuminance

High illuminance

• Fig. 12.

Photometr

Diagram illustrating the laws of illumination. f the obect is er close to the light source and at a straight angle then illumination will be high () whereas as the distance from the light source increases or the angle from the light source increases () the illumination will reduce.

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Clinical Applications

levels are making it di¨cult to see (eg a menu in a restaurant), you can either move closer to a light source or hold the obect at a more suitable angle for an improved visual eperience

Luminance O, so now we know that the light falling onto an obect is the illuminance (or illumination), but what do we see hen we look at a well-lit obect (like in ig ), we can tell that it’s well lit because it will appear to be visibly bright is apparent brightness represents the intensity of light heading from the obect to us (which can be described as the intensity of light emitted in a given direction per proected area of a luminousre¢ecting surface), which is called luminance () (not to be confused with illuminance – see o ) uminance is measured in candelas per square metre (cdm2) and can be calculated using uation , which shows us that as the area () increases, the luminance will decrease lease note however that this euation assumes a ¢at surface (eg part of a table) Equation 12.

L

l SA

Equation 12. (explained) luminance 

luminous intensity surface area

•BO 12.1

lluinance or uinance

ne of the tricier parts of photometr is getting our head around the difference between ‘illuminance’ and ‘luminance’ as the describe er similar things and are almost identical in their spelling (ust to eep things interesting). The tric is to thin about what happens when we see an obect – light will be incident upon the obect some of the light will be absorbed whilst some is reected (see chapter  for reision on this). The reected light will trael towards us (the obserer) and we can process that information and determine what the obect is. The light that falls on the obect is the illuminance (hence wh we describe obects as being ‘illuminated’ or when someone helps to explain something we might use a phrase that suggests the’e ‘illuminated’ the point). oweer the light that reects off the obect is the luminance which means it’s partl determined b the illuminance but also b the surface properties themseles.

DEMO QUESTION 12.2  spherical light source of diameter  m emits 000 lm uniforml in all directions. hat is the aerage illuminance on a surface  m from its centre tep  Determine what we need to calculate illuminance  tep  Dene ariables f 5 000 r 5  m (radius is equal to diameter divided by 2) d5m  5 p (solid angle when emitted in all directions as  5 4p22 and 51)

Luminous flux (lm) perceived power in all directions

Luminance (cd/m2) intensity of light emitted in a given direction per projected area

Luminous intensity (cd) power to emit light in a given direction

Illuminance (lx) luminous flux density on a surface

• Fig. 12.

Diagram designed to help ou remember the difference between luminous ux luminous intensit illuminance and luminance.

CHAPTER 12

DEMO QUESTION 12.2 – cont’d tep  Determine necessar euation l 5 f   (Equation 12.4)  5 l  d (Equation 12.) tep  Calculate – but remember our calculator needs to be in ‘radians’ l5f l 5 000  p l 5 9 cd  5 l  d  5 9    5 9  9  5 .0 lx (don’t orget the units)

Practice Questions 12.2: 12.2.1  spherical light source of diameter  m emits 00 lm uniforml in all directions. hat is the aerage illuminance on a surface  m from its centre 12.2.2  spherical light source of diameter  m emits 00 lm uniforml in all directions. hat is the aerage illuminance on a surface 0 cm from its centre

A Cool Eperiment f you’re interested in learning more about the inverse square la of illumination, then you can do a uick and easy eperiment at home to see it in action ll you need to do is set up two identically powered light sources at a reasonable distance ( cm1) from one another, with them both pointed towards one another (as shown in ig ) ou will also need some oil (the normal kind from the kitchen) and a piece of paper mportantly, you don’t need to know the power of the light sources, as long as they are both the same ow, the inverse suare law of illumination states that as you increase the distance away from these light sources, the illuminance should reduce s these light sources are facing each other – this means that if we put an

11

obect like a piece of paper very close to one of the light sources – the illuminance on the paper from the close source should be higher than from the opposite source t also means that if we put the paper right in the middle of the two light sources, then illuminance levels should be identical on either side ut how do we prove this ell, let’s think about a piece of paper n normal conditions, a standard piece of paper will mostly re¢ect light (as opposed to letting it transmit through to the other side) owever, if the paper is saturated with oil, the patch of oil will re¢ect much less of the light and will instead transmit more light through to the other side ou can test this for yourself by putting a small ( cm) dot of oil onto a piece of paper and holding it up to a light source ’d predict that the oil patch will appear brighter than the rest of the paper because it’s letting the light pass through to your eye is implies that if the light source behind the oil patch is brighter (higher levels of illumination) on one side, then two things will be true () the side of the oil spot facing away from the bright light source will look bright (as it lets the light pass through) and () the side of the oil spot facing the light source will look dark, as the patch of oil is letting more light transmit through in that spot than the rest of the paper, meaning it re¢ects less back in that region relative to the rest of the paper igs ,  and  demonstrate this nicely, showing that with two identical light sources, the oil spot will appear bright if the illuminance level is higher behind it (ig ), and it will appear dark if the illuminance level is higher in front of it (ig ) f the illuminance levels are the same on both sides, then the oil spot will seem to disappear because eual amounts of light are being transmitted through the oil spot and re¢ected o the paper (ig ), and with two identical light sources, the inverse suare law of illumination tells us that this can only occur if the paper is positioned eactly halfway between the two light sources ry it for yourself and see

Paper with oil spot in the centre - closer to torch 1

Observer

Torch 1

Torch 2

Bright oil spot, dark paper Brighter light transmitted through than reaches the paper

•Fig. 12.

Photometr

llustration of experiment with two identical light sources pointing towards one another and a piece of paper with a spot of oil on the front of it. The oil allows light to transmit through the paper (where it would normall be reected) so if iewing from the right when the paper is near torch  the light from torch  reecting off the paper will be less bright than the light from torch  which is transmitting through the oil. o the oil spot will appear bright relatie to the paper.

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Clinical Applications

Observer Paper with oil spot in the centre - closer to torch 2

Torch 1

Torch 2

Dark oil spot, bright paper Brighter light reflected than is transmitting from the other side

•Fig. 12.

llustration of experiment with two identical light sources pointing towards one another and a piece of paper with a spot of oil on the front of it. The oil allows light to transmit through the paper (where it would normall be reected) so if iewing from the right when the paper is near torch  the light from torch  reecting off the paper will be brighter than the light from torch  which is transmitting through the oil. o the oil spot will appear dar relatie to the paper.

Paper with oil spot in the centre - equidistant to both torches

Torch 1

Observer

Torch 2

Invisible oil spot Equal brightness on both sides

• Fig. 12.1

llustration of experiment with two identical light sources pointing towards one another and a piece of paper with a spot of oil on the front of it. The oil allows light to transmit through the paper (where it would normall be reected) so if iewing from the right when the paper is exactl halfwa between the light sources the light from torch  reecting off the paper will be as bright as the light from torch  which is transmitting through the oil. o the oil spot will become inisible relatie to the paper.

Colour Temperature f you’re big into photography, art, or lters for your seles, then you’ll probably be familiar with the idea that light can be described as ‘cool’ or ‘warm’ ‘ool’ light typically describes light that possesses more of a blueish hue (to correlate with cold things, like ice and winter) whereas ‘warm’ light typically describes light that possesses more of a yellowish hue (to correlate with warm things like the sun and the beach) is means that light sources can be categorised depending on their colour temperature which is measured in elvins () ow, fair warning here, elvins are, 

think, a little counterintuitive because in my brain a higher temperature should be warmer than a low temperature, but with colour temperature it’s the opposite way around is means a lower number of elvins corresponds to warm light ( ), whilst a higher number of elvins corresponds to cool light ( ) eutral ‘white’ light would be thought of as around   (ig ) ome of my students have told me that they remember this by associating it with the colour of ¢ames (for eample, on a unsen burner) in this case the higher temperature ¢ames are blue whilst the lower temperature ¢ames are yellow (ust like with elvins)

CHAPTER 12

Photometr

121

Colour temperature

~2700K

~3500K

~7000K

• Fig. 12.11

Colour temperature of light (measured in elins ) is warm at low leels (left) and cool at high leels (right).

Test Your Knowledge ry the uestions below to see if you need to review any sections again ll answers are available in the back of the book .12.1 hat does ‘photometry’ mean .12.2 escribe a ‘solid angle’ .12. hat does ‘luminous ¢u’ mean and what is it measured in

.12. hich of the following statements uses photometry terms correctly and why ‘e cup is poorly illuminated,’ or ‘e cup is poorly luminated’ .12. hat does a high number of elvins ( ) suggest about a light source

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13 ptical nstruments and Low ision Aids C H A P TE R OU TL IE Introduction Cameras Telescopes Galilean Telescope Keplerian Telescope Reecting Telescope

Low Vision Aids Magniers Adapted Lenses Assistive Technology Test Your Knowledge

O   E C T I VE After working through this chapter, you should be able to: Explain what an ‘optical instrument’ is Explain how cameras wor Explain how Galilean Keplerian and reecting telescopes wor

Explain how optical instruments can help people with low vision

Introduction

If we thin about it logically then in the most basic of forms the camera needs to be constructed to allow light into the camera itself without letting too much in that it will affect the quality of the image so it needs to be a box with a hole in the front of it. In optics the hole that lets light into an optical instrument is called the aperture stop or iris and importantly this hole should not aect the eld of view what the instrument can see. If it helps as an example the human eye can be thought of as an optical instrument and our iris is the aperture stop. hen we’re in brightly lit places our pupils the hole in our iris get smaller to reduce the amount of light entering the eye but this only helps with the brightness it doesn’t restrict our eld of view at all – and the same is true for aperture stops in cameras. ur basic camera will also need a screen or equivalent at the bac of the box to proect the image onto so in theory it could ust be an enclosed box with a pinhole on the front ig. .A or it could be an enclosed box with a slightly larger aperture and a simple lens to focus the image ig. .. owever as we’ve hinted at already with our example of the human eye sometimes we need to alter the amount of light that can enter the instrument e.g. if it’s very dar maybe we want our camera to let in more light or we might want to change the focus e.g. if we want to tae an image of something that’s very far away. In this case we

An optical instrument can include any device or equipment which can alter an image for enhancement or viewing purposes. is includes cameras, binoculars, telescopes and monoculars – some of which are clinically relevant as they’re utilised as low vision aids, which are devices that help people with low acuity to be able to see. is chapter will explain the principles of these devices and then discuss their use in practical settings.

Cameras I thin it’s useful to start this chapter by going over cameras because we all use cameras on our mobile devices fairly regularly so we can apply the nowledge that we already have. A camera then is a device that focuses light onto a screen digital sensor or a lm for the purposes of viewing the image. istorically the term originates from the use of camera obscuras which were dar rooms with a pinhole to let the light in this pinhole proected an image onto a at wall which could be viewed by the observer for details on how pinhole cameras wor see ox . and chapter . ince then cameras have developed signicantly to include the use of fancy lenses and digital enhancing software.

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•BOX 13.1

Clinical Applications

How Does a Pinhole Camera Work?

A pinhole camera (or camera obscura) is a name for an imaging device that utilises the theory behind pinhole photography. When light emanating from an object shines through a pinhole, it produces an upside-down image. This is because light rays have to travel in straight lines (to a limit, see chapter  to review how light can bend around corners when diffraction occurs). This

means that light coming off the tip of the object (e.g. the top of the mug in ig. .) will travel in a straight line through the pinhole and end up near the oor, whilst light rays from the base of the object travel in a straight line through the pinhole and end up near the ceiling. When these rays form a focus, it necessarily produces an inverted image.

Pinhole aperture Object

Image

• Fig. B13.1

iagram showing the path of light rays from a mug, producing an upside-down image in a pinhole camera. The image is also ipped left to right, so in the image (on the right), the tet will be facing the other way (which is why it’s missing).

• Fig. 13.1

Aperture stop

B

Aperture stop

A

llustration of ‘basic’ camera constructions for a pinhole camera (A) or a single-lens camera ().

need our camera to be able to alter the sie of the aperture stop depending on our lighting requirements and we need to be able to oom or replace the lenses to those of dierent focal lengths (f ) which as you probably now is how modern-day cameras wor. e focal length of a camera usually reported in millimetres mm is determined by where the lens will focus light that enters with ero vergence from innity. o for example a camera lens with an  mm focal length would focus ero vergence light  mm behind the lens. is means that as the focal length of the lens changes the lens will be closer to or further away from the sensor at the bac of the camera in order to wor eectively. is is why some telephoto lenses are so long e.g.  mm focal length. e focal length also determines the magnication (m) and the eld of view. ong focal lengths will have narrow elds of

view and high magnication whereas short focal lengths will have wide elds of view and lower levels of magnication ig. .. In terms of varying the diameter of the aperture stop to let in more or less light it’s relatively straightforward as we ust need a sie-changing aperture within the camera. owever photographers describe the diameter ø relative to the focal length f  of the lens – called the f-number – as this helps them understand what eect the aperture stop will have on the nal image. e equation for calculating the f-number is shown in quation .. ere you can see that if a camera had a lens with a focal length of  mm with an aperture stop  mm wide the f-number would be   and so it would be written as f which means the aperture stop in this case has a diameter that equates to a quarter of the focal length. is also highlights that

CHAPTER 13

Shorter focal length

Longer focal length

•

Fig. 13.2 iagram showing the relationship between lens focal length (increasing downwards) and eld of view (shown in shades of blue).

changing either the focal length or the diameter of the aperture stop would alter the f-number. Equation 

f Nø

Equation  (explained) focal length aperture diameter Importantly smaller aperture stops which let in less light might need longer exposure times which is another thing to tae into consideration when planning your shots. In general terms the smaller the f-number the more light is let into the instrument and the better the lens will perform in low levels of light. Altering the aperture sie and f-number can also aect something called depth of eld

Optical Instruments and Low Vision Aids

which describes the range in which something can be in focus. or example let’s say I’m taing a photo of my two cats sitting near each other but one is approximately one metre closer to me than the other. A large depth of eld would mean I could tae a photo that has both of the cats in focus at the same time whereas a small depth of eld would mean I’d have to focus my camera on one or the other and the one that wasn’t in focus would be blurry ig. .. In general smaller apertures lead to larger depths of eld and larger apertures produce smaller depths of eld. In summary cameras are very clever and there’s an awful lot to thin about when trying to tae a photograph

Telescopes Telescopes are optical instruments designed with the purpose of helping us see obects that are far away and they do this by magnifying the far-away obect so we can see it more clearly. or example if we use a telescope to loo at the moon the image of the moon is magnied to help us see it in more detail. A ‘basic’ telescope will need to be made of two lenses – an eyepiece lens the one near your eye and an objective lens the one nearer to the obect. ese lenses wor together ust lie the multiple lens systems we discussed in chapter  to produce an image that is in focus for us as the observer. An interesting thing to learn here is that the image we see through the telescope will be visible through what’s called the exit pupil of the instrument. An exit pupil is the view of the aperture stop from the bac of the instrument so it’s only possible to see the exit pupil through the eyepiece which maes sense the view the other way round would be called the entrance pupil but we’re not going to worry about that in this boo. e position and sie of the exit pupil in a telescope will control the eld of view which means it also serves as the eld stop. A eld stop is an aperture in an optical instrument that controls the eld of view how much of the scene is visible at any one time.

Large depth of field

Small depth of field

• Fig. 13.3

iagram showing visual representation of depth of eld. arger depths of eld mean more depth (areas at different distances from the lens) can be in focus at the same time.

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ere are a few dierent types of telescopes so it’s helpful to discuss them separately.

larger. imilarly we now the nal image will be upright as the three image rays are above the optical axis.

alilean Telescope

Keplerian Telescope

A alilean telescope is made using a negatively powered eyepiece lens and a positively powered obective lens ig. .. is produces a virtual upright magnied image for the observer. e issue with a alilean telescope is that the exit pupil doesn’t coincide with the pupil of the observer as you can see in ig. . the exit pupil the clear window that the image forms in is produced in between the two lenses meaning it can be liened to looing through a eyhole. e observer will need to line up their own pupil with the exit pupil in order to get the best view. It also means that if the observer is holding the eyepiece lens too far away from their eye it will restrict what they can see. ig. . also highlights that the eld of view will be reduced as the three rays leaving the telescope are closer together than the three rays entering the telescope. e can also see evidence of the magnication of the image because the angle of the image relative to the optical axis w9 is larger than that of the obect w suggesting the image is

A eplerian (or epler) telescope is made using two positively powered lenses as long as the higher-powered lens is the eyepiece lens ig. .. ese inds of telescopes produce a real inverted magnied image. owever an upsidedown image can be dicult to mae sense of so sometimes these telescopes are adapted by adding a third lens or a prism in the middle to ip the image to mae it form upright instead of upside down in this case they are referred to as a terrestrial telescope. eplerian telescopes produce an exit pupil that coincides with the observer’s pupil ig. . and ust lie with the alilean telescope we can see that the eld of view is small and the image will be magnied. owever this time we can see that the image rays are below the optical axis indicating an inverted image.

Reecting Telescope e problem with refractive telescopes telescopes that use lenses is that lots of aberrations can be introduced

Eyepiece lens

Objective lens

Image at infinity

Object at infinity

• Fig. 13.4 alilean telescopes have a negatively powered eyepiece lens and a positively powered objective lens.

Objective lens

Eyepiece lens

Exit pupil

w

w’ Fep’

fep’

• Fig. 13.5 iagram of how rays from an object far away (innity) are focused by a alilean telescope. The eit pupil is shown in green, at the focal point of the eyepiece lens ( ep9). The angle of the image relative to the optical ais (w9) is larger than the angle of the object relative to the optical ais, highlighting that magnication has occurred.

CHAPTER 13

Objective lens

• Fig. 13.6

Eyepiece lens

eplerian telescopes have two positively powered lenses.

Eyepiece lens

Objective lens

Fep’

w w’

Exit pupil

Image at infinity

Object at infinity

Optical Instruments and Low Vision Aids

fep’

• Fig. 13. iagram of how rays from an object far away (innity) are focused by a eplerian telescope. The eit pupil is shown in green, at the focal point of the eyepiece lens ( ep9). The angle of the image relative to the optical ais (w9) is larger than the angle of the object relative to the optical ais, highlighting that magnication has occurred.

see chapters  and  for more detail on aberrations so to get around this they usually only possess narrow apertures which limit the light that can enter the instrument. is means that obects that are emitting low levels of light e.g. a star in the sy will be dicult to see. owever the good news is that we can get around this problem by using a mirror instead of lenses and installing a reecting telescope ig. .. ese telescopes use a large concave mirror positively powered to collect light across a wide aperture and form a nice magnied image. e mirror is usually chosen to be paraboloidal aspherical because the wide aperture would mae spherical mirrors susceptible to those pesy spherical aberrations we taled about in chapter . Another advantage of a reecting telescope is that they don’t suer from chromatic aberration see chapter  because chromatic aberration can only occur if light is refracted not reected. o all round this type of telescope is excellent. agnication As we stated at the beginning of the telescope section telescopes are designed to produce magnied images of obects and we could see in igs . and . that the magnication is dependent on the angular dierence between the obect relative to the optical axis w and the image relative to the optical axis w9. is means that telescopic magnication is dened as angular magnica tion (m). owever these angles are determined by the ratio between the powers of the eyepiece lens (ep) and

Prime Incoming focus starlight

Plane mirror

To eye

Eyepiece lens

Focal length

Concave mirror

• Fig. 13.8

iagram of a reecting telescope showing how light from a star (blue) is focused by the concave mirror with the wide aperture.

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objective lens (o), so we can determine angular magnication relatively easily by using quation .. emember as well that the sign of the magnication tells us whether the image is upright 1 or inverted 2 so a alilean telescope should produce positive magnication values upright image and eplerian telescopes should produce negative magnication inverted image. Equation 

Fep

m

Fo

Equation  (explained) mag

power of eyepiece lens power of objectiive lens

DEMO QUESTION 13.1 A alilean telescope has an eyepiece lens with a power of ., and it produces a magnication of 1.. What is the power of the objective lens tep  etermine what we need to calculate power of objective lens, o tep  ene variables ep 5 2. m 5 1. (the sign is important here) tep  etermine necessary euation m 5 2 (ep  o) (Equation 13.2) tep  alculate m 5 2 (ep  o) . 5 2 ( .  o) o 5 2 ( .  .) o 5 1. (don’t forget the 6 sign)

Practice Questions: 13.1.1 A alilean telescope has an eyepiece lens with a power of ., and it produces a magnication of 1. What is the power of the objective lens 13.1.2 A eplerian telescope has an eyepiece lens with a power of 1. and an objective lens with a power of 1.. What is the angular magnication of the telescope 13.1.3 A eplerian telescope has an eyepiece lens with a power of 1., and it produces a magnication of .. What is the power of the objective lens

Low Vision Aids It’s relatively easy to imagine that people might need to use telescopes or binoculars to help them see something far away through magnication for example astronomers bird watchers students completing experiments in optics labs at university. . . . ut it may surprise you to learn that optical instruments lie these are also sometimes prescribed to patients who are diagnosed with low vision. ow vision is a classication of poor-quality vision that can’t be corrected with glasses contact lenses or surgery and it will usually mean a patient is unable to do normal day-to-day tass lie read drive watch  or recognise their friends

and family when out and about. ow vision itself can be caused by a myriad of diseases but the ey factor is that it’s permanent so management of low vision involves trying to help improve the patient’s ability to use the sight they have left. is can include improving the brightness of the lighting in their home environments or advising patients to put lamps close to where they read to improve the illumination see chapter  for details of why this is helpful. Alternatively patients can be advised to use low vision aids such as magniers.

agniers ne of the visual diculties associated with low vision is an inability to resolve small detail for example written text. ften when people experience diculty in this way the tas can be made easier if the obect is made bigger magnied. or example if a person was struggling to read the small print of a boo as shown in ig. .A it might be because the distance of the obect from the observer leads to the obect subtending a small visual angle angle formed at the eye by the rays from the obect distant obect  ud. ollowing this idea it seems lie one way to mae the obect appear bigger and therefore easier to see would be to increase the visual angle. ne way this can be achieved is by bringing the obect closer to the observer as shown in ig. .. is would increase the visual angle close obect uc which will mae the obect much larger on the retina however as I’m sure many of you can appreciate it can be dicult to focus at very short distances away from the eye especially as we advance in age which might mean the obect becomes blurry before it can be large enough to be clearly visible. An alternative solution then is to use a lens between the observer and the obect that is able to produce a magnied image of the obect farther away from the observer as shown in ig. .. is wors because the image is distant so the observer can focus on it but the visual angle is large mimicing the scenario when the obect was much closer. verall this means the desired outcome of the magnier is to produce a virtual upright image that is larger than without the magnier. In general magniers describe high-powered lenses that are designed to mae the image of obects bigger through magnication. agniers themselves can comprise a singlelens system this is typical of handheld and atform mag niers or they might comprise a multiple lens system such as a telescopic monocular emember though not all telescope systems are created equally so if a telescopic device is required it should be one that produces an upright virtual image otherwise they won’t be much use to the individual at all verall each type of magnication device will be useful for dierent types of tass summarised in able . for example a handheld magnier can be useful for nearwor such as reading or watching videos on your mobile device especially if it has a light so that it can illuminate the obect but you wouldn’t be able to use one to loo at something far away lie a train timetable screen or a specials

CHAPTER 13

A

Optical Instruments and Low Vision Aids

Object

Observer θd

B

Object

Observer θc

Image

C

Observer θc

Object Magnifier

• Fig. 13.

implistic illustration showing lin between visual angle (u) and distance. When the object is far away from the observer (A), the visual angle is very small (ud, distant). f the object is brought closer to the observer (), then the visual angle becomes much larger (uc, close), but the observer might nd it difcult to focus on the object. The ideal solution in this scenario, then, is to use a magnifying lens () to mae a small object produce a magnied, virtual image.

TABLE agniaion Deies an heir elaie 13.1 Feares

Magnication Device

Useful For

and-held magniers

ear-wor (e.g. reading)

pectacle magniers

ear-wor (e.g. reading)

tand magnier

ear-wor (e.g. reading)

Telescopic monoculars

ear-distance

board in a restaurant. In these cases a telescopic monocular would be far more useful. agniers and monoculars are therefore very useful for anyone who has issues with their vision but unfortunately a little lie glasses and contact lenses they aren’t perfect. or example magnifying lenses can induce aberrations imperfections in the image see chapter  for more information on this which is especially true if we attempt to utilise sections of the lens that are away from the optical centre and it has a more detrimental eect in high-powered lenses typical of those used in magniers. is means that very high-powered magniers will usually need to have a smaller

lens to prevent aberrations so if you need something magnied to quite a high degree it will often have a smaller eld of view. is can then lead to other issues such as difculty using the lens with both eyes binocular viewing which means that sometimes patients will have to occlude one eye in order to mae eective use of the magnier. imilarly many of the spectacle magniers can have quite a limited depth of eld meaning that unless the obect is positioned within the narrow range of appropriate distances it will appear blurry. verall though despite these limitations low vision aids provide a great solution to the everyday diculties that low vision patients can experience. In addition to magniers and monoculars there are a number of alternative low vision aids available so if you’re interested in this then I’d encourage you to do some research to nd out more.

Adapted Lenses ow we now that magniers can be useful for helping patients with low vision but not all vision-related issues are to do with image sie – sometimes patients can experience diculty with glare that aects their ability to see. is is

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often true for patients with cataracts a clouding of the ocular media lens within the eye which lead to an increase in blur a decrease in ability to detect dierences in contrast and an increase in scattering of light within the eye. rucially this scattering can mae light sources appear very bright which can aect vision and be uncomfortable for the patient – this is called glare. In these cases it is benecial to try to update their glasses prescription to help them see as clearly as possible but it will also be benecial to reduce the amount of light that is able to enter the eye for example through recommending tinted lenses or through advising on positioning of light sources in the home to limit the impact of the glare when performing tass.

Assistie Tecnolog If lie me you have a relatively high myopic shortsighted refractive error and need glasses to see you may have found yourself in the situation where you have taen your glasses o possibly to shower or to get dressed and then been unable to nd them again because you can’t see very well without them and they’re quite small so they easily blur into the bacground at least I hope this has happened to people other than ust me. In this scenario one option is to hope that a relative or housemate might be able to oer assistance but if you’re in the house by yourself it can be a bit of a struggle to get close enough to every surface in order to get it into focus within your far point and see if your glasses are sitting there mocing you. owever a few years ago when this happened to me for the th time I realised that I don’t need to get close to all the surfaces because I have a phone with a camera on it and I can easily focus on the phone screen when it is held close to my face. o I grabbed my phone which is large enough to nd and turned on the camera and was very pleased to discover that

I could now see the room very clearly through the screen on my phone – which helped me locate my glasses in such a short amount of time that it became my new personal best. e reason this wors is because the phone camera captures an image of the scene which is then fed in a live view to the screen of the phone. It is then possible for me as a myope to view the image on the phone when it’s within my very limited range of focus and so this gives me a way to view the world without needing any correction. ow there are obvious drawbacs – rstly it’s not convenient to wal around with your phone held very close to your face but also because I have to hold the device so close to my face it taes a long time for me to scan the scene as I have to mae very deliberate eye movements saccades. owever in a pinch this is an eective way to help me see when I’m uncorrected without my spectacles. hilst I don’t have low vision myself I thin this anecdote serves as a relatable segue into the nal section of this chapter which will discuss assistive technology electronic options for patients with low vision. ome of the more readily available assistive technologies include video magniers and they can be destop xed in position or handheld portable. ideo magniers are electronic magniers that wor along the same principles I was describing with my phone previously. ey use a camera to analyse text or photographs and proect a magnied up to 3 increasedcontrast image onto a digital screen for the patient to see. A neat feature of these devices is that they can tae snapshots so patients can move around the screen at their preference. Also because they utilise digital software it is possible to buy video magniers that will read the text aloud which can help hugely with understanding what is written if patients have low vision. inally these video magniers are not susceptible to the same small eld of view and aberrations that handheld lens magniers are but these advantages come at a cost as these video magniers are often far more expensive.

Test Your Knowledge ry the questions below to see if you need to review any sections again. All answers are available in the bac of the boo. T hat is an ‘optical instrument’ T ould the human eye be classed as an optical instrument xplain your answer.

Reerences . upré . Inside the camera obscura epler’s experiment and theory of optical imagery. Early Sci Med. -. . eat  egge  ullimore A. hat is low vision. Optom Vis Sci. -. . . Low Visual Aids. httpswww.hey.nhs.uwpwp-content uploadsow-isual-Aids.pdf. Accessed ecember  .

T If we wanted to focus our camera on something very far away would we choose a  mm or  mm lens xplain your answer. T ould you expect a alilean telescope to have a positive or negative magnication xplain your answer.

. asa  odgor  atiles  aruso  agno . lare sensitivity in early cataracts. Br J Ophthalmol.  -. . I. Assistive Technology. httpswww.rnib.org.usight-lossadviceequality-rights-andemploymentstaying-worassistivetechnology. Accessed ebruary. Accessed ebruary  .

14 Polarisation   A P T  R   TL I  Introduction

Polarisation by Scattering

Some Light Revision

Applications of Polarisation Photography Sunglasses 3D Films

Types of Polarisation Polarisation by Transmission Polarisation by Reection Polarisation by Refraction

Test Your Knoledge

     T I S After working through this chapter, you should be able to: Explain the dierence between polarised and unpolarised light Explain the process of polarisation by transmission

Explain the process of polarisation by refraction Explain the process of polarisation by reection Explain the process of polarisation by scattering

Introduction

in other electrons forms part of the basic explanation of how polarisation works, but we don’t need to get too bogged down in this right now. f we now go back to thinking about our single light source,  think we can agree that it’s relatively easy to imagine light as a single waveform with a vertically oriented electric eld (and much easier to draw that way too) however, most light sources (the sun,  bulbs) produce light that con tains vibrations in all possible meridians (otherwise known as orientationsplanes), as shown in ig. . (which unfortu nately is much harder to draw and visualise). n this case, the light is formed of many orientations of light, so the electric eld changes orientation randomly over time. When light vibrates in all directions like this (see ig. .), it’s considered to be unpolarised, be cause it has more than one orientation of vibration (and a slightly unpredictable electric eld). owever, it is pos sible to take unpolarised light and reduce it to a beam of light comprising vibrations that occur (mostly) within a single meridian, which we would then describe as polarised light. is process of transforming the unpolarised light into a polarised state is called polarisation, and polarisation can occur through a number of dierent methods, including () transmission, () reflection, () refraction and () scattering. irst, however, let’s discuss types of polarisation.

We’ve already discussed a lot about light and how it can be emitted from a source in all directions (e.g. like the sun), and more specically we’ve talked about how light can be expressed as waves, travelling outwards from the source. is chapter will focus on how we can alter the orientation of those waves, through a process called polarisation.

Some Light Revision (un intended.) When light is emitted from a light source, it can be described in terms of its wavefronts, light rays or light waves. o understand polarisation, we need to focus on light as a wave. magine a single light wave is being emitted from a light source (in this case, possibly a laser), as shown in ig. .. n this example, the light is oscillating vertically, which means its electric eld (or electric vector) is also vertically oriented. We learned in chapter  that visible light is part of the electromagnetic spectrum, but we never specically dis cussed that it is technically an oscillating electric and magnetic eld (both of which are oriented perpendicular to one another). e oscillating electric eld that forms the light has the potential to be able to aect electrons in other mate rials by causing them to start oscillating too. is oscillation

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linical Applications

A

B •Fig. 14.1

Diagram showing how we usually think of light (a wave vibrating or oscillating in one orientation, in this case vertical (A)), and how natural light truly behaves (vibrating in all directions (B)).

Types of Polarisation or light to become polarised, we need to limit the orienta tions (or conne the direction of the electric eld vector). is can be done linearly, circularly or elliptically. Linear polarisation occurs if the electric eld of the light (orientation of the wave) is restricted to a single plane orientation (ig. .). is type of polarisation will be discussed most often within this chapter as it’s the ‘easiest’ to understand. n contrast to this, circular polarisation com prises two linearly polarised (perpendicular) waves that possess a phase dierence of ° whilst possessing the same amplitude (ig. .). n this case, the electric eld pro duced by these waves will rotate as the waves move forward, in a circular shape. is can be clockwise or anticlockwise and is usually referred to as righthand or lefthand circular polarisation, depending on which of the waves is ‘ahead’ (in terms of phase) of the other. f two linearly polarised (per pendicular) waves possess dierent amplitudes or a phase dierence that varies from °, then this produces elliptical polarisation (ig. .). ow polarised light can also be classied as either p-polarised or s-polarised depending on how it is polarised relative to what’s referred to as the plane of incidence. e plane of incidence can be thought of as a at, completely imaginary surface on which the incident (and reected) light exists – so, often, the plane of incidence is drawn per pendicular to the surface (ig. .). P-polarised light de scribes light that possesses an electric eld that is parallel to the plane of incidence (‘p’ for parallel), whereas s-polarised light describes light with an electric eld that is perpendicular to the plane of incidence (‘s’ for s’not parallel at all . . . or, technically, s for ‘senkrecht’, which is erman for per pendicular). o in ig. . if the reected light was parallel to the plane of incidence (vertical) then it would be ppolarised, and if it was perpendicular to the plane of incidence (horion tal), it would be spolarised.

vibrations of light parallel to the transmission axis (or polarisation axis) of the polariser, this method is often called polarisation by transmission. n ig. ., the po larising lter shown has a vertical transmission axis, so only vibrations oriented vertically will be transmitted through. ll other vibration directions will be stopped by the lter, par ticularly those oriented perpendicularly to the transmission axis. n easy way to remember this is to liken it to the ‘picket

A

B

C

Linear

Circular

Elliptical

Polarisation by Transmission ne of the most common manmade methods of polarising light is to use a material as a polariser (a system that can turn unpolarised light into a polarised state). ese materi als are specially designed to only transmit waves vibrating in a single direction (much like in the example in ig. .). ow, because this method of polarisation will transmit

•Fig. 14.2

Diagram showing linear polarised light (A), circular polarised light (B), and elliptical polarised light (). ere the light is propagating from left to right as indicated by the black dashed line. ight waves are shown in blue, the electric eld is shown in solid black and the type of polarisation is indicated in a bright pink.

CHAPTER 14

Plane of incidence

Polarisation by Reection

•Fig. 14.3

Diagram showing plane of incidence (light blue) as eisting along the line of the incident and reecting light (red). n this case it’s perpendicular to the surface (grey).

fence’ analogy. is analogy assumes that a dog (for exam ple) with a stick in its mouth will only be able to pass through a gap in a picket fence if the stick is vertical (meaning the dog will need to turn its head). olding the stick horiontally will not work – ust like with this example of polarisation is also means that if you put two perpendicularly aligned polarisers one after another, the light would become polarised as it transmits through the rst polariser, but it would then get stopped by the second polariser. ese are called crossed polarisers (ig. .). mportantly, as these polarisers are technically stopping uite a good amount of light from passing through to the other side, there is a reduction in brightness () of the light after transmission. is helps form part of the

nother method of polarising light can be to reect it o a surface in proper scientic terms this is due to the relation ship between the electric eld of the light relative to the plane of incidence upon the surface. ssentially, if the electric vector is perpendicular to the plane of incidence, then the light will be polarised in the same orientation as the electric vector. n a simple example, if we have incident light upon a horiontal surface (e.g. a lake), then the plane of incidence will be vertical (as shown in ig. .). is means that the light will ultimately end up being polarised to be parallel to the surface of the road (also horiontal), which will be perpendicular to the plane of incidence, mak ing it spolarised. n simpler, more general terms, this all means that given the right conditions, the reected light will be polarised par allel to the surface from which it was reected, so for the maority of examples (snow, roads, lakes) the reected light will be polarised in the horiontal plane, but in some cases (e.g. light reecting o a window on a house) it will be po larised in the vertical plane.

Unpolarised light Polarising filter

Polarised light

olarisation by transmission shown with a lter that possesses a vertical transmission ais.

Unpolarised light Polarising filter (vertical TA) Polarised light

Polarising filter (horizontal TA)

No light

• Fig. 14.5

133

explanation as to why polarising lters make great lenses for sunglasses (but it isn’t the whole story). e benet of having polarised lenses is discussed towards the end of this chapter.

Laser pointer

•Fig. 14.4

Polarisation

rossed polarisers (angled at ° to each other) prevent light from transmitting through the system, because after passing through the rst polariser with a vertical transmission ais (A), the light is inappropriately oriented for transmitting through the second polariser with a horiontal A.

linical Applications

Incident unpolarised light

Normal

134 S EC TI ON 3

Pla ne o f

inci den ce

Brewster angle () alle Par

l po

l

e ect refl d e aris

ht d lig

Partially polarised refracted light

• Fig. 14.6

Diagram showing Brewster angle (u) with the eample of unpolarised light incident upon a glass block. he angle between the refracted and reected rays is °, and the reected light is polarised parallel to the surface of the glass block.

n interesting feature of polarisation by reection is that it can occur when some of the light is transmitted through the material too (like in the examples of lakes and windows). owever, there are some special circumstances, where the in cident light is at the Brewster angle (sometimes called the polarising angle) relative to the normal. e rewster angle is related to the wavelength of light but also to the refractive index dierence between the two materials. or example, if light starts in air (n 5 .) and travels into a glass block (n 5 .), then uation . shows that the rewster angle will be roughly .°. mportantly, if ppolarised light is inci dent upon a surface at the rewster angle, no light will be re ected, whereas if unpolarised light is incident upon a surface at the rewster angle, then the reected rays will be polarised par allel to the plane of the surface – in which case the remaining refracted light will be partially polarised (see ig. .). n all cases, when the incident light is at the rewster angle, the angle between the refracted and reected rays will always be °. Euation 

tan

1

⎛ n2 ⎞ ⎝ n1 ⎠

Euation  (explained) Brewster angle

tan

1

second ref. index rst ref. index

owever, the problem with polarisation by reection is that in the real world it can cause a great deal of glare, which is something we’ll pick up on in the section pplications of olarisation’.

Polarisation by Refraction n some cases, light can become polarised through refraction. e term ‘refraction’ describes the change in direction of light when moving from one material into another. is directional change will occur at the boundaries of the mate rials, so if light entered the front surface of a glass block (from air) and then left the block at the back surface, refrac tion would occur at both the front and back surfaces. ow, most materials that refract light are what we call ‘optically isotropic’, meaning they possess the same optical properties in all directions, but some special materials (e.g. calcite crystal) can have a degree of optical symmetry, meaning the natural freuency of the material (and its propensity for transmitting light) will be dierent depending on the axis. When light enters this type of material, two linearly polar ised rays are produced  like to think of this as double refraction, but the technical term is birefringence (ig. .). n this example, one of the polarised rays will be linearly polarised parallel to the surface, and one will be polarised perpendicular to the surface, which ultimately results in two distinct images being produced.  nice, realworld example of this is the image formed when viewed through a calcite crystal (sometimes called celand spar).  calcite crystal is transparenttranslucent and is a clear (glassy) type of crystal, meaning that it can transmit light through it uite easily. owever, because cal cite crystals are optically symmetrical, they produce bire fringence which produces two images of any obect when viewed through the crystal. We can see a nice example of this in ig. .

CHAPTER 14

Unpolarised light

Polarisation

135

Perpendicular to surface Parallel to surface

Glass block

•Fig. 14.7

llustration showing incident light (blue ray) upon a glass block. At the front surface of the block, the light is split into two perpendicularly, linearly polarised rays, which will produce two images.

n clinical applications, the retinal nerve bre layer () in the eye is a birefringent material (meaning it splits light into two rays). f light is sent into the eye (let’s say from a camera), and it passes through the , then the dierence between the subseuent two rays will reveal a lot about the thickness of the , so this feature is used in some optical imaging for monitoring the health of the back of the eye.

light is scattered as it travels through a medium. or exam ple, when unpolarised white light from the sun travels through the arth’s atmosphere, it will come into contact with the various atoms that make up the atmosphere. is will set the electrons in the atoms into vibration, which causes them to produce their own electromagnetic wave that radiates out in all directions (ig. .). is newly gener ated wave then strikes neighbouring atoms, which causes a knockon eect of the same vibration and production of electromagnetic waves, which are once again radiated out in all directions. is absorption and reemission of light waves is what causes the light to be ‘scattered’ within the medium. or a realworld example, let’s consider that the sun looks yellow at noon and red at sunset, and why the sky looks blue. When this absorption and reemission process happens with unpolarised sunlight, shorter (more blue) wavelengths within the sunlight are more easily polarisedscattered than the longer wavelengths. t noon, the light from the sun only has a relatively small amount of atmosphere to travel through, which limits the amount of polarisation that can happen (ig. .). owever, the scattering of the blue wavelengths into the atmosphere makes the sky look blue, and as a conseuence of them being removed from the white light of the sun, the sun appears shifted towards the longer side of the spectrum – making it appear yellow instead of white. imilarly, at sunrise and sunset, the light from the sun has much more atmosphere to travel through, which causes some of the slightly longer (more yellow) wavelengths to be scattered as well (see ig. .). is means that by the time the light from the sun reaches us at this time of day, the blue and yellow light will be scattered into the atmo sphere (making the sky appear orange), and the appearance of the sun will be shifted further towards the long wave length side of the spectrum, making it appear red.

Polarisation by Scattering

Applications of Polarisation

e nal method of polarisation we’re going to discuss in this chapter is polarisation by scattering. is method (ust like it says on the tin) describes polarisation that occurs as

olarisation and polarising lters can be utilised eectively for a number of applications, including photography, glare reducing sunglasses, and dimensional () lms.

A

B

•Fig. 14.8

An eample of birefringence in a calcite crystal (also known as celand spar). ou can see that without the crystal, the writing is neat(ish) and very clear (A), whereas when the crystal is placed on top of the writing, it produces two images of the writing and the suare lines on the page (B). hese two images will be produced by perpendicularly polarised light.

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linical Applications

Unpolarised sunlight

Linearly polarised Molecule

Linearly polarised Partially polarised

• Fig. 14.9

A diagram illustrating how unpolarised sunlight can be polarised in a number of directions when making contact with a molecule in the atmosphere.

Sun at noon

Sun at sunrise/sunset

Atm os ph er

e

• Fig. 14.10

An illustration of a person on arth (not to scale) viewing the sun at noon and at sunrise sunset. At sunrisesunset the sunlight has farther to travel to reach the person, which means more of the light is scattered. his makes the sun appear redder as it gets nearer the horion.

Photography When taking photographs, polarised light (particularly that caused by reection) can create a lot of glare and can aect the colour of certain obects within the image. hotogra phers then, can use a polarising lter on the front of their camera lens, which will reduce glare, darken the colour of the sky, and manage unwanted light reections. odern polarising lters are typically circular polarisers to make them maximally eective, as linear polarisers can aect the autofocus capabilities of some cameras.

Sunglasses olarised sunglasses are often seen as a way for people sell ing sunglasses to charge you extra for the lenses, but they really do serve a purpose ese lenses are typically designed to remove the glare and reections from polarised light

reected o water, roads, snow, etc. is means that most of the light they are trying to block out will be polarised parallel to the surface it’s reected from, which should (in most cases) be horiontal relative to the observer. is means that polarised lenses in sunglasses can be linearly polarised to only permit vertical light through (to remove all horiontal glare light from the scene). s discussed ear lier in the chapter, this will also reduce the brightness of the scene, which highlights that they’re eective for use as sun glasses. e advantage of polarised lenses is that the removal of the reected light allows people to see into the water (so great for shermen and sailors), and they can improve con trast of the scene (so also great for sportspeople like golfers). owever, these lenses can also remove the light information from certain digital displays that emit polarised light, so if you’ve ever noticed that a digital sign appears to vanish at certain angles through your polarised lenses, it’s because the light emitted is also polarised

CHAPTER 14

 ilms odern  lms that reuire the cool, blacktinted specs (as opposed to the slightly less cool red and green specs) utilise principles of polarisation to make their scenes ‘pop out’ of the screen. ey do this by using two  cameras to lm two sidebyside, simultaneous shots of every scene which identies the disparity (relative dierence in position) between obects. f an obect is closer to the cameras, it will appear to have larger disparity than an obect far away. ese scenes are then superimposed on top of one another using two proectors (which makes them look a little blurry – par ticularly for obects close to the camera – if viewing without the appropriate specs on). or linearly polarised lms, these proectors will be set to emit light that is polarised in a

Polarisation

particular orientation, but crucially, they need to be set dif ferently to one another. or example, one proector might emit light linearly polarised vertically, and one might emit light linearly polarised horiontally. our special glasses will have two dierent lenses – one set to allow through each type of polarised light so for example, perhaps only verti cally polarised light (from the rst proector) can reach the left eye, and the opposite for the right eye. ur brains can then interpret and merge the two disparate images to achieve that sense of depth. owever, an issue arises if we tilt our heads whilst in the cinema (or snuggle up to a loved one), because it changes the orientation of the polarising lter in the lenses and can prevent the light getting through prop erly, which is less than ideal. o x this, proectors can utilise circular polarisation, which means headtilting is  again.

Test Your Knowledge ry the uestions below to see if you need to review any sections again. ll answers are available in the back of the book.  ene ‘unpolarised light’.  xplain the dierence between circular and el liptical polarisation.

Reference . hysics lassroom. olarisation. n.d. httpswww.physicsclass room.comclasslightessonolariationtext5 picketfenceanalogyis,vibratesina singleplane. ccessed ecember , .

 What type of light would be emitted through two identically oriented polarising lters  s light reected o a lake more likely to be po larised in the horiontal or vertical plane  xplain why the sun appears red at sunset.

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15 Imain the Eye and easurin eractie Error   A  T E R O  T I E Introduction Imaging the Eye Why Can’t We See Inside the Eye Anyway? Ophthalmoscopy Direct Ophthalmoscopy Indirect Ophthalmoscopy

Gonioscopy Direct Gonioscopy Indirect Gonioscopy Measuring Refractive Error Applanation Tonometry Test Your Knowledge

O   E  T I E After working through this chapter, you should be able to: Explain why it is dicult to see the ac o the eye and the anterior anle without the help o special deices Explain how direct and indirect ophthalmoscopy wors

Explain how direct and indirect onioscopy wors Explain retinoscopy and how it wors Explain how applanation tonometry wors

Introduction

(the aperture in the iris leading to the back of the eye) appears black.

In this chapter, we’re going to use some of the knowledge we gained in previous chapters in order to start to understand some of the basic principles behind imaging the inside of the eye (which as you might imagine, requires special optical systems), before moving on to discuss how we can use optical physics to determine a person’s refractive error and how applanation tonometry works. lease note that this book is not designed to be a guide for how to perform these techniques safely in clinical practice instead consider this an interesting summary of the underlying principles to help you understand the techniques.

Imaging the Eye o start with, let’s think about imaging the inside of the eye. ow, when I say ‘imaging’ here, I’m referring to the ability to produce an optical image of the inside of the eye. e reason for needing to do this would be to assess the health of a patient’s eye, so it is an essential part of an optometrist’s daily practice. owever, if you’ve ever met another human being before you’ll know that we can’t ust look directly into a person’s eye, because in normal circumstances the pupil

hy an’t e ee Inside the Eye Anyway et’s begin by considering the reduced human eye that we learned about in chapter  (ig. . for a refresher). is reduced eye assumes that the overall power of the eye will be 1., and the retina (the sensory tissue at the back of the inside of the eye) will eist at the secondary focal point of the eye (1. mm). is means that we can assume that light reecting from the retina (forming an image of the retina) will possess a vergence of 2. when it reaches the front surface of the cornea, and that therefore the light leaving the patients’ eye will have parallel vergence. ollowing this, we can start to understand that an in-focus image of the retina will be ‘at innity’ (we can conrm this using vergence equations from chapter ). is means that in fact, we absolutely should be able to see the retina from outside the eye, providing we can line ourselves up with the eit pupil of the eye (see chapter  for a reminder of this idea). owever, there are two issues . e inside of the eye is very dark, and we can only see illuminated obects (see chapter  for a review of this). . e human pupil is usually very small (. mm), which will severely restrict our eld of view. 139

140 S EC TI ON 3

linical Applications

F = +60.00 n = 1.00

Reduced eye

n’ = 1.  Apex (cornea)

F

–16.67 mm (f)

F’ (fovea)

+5.55 mm (r)

+22.22 mm (f’)

• Fig. 15.1

Reduced model eye with power, distances, and refractive indices labelled appropriately.

ogically, then, if we could illuminate the inside of the eye somehow, and overcome the issue of the eld of view, then it should be possible to see inside the eye.

Ophthalmoscopy ne method for viewing the back of the eye is called ophthalmoscopy (ophthalmo-, eye; -scopy, view), and there are two main types direct ophthalmoscopy and indirect ophthalmoscopy

irect Ophthalmoscopy irect ophthalmoscopy requires the use of a hand-held device called an ophthalmoscope. is device sends parallel (ero vergence) light into the eye, which will then focus on

the retina, thanks to the focusing power of the cornea and lens. e light then travels along the same path back out of the eye, which means it leaves the eye as parallel light again. is parallel light can then approach the clinician’s eye, which will form a lovely, focused image of the patient’s retina on the clinician’s retina and allow them to see eactly what’s going on inside the patient’s eye (ig. .). e nice thing about direct ophthalmoscopy is that it produces a virtual (upright) image, so if the top of the image looks a bit suspect, then it means the top of the retina looks a bit suspect, and it’s easy to relate the two (unlike with indirect ophthalmoscopy which produces a real, inverted image – see section ‘Indirect phthalmoscopy’ for more details). ssuming that the clinician has corrected vision, and the patient is emmetropic (no refractive error), then the clinician will not need a lens of any power in order to see the patient’s retina. Instead, the clinician will ust need to get very close to the patient – we’re talking  to  cm away – in order to be able to see through the patient’s pupil. owever, due to the sie of the pupil, the clinician will still only be able to see a small amount of the retina at any one time (small eld of view), which means they will need to change their viewing angle to see other parts of the retina (a bit like viewing through a keyhole – ig. .). is is why, if you’ve ever had this done, clinicians will wiggle around and do the ophthalmoscopy dance in front of you – they’re attempting to see all the parts of your retina. e clinician will also usually ask the patient to move their gae (upwards, up and to the right, rightwards, down and to the right, etc.) to help them view as much of the back of the eye as possible. owever, this method only allows the clinician to use one eye at a time (due to how close the clinician needs to be to the patient), which can be an issue if the clinician has

Clinician

Patient Compensation lens wheel

Mirror/ prism

Lens

Aperture wheel Lens

Bulb

•Fig. 15.2

Breakdown of how direct ophthalmoscopy images the patient’s retina (right) to allow the clinician to see what it looks like (left).

CHAPTER 15

A

Imaging the Eye and Measuring Refractive Error

141

B

• Fig. 15.3

igure showing clinician using a direct ophthalmoscope to look at the patient’s fovea () and then adusting their position to look at the inferior (bottom) part of the patient’s retina (B).

amblyopia or any other condition aecting the vision of a single eye. e other, slightly complicated factor to think about is that if the patient (or clinician) has a refractive error, then the light from the ophthalmoscope will be focused incorrectly (either behind or in front) of the patient or clinician’s retina, and so the ophthalmoscope will need to be adusted to account for that.

Indirect Ophthalmoscopy In contrast to direct ophthalmoscopy, indirect ophthalmoscopy allows the clinician to use two eyes (stereoscopic viewing) and utilises a high-power plus lens (e.g. 1., 1.) called a condensing lens  ight is shone through the condensing lens and focused on the patient’s

retina, and, ust like before, the light will reect back out of the eye following along the same path and passing back through the condensing lens (ig. .). is lens then forms an image of the retina which is magnied but inverted (upside down and ipped left to right), which is passed to the clinician’s retina in order for them to see the image. e eld of view will be determined by the power of the condensing lens – so clinicians will usually have a preferred type of lens. e advantages of the indirect ophthalmoscopy technique (compared to direct) are that the clinician can view the patient’s retina using both eyes, and thanks to the power of the condensing lens, the eld of view will be much greater. is method can be utilised using either a head-mounted binocular indirect ophthalmoscope (abbreviated as I) or with a slit-lamp biomicroscope (abbreviated as ).

Clinician

Handheld condensing lens

Patient

Bulb

Mirrors

• Fig. 15.4

Breakdown of how indirect ophthalmoscopy images the patient’s retina (right) to allow the clinician to see what it looks like (left).

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If using a slit-lamp biomicroscope, then another thing clinicians need to think about is that they can’t move themselves very easily to view dierent parts of the patient’s retina. o get around this, they will ask the patient to move their gae (upwards, up and to the right, rightwards, down and to the right, etc.) to help them view as much of the fundus as possible.

pressure goes too high then this can lead to damage which can ultimately lead to sight loss – so it’s useful for clinicians to view the anterior angle and check that it’s healthy in order to determine whether the aqueous is able to drain eectively. owever, the problem is that it’s not possible to see the anterior angle from outside the eye, as total internal reection occurs. If you recall back in chapter  we discussed the idea that when moving from a medium with a higher refractive inde into a medium with a lower refractive inde, if the angle of incidence of light leaving the obect eceeds the critical angle for that medium, then all the light from the obect will be reected back into the higher refractive inde medium (which is appropriately called total internal reection). is is what happens with the anterior angle in the eye (as seen in ig. .), where due to the refractive inde dierence between the aqueous and cornea (. and .), relative to the air (.), the light from the anterior angle meets the corneal surface at an angle that eceeds the critical angle (ic). is causes it to reect back into the anterior chamber – making it seemingly invisible to anyone outside of the eye.

Gonioscopy ow, let’s imagine that we’re not as interested in the back of the eye, but instead we’d be interested to see the anterior angle. e haven’t discussed much anatomy and physiology in this book, so you’ll ust have to trust me when I say that the anterior angle (the point where the inside surface of the cornea becomes the limbus and connects to the iris – shown by an arrow in ig. .) is an important site for aqueous drainage. queous humour is the name for the uid that eists in the front part of our eye, and it helps to nourish certain structures and also helps to maintain a certain intraocular pressure (IOP). I is important because if the

Zoomed in anterior eye Cross-section of an eye

• Fig. 15.5

iagram showing cross-section of an eye (left) with oomed-in section (red dashed line) to show anterior angle (black arrows).

n’ = 1,00 Front surface of eye

n=

1.3 76

n = 1.336

• Fig. 15.6

llustration showing cross-section of the anterior eye. ight from the anterior angle (see ig. .) suffers from total internal reection due to the refractive inde difference between the inside of the eye (relative to the outside).

CHAPTER 15

Lens

•Fig. 15.7

143

B r Mirro

A

Imaging the Eye and Measuring Refractive Error

Ocular lubricant

iagram illustrating how direct gonioscopy () and indirect gonioscopy (B) work.

In order to get around this, clinicians need to utilise a goniolens – a special lens for viewing the anterior angle. In general terms, these lenses come in two forms to allow clinicians to perform direct gonioscopy (ig. .) and indirect gonioscopy (ig. .). oth techniques require a lens to be placed directly onto the front surface of the eye (so you’ll be relieved to know that the eye will be anaesthetised rst).

A

Superior view

Lateral view

B

Superior view

Lateral view

irect Gonioscopy irect gonioscopy involves placing a special lens onto the front surface of the eye. is lens will be manufactured at a refractive inde more similar to the eye than air, which means light from the anterior angle can pass into the lens itself. en, the special (very steep) curve of the outer surface of the lens allows the light to leave the lens as it no longer eceeds the critical angle and therefore isn’t susceptible to total internal reection. is is a useful technique because it provides a ‘direct’ view of the angle, from all possible angles of viewing. owever, the problem with direct gonioscopy is that it needs to be placed on the patient’s eye whilst they are lying down, so a traditional optometric practice wouldn’t be able to perform this technique.

•Fig. 15.8

olour-coded illustration view of three-mirror goniolens () and four-mirror goniolens (B) from the superior view and the lateral view. ach mirror is outlined in a colour which is proected onto the cross-section view of the eye to show the area it makes visible.

Indirect Gonioscopy Indirect gonioscopy relies on similar principles to the direct gonioscopy technique, but this time, the clinician will use powered mirrors instead of a lens. Indirect gonioscopy can involve either a three-mirror goniolens or a fourmirror goniolens, depending on what the clinician is looking for, and (to some etent) personal preference. s shown in ig. ., three-mirror lenses can view the posterior pole of the eye by looking straight through (blue), or they can help the clinician view the equatorial section (yellow) using the trapeoid mirror (°), or they can allow the clinician to view the ora serrata (green) using the rectangular mirror (°). inally, the clinician can use the -shape mirror (°) to view the anterior angle (pink). In contrast to this, a four-mirror goniolens (ig. .) can only view the posterior pole (blue) and the anterior angle (pink) as it has four -shape mirrors (°) placed around the inside of the lens. s these lenses utilise a mirror reection, they

produce an indirect view of the angle and require repositioning of the lens depending on which part of the angle the clinician wants to view.

Measuring Refractive Error t this point in the chapter, we’re going to slightly deviate away from health-related imaging and instead start to consider the refractive power (and associated refractive error) of the eye. e learned in chapter  that the human eye can possess spherical or cylindrical refractive errors (or both), with spherical refractive errors being dened as either too much power in all planes (myopia), requiring negative spherical lenses, or too little power across all planes (hyperopia), requiring positive spherical lenses. ylindrical error, on the other hand, is dened as a dierence in refractive

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power along a single plane (or meridian) and requires specialised cylindrical (or toric) lenses. part from using an automated machine to calculate refractive errors, there are two main ways of assessing a patient’s refractive error () subective refraction (asking patients what they can see with a sequence of sensibly chosen lenses) and () obective refraction (shining a light into the patient’s eye and letting the laws of physics tell you what the error is likely to be, whilst utilising a sequence of sensibly chosen lenses). s you might have guessed, we’re going to focus on obective refraction in this book, and in particular we’re going to focus on a method called retinoscopy (or sometimes skiascopy), which is a technique for measuring refractive error that dates back to the s eutralisation retinoscopy relies on the principles of the oucault nife test in which the shadow formed by obects

appear to move dierently depending on where the obect is relative to the focal plane. or eample, look at the diagrams shown in igs. . to .. ere, you can see that a light source (bulb) is focusing light through a condensing lens at a point labelled the focal plane (shown with dashed black lines). ere is also a knife (grey, sharp obect) which is either located at the focal plane (see ig. .) behind the focal plane, farther from the observer (see ig. .) or in front of the focal plane, closer to the observer (see ig. .). In all these gures you can see the shadows formed by the knife in each eample. hen the knife is at the focal plane but not obstructing the light rays (see ig. .), no shadow is produced however, if the knife is moved slightly upwards, it will quickly block the light (see ig. .). is means that if the shadow quickly alternates between noneistent and ‘full coverage’ when the knife is moved (as shown in ig. .), we

A Focal plane

B

Focal plane

• Fig. 15.9

 diagram demonstrating the oucault knife test  with the knife placed at the focal plane. n this setup, light is shone through a positively powered lens to form a focus (focal plane). f the knife (shown in grey) is placed at the focal plane () but without obscuring the light, then it will allow all light to pass (not obstructing). f, however, it is moved slightly upwards (B), it will uickly block all the light completely, meaning the shadow is either ‘full’ or absent.

A Focal plane

B Focal plane

• Fig. 15.10

 diagram demonstrating the oucault knife test  with the knife placed farther from the observer than the focal plane. n this setup, light is shone through a positively powered lens to form a focus (focal plane). f the knife (shown in grey) is placed farther from the observer than the focal plane (), then it will slightly obscure the light and produce an inverted shadow (after the light crosses following convergence). f the knife is then moved slightly upwards (B), it will move the shadow (and the remaining light) downwards in the opposite direction to the movement of the knife (against movement).

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145

A Focal plane

B Focal plane

•Fig. 15.11  diagram demonstrating the oucault knife test with the knife placed closer to the observer than the focal plane. n this setup, light is shone through a positively powered lens to form a focus (focal plane). f the knife (shown in grey) is placed closer to the observer than the focal plane (), then it will slightly obscure the light and produce an upright shadow (as the light continues in a straight line). f the knife is then moved slightly upwards (B) it will move the shadow (and the remaining light) upwards, in the same direction to the movement of the knife (with movement).

can assume the obect is at the focal plane. lternatively, when the knife is behind the focal plane, farther from the observer (see ig. .), it produces an inverted shadow because the light converges at the focal plane and crosses to the opposite side. If this knife is then moved upwards (further into the light), then we can see from ig. . that the shadow will move in the opposite direction. is is called an against movement because the shadow is moving in the opposite direction to the obect, which suggests the obect is farther from the observer than the focal plane. inally, when the knife is in front of the focal plane, closer to the observer (see ig. .), it produces a shadow that falls on the same “Peephole” Mirror

Clinician

side as the knife. If this knife is then moved upwards (farther into the light), then ig. . shows us that the corresponding shadow will move in the same direction as the obect. is is called a with movement because the shadow is moving in the same direction as the obect, which suggests the obect is closer to the observer than the focal plane. is is informative to us because we can estimate the position of the knife relative to the focal plane (and vice versa) using the direction and appearance of the shadows. ow, you may be thinking this is all well and good, but how does this relate to retinoscopy ell, with retinoscopy a clinician uses a device called a (self-illuminated) retinoscope Working distance

Patient

Streak aperture

Lens

Bulb

• Fig. 15.12

implied illustration of how a self-illuminating retinoscope works, with light from the bulb passing through a positive lens before being reected to the patient by a mirror. he mirror contains a ‘peephole’ so that the clinician can see the patient through the device.

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which utilises an angled mirror (ig. . for an eample) to shine a bright light at the patient’s eye whilst the patient is focusing on a ‘distant’ target. rucially, this mirror will have an aperture (‘peephole’) or be ‘half silvered’ (see-through), which allows the clinician to see the reection of the light (from the patient’s retina). is peephole in the mirror is designed to align the view of the clinician with the light leaving the retinoscope, and the bright light can appear as a spot (spot retinoscopy) or a streak (streak retinoscopy), but the tet in this chapter will focus on the streak retinoscope and a plane (unpowered) mirror. is streak of light will sit partially across the iris, and some of it will enter the patient’s pupil. e clinician then moves the streak side-to-side and monitors the direction and shape of the light (and the shadow) visible through the pupil, which is shining o the back of the eye (this image of the light is called the ree). e goal of retinoscopy is to line up the focal plane (far point see chapter  for a reminder of this) of the eye with the ‘entrance pupil’ of the retinoscope – which, because the clinician will be  or  cm away from the patient, assumes that light from the patient’s eye will need to converge to nd the retinoscope’s entrance pupil appropriately. owever, remember that the patient will be focusing on a distant target (parallel vergence) because the clinician is trying to determine their distance refractive error, which means clinicians need to be very careful to make a note of eactly how far they are away from the patient. s we learned in chapter , vergence is related to distance, and as the patient is focusing on the distant target (innity), the clinician will need to account for their relative ‘closeness’, as they will nd that they need to add too much positive power (relative to

With

innity) to get the focal plane lined up with their retinoscope. is means that once the ree is ‘neutralised’, the clinician will need to account for their distance from the patient – a distance referred to as woring distance – which can be done at the start of the procedure, or at the end. e eact mechanics of working distance calculations will be discussed after we consider how to decide which lenses to add (or take away). If the patient’s eye is emmetropic or corrected appropriately, then (assuming working distance has been accounted for) the far point will nicely line up with the retinoscope, and the ree will appear bright and ll the pupil and disappear instantaneously as the streak moves away from the pupil this is considered ‘neutrality’ – that is, the light is focusing appropriately on the retina (ig. ., right). owever, if the ree has a visible movement, or is dull (less bright), or if the ree moves at a dierent speed to the streak, then a refractive error is present – that is, the light is not focusing appropriately on the retina and is instead focusing too far in front (myopia) or too far behind (hyperopia). e goal of retinoscopy is to place corrective lenses in front of the patient’s eye until the ree shows a ‘reversal’ (neutrality). e way to determine whether a positive lens or a negative lens is needed to correct the error is to look at the movement of the light (and shadow, like with the knife edge). If the ree has a with movement, then it means the far point of the eye is virtual (or behind the clinician) – which indicates hyperopia – so a positive corrective lens is needed (ig. ., left). lternatively, if the ree has an against movement, then it means the far point of the eye is

Against

Neutral

Sweep streak from right to left

Direction of streak Direction of reflex

•Fig. 15.13

llustration showing retinoscopy and associated ree as the streak moves from right to left. iagrams show what ree looks like when positive lenses need to be added (left), when negative lenses need to be added (middle), and when no lenses need to be added (right).

CHAPTER 15

between the patient and the clinician – which indicates myopia – and so a negative corrective lens is needed (ig. ., middle). If the ree moves at an angle that diers to that of the streak, this would indicate the presence of astigmatism (cylindrical error see chapter ). e amount of refractive error present is determined by adding lenses (appropriately) until neutrality is reached however, as we indicated a moment ago, this is inuenced by the clinician’s working distance. is means that to nd the true refractive error, clinicians need to calculate the relative vergence assumed by their distance from the patient for eample, if they are sitting  cm away from the patient then they will be adding approimately 1. of vergence to the light leaving the patients’ eye ( 5 nl 51. 5 1.) in order to focus it at their own eye. is means that at some point in the refraction process, the clinician will need to subtract this vergence from the estimate before determining the nal prescription. or eample, if an eperienced clinician was performing retinoscopy on me with my 2. eyes, and they were sitting  cm away – they should nd that they need to initially add 2. of power to reach neutrality, but then subtract their working distance vergence (2. 2 (1.)) to determine that my true prescription is 2..

Applanation Tonometry In this nal section, we will briey discuss how split-image prisms can be used to measure I. s we mentioned in the section on gonioscopy, it’s important to monitor a patient’s I because I that is too low or too high can lead to eye problems and vision loss. easurement of pressure is called tonometry, and the technique which we will discuss here is called applanation tonometry (or contact tonometry), which kind of means ‘measure pressure by attening’. In this case, an applanation tonometer can measure I by attening the cornea over a set area. is relies on the Imbert-ic principle, which states that the pressure inside a sphere is related to the force required to atten the surface of the sphere over a particular area (quation .). quation 

P5

F A

Top view

•Fig. 15.15

Imaging the Eye and Measuring Refractive Error

147

Tonometer prism Patient

• Fig. 15.14

iagram showing applanation tonometry. he tonometer (with prisms inside) is in contact with the patient’s eye, which is stained with uorescein (yellow dye) to help show the meniscus of the tear lm.

quation  (eplained)

Pressure 

force area

In applanation tonometry, the tonometer will atten an area of the cornea that corresponds to . mm in diameter, and the I (mmg) can be determined by the amount of force (grams) needed to atten the cornea at this area, multiplied by  (ig. .). is means that the tonometer not only needs to be in contact with the front of the eye (but don’t worry, the eye will have anaesthetic so it won’t feel anything), but it also needs to apply a small amount of force. o, how does the clinician know when the cornea is truly and appropriately attened In order to correctly atten the cornea, the clinician will need to stain the front surface of the eye using a yellow dye called uorescein. is dye is useful because it highlights the tear lm, so when the tonometer is positioned on the eye (in contact with the eye and tear lm), the tear lm produces a meniscus (curve in the surface) around the edge of the tonometer. is meniscus will be viewed through a split-image prism (two cylindrical lenses cut at opposite, diagonal angles to one another – ig. .) which (as its name suggests) splits the image of the spherical meniscus into two semi-circle halves called mires. e way the splitprisms are constructed means that the clinician can see how the mires align with one another (ig. .) to decide whether to add more force or less force until the balance is appropriate to record the I.

Rotated view

Side view

llustration showing how the tonometer works from three cross-sectional viewpoints. nside, there is a split-image prism (two cylindrical lenses which have been cut to be oppositely angled).

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Low dial reading

High dial reading

Correct alignment

•Fig. 15.16

iagram showing clinician’s view of the mires (yellow lines) which are apparent when pressure is made on the patient’s eye. f the clinician is applying too little force, the mires will be too far apart, whereas if the clinician is applying too much force, the mires will be overlapping too much. hen the clinician has applied the correct amount of force, the mires will be aligned as shown on the right with the inner edges in contact.

Test Your Knowledge ry the questions below to see if you need to review any sections again. ll answers are available in the back of the book.  hy is it not possible to see the back of a patient’s eye without the help of a special device  Is a slit-lamp biomicroscope a form of direct or indirect ophthalmoscopy  hy is it advantageous to the clinician to ask the patient to move their gae when performing ophthalmoscopy  plain why the anterior angle of the eye is not visible without the help of a special lens.

References . minihaibashi , agen , oldal , aeng , speseth . Individual dierences in resting-state pupil sie evidence for association between working memory capacity and pupil sie variability. Int  sychophysiol. -. . herman   e history of the ophthalmoscope. In enkes , renner , eds. istory of phthalmology (ub auspiciis cademiae phthalmologicae Internationalis). ordrecht pringer -.

 n a three-mirror goniolens, which mirror would allow the clinician to view the anterior angle  plain what an against movement would look like during retinoscopy.  If a clinician performed retinoscopy on a patient at a working distance of  cm, what spherical power would they need to account for in the nal refractive error  hat equation does applanation tonometry rely on for calculating intraocular pressure (I)

. orboy . e Retinoscopy Book: An Introdctory anal for ye are rofessionals th ed.   Incorporated . . allak . eections on retinoscopy. Am  ptom hysiol pt. ()-. . lliott . linical rocedres in rimary ye are. th ed. msterdam lsevier ealth ciences . . hee . olor Atlas  ynopsis of linical phthalmology: lacoma. rd ed. e etherlands olters luwer .

16 avefront berrations and daptive ptics  H A  T E R   TL I E Introduction

Aberrations in the Human Eye

Wavefront Aberrations

Measuring Aberrations in the Eye

Types of Aberrations

Removing Aberrations

Aberrations in Lenses

Test Your Knoedge

  E  TI E After working through this chapter, you should be able to: Dene a wavefront aberration Explain how aberrations can aect the resultant image Explain a Zernike polynomial

Understand how aberrometers can quantify aberrations Explain how adaptive optics systems can improve image resolution

Introduction

be blurring (less focused ig. .) or distorting (warped in shape somehow ig. .) the image. nd, slightly frustratingly, these aberrations exist in almost all optical systems (lenses, imaging systems, eyeballs). is can lead to issues such as blurry vision for people in the real world, or blurry images being taen of the bac of a patient’s eye. We will go into more detail on these aberrations and their eects throughout this chapter.

We’ve already discussed aberrations (imperfections in the image) in previous chapters when we’ve considered spherical aberrations and chromatic aberrations. is chapter will focus on aberrations more generally before explaining why they’re such a nuisance, and how to get rid of them.

Wavefront Aberrations n a perfect optical system that produces an image, all the light rays from the obect would be perfectly focused to form a complete, comprehensive and accurate representation of the obect (e.g. ig. .). or example, if we thin of our eyes as optical systems, the goal of the eye is to focus a clear image of the world onto the bac of our eyes so that we can see in perfect clarity. owever, as you’ll be familiar with your own eyes, you now that this isn’t the case. f we need glasses, then we’ll see blurry images without our correction, and even if we don’t need glasses, we might struggle to focus on things too close or too far away. ese are examples of imperfections in the image which are called wavefront aberrations (because these aberrations exist in the wavefront). n its most basic form, the term ‘aberration’ describes a failure of light rays to converge sensibly at a point of focus. is means that the rays will, in some way,

Types of Aberrations efore we go into detail of how aberrations can cause problems, let’s start by discussing the type of aberrations that can be induced. n the previous section we started to distinguish between blurring relative to distorting, but in actuality, we can brea it down a lot more by not only classifying the type of aberration but also classifying how impactful the aberration is. ne way to do this is to use Zernike polynomials, which turn the induced aberrations into a mathematical construct so we can tell how disruptive one aberration might be relative to another. n this case, each aberration can be assigned a value that is either positive or negative, and these values will predict alterations in the shape and uality of the image. ernie polynomials are expressed in the form Z nm , where the subscript n denes the order of the aberration, and the superscript m denes 149

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inica Appications

A

Object F’

‘Perfect’ optical system

B

Image

Object F’

Blur-inducing optical system Image

C

Object F’

Distortion-inducing optical system

Image

• Fig. 16.1 Diagram showing examples of how optical systems should focus light to form a clear image (A), and how they can induce blur (B) and distortion (C).

the angular frequency (how many times the wavefront pattern repeats itself every p radians). ig. . shows a view of the polynomials labelled and described. n optometric practice, the angular freuency aligns with the number of planes of the cornea that the aberration aects – for example, you can imagine that a ‘tilt’ aberration would only aect one plane (horiontal or vertical), so its angular freuency is  (2 or 1 depending on the direction). lso, typically, aberrations with a negative angular freuency are aligned in the vertical plane, whereas aberrations with a positive angular freuency are aligned in the horiontal plane. o provide a clinically relevant example, in chapter  we discussed that people can have astigmatism (a refractive error along a particular plane within the eye). n ig. . we can see that astigmatism is a second-order wavefront aberration with either a 1 or 2 angular freuency.

Aberrations in Lenses enses are used in optical instruments (and glasses) in order to focus light in a set way. is means that you could be forgiven for thining that all light is refracted through the lens eually, but in reality this isn’t the case. e optical centre of the lens will possess the clearest image, but as light moves away from the optical centre, more and more aberrations (imperfections)

will be induced (ig. .). is is particularly true for highpowered lenses where the lenses need to be thicer in order to produce the power reuired. f we wear glasses, we can demo this ourselves, otherwise we can use ig. . to help thin about it. enses need to have curved surfaces in order to focus light correctly, but with two curved surfaces, the shape of the lens will vary from the centre relative to the edges. n ig. . you can see that the edges of the convex and concave lenses begin to resemble prisms, which is a clear indicator that they will start to introduce prismatic eects. s light moves away from the optical centre of a glasses lens, the light will undergo prismatic changes, which will aect the uality of the image seen by the wearer. f you have a pair of glasses, try looing through the edges of the lens – it should be less clear than when you wear them normally, because it’s introducing these pesy aberrations.

Aberrations in the Human Eye f we now tae a moment to thin of some common-sense biology, we now that the goal of the eye is to refract light in such a way that incoming light focuses on the bac of the eye, and we also now that if a person has a particular refractive error, then there may be some blurring (e.g. defocus).

CHAPTER 16

Wavefront Aberrations and Adaptive ptics

Angular frequency(superscript) More negative

More positive

0

Z0

Order(subscript)

Piston ertical tilt

Z

1 Z 1 Horizontal tilt

-1 1

Defocus Oblique astigmatism

2

0

-2

Z 2 Astigmatism

Z2

Z2

Coma Trefoil

Z

-3 3

Z

-1 3

1

3 Z 3 Trefoil

Z3

Spherical Quadrafoil

Z

0

-2

-4 4

Secondary astigmatism Pentafoil

Z

-5 5

Z Secondary trefoil

-3 5

2

Z4

Z4

4 Z 4 Quadrafoil

Z4

Secondary astigmatism Z

-1 5

1

3

Z5

Secondary coma

Z5

5

Z5 Pentafoil

Secondary trefoil

• Fig. 16.2

llustration of the rst e orders of ernike polynomial aberrations. Black text describes the type of aberration (e.g. ‘defocus’) whilst green text identies the polynomial. rder increases linearly in a downward direction (e.g. ‘iston’ at the top has an order of , whereas ‘entafoil’ at the bottom has an order of ). Angular freuency increases outwards from the middle, with the left side being negatie and the right side being positie.

A

• Fig. 16.4 • Fig. 16.3

Diagram showing a lens on top of a grid of black circles. hen the circles are refracting through the edges of the lens, they ex perience greater amounts of distortion compared to the central region.

B

Diagram showing light (blue) refracting through the edges of a conex (A) and concae (B) lens. Black dashed lines illustrate that the edges of a lens become prism shaped, whereas blue dashed lines indicate that the image will be displaced upwards (conex, A) or down wards (concae, B) due to the prismatic effects at the lens edges.

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owever, we can hopefully also appreciate that it’s unliely that all human eyes would produce a perfect image, so it won’t be a surprise then to learn that the various refractive components of the eye (cornea, lens) can induce lower-order aberrations. or example, if the lens and cornea of the eye do not line up perfectly along the optical axis of the eye, it could induce coma or spherical aberrations in the image at the bac of the eye. is means that it’s fair to say most eyes will experience at least some lower-order aberrations in the image that they transduce into a signal for the brain. owever, we often don’t notice these aberrations because the human brain is very good at getting rid of low-level ‘noise’ in the incoming signal. is means the brain can compensate for lower-order aberrations so that we don’t notice them in our perceptual experience. igher-order aberrations, however, are more dicult for the brain to compensate for, so they can have a relatively large impact on the perceptual experience of the observer. ese types of aberrations can occur in dimly lit environments (e.g. dim – mesopic dar – scotopic) where the pupil needs to dilate (mae itself larger) in order to let in more light. s discussed in the previous section, the farther away from the central region the light is, the more liely it is to experience aberrations, and this is true in the cornea as well. s the pupil enlarges, more light can pass through the outer edges of the cornea and lens, which increases the amount of aberrations present in the image.

Measuring Aberrations in the Eye ome wavefront aberrations can be identied in clinical practice by performing a refraction to determine which lenses help the patient see more clearly however, higherorder aberrations are more dicult to identify, so for these we need to utilise a device called an aberrometer (aberr-, aberration -ometer, measurement). istorically, one of the rst aberrometers was devised by hristoph cheiner in , when he realised that a twopinhole occluder could identify if a person was able to see clearly or not. is device is called a Scheiner disc and wors under the principles that () light can only travel in straight lines, and () the human eye should focus distant light onto the bac of the eye. n theory, then, if you put two small, adacent pinholes in an occluder and as the

A

patient to loo at a distant light, the light should be approaching the eye with ero vergence (parallel rays see chapter  for a refresher on vergence). is means the pinholes should only let through two small spots of light, which, if no refractive error is present, should focus on the bac of the eye to form an image of a single spot (ig. .). erefore, an emmetropic patient should report seeing one spot of light when looing through a cheiner disc. owever, if the patient’s eye is inducing aberrations (e.g. defocus, spherical), then the light will focus incorrectly and will appear as two separate spots of light (ig. .). emember, though, these two spots should still be in focus because pinholes reduce the aberrations experienced by the eye, so the ey factor for determining a refractive error would be if the spots appeared twice rather than being blurry. ence, it was relatively easy to identify patients who could see clearly and those who couldn’t. t also means that it was possible to estimate how large the aberrations were, because if the two spots were very far apart, then more refractive error is present. owever, an estimation of how far away a patient describes seeing two spots is hardly the rigorous, scientic approach we lie to use these days, so you’ll be pleased to now that we now have more accurate ways of uantifying (mathematically determining) aberrations. ssentially, then, if we had a patient who reported seeing two spots of light, we could potentially alter the incoming light until it formed a single spot of light. e amount of adustment needed to correct the aberration would indicate the amount of error in the eye. is measurement of displacement of light is essentially how a modern-day wavefront aberrometer wors. ese devices are comprised of many smaller lenses called lenslets, all possessing identical focal lengths (meaning they should individually focus light in the same way). ese lenslets are designed to focus incoming light onto a sensor that can measure how much light deviates from what would be expected if the incoming light was an aberration-free wavefront. n ig. . you can see that a perfect wavefront would cause the lenslets to focus light on all the intersections of the sensor – meaning they are all evenly distributed and have undergone no aberrations. owever, if aberrations are present (ig. .), then the light will be focused inappropriately on the sensor, and the sensor can determine the degree (and shape) of the aberration based on these data.

B

•Fig. 16.5

In focus (one spot)

Out of focus (two spots)

Diagram showing how a cheiner disc would work in an eye with no refractie error (A) which would experience a single spot of light, relatie to an eye with refractie error (B) which would experience two spots of light.

CHAPTER 16

Wavefront Aberrations and Adaptive ptics

Side view of aberrometer Sensor (front view)

Sensor (side view)

Incoming light

Lenslets

Perfect wavefront

Subdivided wavefront

• Fig. 16.6

Diagram of aberrometer focusing light through the lenslets onto the sensor. ncoming light is free of aberra tions, so the resulting image is free of distortions and blur.

Side view of aberrometer Sensor (front view)

Sensor (side view)

Incoming light

Lenslets

Aberrated wavefront

Subdivided wavefront

• Fig. 16.7

Diagram of aberrometer focusing light through the lenslets onto the sensor. ncoming light contains aberra tions, so the resulting image is distorted.

Removing Aberrations p to this point in the chapter, we’ve considered what an aberration is and what impact it can have on vision, and we’ve discussed how to uantify the degree of aberration present in an image. owever, the most useful reason for measuring aberrations is to remove them from the image. is can be achieved with refractive errors in the eye through prescribing corrective lenses or contact lenses, but what about aberrations present in medical images n clinical terms, the problem with aberrations in the eye is that they distort images that we can tae of the bac of the eye itself. ou can imagine that if light focuses on the retina with some lower-order aberrations present and we wanted to tae a photo of the bac of the eye, the resulting image would also contain these lower-order aberrations because the light from our camera (or microscope) will experience the aberrations of the eye as it passes into the eye and as it re¡ects bac out again. is is troubling because it limits the resolution (clarity) of the images we can tae of the bac of the eye, which means clinicians might miss small changes in health. t is therefore in our interest to be able to () detect the aberrations in the rst place, () uantify the aberrations, () localise the aberrations and () be able to rapidly adust adapt the optical system to compensate for any aberrations

in the image. anfully imaging systems can achieve this very successfully through the principles of adaptive optics. e term ‘adaptive optics’ refers to any optical system (or imaging system) that can adapt to compensate for any aberrations introduced between the obect (e.g. the bac of the eye) and the image. e goal is to remove the aberrations to improve the overall uality and resolution of the nal image. When imaging the human eye, adaptive optics systems compensate for the eye’s aberrations by utilising a rapidly deformable mirror (. ) which is constantly adusted by monitoring the incoming light and changing (deforming) the shape of the mirror in order to undo the aberrations (ig. .). Incoming aberrated wavefronts

Outgoing corrected wavefronts

Adaptive/deformable mirror

• Fig. 16.8

Diagram showing how a deformable mirror could change its shape to correct (and remoe) aberrations in an image.

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Light from back of eye Beam splitter Light Source

Aberrated wavefront

Adaptive mirror

Corrected wavefront

Beam splitter Control system

High resolution retinal image

Wavefront sensor

• Fig. 16.9

Diagram showing how an adaptie optics imaging system could photograph the back of the eye. ee text for details.

ltimately this means that adaptive optics systems need to have a detection system to measure the aberrations and an adaptive optical element (deformable mirror) to correct the aberrated incoming wavefront using an opposite cancelling distortion. or example, in ig. ., a spot of light from the light source is being focused on the bac of the eye. is is then re¡ected o the bac of the eye and leaves out of the front of the eye to pass through the beam splitter, but at this point it now contains aberrations from the eye. e light is then re¡ected o the deformable mirror (though

initially the mirror won’t be able to cancel out the aberrations as it doesn’t yet now what to do) and re¡ected o a beam splitter towards a wavefront sensor which will uantify the aberrations in the image. e control system then tells the deformable mirror how to compensate for the aberrations, and the deformable mirror wors its magic to produce a corrected wavefront and a high-resolution nal image. e tric is that this system is constantly updating and adapting so it can even account for small eye movements and changes in accommodation in the eye itself.

Test Your Knowledge ry the uestions below to see if you need to review any sections again. ll answers are available in the bac of the boo. TYK What is an aberration TYK Which part of a lens induces the greatest amount of aberration

TYK n terms of ernie polynomials, what ‘order’ of aberration is defocus TYK xplain how a cheiner disc can identify the presence of a refractive error. TYK xplain how adaptive optics systems can tae high-resolution images.

References . orn , Wolf , hatia , et al. Principles of Optics: Electromagnetic eory of Propagation, Interference and Diraction of Light. th ed.  ambridge niversity ress .

. orris W. e cheiner optometer. lin Ep Optom. ()–.

17 Optical Coerence Tomorap C  A  T  R O  TL I  Introduction Some Light Revision The Interferometer

Swept-Source OCT (SS-OCT) Conventional Versus En Face OCT Aniorap (OCT-A)

What Is OCT?

Clinical Applications

Interferometry and OCT Fibre-Based Time-Domain OCT (TD-OCT) Fibre-Based Fourier-Domain OCT (FD-OCT)

Test Your noledge

O    C T I S After working through this chapter, you should be able to: Eplain wat intererence is and ow it can be used to measure distances Eplain wat OCT stands or Eplain (in simple terms) ow time-domain OCT wors

Eplain (in simple terms) ow Fourier-domain OCT wors Eplain (in simple terms) ow swept-source OCT wors Describe clinical applications o OCT

Introduction

destructive interference a decrease in amplitude. ig. . shows a reminder of this. n chapter  we also discussed that the path dierence travelled y two eams of light can help us measure distances. is reuires two identical waves which can e from a single light source split into two to travel separate distances efore meeting again at a detector which can uantify the resultant amplitude and determine the relative path dierence. is has een discussed efore in the context of the Michelson interferometer, which we will review in the next section.

is chapter will focus on the exciting world of optical coherence tomography (OCT) imaging, which is currently considered the ‘gold standard’ of imaging the eye. However, OC relies on the wave theory of light, so we need to start with some light revision as always, pun intended.

Some Light Revision n chapter  we discussed that light can e thought of as a wave, and that multiple waves can superimpose on top of one another y arriving at a point at the same time and comine to produce a resultant wave. is resultant wave may have a smaller or larger amplitude rightness or intensity depending on how the waves interact. ememer, as well, that if the waves are coherent meaning they possess the same wavelength and freuency, then they will produce either constructive or destructive interference. e type and extent of the interference will depend on whether the waves arrive in-phase path dierence a whole multiple of the wavelength to produce constructive interference an increase in amplitude or if they arrive out-ofphase path dierence half a wavelength out to produce

The Interferometer e simplest interferometer to start with is the ichelson interferometer, which utilises light from a coherent light source and splits it into two ig. .. One of the paths of light travels to a moveale mirror a nown distance away, whilst the other path of light travels to something we want to measure the distance of. n the case of the ichelson interferometer, the second path of light travels to a xed mirror. s discussed in chapter , the idea of this setup is that y moving the moveale mirror set amounts, the interference pattern at the detector will cycle through constructive 155

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Clinical Applications

layers at the ac of the eye, ut it can also e used to image the anterior eye as well. f we use the example of imaging the retinal layers, OC utilises the principles of interferometry to measure how light reects from each of the individual layers in order to determine the distance they are away from the detector ig. .. is distance information can then e transformed into a lac-and-white image of the layers, which can help to monitor health and disease.



Path difference N completely in phase constructive interference bright light!

Interferometry and OCT s discussed, OC systems utilise these principles of interferometry in order to image the layers of the eye, ut there are several dierent types of OC system.

ireased Timeomain OCT TOCT

Path difference N+0.5 completely out of phase destructive interference no light

•Fig. 17.1

Diagram showing difference between constructive interference (top) and destructive interference (bottom). The resultant amplitudes are shown in black on the right-hand side.

Fixed mirror (or something we want to measure)

Coherent light source Moveable mirror Beam splitter

e rst type of OC is called bre-based time-domain OCT or TD-OCT for short. is method of OC utilises a moveable mirror ust lie in the ichelson interferometer which means it measures the interference patterns over time as the mirror is moved hence why it’s called a time-domain. ey usually use a low-coherence nearinfrared light source which is produced using a superluminescent diode. ig. . shows roughly how the system wors you’ll notice it’s very similar to that of the ichelson interferometer, ut with a lens system for helping to image side to side laterally across the ac of the eye. One of the paths of light is incident upon the moveale mirror reference beam, whilst the other is incident upon the eye measurement beam, and the measurement eam is reected, or acscattered, from the ac of the eye with dierent delay times which are dependent on the optical properties of the tissue and the distance away from the light source. e software within the system can then interpret the interference fringe or reectance proles as they pass through the detector in order to determine the depth of the tissue

Detector

•Fig. 17.2

Diagram of a Michelson interferometer.

Incoming light

interference increase in amplitude and destructive interference decrease in amplitude. or example, if we move the mirror y a distance that euates to . of the wavelength of the light, then the path dierence will euate to . of the wavelength as the light will experience 1. approaching the mirror and 1. after reecting, totalling 1.. is means that, providing we now the wavelength of the light, y measuring the interference at the detector we can determine the distance the light has travelled. is is, in relatively simple terms, how OC wors.

Reflected light

What Is OCT? OC is a method of structurally imaging the individual layers of the eye. ypically this is used to image the retinal

• Fig.

17.3 Illustration of an optical coherence tomography (T) image of retinal layers (shown in shades of grey) with representative light reecting off the layers back to the detector (shown in blue).

CHAPTER 17

This mirror controls depth of image

Optical Coherence Tomography

Mirror (moveable) d Lenses

Low coherence light source

Part of the eye Beam splitter

Detector

This system controls the lateral position of the image

•Fig. 17.4

Diagram of bre-based time-domain optical coherence tomography (TD-T) system imaging part of the eye (shown in purple on the right). otice that the system uses a moveable mirror to image in depth.

and produce the nice lac-and-white image of conventional OC as shown in ig. .. mportantly, ecause these images reveal cross-sectional views of the retina, they are imaging the -plane as opposed to the x-y, transverse plane ig. .. However, of all the types of OC, this system is slow and of the lowest resolution lowest-uality images.

x-y plane

z plane

• Fig. 17.5

Illustration showing the difference between imaging in the -y (face-on) plane and the  (depth) plane.

ireased ourieromain OCT OCT e second type of OC is called bre-based ourierdomain OCT or D-OCT for short also sometimes called spectral domain or D-OCT. n this case, instead of recording interference and intensity at dierent locations of the reference mirror, the interference is recorded as a function of wavelengths or freuencies of light. is is achieved y using a roadand light source containing multiple wavelengths centred on ∼ nm and then inserting a diraction grating or spectrometer in the system ust efore the detector. e detector then loos at the interference proles of each of the wavelengthsfreuencies hence the name Fourier-domain which is called spectral interference ig. .. e advantage of this techniue is that the depth information from the eye can e acuired simultaneously without moving the mirror over time, which allows it to tae images much faster than -OC therefore, we can descrie it as having a faster acuisition time. ast This system controls the lateral position of the image

Mirror (stationary)

Lenses Broadband light source

Part of the eye Beam splitter Diffraction grating

Detector

• Fig. 17.6 Diagram of a bre-based ourier-domain optical coherence tomography (D-T) system imaging part of the eye (shown in purple on the right). otice that the system uses a diffraction grating to measure spectral interference.

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acuisition times are useful in clinical practice ecause patients usually don’t lie to sit for long periods of time, so if high-resolution images can e taen in shorter amounts of time, then this is very advantageous. -OC can also acuire images at a higher spatial resolution than OC ox ..

•Box 17.1

SeptSource OCT SSOCT wept-source OC or -OC for short is a type of OC. ith -OC the freuency-wavelength-dependent intensity is not acuired simultaneously using a roadand light source, ut instead the wavelength of the light source is

Spatial Resolution

The concept of spatial resolution is important because this denes the level of structural detail that an imaging system is capable of capturing. In simple terms this can be likened to the number of megapiels in a digital camera. This means a system with low spatial resolution would blur the image a lot more than a system capable of producing a higher spatial resolution. ig. . shows an obect (the eye) which is going to be imaged by two systems. The one on the left has a lower spatial resolution (shown by the larger sie of the imaging ‘units’) which means the resultant image

will be almost unrecogniable and a lot of details will be missing. owever the imaging system on the right has a higher spatial resolution (shown by the smaller sie of the imaging ‘units’) which means it becomes recognisable as an eye again and you can even start to make out the shape and the position of the pupil. The potential clinical conseuence of using a system with a low spatial resolution is that a clinician may be unable to detect small changes or subtle indications of pathology so in general terms the higher the resolution (and smaller the imaging ‘units’) the better.

Object

Imaging system 1 Lower spatial resolution (unit size ~4mm)

• Fig. B17.1

Imaging system 2 Higher spatial resolution (unit size ~1.4mm)

 diagram showing two hypothetical imaging systems ( and  left and right respectively) photographing an eye (the obect). ach pink bo represents an imaging ‘unit’ of the system. Imaging system  possesses a low spatial-resolution system (fewer larger ‘units’ indicated by the pink boes) which produces a blurry unresolvable image whereas imaging system  possesses a higher spatialresolution system (many smaller ‘units’ indicated by the pink boes) which produces an image that is beginning to resemble the original.

CHAPTER 17

Optical Coherence Tomography

Mirror (stationary)

Lenses

Tunable laser light source

Part of the eye Beam splitter

Detector

This system controls the lateral position of the image

•Fig. 17.7

Diagram of -T system imaging part of the eye (shown in purple on the right). otice that the system uses a tunable laser as the light source.

tuned to sweep hence swept . . . through a narrow range of wavelengths seuentially, as shown in ig. .. Crucially, however, this range will e centred on approximately  nm, indicating that -OC utilises a longer wavelength of light relative to -OC. e advantage of using longer wavelengths is that the tissue penetration is greater, meaning it is possile to image deeper into the eye even down to the level of the choroid – the vascular layer ehind the retina, ut the pitfalls of using a longer wavelength light source is that the spatial resolution can e reduced. e good news, however, is that it is possile to slightly improve the spatial resolution through enhancements uilt into the imaging system’s software. ltimately, the advantages of -OC relative to -OC and -OC are that it maintains a high resolution whilst also possessing the fastest acuisition time, and it allows imaging of deeper tissue structures.

Conventional ersus n ace Conventional OC scans will typically image tissue structures in depth, meaning that data from these scans will show cross-sectional views of the part of the eye eing imaged, which is very useful in a clinical environment, as optometrists and ophthalmologists can scroll through sections of the retina to assess health and monitor changes over time. However, it is also now possile to use OC systems to produce en face face-up scans also called C-scans. ese scans can uild on the OC system’s aility to image in depth y adding confocal analysis of the data. is produces a high-resolution image of the retina in the x-y transverse plane, meaning they can reveal at, face-on images of the eye at any specied depth, for example, at the outer retinal level or the choroidal level. Overall, these additional data can provide further information of sutle, very smallscale changes in retinal tissue, which is helping researchers learn more aout the anatomical changes associated with disease.

OCT Angiography OCTA One nal type of OC imaging that we will consider here is OCT angiography (OCT-) which utilises en face scanning methods. e word angiography means ‘vessel’ angio- measurement -graphy, highlighting that this type of OC serves as a noninvasive method of imaging lood vessels in the eye. ore specically, it wors y comparing the reectance of the light from red lood cells that are moving within the retinal and choroidal vessels therefore, it is imaging changes in the reectance prole in the same location over time. n very simple terms, this wors ecause moving red lood cells will induce more of a change in the reectance prole than stationary layers of tissue, and it’s possile for the software to analyse this dierence etween moving and static tissue to estimate where the lood vessels are and how fast the lood is moving through them.

Clinical Applications OC is a system that allows us to see high-resolution images of individual layers in structures of the eye – which is advantageous to standard fundus photography ecause it has higher resolution and has the aility to image in depth. erefore, it’s incredily useful as a techniue for imaging the health of the eye ecause clinicians can clearly see if any pathology is present in deep layers of the tissue and ecause it’s so high resolution it can also give an incredily accurate representation of any anatomical changes associated with pathology over time, at the level of a few micrometres. t also allows precise measurements of distance e.g. how thic the retinal nerve re layer is, or how long the axial depth of the eyeall is, and it can help to measure the shape of the cornea e.g. in eratoconus or to see how well a contact lens is sitting on the front surface of the eye.

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Clinical Applications

Test Your Knowledge ry the uestions elow to see if you need to review any sections again. ll answers are availale in the ac of the oo. T hat does OC stand for T xplain how far the mirror in a ichelson interferometer would need to move to produce a path dierence of a whole wavelength.

T xplain how a -OC system wors. T ame one dierence etween a -OC and an -OC system. T ame one clinical application of OC.

SECTION

4

erments to o at Home 18. Create Your Own Camera Obscura, 163

21. easure te See o t, 17 22. Create a ‘Cornea’, 179

19. Create a Blue Sky at Home, 167 23. tcen n lm ntererence, 183 2. Create a rsm, 171

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18 Create Your Own Camera Obscura C H A P T E R O U T L INE Introduction The Experiment ument eure

eto esults Test Your Knowlede

O   E C T I E After working through this chapter, you should be able to: lan ow a nole rouces an mae Create your own camera obscura

Introduction As we discussed very briey in chapter 13, a ‘camera obscura’ is an imaging device that utilises the theory behind pinhole photography. When light emanating from an obect shines through a pinhole, it produces an upsidedown image. is is because light rays from the tip of the obect e.g. the top of the mug in ig. 1.1 travel in a straight line through the pinhole and end up near the oor, whilst light rays from the base of the obect travel in a straight line through the pinhole and end up near the ceiling. When these rays form a focus, it necessarily produces an inverted image.

The Experiment n order to see how this wors, it’s best to mae our own. o do this, we’ll need a pinhole aperture and something to proect the image onto ig. 1..

Euipment Reuired • A pencil • A drawing pinneedlesomething with which to mae a very small hole • wo pieces of Asied blac card if you don’t have blac card, a cereal bo might do • ape of some ind • cissors • racing paper or euivalently translucent material that, when held up to a lamp, is diusely illuminated but does

not show a clear image of the lamp through the material e.g. a translucent sandwich bag or the bag from inside a cereal bo – if these are too transparent, you can double up the layers

ethod 1. ut an  cm strip o the rst piece of blac card as shown in ig. 1.3. et the  cm strip to one side for now. . ae the remaining part of the rst piece of blac card and roll it up to mae a tube tube 1 in ig. 1.. ape to secure. e diameter of the tube should be roughly  cm, so overlapping is allowed. 3. oll up the second uncut piece of blac card to t inside the rst tube. is needs to t very snugly, so  recommend rolling it up small, placing it inside the rst tube, and then letting it epand inside. arefully remove this narrower tube and secure with tape tube  in ig. 1.. ou should now have two tubes of slightly dierent length and diameter with the wider tube being shorter in length. . ow secure the translucent element e.g. tracing paper over the edge of tube  as shown in ig. 1.. is is the screen on which the image will appear, so mae sure it’s tight and at. . ollect the remining  cm strip of blac card and tube 1 the shorter tube. raw around the tube to create a circular template on the card, then draw a slightly bigger circle and connect the two circles with radial lines as shown in ig. 1.. 163

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Experiments to o t Home

Pinhole aperture Object

Image

•Fig. 18.1

Diagram showing path of light rays from a mug, producing an upside-down image in a pinhole camera. The image is also ipped left to right, so in the image (on the right), the text will be facing the other way (which is why it’s missing).

•Fig. 18.2

•Fig. 18.3

Diagram of step  of method – cut an  cm strip off one of the pieces of card.

Tube 1 (shorter, wider)

•Fig. 18.4

llustration of reuired materials.

Tube 2 (longer, fits snugly inside tube 1)

There should be two tubes – a short, wide one (tube ) and a narrow, long one (tube ).

Crete Your Own Cmer Oscur

CHAPTER 18

Tube 2

•Fig. 18.5 The translucent material needs to be taped nice and tight on the end of the narrow, long tube (tube ).

Tube 1

g

Sl i

ly arger cir

l

•Fig. 18.6

se the short, wide tube (tube ) to draw a circle on the spare  cm strip of card (left). Then draw a larger circle around it and connect the two with radial lines (right).

Tube 1

•Fig. 18.7

hen cut and pierced, the ‘toothy circle’ (left) should then be attached to one end of the short, wide tube (tube ).

. ut the larger circle out of the card, and then cut the radial lines to mae ‘teeth’ around the inside circle. . se your drawing pin or euivalent to mae a pinhole in the centre of the inner circle you should now have something lie that shown in ig. 1. . Attach the toothy circle to tube 1 using tape see ig. 1.. . lace the ‘screen’ end of tube  inside tube 1 the screen needs to be inside the tube to mae sure the image isn’t aected by other light sources. ou’ve now made your camera obscura ig. 1. 1. oint the pinhole end towards something well lit this wors best outdoors on a sunny day to see an upside down image form on the screen. ou can adust the distance of the screen from the pinhole by pushing the inner tube closer and farther away. is should change the sie of the image seen on the screen which should be inside the tube.

e1

b Tu

e2

b Tu

Results opefully your camera obscura was able to produce a clear, upsidedown image of the world e.g. ig. 1.. e reason

•Fig. 18.8 The narrow, long tube (tube ) goes screen-side rst inside the short, wide tube (tube ).

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TABLE Trouble-Shooting the Camera Obura 18.1

Reported Issue

Potential Problem

o image seen (dar)

. inhole too small . bect not bright enough

. ae pinhole larger

o image seen (bright)

. inhole too large

. emae a smaller pinhole and stic oer the original . Temporarily remoe the screen tube and loo at a light source through it – it should produce diffuse illumination of the screen without producing a isible image of the light

. creen too opaue translucent

•Fig. 18.9

 photograph of the author’s cat looing ery thoughtful at the window. The photograph was taen through the author’s camera obscura.

this wors is because the pinhole on the front aects how light can travel inside the camera, which produces the ipped image on the screen inside. f you nd that it isn’t woring very well, you can troubleshoot the issue using able 1.1 as a guide. As a rough guide, however, mae sure to point the camera at something very well lit e.g. a tree in the sunshine and maybe eperiment with the sie of the pinhole – if it’s too small then it won’t let enough light in, but if it’s too big then the image will be very blurry. As with all eperiments, if it didn’t wor out even after troubleshooting then you should thin about what you could do to improve the method net time.

mage seen but not upside down

. inhole too large . ooing through camera the wrong-way round

Potential Solution

. hine a torch on an obect and try again

. emae a smaller pinhole and stic oer the original . ae sure you’re looing through the empty end of the screen tube, with the pinhole closest to the obect

Test Your Knowledge ry the uestions below to see if you need to review any sections again. All answers are available in the bac of the boo. TYK.18.1 plain how a pinhole produces an upsidedown image.

TYK.18.2 plain why moving the screen away from the pinhole produces a larger image.

19 Create a Blue Sky at Home C H A P T E R O U T L INE Introduction The Experiment Equipment Required

Method Results Test Your Knowlede

O   E C T I E After working through this chapter, you should be able to: Create a demonstration of polarisation by scattering Explain how polarisation by scattering works

Introduction As a brief recap, in chapter 14 we discussed how light can be made to only have one (or at least very few) orientations of electric eld if it undergoes a process of polarisation. is is important because light from the sun is unpolarised, mean ing that the light is vibrating in all orientations, and there fore the orientation of the electric eld varies randomly over time. ne way in which this natural sunlight can become polarised is through polarisation by scattering in which the light from the sun will be scattered (and polarised) when it comes into contact with molecules in the atmosphere. n particular, shorter wavelengths (within the sunlight) are scat tered more easily than the longer wavelengths, which means that as light travels through the atmosphere, it scatters shorter (blue) wavelengths into the atmosphere (maing the sy loo blue), which subseuently maes the light from the sun appear to be shifted slightly towards the longer wavelengths (maing the sun loo more yellow). n this eperiment we are going to investigate this principle by maing our very own atmosphere in a glass and using a torch as our ‘sun’.

The Experiment or this eperiment we are going to attempt to prove that when light travels through a busy medium (lie the atmo sphere), it will scatter shorter wavelengths more than longer wavelengths. is will be achieved by creating an atmosphere (mily water in a glass) and seeing what the sun (a torch light) loos lie as it passes through the atmosphere.

efore starting the eperiment, please mae sure you have all the euipment you need (and a mobile device to record your results).

Euipment Reuired • A clear glass (a reasonable sie glass, no pattern if possible) • A whitelight torch (mobile phone lights wor , but torches wor better) • ome cold tap water • A small amount of mil • omething to stir the miture with • ptional (but advised) a towel or itchen roll to mop up watermil spillages

ethod 1. ill the glass with cold water. . hine the torch from behind the glass, directly towards yourself. . ae a note of the colour of the torch light – hopefully it loos uite white (unaected by travelling through the water). 4. ae a note of the colour of the water – hopefully it loos uite clear. . ae a note of the colour of the mil (before we put it into the water) – hopefully it loos white. . ow add a very tiny amount of mil into the water (if you add too much then the eperiment won’t wor, so it’s better to add too little than too much). 167

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. . . 1. 11.

Experiments To o t Home

ou may need to stir the miture to mae it wor appropriately. hine the torch from behind the glass, directly towards yourself. ae a note of the colour of the torch light – hopefully it loos more yellow. eep adding mil until the torch light begins to become convincingly yellow in appearance. ae a note of the colour of the mily water – hopefully it loos to be bluer than without the light. ow eep the torch still but view the mily water from above the glass tae notice of what’s happened to the colour – does it loo bluer with the torch than without

Results opefully you could see that the torch light magically (or scientically and predictably) changed colour to become

more yelloworange when the mil was added to the water, and that the mily water became blue. e reason this wors is because the mil in the water scatters the light from the torch through polarisation by scattering (ig. 1.1). As we learned in chapter 14, shorter wavelength (blue) light scatters more easily than longer wavelength light, so as a result of this the blue light from the torch ends up diusing through the mil, maing it loo bluer, and the light source will loo more yellow (due to the missing blue). owever, if you nd that the eperiment isn’t woring very well, you can troubleshoot the issue using able 1.1 as a guide. As with all eperiments, if it didn’t wor out (even after troubleshooting), then you should thin about what you could do to improve the method net time. ou can also watch me do a demonstration of the eperiment on the associated lsevier website (ig. 1.).

Torch (behind glass)

Front view

Water

Milky water

Even milkier water

Milky water

Even milkier water

Glass

Torch

Top view

Water

Glass • Fig. 19.1

Illustration of the effect viewed from the front of the glass (top) and the top of the glass (bottom). As the water becomes milkier (right), the light from the torch will experience polarisation by scattering, which will make the torch light seem more yellow (top) and the milky water look bluer (bottom).

CHAPTER 19

Crete  lue  t Home

TABLE Troubleshooting the Polarisation Experiment 19.1

Reported Issue

Potential Problem

Potential Solution

ight source does not turn orange (but milk turns blue)

. ight source producing diffuse light . ot enough milk added

. Angle the light source slightly downwards towards the table – this will mean the light that reaches you is less diffuse (alternatively, try a dif ferent light source) . Add more milk and try again

ight source disappears when milk is added

. Too much milk added . ight source too low intensity

. eplace the water and try again – this time add the milk a tiny drop at a time . Try to nd a brighter light source

• Fig. 19.2

Two stills taken from the associated video version of the experiment being completed by the author. The left image shows the bright white light being transmitted through the clear water in the glass. The right image shows the torchlight becoming more yelloworange once the milk is introduced. It is also clear that the milk has taken on a blueish hue.

Test Your Knowledge ry the uestions below to see if you need to review any sections again. All answers are available in the bac of the boo. TYK.19.1 plain why the torch loos more yellow as we add more mil.

TYK.19.2 plain why using a red light source wouldn’t wor.

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20 Create a Prism C H A P T ER O U TL INE Introduction The Experiment Equipment Required

Method Results Test Your Knowlede

O  E C TI E After working through this chapter, you should be able to: Explain why light can disperse through water if a mirror is introduced Explain dispersion

Introduction Looking back over the book so far, in chapter 1 we learned that white light (e.g. sunlight) contains all wavelengths of visible light, and in chapter 8 we discussed that these wavelengths of light can be dispersed to be independently visible (e.g. in the case of a rainbow where white light disperses through a raindrop, or when light travels through a prism). Dispersion is soething that we’ll have eperienced ourselves in real life if we’ve ever seen light pass through a crystal ornaent in a window – it akes a rainbow pattern appear. is rainbow pattern is siply the white (incident) light fro the sun dispersing (splitting out into its constituent wavelengths), which akes it appear to be lots of colours instead of ust white. is phenoenon occurs because the wavelength of the light is directly related to how uch refraction it undergoes as it travels fro one aterial to another. ere is an old adage ‘blue bends best’ which is supposed to help us reeber that shorter wavelength light (e.g. blue) will refract to a greater degree than longer wavelength light (e.g. red) when dispersion takes place. s any science teacher will probably tell you, the best way to understand this process is to get your hands on a pris and see the eect rst-hand. nfortunately, optical priss aren’t that coon around the hoe, so instead ’d like to encourage you to build your own pris. or this eperient we are going to create our own pris at hoe in order to split white light into the colours of the rainbow – which will also allow us to see rst-hand how

changing the apical angle will aect the deviation (and the aount of dispersion).

The Experiment ur ‘at-hoe’ pris eperient works best (and is ost ipressive) when constructed outside on a sunny day, but it can also work very eectively if you have a torch that produces white light. e only thing to note is that the torch light ight ake a slightly less-convincing rainbow, depending on its particular spectru, so please bear this in ind when copleting the eperient outlined here. efore starting the eperient, please ake sure you have all the euipent you need (and a obile device to record your results).

Euipment Reuired •  tray or a bowl that can be lled with water •  plane (at) irror that can be placed in the traybowl • oething to see the dispersed light on (e.g. piece of paper or a wall) • Optional: something to secure the mirror in place, for example, a stone to rest it on or some tape (just in case it makes setting up the experiment easier)

ethod 1. Fill the tray/bowl with water – ake sure there is enough water that the irror would be at least partially 171

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.

. . .

Experiments to o t Home

suberged if placed in the traybowl at an angle. ’ve found that this works best with at least  inches of water, so please bear this in ind when selecting your tray. Place the mirror into the tray/ bowl – it ust be at an obliue angle (e.g. °) and ust be at least partially suberged. is is ade easiest by taping the edge of the irror to the edge of the tray or by resting it on soething (or if you have soeone helping you, one of you can hold the irror in place). Place the tray/bowl in a location where the sun will shine directly onto the water, or angle the torch so that it is shining into the water, towards the mirror. Hold the paper parallel to the tray/bowl above the water and look to see if you’ve produced dispersion (see igs. .1 and . for a deonstration of this). ow change the angle of the irror (or increase the aount of water) and note what happens to the rainbow – does it change, or does it stay the sae

Results is works because as the light is refracted through the water, then reected at the irror, and then refracted as it leaves the water again, it behaves as raindrops do in the sky to produce a rainbow ll the wavelengths that ake up white light will refract at dierent angles, so by refracting the twice, we are refracting the to such an etent that they becoe separated – this is dispersion. owever, if you nd that it isn’t working very well, you can troubleshoot the issue using able .1 as a guide. s with all eperients, if it didn’t work out (even after troubleshooting), then you should think about what you could do to iprove the ethod net tie. ou can also watch e do a deonstration of the eperient on the associated lsevier website.

Paper

Sunlight

r

Mirro

Water

Tray/bowl

•Fig. 20.1

An illustration of the setup for this experiment. Please note that the sun can be replaced with a torch if necessary.

•Fig. 20.2

As the white light from the sun (or torch) refracts through the water and reects back out again, the wavelengths that make up white light are dispersed to form a rainbow.

CHAPTER 20

Crete  Prism

TABLE Troubleshooting the Homemade Prism 20.1

Reported Issue

Potential Problem

Potential Solution

o rainbow visible

. irror not angled properly . ight source not angled properly . ater not deep enough

. ry altering the angle of the mirror as shown in the video demo on the lsevier website . ry altering the angle of the light source as shown in the video demo on the lsevier website . Add more water (you may need a deeper bowl)

Test Your Knowledge ry the uestions below to see if you need to review any sections again. ll answers are available in the back of the book. .. plain why the white light produces a rainbow when it passes through the setup described in this chapter.

.. plain what you think would happen if you changed the angle of the light approaching our hoeade priss.

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21 Measure the Speed of Light C H A P T E R O U T L INE Introduction The Experiment Notes of Caution Equipment Required

Method Results Test Your Knowlede

O   E C T I E After working through this chapter, you should be able to: Explain how microwaves heat food using interference Explain how we can use chocolate to measure the speed of light

Introduction In chapter 1 we briey mentioned dierent types of electromagnetic radiation, one of which involved a type of energy you can nd in most kitchens – microwaves s a uick recap, microwaves are a form of electromagnetic energy comprising wavelengths that fall between infrared and radio waves, meaning that in general terms their wavelengths are relatively long, and they eert relatively low energy compared to other types of energy eg  ig 11 ow, it’s probably no surprise to hear that one of the most easy-to-understand practical uses of microwave radiation is for the generation of heat thermal energy in a microwave oven is means that every time you convert an item of food from a chilled substance into its molten lava counterpart in the microwave, you’re technically doing a scientic eperiment that utilises interference properties of electromagnetic radiation see chapter 1 is is possible because the microwave energy is absorbed by fats, sugars, proteins and water, causing them to heat up e microwaves also interfere with each other causing the formation of hot spots in the food in locations where constructive interference takes place is can be likened to the ‘bright maima’ we discussed in chapter 1, but instead of bright light, it produces hot spots If your microwave contains a turntable, then the turntable serves to dissipate these hot spots, enabling food to be cooked more evenly, and if your microwave doesn’t contain a turntable, then instead it will contain a built-in, rotating motor which serves the same purpose

e hot spots generated by the constructive interference are related to the wavelength of the microwaves, as anywhere two peaks or two troughs align, there will be maimum heat produced ig 1 is means that the distance between the hot spots will euate to half of the wavelength

The Experiment In order to nd the wavelength of the microwaves in our microwave ovens, we purposely want to create hot spots within the oven – thereby cooking something incredibly unevenly    is means we will need to remove the turntable from the microwave before we begin lease note if your microwave does not contain a turntable or if the turntable can’t be removed, then unfortunately you won’t be able to do this eperiment unless you borrow someone else’s microwave

Notes o Cution • icrowaves produce heat e careful not to burn yourself, particularly as the turntable used to distribute the heat is not being used • o not place metal obects in the microwave oven icrowaves reect o metal and will damage the internal workings of the microwave oven • o not use the microwave without food inside if there is nowhere for the microwaves to go, they will damage the microwave oven 175

176 S EC TI ON 4

Experiments to o t Home

Shorter wavelength higher frequency higher energy

Cosmic rays

Longer wavelength lower frequency lower energy

Visible

X-rays

UV

Infrared

Microwaves

adiowaves

•

Fig. 21.1 Diagram of the electromagnetic spectrum () showing the relationship between wavelength, freuency and energy.

Hot spots

Temperature profile Microwave

 Once visible, measure the distance between the two hot spots with a ruler onvert this measurement into metres and multiply by  to get the wavelength   alculate the speed of light using the euation below speed of light 5 wavelength × freuency is is often written as c 5 

Microwave wavelength, 

• Fig. 21.2

Diagram showing the amplitude prole and wavelength of a microwave (black line). As the amplitude increases, the energy also increases, meaning that both the peak and the trough of the microwave prole will produce the highest amount of intensity (heat). This means that the microwave will produce two hot spots (shown in red dashed lines) whose distance will euate to half a wavelength (distance between peak to trough).

Euipment Reuired • icrowave oven that utilises a removable turntable •  1 cm1 bar of chocolate select one that’s tasty and then you can eat it afterwards •  timer •  ruler

ethod 1 Prepare the chocolate to be as at as possible eg grating the bumps o the chocolate  ead the freuency  of the microwave o the label on the back of the microwave oven or by consulting the manual, and make a note of it somewhere It is usually given in megahert , which must be converted to hert  onvert from  to  by multiplying by 1 eg ,  5 ,,,   Remove the turntable from the microwave oven  Place the prepared chocolate onto a microwave-proof tray and then place in the microwave  Microwave the chocolate in short, 5-second bursts and eamine the food after each burst ou are looking for the point at which the chocolate just begins to melt wo or more areas of melting will occur

Results e actual speed of light is ,,5 m s  ow do your results compare If you nd that the results don’t match up with the real speed of light, or if you’re struggling to locate the hot spots, you can troubleshoot the issue using able 11 as a guide TABLE Troubleshooting the Speed of Light 21.1 Experiment

Reported Issue

Potential Problem

hocolate melts

. eft in microwave too long

ot spots are really large

. eft in micro . Try shorter bursts (e.g. wave too long  seconds) . hocolate . ake sure not to dis was moved turb the chocolate whilst doing the eperiment – during can you check for hot process spots without removing it from the microwave

Distance . hocolate between was moved hot during spots is process too smalltoo large . Distance was not measured from euivalent points

Potential Solution . Try shorter bursts (e.g.  seconds)

. ake sure not to dis turb the chocolate whilst doing the eperiment – can you check for hot spots without removing it from the microwave . ake sure to measure from one edge to the euivalent edge (e.g. left of one hot spot to left of other hot spot) for maimum accuracy

CHAPTER 21

s with all eperiments, if it didn’t work out then you should think about what you could do to improve the method net time

esure the peed o Liht

ou can also watch me do a demonstration of the eperiment on the associated lsevier website

Test Your Knowledge ry the uestions below to see if you need to review any sections again ll answers are available in the back of the book  plain why the distance between the hot spots is eual to half the wavelength of a microwave

 plain why we needed to take the turntable out of the microwave for it to work

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22 Create a ‘Cornea’ C H A P T ER O U TL INE Introduction The Experiment Equipment Required

Method Results Test Your Knowlede

O  E C TI E After working through this chapter, you should be able to: Explain how the human eye focuses distant light on the retina

Explain the dierent contributions of refractive index and curvature on an image

Introduction

The Experiment

In this book we’ve learned a lot about light and how it interacts with both at (plane) and curved surfaces. We’ve also learned that when light moves from a material of one refractive index to another, it will undergo refraction (a change in direction). We’ve also touched on the idea that the human ee (see chapter  for revision on this) is able to focus light on the back of the ee through two ke characteristics . e cornea (front of the ee) is a dierent refractive index to that of air (cornea . air .), which al lows refraction to take place. Importantl the uid in side the ee that sits behind the cornea (the aqueous) is a similar refractive index to the cornea. . e cornea is a convex, curved shape, which means it has dioptric power. is allows the cornea to add convergence to the incoming light (ig. .). is is important because it shows us that the basic focusing of distant light onto the back of the ee has nothing to do with an muscles (which is a common misconception) and can therefore not be ‘trained’. Instead, if a refractive error is present and light focuses too earl (mopia) or too late (hperopia), it can onl be corrected b altering the vergence of the light before it enters the ee (glasses or contact lenses), or b changing the shape of the cornea (refractive surger). o prove this, this chapter will explain how we can make a mock (relativel huge) cornea at home using common household items.

e goal of this experiment is to show that the refractive index change alone would not be enough for the ee to focus light over such a short distance (corresponding to the length of the eeball) instead we’re going to prove that it’s the curvature of the cornea that adds all the power. o that end we need to produce a convex surface that’s a dierent refractive index to air and see how the image compares to a at surface that’s a dierent refractive index to air. lease note, however, that there will be uite a sie dierence be tween our homemade ‘cornea’ and a real cornea, so the ra dius of curvature is not euivalent and so our homemade cornea will not have a power euivalent to a real cornea. Instead, it will serve to demonstrate the basic principles. efore starting the experiment, please make sure ou have all the euipment ou need (and a mobile device to record our results).

Euipment Reuired • n empt, clear plastic    drink bottle – note the shape of the bottle is uite important so tr to nd one that looks like the one in ig. .. •  pair of scissors •  felttip pen or marker of some kind • n empt, clear plastic at tub this could be a food storage container or an empt packet of sliced cheese – anthing that’s at, plastic and completel clear. 179

180 S EC TI ON 4

Experiments to o t Home

Air n = 1.00 Incident light

Retina Fovea

Cornea

Aqueous

n’ = 1.376

n’ = 1.336

•Fig. 22.1

• Fig. 22.3

A diagram showing how to produce the disc from the 2 L bottle, which will act as the front surface of our ‘cornea’.

An illustration of a crosssection of a human ee with light blue lines entering through the cornea labelled and being focused onto the foea in the retina labelled. he refractie index of air n 5 . and the refractie index of the cornea n9 5 . and aqueous n9 5 . are labelled for information.

• Fig. 22.4

After cutting out the disc, this is the equipment we will be using for the experiment itself.

.

•Fig. 22.2

A diagram of all the equipment needed for this experiment, including: a pen, scissors, water, a clear at plastic tub, a 2 L empt plastic bottle and a doodle or some writing on a piece of paper. lease note that the 2 L bottle needs to be clear plastic and needs to be this shape round ‘shoulders’ near the top.

• Water • n obect that can be laid at on the table, which we will look at through our homemade cornea this could be a drawing on a piece of paper (like in ig. .) or possibl something written down, but please don’t use anthing electrical, valuable or important, because there is a small risk of spilling the water onto the obect. • Optional (but advised): a towel or kitchen roll to mop up water spillages

ethod . Using the pen, draw a circle shape on the top ‘shoulder’ area of the 2 L bottle (like shown in ig. .). ake

.

. . .  .

sure the circle is a reasonable sie (at least  cm in diameter). It does not have to be perfectl spherical. Using the scissors, cut out the plastic disc from the top of the 2 L bottle – ou should now have the disc, the at plastic tub, the water and the obect remaining (ig. .). Create (or source) the obect – for me, I like to do a little doodle on a piece of paper (see ig. .), but ou can ust write a single word if ou feel more com fortable doing that (please see the associated video content on lsevier’s website for a demonstration of this experiment). lace the obect at onto the table Loo at the obect through the empt, at plastic tub It shouldn’t change. ow loo at the obect through the empt, curved plastic disc It still shouldn’t change. our some of the water (at least  cm deep) into the clear, at plastic tub and hold it  cm above the obect Loo through the water (lie shown in ig 22) to see what the obect loos lie ou should be able to see that, as ou move the at tub (lled with water) over the obect, the image of the obect looks the same (though possibl slightl wobbl due to the water). It should not be magnied, or upside down. It will ust be a virtual (upright), samesie image.

CHAPTER 22

Crete  ‘Corne’

Refracted image of object Observer Flat plastic tub filled with water (held above object)

Object

Paper with object on

• Fig. 22.5

he rst part of the experiment inoles looing at the image of the obect through the ‘at lens’ water inside the at plastic tub. his diagram is a side iew of how the experiment should be done and illustrates the best setup for optimal iewing.

Refracted and magnified image of object Observer

Homemade ‘cornea’ filed with water (held above object) Paper with object on

Object

• Fig. 22.6

he second part of the experiment inoles looing at the image of the obect through the ‘cornea’ water inside the conex disc. his diagram is a side iew of how the experiment should be done and illustrates the best setup for optimal iewing.

. ow put the tub to one side and pour some water into the plastic disc we cut out of the 2 L bottle – this is our fake cornea. . Carefull hold the disc  cm above the obect (this is where most water spillage happens). e disc should be convex, with its apex pointing towards the table. . Loo through the water in the disc (lie shown in ig 22) to see what the obect loos lie ou should be able to see that, as ou move the disc (lled with water) over the obect, the image of the obect looks dierent (though still probabl slightl wobbl due to the water). is time the image should be magnied and virtual (upright). . r changing the distance from the obect to the ‘cornea’ – what does it do to the image

Results ig. . shows some screenshots from m own attempt at this experiment (the full version can be viewed on

lsevier’s website), and hopefull our experiment found the same thing. Ideall, when ou looked at the obect through the ‘at lens’ (the water in the at plastic tub), there should have been no real dierence in the image of the obect and the obect itself. is is because when light travels through a atsurface material that has a dierent refractive index to the material in which the obect exists (in our case, this is air), it might change the position of the image, but it wouldn’t change the image characteris tics. owever, when we hold our convex ‘cornea’ over the image, the combination of the curved surface and the refractive index dierence means that our ‘cornea’ is positivel powered. is is evidence of the fact that the image appears to be magnied (something that is impos sible with a negativel powered lens – see chapters  and  for review on image formation). is shows that it is simpl the combination of the cur vature of the cornea and the refractive index dierence (of the cornea relative to the air) that enables our ees to focus light over such a short distance.

181

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No ‘lens’

Flat ‘lens’

‘Cornea’

•Fig. 22.7

creenshots from the ideo ersion of me completing the experiment at home. he left image shows m ‘obect’ drawing of a cat without an inolement from m water lenses. he middle image shows the image of the obect as seen through the ‘at lens’ water in the at plastic tub. he image on the right shows the image of the obect as seen through the conex ‘cornea’, and it is clearl magnied.

TABLE Troubleshooting the ‘Create a Cornea’ Experient 22.1

Reported Issue

Potential Problem

Potential Solution

‘ornea’ does not magnif the image

. ot enough water is being used

. ou need to mae sure there’s enough water in the ‘cornea’ that the ergence of the light will be altered chec to see if there’s at least . cm 2. ae sure to use the er cured part up at the top ‘shoulder’ of the 2 L bottle . ut it out a little larger in our next attempt . old the ‘cornea’ a little higher, and iew from aboe

2. he ‘cornea’ isn’t cured enough . he ‘cornea’ is too small . he ‘cornea’ is being held too close to the doodle he at tra also affects the image

. he tra is too small so cured edges confound the results 2. he tra is not trul at, or the plastic is too thic

onus – did ou happen to notice that the image was sometimes a little distorted near the edges of the ‘cornea’ an ou use the information from chapter  to explain what’s happening here lso, as ever, if ou nd that the experiment isn’t working ver well, ou can troubleshoot the issue using able . as a guide.

. wap the tra for a larger one 2. wap the tra for a atter, thinner plastic one

s with all experiments, if it didn’t work out (even after troubleshooting), then ou should think about what ou could do to improve the method next time. ou can also watch me do a demonstration of the experiment on the associated lsevier website.

Test Your Knowledge r the uestions below to see if ou need to review an sections again. ll answers are available in the back of the book. 22 xplain how the human ee can focus distant light onto the back of the ee. 222 Wh did we use water for the refractive index dierence

22 xplain wh the image is sometimes a little distorted near the edges of the ‘cornea’ in our demonstra tion (see ig. . for an example of this distortion). 22 If we had lled our lenses with glcerine (a viscous, clear liuid of refractive index .), do ou think the results would have been the same iscuss our thoughts.

23 Kitchen Thin Film nterference C H A P T ER O U TL INE Introduction The Experiment Equipment Required

Method Results Test Your Knowlede

O  E C TI E After working through this chapter, you should be able to: Explain ‘thin lm interference’ Explain why the interference pattern is colourful Produce a thin lm demonstration of your own

Introduction Before we delve deep into the wonderful world of thin lm interference, let’s take a moment to remind ourselves what interference is in the rst place (you can also review chapter 10 for additional revision on this). Interference (in optics) descries the variation in wave amplitude that occurs when multiple waves (e.. of liht) interact with one another. e type of interference (and amplitude of the resultant, comined wave) is determined y the relative phase dierence or path dierence of the individual waves. i. .1 shows an eample of liht interferin construc tively, with two inphase lue waves shown (phase dier ence 0°) and a path dierence that euates to a whole wavelenth (n). is leads to an increase in amplitude in the resultant wave (shown in orane). ontrarily, i. .1B shows an eample of liht interferin destruc tively, with the two outofphase lue waves (phase dier ence 10°) and a path dierence that euates to half a wavelenth (n 1 0.). is leads to a decrease in ampli tude in the resultant wave (shown in orane the eample in i. .1B is complete destructive interference, meanin the resultant amplitude is ero). e important part of this is that the dierence in path lenth of the waves will deter mine the interference.  reat eample of interference in the real world is thin lm interference, where liht will partially reect o the front and ack surfaces of a very thin lm – which essen tially reects two versions of the liht ack towards you. is allows the liht to produce interference which will either

produce a reection (constructive) or not (destructive). Interestinly, if white liht (comprisin lots of wavelenths) is incident upon one of these lms, the individual wavelenths of liht will interfere dierently with one another, which pro duces a nice rainow pattern of interference. ain, the key factor determinin this is the path dierence, so the anle of the approachin liht plays a role, ut crucially the thickness of the lm does too (see i. .). is chapter aims to allow you to see this for yourself whilst also demonstratin that the thickness of the lm plays a key role in the oserved interference pattern. is means we’re oin to make our own thin lms

The Experiment e oal of this eperiment is to show that the path dier ences in liht can e introduced when it reects o the front and ack surfaces of a thin lm and to show that the thickness determines the shape of the pattern. o that end we need to make our own thin lm that varies in thickness, and we need to shine some white liht throuh it. Before startin the eperiment, please make sure you have all the euipment you need (and a moile device to record your results).

Euipment Reuired • ome cold water in a dish or tu of some kind • ome liuid soap for washin dishes (or ulelowin miture, if you have it) 183

184 S EC TI ON 4

Experiments to o t Home

A

Phase difference: 0°

B

Phase difference: 180°

Path difference:

Path difference:

n

n + 0.5

• Fig. 23.1

In cid en Side view t li gh tr ay

First reflected ray

A

Second reflected ray

Illustration showing blue waves producing constructive (A) and destructive (B) interference. Orange waves show approximate resultant waves.

In cid en Side view t li gh tr ay

Second reflected ray

B

First reflected ray

Thinner film

•Fig. 23.3

An illustration of the materials reuired for this experiment.

Thicker film

• Fig. 23.2

Diagram showing that thickness of the lm will alter the path difference between the two reected ras of light. If the lm is thinner (A) the waves have less of a path difference than if the lm is thicker (B).

•  liht source (torch or moile phone liht) • omethin with an aperture (e.. a metal anle that you wouldn’t mind ettin soapy and wet) or some craft wire (the thinner the wire the etter) and wire cutters (as shown in i. .) • Optional (but advised): a towel or kitchen roll to mop up water spillages

ethod 1. If you have an oect with an aperture then skip this step otherwise, cut approimately 10 to 1 cm of wire with the wire cutters and shape to form a circle (it doesn’t have to e a perfect circle). wist the wire ends

•Fig. 23.4

Illustration showing the reuired shape of the wire (left) that needs to be placed in the soap mixture (right) to produce the lm.

toether to keep it nice (and ive you somethin to hold on to) like shown in i. .. . Add one part liquid soap with ve parts water and mix together enthusiastically (to make ules and have a fun time). If usin ulelowin miture, sim ply pour the ule miture into the owl. lso, look at the ules in the miture – are they rainow coloured If so, this is thin lm interference . Dip the wire ring (or object with aperture) into the soapy liquid and carefully remove it so that it retains a thin lm of the soap mixture (like a ulelower wand see i. .). If the lm disappears at any point whilst completin the eperiment, simply repeat this step. . Hold the ring (or object with aperture) vertically and shine the torch on it from whichever angle gives you the best view of the interference pattern

CHAPTER 23

Front view

Side view Thickness increases due to gravity

. ae a note of what you can see (you may need to adust the torch to et the est view, and sometimes the pattern takes a few seconds to materialise) – there should be rainbow stripes present in the lm . ae a note of how the stripes vary as you go down the lm – do they change at all

Results i. . shows a photoraph I took when I attempted this eperiment, and althouh it turned out to e etremely dicult to photoraph, I’m hopeful that you can see the interference patterns comin throuh nicely in the picture (particularly near the top of the rin). ere is also a patch of destructive interfer ence visile at the very top of the lm, which is where the lm is so thin that all visile wavelenths destructively interfere with one another. ere is also evidence that the interference stripes chane in thickness as they et nearer the ottom of the lm, thouh this is likely more visile in your own version which you can see live at home. et’s eplore why this chane in stripe thickness happens with the help of a diaram (i. .). In i. . you can see a front view and a side (crosssectional) view of the lm. ememer in the introduction of the chapter we said that the anle of the incident liht and the thickness of the lm would impact the interference pattern ell, what you miht not have realised is how relatively small chanes in thickness can have a relatively lare impact on the interfer ence pattern. In our eperiment, we held the lm vertically, which meant that the thickness was uneven across the lm ecause ravity is applyin its forces to the lm, makin it thicker towards the ottom of the lm itself. is means that when liht reects from the top part of the lm, it will

Interference pattern

•Fig. 23.6 A diagram of a cross section of the lm (right) showing how the thickness varies verticall.

eperience a dierent pattern of path dierences to the liht at the ottom of the lm. ou should try this eperiment aain, ut this time hold the lm horiontally. hat would you epect to see lso, as ever, if you nd that the eperiment isn’t workin very well, you can trouleshoot the issue usin ale .1 as a uide. s with all eperiments, if it didn’t work out (even after trouleshootin), then you should think aout what you could do to improve the method net time. TABLE Troubleshooting the Thin Film Interference 23.1 Experiment

Reported Issue

Potential Problem

. ilm is too thick

A photograph of the interference pattern I made when I tried this experiment at home.

Potential Solution

oap lm . ixture is . Add more washing pops up liuid or leave to too water before stand (covered) interference overnight – this can be makes it bubblier viewed . he ring is . r using a smaller too large so ring the weight of the soap is too great Interference is . ighting not visible is not maximall effective

• Fig. 23.5

185

Kitchen Thin ilm Intererence

. emember that inter ference can onl occur if light is present I found this worked best in well lit rooms against a dark background with a torch lighting the ring from the side . his can happen if the mixture is too soap or if the ring is too large – tr a smaller ring and a more water mixture

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Experiments to o t Home

Test Your Knowledge ry the uestions elow to see if you need to review any sections aain. ll answers are availale in the ack of the ook.  plain how constructive interference is produced with two coherent liht sources.  If two coherent waves arrive at a detector ‘in phase’, what would the path dierence e

 plain why the interference pattern varies vertically in our eperiment.  hy do you think there’s no visile interference pattern towards the very top of the lm (see i. .)

SECTION

5

Question Answers Answers to Practice Questions, 189

Answers to Test Your Knowledge Questions, 201

187

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Answers to Practice Questions Below are some representative answers to the ‘practice questions’ that appear within any chapters where we need to learn to apply some maths. ey’re designed to help you focus your learning, so make sure you have a go at answering them yourself before you take a peek at the answers

Chapter 2 2.1.1 f an obect is placed  cm in front of a surface, what is the obect’s vergence at the point where light from the obect meets the surface l 5 2. n 5 .  5 nl  5 . 2.  5 2. 2.1.2 f an obect is placed  cm in front of a surface, what is the obect’s vergence at the point where light from the obect meets the surface l 5 2. n 5 .  5 nl  5 .  2.  5 2. 2.2.1 f a light ray is incident on a glass block refractive inde . at an angle of °, what is the angle of refraction i 5 ° n 5 . n9 5 . n sin i 5 n9 sin i9 . sin  5 . sin i9 ... . 5 sin i9 ... 5 sin i9 sin2 ... 5 i9 i9 5 .° 2.2.2 f a light ray is incident on a piece of plastic refractive inde . at an angle of °, what is the angle of refraction i 5 ° n 5 . n9 5 . n sin i 5 n9 sin i9 . sin  5 . sin i9 ...  . 5 sin i9 ... 5 sin i9 sin2 ... 5 i9 i9 5 .°

2.2.3 f a light ray refracts out of a glass block refractive inde . at an angle of °, what was the angle of incidence at the back surface i9 5 ° n 5 . because the light is leaving the glass block n9 5 . because the light is moving into air n sin i 5 n9 sin i9 . sin i 5 . sin  . sin i 5 ... sin i 5 ...  . sin i 5 ... i 5 sin2 ... i 5 .° 2.3.1 alculate the lateral displacement of a light ray that enters a -cm-wide glass block refractive inde . at an angle of °. i 5 ° n 5 . n9 5 . d 5 . n sin i 5 n9 sin i9 . sin  5 . sin i9 ... . 5 sin i9 ... 5 sin i9 sin2 ... 5 i9 i9 5 ...° s 5 d sin i – i9  cos i9 s 5 . sin 2 ...  cos ... s 5 . m s 5 . cm 2.3.2 alculate the lateral displacement of a light ray that enters a -cm-wide glass block refractive inde . at an angle of °. i 5 ° n 5 . n9 5 . d 5 . n sin i 5 n9 sin i9 . sin  5 . sin i9 ...  . 5 sin i9 ... 5 sin i9 sin2 ... 5 i9 i9 5 ...° s 5 d sin i – i9  cos i9 s 5 . sin 2 ...  cos ... s 5 . m s 5 . cm 189

190

Answers to Practice Questions

2.4.1 etermine the power of a conve spherical glass surface refractive inde . with a radius of curvature of  cm. r 5 1. m n 5 . n9 5 .  5 n9 – n  r  5 . – .  .  5 1. 2.4.2 etermine the power of a concave spherical glass surface refractive inde . with a radius of curvature of  cm. r 5 2. m n 5 . n9 5 .  5 n9 – n  r  5 . – .  2.  5 2. 2.4.3 etermine the power of a conve spherical glass surface refractive inde . with a radius of curvature of  cm. r 5 1. m n 5 . n9 5 .  5 n9 – n  r  5 . – .  .  5 1. 2.5.1 n obect is placed  cm in front of a conve spherical glass surface refractive inde . with a radius of curvature of  cm. here does the image form l 5 2. m r 5 1. m n 5 . n9 5 .  5 n9 – n  r  5 . – .  .  5 1... 5nl  5 .  2.  5 2 9 5  1  9 5 2 1 ... 9 5 2... l9 5 n9  9 l9 5 .  2... l9 5 2. m l9 5 2. cm 2.5.2 n obect is placed  cm in front of a concave spherical glass surface refractive inde . with a radius of curvature of  cm. here does the image form l 5 2. m r 5 2. m n 5 . n9 5 .

 5 n9 – n  r  5 . – .  2.  5 2... 5nl  5 .  2.  5 2... 9 5  1  9 5 2... 1 2... 9 5 2. l9 5 n9  9 l9 5 .  2. l9 5 2. m l9 5 2. cm

Chapter 3 3.1.1  biconcave thin lens has a front surface power of 2. and a back surface power of 2.. hat is the overall power of this lens  5 2.  5 2.  5  1   5 2. 1 2.  5 2. 3.1.2  plus meniscus thin lens has a front surface power of 1. and a back surface power of 2.. hat is the overall power of this lens  5 1.  5 2.  5  1   5 . 1 2.  5 1. 3.2.1  biconve thin lens has a focal length of  cm. hat is its power n9 5 . f9 5 1.  5 n9  f9  5 .  1.  5 1. 3.2.2  biconcave thin lens has a focal length of  cm. hat is its power n9 5 . f9 5 2.  5 n9  f9  5 .  2.  5 2. 3.2.3  biconve thin lens has a focal length of  cm. hat is its power in water refractive inde . n9 5 . f9 5 1.  5 n9  f9  5 .  1.  5 1.

Answers to Practice Questions

3.2.4  biconcave thin lens has a power of 2.. hat is its focal length n9 5 .  5 2. f9 5 n9   f9 5 .  2. f9 5 2. m or 2. cm 3.3.1 n obect is placed  cm in front of a biconve thin lens with a power of 1.. here does the image form l 5 2. m  5 1. n 5 . n9 5 . 5nl  5 .  2.  5 2. 9 5  1  9 5 2. 1 1. 9 5 1. l9 5 n9  9 l9 5 .  1. l9 5 1. m or 1. cm right of lens 3.3.2 n obect is placed  cm in front of a biconcave thin lens with a power of 2.. hat is the magnication of the image l 5 2. m  5 2. n 5 . n9 5 . 5nl  5 .  2.  5 2... 9 5  1  9 5 2... 1 2. 9 5 2... m 5   9 m 5 2...  2... m 5 1. 3.3.3 n obect is placed  cm in front of a biconcave thin lens with a focal length of  cm. here does the image form l 5 2. m f9 5 2. m n 5 . n9 5 .  5 n9  f9  5 .  2.  5 2. 5nl  5 .  2.  5 2... 9 5  1  9 5 2... 1 2. 9 5 2...

l9 5 n9 9 l9 5 .  2. l9 5 2. m or 2. cm left of lens 3.4.1 wo thin lenses of powers 1. and 1. are in contact with each other. f an obect is placed  cm in front of the rst lens, where does the image form l 5 2. m  5 1.  5 1. n 5 . n9 5 . e 5  1  e 5 1. 1 1. e 5 1. 5nl  5 .  2.  5 2. 9 5  1 e 9 5 2 1 . 9 5 2. l9 5 n9  9 l9 5 .  2. l9 5 2. m or 2. cm left of lenses 3.4.2 wo thin lenses of powers 1. and 2. are in contact with each other. f an obect is placed  cm in front of the rst lens, what is the linear magnication l 5 2. m  5 1.  5 2. n 5 . n9 5 . e 5  1  e 5 1. 1 2. e 5 1. 5nl  5 .  2.  5 2. 9 5  1 e 9 5 2. 1 . 9 5 2. m 5   9 m 5 2.  2. m 5 1. 3.4.3 ree thin lenses of powers 1., 2. and 1. are in contact with each other. f an obect is placed  cm in front of the rst lens, where does the image form l 5 2. m  5 1.  5 2.  5 1. n 5 . n9 5 . e 5  1  1  e 5 1. 1 2. 1 1.

191

192

Answers to Practice Questions

e 5 1. 5nl  5 .  2.  5 2. 9 5  1 e 9 5 21 . 9 5 1. l9 5 n9  9 l9 5 .  1. l9 5 1. m or 1. cm right of lenses 3.5.1 wo thin lenses of powers 1. and 2. are separated by a distance of  cm. hat is the back verte power of the system  5 1.  5 2. d 5 . m v9 5  1  2 d   2 d v9 5 1. 1 2. – .1.2.   2 .1. v9 5 1. 3.5.2 wo thin lenses of powers 1. and 2. are separated by a distance of  cm. hat is the front verte power of the system  5 1.  5 2. d 5 . m v 5  1  2 d   – d v 5 1. 1 2. – .1.2.   2 .2. v 5 1. 3.6.1 wo thin lenses of powers 1. and 2. are separated by a distance of  cm. hat is the back verte focal length of the system  5 1.  5 2. d 5 . m v9 5  1  2 d   2 d v9 5 1. 1 2. – .1.2.   2 .1. v9 5 1... f v9 5 n   v 9 fv9 5   1... fv9 5 1. m or 1. cm 3.6.2 wo thin lenses of powers 2. and 2. are separated by a distance of  cm. hat is the front verte focal length of the system  5 2.  5 2. d 5 . m v 5  1  2 d   – d v 5 2. 1 2. – .2.2.   2 .2.

v 5 2... fv 5 2n  v fv 5 2  2.… fv 5 2 . m or 2. cm 3.7.1 wo thin lenses of powers 1. and 1. are separated by a distance of  cm. f an obect is placed  cm in front of the rst lens, where will the image form  5 1.  5 1. d 5 . m l 5 2. m n 5 . n9 5 .  5 n  l   5 .  2.  5 2.  9 5   1   9 5 2. 1 1. 9 5 1.  5 9   – d9  5 1.   – . 3 .  5 1.…  9 5   1   9 5 1.… 1 . 9 5 1.… l9 5 n9  9 l9 5 .  1.… l9 5 1. m or 1. cm 3.7.2 wo thin lenses of powers 2. and 1. are separated by a distance of  cm. f an obect is placed  cm in front of the rst lens, where will the image form  5 2.  5 1. d 5 . m l 5 2. m n 5 . n9 5 .  5 n  l   5 .  2.  5 2.  9 5   1   9 5 2. 1 2. 9 5 2.  5 9   – d9  5 2.   – .32.  5 2.  9 5   1   9 5 2. 1 . 9 5 1. l9 5 n9  9 l9 5 .  1. l9 5 1. m or 1. cm

Answers to Practice Questions

3.8.1 wo thin lenses of powers 1. and 1. are separated by a distance of  cm. hat is the equivalent power  5 1.  5 1. d 5 . m e 5  1  – d e 5 1. 1 1. – . 3 1. 3 1. e 5 1. 3.8.2 wo thin lenses of powers 2. and 1. are separated by a distance of . cm. hat is the equivalent power  5 2.  5 1. d 5 . m e 5  1  – d e 5 2. 1 1. – . 3 2. 3 1. e 5 1. 3.9.1 wo thin lenses of powers 2. and 1. are separated by a distance of  cm. hat is the secondary equivalent focal length of the system  5 2.  5 1. d 5 . m n 5 . e 5  1  – d e 5 2. 1 1. – . 3 2. 3 1. e 5 1... fe9 5 n  e fe9 5   1... fe9 5 1. m or 1. cm 3.9.2 wo thin lenses of powers 2. and 1. are separated by a distance of  cm. hat is the primary equivalent focal length of the system  5 2.  5 1. d 5 . m n 5 . e 5  1  – d e 5 2. 1 1. – . 3 2. 3 1 e 5 1. fe 5 2n  e fe 5 2  .. fe 5 2. m or 2. cm 3.9.3  multiple lens system has a secondary equivalent focal length of 1 cm. hat is the primary equivalent focal length of the system fe9 5 1. m f e 5 2 f e9 fe 5 2. m or 2. cm

3.10.1 wo thin lenses of powers 2. and 1. are separated by a distance of  cm. f an obect is placed  cm in front of the primary focal point of the system, where will the image form  5 2.  5 1. d 5 . m  5 2. m n 5 . n9 5 . e 5  1  – d e 5 2. 1 1. – . 3 2. 3 1. e 5 1 fe9 5 n  e fe9 5   1 fe9 5 1. m fe9 5 29 . 5 2 2. 9 ...  2. 5 29 2. 5 29 9 5 1. m or 1. cm 3.10.2 wo thin lenses of powers 1. and 1. are separated by a distance of  cm. f an obect is placed  cm in front of the primary focal point of the system, where will the image form  5 1.  5 1. d 5 . m  5 2. m n 5 . n9 5 . e 5  1  – d e 5 1. 1 1. – . 3 1. 3 1. e 5 1. fe9 5 n  e fe9 5   1. fe9 5 1 ...m fe9 5 29 ... 5 2 2. 9 ...  2. 5 29 2. 5 29 9 5 1. m or 1. cm 3.10.3  multiple lens system has a secondary equivalent focal length of 1. cm. f an obect is placed  cm in front of the primary focal point of the system, where will the image form fe 5 1. m  5 2. m n 5 . n9 5 . fe9 5 29

193

194

Answers to Practice Questions

. 5 2 2. 9 ...  2. 5 29 2. 5 29 9 5 1. m or 1. cm 3.11.1 wo thin lenses of powers 2. and 1. are separated by a distance of  cm. f an obect is placed  cm in front of the primary focal point of the system, what is the linear magnication of the image  5 2.  5 1. d 5 . m  5 2. m n 5 . n9 5 . e 5  1  – d e 5 2. 1 1. – . 3 2. 3 1. e 5 1. fe9 5 n  e fe9 5   1. fe9 5 1...m fe9 5 29 ... 5 2 2. 9 ...  2. 5 29 2... 5 29 9 5 1...m m 5 29e m 5 2 1...1. m 5 2. 3.11.2 wo thin lenses of powers 1. and 1. are separated by a distance of  cm. f an obect is placed  cm in front of the primary focal point of the system, what is the linear magnication of the image  5 1.  5 1. d 5 . m  5 2. m n 5 . n9 5 . e 5  1  – d e 5 1. 1 1 – . 3 1. 3 1. e 5 1... fe9 5 n  e fe9 5   1... fe9 5 1...m fe9 5 29 ... 5 2 2. 9 ...  2. 5 29 2... 5 29 9 5 1...m m 5 29e m 5 2 1...1... m5 2.

Chapter 4 4.1.1  -cm-wide planoconve lens has a radius of curvature of  cm. hat is the sag of the lens y 5 . m r 5 . s 5 r 2 √r – y s 5 . 2 √. – . s 5 . m s 5 . cm 4.1.2  -cm-wide planoconve lens has a radius of curvature of  cm. hat is the sag of the lens y 5 . m r 5 . s 5 r 2 √r – y s 5 . 2 √. – . s 5 . m s 5 . cm 4.1.3  -cm-wide planoconve lens has a radius of curvature of  cm. hat is the sag of the lens y 5 . m r 5 . s 5 r 2 √r – y s 5 . 2 √. – . s 5 . m s 5 . cm 4.2.1  thick lens refractive inde . has a conve front surface with a radius of curvature of  cm. hat is the power of the front surface n9 5 . r 5 1. m n 5 .  5 n9 – n  r  5 . – .  .  5 1. 4.2.2  thick lens refractive inde . has a concave back surface with a radius of curvature of  cm. hat is the power of the back surface n9 5 . r 5 1. m n 5 .  5 n9 – n  r  5 . – .  1.  5 2. 4.2.3  -cm-thick lens refractive inde . has a front surface power of 2. and a back surface power of 1.. hat is the power of the lens t 5 . m ng 5 .  5 2.

Answers to Practice Questions

 5 1. n 5 . ˉt 5 t  ng ˉt 5 .  . ˉt 5 ... e 5  1  – ˉt  e 5 2. 1 . – .… 3 2. 3 . e 5 1. 4.3.1 magine a . cm thick biconve lens surface powers 1. and 1. refractive inde .. here is the secondary principal plane, relative to the back surface of the lens  5 1.  5 1. t 5 . m n 5 . ng 5 . ˉt 5 t  ng ˉt 5 .  . ˉt 5 ... v9 5  1  2 ˉt   2 ˉt  v9 5 . 1 . 2 ...3.3.   2 ...3 . v9 5 1... f v9 5 n   v 9 fv9 5 .  1... fv9 5 1... e 5  1  2 ˉt  e 5 .1. 2 ...3.3. e 5 1... fe9 5 n  e fe9 5 .  1... fe9 5 1... 9 5 e9 5 fv9 – fe9 9 5 e9 5 1... 2 1... 9 5 e9 5 2. m r . cm left of the back lens 4.3.2 magine a -cm-thick biconve lens surface powers 1. and 1. refractive inde .. here is the primary principal plane, relative to the back surface of the lens  5 1.  5 1. t 5 . m n 5 . ng 5 . ˉt 5 t  ng ˉt 5 .  . ˉt 5 ... v 5 12 ˉt    2 ˉt  v 5 . 1 . 2 ...33.   2 ...3. v 5 1... fv 5 2n  v fv 5 2.  1... fv 5 2... e 5 1 2 ˉt  e 5 . 1 . 2 ...3.3.

e 5 1... fe 5 2n  e  fe 5 2.  1... fe 5 2...    5 e 5 fv – f e  5 e 5 2... 2 2...  5 e 5 1. m rimary principal plane is 1. cm right of front surface and is therefore 2. cm left of the back surface of the lens. 4.4.1 n obect is placed  cm in front of a -cm-thick biconve lens refractive inde . with a front surface power of 1. and a back surface power of 1.. here does the image form relative to the back surface of the lens l 5 2. m  5 1.  5 1. t 5 . m n 5 . n9 5 .  5 n   l   5 .  2.  5 2.  9 5   1   9 5 2. 1 . 9 5 2. l 9 5 n  9    9 l9 5 .  2. l9 5 2... l 5 l  9 – t l 5 2.… – . l 5 2.…  5 n   l   5 .  2.…  5 2.…  9 5   1   9 5 2.… 1 . 9 5 2.… l 9 5 n  9    9 l9 5 .  2.… l9 5 2. m or 2. cm 4.4.2 n obect is placed  cm in front of a .-cm-thick biconcave lens refractive inde . with a front surface power of 2. and a back surface power of 2.. here does the image form relative to the back surface of the lens l 5 2. m  5 2.  5 2. t 5 . m n 5 . n9 5 .  5 n   l   5 .  2.  5 2.  9 5   1  

195

196

Answers to Practice Questions

9 5 2. 1 2. 9 5 2. l 9 5 n  9    9 l9 5 .  2. l9 5 2... l 5 l  9 – t l 5 2.… – . l 5 2.…   5 n  l   5 .  2.…  5 2.…  9 5   1   9 5 2.… 1 2. 9 5 2.… l 9 5 n  9    9 l9 5 .  2.… l9 5 2. m or 2. cm 4.5.1 n obect is placed  cm in front of a -cm-thick biconve lens refractive inde . with a front surface power of 1. and a back surface power of 1.. here does the image form relative to the back surface of the lens l 5 2. m  5 1.  5 1. t 5 . m n 5 . n9 5 .   5 n  l   5 .  2.  5 2...  9 5   1   9 5 2... 1  9 5 2... ˉt 5 t  ng ˉt 5 .  . ˉt 5 ...  5 9   – ˉt 9  5 2...  – ... 3 2...  5 2...  9 5   1   9 5 2.… 1 . 9 5 2.… l 9 5 n  9    9 l9 5 .  2.… l9 5 2. m or 2. cm 4.5.2 n obect is placed  cm in front of a -cm-thick biconcave lens refractive inde . with a front surface power of 2. and a back surface power of 2.. here does the image form relative to the back surface of the lens l 5 2. m  5 2.  5 2. t 5 . m

n 5 . n9 5 .  5 n   l   5 .  2.  5 2.  9 5   1   9 5 2. 1 2. 9 5 2. ˉt 5 t  ng ˉt 5 .  . ˉt 5 ...  5 9   – ˉt 9  5 2.  – ...32.  5 2...  9 5   1   9 5 2.… 1 2. 9 5 2.… l 9 5 n  9    9 l9 5 .  2.… l9 5 2. m or 2. cm

Chapter 6 6.1.1 wo plane mirrors are inclined at an angle of ° towards one another. hat will the angle of deviation be a 5 ° d 5  2 a d 5  –  d 5  –  d 5 ° 6.1.2 wo plane mirrors are inclined at an angle of .° towards one another. hat will the angle of deviation be a 5 .° d 5  2 a d 5  – . d 5  –  d 5 ° 6.2.1  conve spherical mirror has a radius of curvature of  cm. hat is its power r 5 1 m n 5 . f5r f 5   f 5 1. m  5 2 n  f   5 2 .  1.  5 2. 6.2.2  concave spherical mirror has a focal length of  cm. hat is its power f 5 2. m n 5 .  5 2 n  f   5 2 .  2.  5 1.

Answers to Practice Questions

6.3.1 n obect is placed  cm in front of a conve spherical mirror with a radius of curvature of  cm. here does the image form l 5 2. m r 5 1. m n 5 . f5r f 5 1.   f 5 1.  5 2 n  f   5 2 . 1.  5 2. 5nl  5 .  2.  5 2 9 5  1  9 5 2 1 2 9 5 2 l9 5 2 n9  9 l9 5 2 .  2 l9 5 1. m or 1. cm

f 5 2.   l9 1   l 5   f    l9 1   2. 5   2.   l9 5 2 l9 5   2 l9 5 2. m or 2. cm

6.3.2 n obect is placed  cm in front of a concave spherical mirror with a power of 1.. here does the image form l 5 2. m  5 1. n 5 . 5nl  5 .  2.  5 2... 9 5  1  9 5 2... 1 2 9 5 1... l9 5 2 n9  9 l9 5 2 .  ... l9 5 2. m or 2. cm

6.5.2 n obect is placed  cm in front of a concave spherical mirror of power 1.. hat is the magnication of the image l 5 2. m  5 1. n 5 . f 5 2 n   f 5 2 .  1. f 5 2.   l9 1   l 5   f    l9 1   2. 5   2.   l9 5 2... l9 5   2... l9 5 2. m m 5 2 l9  l m 5 2 2.  2. m 5 2. minied, inverted, real

6.4.1 n obect is placed  cm in front of a conve spherical mirror with a radius of curvature of  cm. here does the image form l 5 2. m r 5 1. m n 5 .   l9 1   l 5   r   l9 1   2. 5   .   l9 5 ... l9 5   ... l9 5 1. m or 1. cm 6.4.2 n obect is placed  cm in front of a concave spherical mirror with a power of 1.. here does the image form l 5 2. m  5 1. n 5 . f 5 2 n   f 5 2 .  1.

6.5.1 n obect is placed  cm in front of a concave spherical mirror with a radius of curvature of  cm. hat is the magnication of the image l 5 2. m r 5 2. m n 5 .   l9 1   l 5   r   l9 1   2. 5   2.   l9 5 ... l9 5   ... l9 5 1. m m 5 2 l9  l m 5 2 1.  2. m 5 1. magnied, upright, virtual

Chapter 8 8.1.1 hat is the chromatic aberration of a crown glass lens with a power of 1.  5   5 1.  5     5 .    5 . 8.1.2 hat is the chromatic aberration of a ¢int glass lens with a power of 1.  5   5 1.  5     5 .    5 .

197

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Answers to Practice Questions

8.2.1 hat power of ¢int glass would be required to remove chromatic aberration from a lens made partly of crown glass 1. c 5  f 5  c 5 1.  5   5 c  c 1 f  f   5 .   1 f   2... 5 f   2... 3  5 f 2. 5 f 8.2.2 hat power of crown glass would be required to remove chromatic aberration from a lens made partly of ¢int glass 1. c 5  f 5  f 5 1.  5   5 c  c 1 f  f   5 c   1 .   2... 5 c   2... 3  5 c 2. 5 f

Chapter 9 9.1.1  ray of light is incident upon a prism at an angle of ° and leaves the prism at an angle of °. f the prism has an apical angle of °, what is the angle of deviation of the light i 5 ° i9 5 ° a 5 ° d 5 i 1 i9 2 a d 5  1  2  d 5 ° 9.1.2  ray of light is incident upon a prism at an angle of .° and leaves the prism at an angle of .°. f the prism has an apical angle of °, what is the angle of deviation of the light i 5 .° i9 5 .° a 5 ° d 5 i 1 i9 2 a d 5 . 1 . 2  d 5 .° 9.2.1 alculate the minimum angle of deviation of a prism constructed of refractive inde . of apical angle °. np 5 . ns 5 . a 5  ns sin.a 1 dmin 5 np sin .3a . sin. 1 dmin 5 . sin .3

sin. 1 dmin 5 . sin .3 sin. 1 dmin 5 . . 1 dmin 5 sin2. . 1 dmin 5 ...  1 .dmin 5 ... .dmin 5 ... – .dmin 5 ... dmin 5 ...  . dmin 5 .° 9.2.2 alculate the minimum angle of deviation of a prism constructed of refractive inde . of apical angle °. np 5 . ns 5 . a 5  ns sin.a 1 dmin 5 np sin .3a . sin. 1 dmin 5 . sin .3 sin. 1 dmin 5 . sin .3 sin. 1 dmin 5 ... . 1 dmin 5 sin2... . 1 dmin 5 ...  1 .dmin 5 ... .dmin 5 ... – .dmin 5 ... dmin 5 ...  . dmin 5 .° 9.3.1 etermine the smallest angle of incidence at the rst face of an equilateral triangular prism of refractive inde . for light to ust pass through the second face. e apical angle of the prism is °. np 5 . ns 5 . a 5  sinic 5   np sinic 5   . ic 5 sin2   . ic 5 ...° i 5 ic i9 5 a 2 i i9 5  – ... i9 5 ...° ns sin i 5 np sin i9 . sin i 5 . sin ... sin i 5 . sin ...   i 5 sin2 . sin ... i 5 .° 9.3.2 etermine the smallest angle of incidence at the rst face of an equilateral triangular prism of refractive inde . for light to ust pass through the second face. e apical angle of the prism is °. np 5 . ns 5 . a 5  sinic 5   np sinic 5   .

Answers to Practice Questions

ic 5 sin    . ic 5 ...° i 5 i c i9 5 a 2 i  i9 5  – ... i9 5 ...° ns sin i 5 np sin i9 . sin i 5 . sin ... sin i 5 . sin ...   i 5 sin  . sin ... i 5 .°

 5 . 3 2.  5 2.∆

9.3.3 etermine the smallest angle of incidence at the rst face of an equilateral triangular prism of refractive inde . for light to ust pass through the second face. e apical angle of the prism is °. np 5 . ns 5 . a 5  sinic 5   np sinic 5   . ic 5 sin    . ic 5 ...° i 5 i c i9 5 a 2 i  i9 5  – ... i9 5 ...° ns sin i 5 np sin i9 . sin i 5 . sin ... sin i 5 . sin ...   i 5 sin  . sin ... i 5 .°

Chapter 10

9.4.1  prism produces  m of displacement at  m. hat is its power  5  cm y 5 , cm  5 y  5   ,  5 ∆ 9.4.2  prism produces  cm of displacement at  cm. hat is its power  5  cm y 5  cm  5 y  5     5 .∆ 9.5.1 ow much prism power will be induced if a patient looks through their 2. lens  mm left of the optical centre  5 2. c 5 . cm  5 c

9.5.2 ow much prism power will be induced if a patient looks through their 1. lens  mm right of the optical centre  5 1. c 5 . cm  5 c  5 . 3 .  5 .∆

10.1.1  single slit is placed  cm in front of a wall. f the rst and second maima are  cm apart from one another, what is the angle of di¦raction  5 . m y 5 . m sin f 5 y  √ 1 y sin f 5 .  √. 1 . f 5 sin ... f 5 .° 10.1.2  single slit is placed  cm in front of a wall. f the rst and second maima are  cm apart from one another, what is the angle of di¦raction  5 . m y 5 . m sin f 5 y  √ 1 y sin f 5 .  √. 1 . f 5 sin ... f 5 .°

Chapter 12 12.1.1 alculate the planar angle shown as ‘’ in the following image 35cm ? 20cm

c 5 . m  5 . m u5c u 5 .  . u 5 . (p radians) u 5 .  p u 5 . p rad (degrees)

199

200

Answers to Practice Questions

u 5 .    p u 5 .° 12.1.2 alculate the planar angle shown as ‘’ in the following image 60cm 10cm ?

d 5 . m  5 p l5f l 5   p l 5 ...cd  5 l  d  5 ...  .  5 ...  .  5 . l

Chapter 13 c 5 . m  5 . m u5c u 5 .  . u5 (p radians) u5p u 5 . p rad (degrees) u 5     p u 5 .° 12.2.1  spherical light source of diameter  m emits  lm uniformly in all directions. hat is the average illuminance on a surface  m from its centre f 5  5m d5m  5 p l5f l 5   p l 5 . cd  5 l  d  5 ...    5     5 . l 12.2.2  spherical light source of diameter  m emits  lm uniformly in all directions. hat is the average illuminance on a surface  cm from its centre f 5  5m

13.1.1  alilean telescope has an eyepiece lens with a power of 2., and it produces a magnication of 1. hat is the power of the obective lens ep 5 2. m 5 1 m 5 2 ep  o o 5 2 2.   o 5 1. 13.1.2  eplerian telescope has an eyepiece lens with a power of 1. and an obective lens with a power of 1.. hat is the angular magnication of the telescope ep 5 1. o 5 1. m 5 2 ep  o m 5 2   . m 5 2. 13.1.3  eplerian telescope has an eyepiece lens with a power of 1. and it produces a magnication of 2.. hat is the power of the obective lens ep 5 . m 5 2. m 5 2 ep  o o 5 2 .  2. o 5 1.

Answers to Test Your Knowledge Questions

Below are some representative answers to the ‘Test Your Knowledge’ questions that appear at the end of every chapter. ey’re designed to help you focus your learning so mae sure you have a go at answering them efore you tae a pee at the answers

Chapter 1 TYK.1.1 ow do we measure ‘wavelength’ rom two equal points on the wave for eample from pea to pea or trough to trough. TYK.1.2 oes ultraviolet light have higher or lower energy than visile light ltraviolet light has higher energy than visile light ecause it has a shorter wavelength and higher frequency. TYK.1.3 oes ultraviolet light have a longer or shorter wavelength than visile light ltraviolet light has a smaller wavelength than visile light. TYK.1.4 in aout a regular tale. o you thin light approaching the tale is asored reected transmitted or a comination of a few of these ight needs to e reected for us to e ale to see it ut it will e asoring some of the light as well. is is particularly true for lac or rown tales. TYK.1.5 hat colour would e produced if we comined green and red wavelengths Yellow. TYK.1.6 hat is a shadow  shadow is an asence of at least some light produced when an ostacle locs the path of the light source.

Chapter 2 TYK.2.1 hat is a collection of light rays called  pencil. TYK.2.2 hat does parallel vergence tell us aout the origin of the light rays e rays have come from far away innity ecause the wavefronts are at relative to one another ero vergence. TYK.2.3 ow do we decide whether an oect distance will e negative or positive ect distance will always e negative ecause light starts at the oect and our convention states that light travels

from left where the oect is to right. e also always measure from the surface so oect distances will always e measured against the direction of light and therefore always e negative. TYK.2.4 f a light ray moved from a medium with a refractive inde of . into a medium with a refractive inde of . would it end towards or away from the normal t would end towards the normal. TYK.2.5 f you wor out that following refraction an image will have a vergence of 1. are the rays converging or diverging onverging positive vergence suggests they are converging. TYK.2.6 ould a concave spherical surface possess a negative or positive power hy  concave spherical surface would possess a negative power ecause the centre of curvature will e on the left meaning the radius of curvature r will e negative.

Chapter 3 TYK.3.1 hat is the denition of a thin lens  lens in which the thicness is small enough relative to the radius of curvature of each surface that it’s assumed that the refractive inde of the lens material has a negligile eect on the power so it can e ignored. TYK.3.2 hat would a magnication of 2. tell us aout the nature of the image t’s negative so that tells us it’s inverted real and it’s etween  and 2 so that tells us it’s minied. TYK.3.3 ow would you calculate equivalent power of two thin lenses in contact with one another dd the two powers together e 5  1 . TYK.3.4 hat is a principal plane  principal plane is the location where an equivalently powered lens would need to e placed within a multiple lens system in order to coincide with the secondary or primary focal point of the system. TYK.3.5 hat is the dierence etween ac verte focal length and secondary equivalent focal length e ac verte focal length fe9 is the distance etween the ac lens and the secondary focal point 9 whereas the secondary equivalent focal length is the distance etween the secondary principal plane 99 and the secondary focal point 9. 201

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Answers to Test Your Knowledge Questions

TYK.3.6 hat determines whether we use step-along vergence or ewton’s formulae to nd image distance with a multiple lens system e variale we’ve een given for oect distance determines which formulae to use. f we’ve een given the oect distance relative to the rst lens l then we use step along ut if we’ve een given the oect distance relative to the primary focal point  then we use ewton’s formulae.

Chapter 4 TYK.4.1 hat is the denition of a thic lens relative to a thin lens  thic lens taes the refractive inde of the lens into account when solving the equations whereas a thin lens does not. TYK.4.2 hat is the dierence etween the sag and the edge thicness e sag is the height of the segment of the sphere that maes a lens surface whereas the edge thicness is the literal thicness at the edges of the lens. TYK.4.3 n an equiconcave lens does the ac surface have a positive or negative radius of curvature n an equiconcave lens the ac surface would have a positive radius of curvature. TYK.4.4 hat is a virtual oect  virtual oect is formed when an image acts as a new oect for another surface or lens. TYK.4.5 hat causes the resolution of a resnel lens to e reduced relative to a ‘regular’ lens mage quality will e slightly reduced due to diraction occurring at the ridges etween each curved ring that maes up the resnel lens.

Chapter 5 TYK.5.1 hat is the presumed power of the reduced eye 1. TYK.5.2 hat is the dierence etween spherical and cylindrical refractive error pherical refractive errors are the same in all orientations whereas cylindrical refractive errors have a dierent error along a specic ais. TYK.5.3 f the far-point of a patient’s eye is 2. cm what is their refractive error eir refractive error is 2.  5   2.m.

TYK.5.4 hich meridian in a cylindrical lens is written as the ‘ais’ e ais meridian the ais with no power.

Chapter 6 TYK.6.1 hat are the two laws of reection  e incident light ray and the reected light ray lie in one plane.  e angle of incidence i is equal to the angle of reection i9. TYK.6.2 f an oect is  cm in front of a plane mirror where will the image form  cm within the mirror image distance is equal to oect distance in a plane mirror. TYK.6.3 escrie the nature of an image formed in a plane mirror. e image is the same sie as the oect virtual upright reversed and laterally inversed. TYK.6.4 plain why the formula for calculating the power  of a spherical mirror has a ‘minus sign’ in it. nlie with lenses mirrors reect light ac in the direction it came meaning that after reection without including the minus sign all variales would e assigned the opposite incorrect sign. TYK.6.5 f a spherical mirror has a radius of curvature of 1 cm what is the focal length f  1 cm f 5 r

Chapter 7 TYK.7.1 sing your nowledge of optical systems can you eplain why a -ray will pass through 9 Because the incident light is parallel to the optical ais and therefore suggesting it has ero vergence and light with ero vergence focuses at the secondary focal point 9. TYK.7.2 plain how you could use a ray diagram to decide whether an image distance would e positive or negative. f the ray diagram shows that the image forms on the right of the lens or mirror then the distance will e positive whereas if the ray diagram shows that the image forms on the left of the lens or mirror then the distance will e negative.

Answers to Test Your Knowledge Questions

TYK.7.3 n the diagram elow which image is drawn correctly plain your answer.

2F’

F’ Object

F

2F

A B

‘’ is drawn correctly ecause it’s formed at the intersection of the acwards-proected refracted rays. e ‘B’ intersection is incorrect ecause it is intersecting a refracted ray with an incident ray. TYK.7.4 raw a ray diagram to show where an image would form if an oect was placed at  in front of a positively powered lens.

Object 2F

F’

2F’

F Image

TYK.7.5 raw a ray diagram to show where an image would form if an oect was placed etween 9 and the lens in front of a negatively powered lens.

2F’

F’ Object

F Image

2F

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Answers to Test Your Knowledge Questions

TYK.7.6 raw a ray diagram to show where an image would form if an oect was placed left of  in front of a positively powered mirror.

F Object

C Image

TYK.7.7 raw a ray diagram to show where an image would form if an oect was placed at any location in front of a negatively powered mirror.

C Object

TYK.7.8 To get an inverted minied image what ind of lens would e needed positive or negative and where would the oect need to e placed n inverted image can only e produced y a positively powered lens negative lenses always produce upright images and minied inverted images can only e produced when the oect is etween  and innity.

Chapter 8 TYK.8.1 hat wavelength refracts the most – red or lue Blue ecause it’s a shorter wavelength lue ends est. TYK.8.2 plain how raindrops produce a rainow. hen white light from the sun encounters a raindrop each constituent wavelength is refracted slightly dierent amounts. e light egins to disperse ut when it reaches the other side of the raindrop some of the light is reected ac again and so it is then refracted a second time as it leaves the raindrop. t is the comination

Image F

of the two refraction opportunities that disperse the light to produce the rainow. TYK.8.3 ould dispersion occur if we shone a red laser through a prism plain your answer. o – dispersion descries the splitting of light into constituent wavelengths.  laser light only possesses one single wavelength so although the light will refract and change direction it will not disperse. TYK.8.4 hat is chromatic aerration n ‘aerration’ is a reduction in quality of an image. e ‘chromatic’ description suggests it is related to colour and so ‘chromatic aerration’ descries imperfections in an image related to the splitting of colours. TYK8.5 f a person is slightly under-minussed in their glasses prescription will green or red wavelengths e more liely to focus on the retina f a person is under-minussed their eye is focusing light with too much power. is means that it is liely that all

Answers to Test Your Knowledge Questions

the incoming wavelengths of light will e focused earlier in the eye than epected and therefore the red wavelengths which usually focus ehind the eye will e more liely to focus on the retina in this case.

Chapter 9 TYK.9.1 ill light deviate towards the ase or the ape of the prism ight deviates towards the ase the image deviates towards the ape. TYK.9.2 hat is the ‘critical angle’  ‘critical angle’ of incidence descries conditions in which light will leave the surface along the surface itself indicating an angle of refraction of °. t can only occur when moving from a material with a high refractive inde to a material with a lower refractive inde. TYK.9.3 f a prism was sumerged in water would it deviate light dierently than if it was in air plain your answer. Yes – when light leaves the prism the amount it is refracted is partly dictated y the refractive inde dierence etween the prism and the surrounding medium. f we alter the refractive inde of the surrounding medium it will change how the light leaves the prism.

some of the light reects at the front surface of the lm ut some passes through to the ac surface. en at the ac surface some light transmits through ut some is reected which means that even though we started with one light source there will e in this case two waves that reect towards us the oserver. is means that we’ll have two waves reaching us one of which has now travelled a slightly greater distance than the rst. ese light rays will produce constructive interference if in phase and destructive interference if out of phase for each of the wavelengths that mae up the white light.

Chapter 11 TYK.11.1 hat does a focimeter do  focimeter determines the spherical power cylindrical power and corresponding ais prismatic power and the optical centre of a lens. TYK.11.2 hat is the graticule  graticule is the networ of lines in the eyepiece of the focimeter shown in lac in the diagrams that act as a measuring scale. n this case the lines represent meridians and degree of prismatic eects.

Chapter 1

TYK.11.3 hy is it important to focus the eyepiece efore attempting to use a focimeter ach oserver may have a small amount of uncorrected refractive error which might mae the target appear lurry even though it is actually in focus. is could lead to errors.

TYK.10.1 hat is the ‘phase’ of a wave e phase of a wave is dened as the location of a point on the wave within a cycle measured in degrees.

TYK.11.4 f the target falls elow the centre of the graticule would this indicate ase-down or ase-up prism is would indicate ase-down prism.

TYK.9.4 ould increasing the apical angle increase the power of the prism or decrease it ncrease.

TYK.10.2 f two identical waves are ° ‘out of phase’ what will happen estructive interference will occur with a net amplitude of ero. is results in no light eing produced. TYK.10.3 hat does uygen’s principle predict aout light when it gets loced y an ostacle uygen’s principle predicts that the wavelets on the primary wavefronts will permit the light to ‘end’ slightly around the ostacle to produce light in the geometric shadow of the ostacle – this is diraction. TYK.10.4 f we increased the numer of slits in a diraction eperiment from one slit to ve slits what do you thin would happen to the diraction pattern e diraction pattern would have more interference patterns present as it would e lie having ve light sources and so the right spots within the maima would get smaller. TYK.10.5 hy do soap ules loo multicoloured sometimes oap ules are made of a thin lm which means that as incident white light reaches the ‘lm’ of the ule

Chapter 12 TYK.12.1 hat does ‘photometry’ mean easurement of light. TYK.12.2 escrie a ‘solid angle’.  solid angle is a -dimensional angle which descries the eld of view from a particular point or ape liened to the angle at the top of a circular cone. TYK.12.3 hat does ‘luminous u’ mean and what is it measured in uminous u denes the measure of power or perceived power emitted y a light source and is measured in lumens lm. TYK.12.4 hich of the following statements uses photometry terms correctly and why ‘e cup is poorly illuminated’ or ‘e cup is poorly luminated’ ‘e cup is poorly illuminated’ – ecause illumination descries light falling onto the cup whereas luminance is the perceived rightness of light coming o the cup.

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Answers to Test Your Knowledge Questions

TYK.12.5 hat does a high numer of Kelvins  K suggest aout a light source t will appear to loo ‘cold’ towards the lue end of the colour spectrum.

Chapter 13 TYK.13.1 hat is an ‘optical instrument’ n optical instrument is any device or equipment which can alter an image for enhancement or viewing purposes. TYK.13.2 ould the human eye e classed as an optical instrument plain your answer. Yes – ecause it refracts light to focus it in a specic place the ac of the eye and can adust the power accommodation of the lens in circumstances where needed. TYK.13.3 f we wanted to focus our camera on something very far away would we choose a  mm or  mm lens plain your answer. e would choose the  mm longer focal length lens ecause it will produce a higher magnication and allow us to see distant oects more clearly. TYK.13.4 ould you epect a alilean telescope to have a positive or negative magnication plain your answer. ositive magnication. alilean telescopes produce an upright image so the magnication should always e positive see chapter 3 for revision on this.

Chapter 14 TYK.14.1 ene ‘unpolarised light’. npolarised light is light that is oscillatingvirating in all orientations or a large range of orientations which means the electric eld changes randomly over time. TYK.14.2 plain the dierence etween circular and elliptical polarisation. Both types of polarisation require two perpendicular linearly polarised waves ut for circular they need to have the same amplitude and e ° out of phase whereas elliptical polarisation can e produced with dierent amplitudes or dierent degrees-of-phase dierence. TYK.14.3 hat type of light would e emitted through two identically oriented polarising lters olarised light which is polarised according to the orientation of the lters. f oth lters are at the same angle then they will let through the same light. TYK.14.4 s light reected o a lae more liely to e polarised in the horiontal or vertical plane oriontal – when light is polarised y reection it is polarised parallel to the surface from which it is reected. aes are usually horiontal TYK.14.5 plain why the sun appears red at sunset. s the unpolarised sunlight passes through the molecules in the atmosphere the wavelengths are scattered. is scattering

aects shorter wavelengths rst so when the sunlight travels a large distance through the atmosphere as it does at sunrisesunset then more of the wavelengths will e scattered out of the light and sent into the atmosphere. is means that only the longer wavelength light is left which will mae the sun appear red.

Chapter 15 TYK.15.1 hy is it not possile to see the ac of a patient’s eye without the help of a special device e inside of the eye is very dar meaning you’d need a light to illuminate the inside of the eye to see it and the pupil is very small which restricts the eld of view. TYK.15.2 s a slit-lamp iomicroscope a form of direct or indirect ophthalmoscopy ndirect. TYK.15.3 hy is it advantageous to the clinician to as the patient to move their gae when performing ophthalmoscopy e eld of view when performing ophthalmoscopy is relatively small even with indirect so clinicians can as the patient to change the position of their eyes to allow them to see dierent parts of the ac of the eye. TYK.15.4 plain why the anterior angle of the eye is not visile without the help of a special lens. Because of the refractive inde dierence etween the front surface of the cornea and the air light from the anterior angle eceeds the critical angle meaning that it eperiences total internal reection all the light reects ac into the eye maing it invisile to the eternal viewer. TYK.15.5 n a three-mirror goniolens which mirror would allow the clinician to view the anterior angle e -shape ° lens. TYK.15.6 plain what an against movement would loo lie during retinoscopy. f an against movement is present the ree light from inside the eye would move in the opposite direction to that of the strea. or eample if the strea was moving from right to left the ree would e seen to e moving from left to right. TYK.15.7 f a clinician performed retinoscopy on a patient at a woring distance of  cm what spherical power would they need to account for in the nal refractive error ey would need to account for 2. when determining their nal refractive error  5 nl 5 . 5 . TYK.15.8 hat equation does applanation tonometry rely on for calculating intraocular pressure  ressure 5 force  area

Chapter 16 TYK.16.1 hat is an aerration n imperfection in the quality of an image – could e from distortion lurring or oth.

Answers to Test Your Knowledge Questions

TYK.16.2 hich part of a lens induces the greatest amount of aerration e periphery edges – as distance increases away from the optical centre aerrations ecome larger. TYK.16.3 n terms of ernie polynomials what ‘order’ of aerration is defocus  ecause it’s Z 20. TYK.16.4 plain how a cheiner disc can identify the presence of a refractive error. e cheiner disc is an occluder with two small adacent pinholes. e occluder is placed in front of the patient’s eye and a distant parallel vergence light is shone at the occlude. e pinholes should then only let through two small spots of light which if no refractive error is present should focus on the ac of the eye to form an image of a single spot. owever if the patient’s eye is inducing aerrations then the light will focus incorrectly and will appear as two separate spots of light. TYK.16.5 plain how adaptive optics systems can tae high-resolution images. daptive optics systems utilise a wavefront sensor and a rapidly deformale mirror to constantly monitor aerrations sensor and compensate for them deformale mirror.

Chapter 17 TYK.17.1 hat does T stand for ptical coherence tomography. TYK.17.2 plain how far the mirror in a ichelson interferometer would need to move to produce a path dierence of a whole wavelength. e mirror would need to move a distance equivalent to half a wavelength in order to produce a full wavelength of path dierence ecause moving the mirror half a wavelength will add or remove half from the incident ray and also the reected ray totalling a whole wavelength. TYK.17.3 plain how a T-T system wors. T-T utilises a low-coherence near-infrared light source which is split into two eams a reference eam and a measurement eam y a eam splitter. ne eam measurement will enter the patient’s eye whilst the other eam reference will reect o a moveale mirror. e measurement eam is reected from the ac of the eye with dierent delay times which are dependent on the optical properties of the tissue and the distance away from the light source. e software within the system can then interpret the interference fringe proles as they pass through the detector in order to determine the depth of the tissue and produce the nice lac-and-white image. TYK.17.4 ame one dierence etween a T-T and an -T system. oveale mirror T-T versus diraction grating -T.

ow-coherence near-infrared light course T-T versus roadand light source -T. TYK.17.5 ame one clinical application of T. hecing the health of the layers of the retina measuring distances thicness of cornea aial length thicness of retinal nerve re layer  measuring the shape of the cornea checing the t of a contact lens.

Chapter 18 TYK.18.1 plain how a pinhole produces an upside-down image. ight rays have to travel in straight lines ut a pinhole is very small therefore light coming o the tip of an oect will travel in a straight line through the pinhole and end up near the oor whilst light rays from the ase of the oect travel in a straight line through the pinhole and end up near the ceiling. TYK.18.2 plain why moving the screen away from the pinhole produces a larger image. f the oect remains still then the angle of the rays on the other side of the pinhole should remain constant. o if the screen is closer to the pinhole it will see more of the scene than if it is farther away.

Chapter 19 TYK.19.1 plain why the torch loos more yellow as we add more mil. dding more mil causes more scattering of the light within the glass. is maes the shorter lue wavelengths of light scatter into the mil causing the torchlight to loo more yellow towards the longer wavelengths and maing the mil loo more lue. TYK.19.2 plain why using a red light source wouldn’t wor.  red light source wouldn’t wor ecause it only contains red long wavelengths. onger wavelengths are more di«cult to scatter so would require so much mil the light wouldn’t e visile through the muriness ut also lac the full range of wavelengths as with white light so there are no lue wavelengths to scatter and mae the mil lue.

Chapter 2 TYK.20.1 plain why the white light produces a rainow when it passes through the setup descried in this chapter. hen white light passes through a prism a material with two or more refracting surfaces the individual wavelengths within the white light refract to greater or lesser amounts depending on their wavelength e.g. red wavelength refracts the least ut lue refracts the most. hen the mirror is placed in the tray of water it turns the tray of water into a prism and splits the wavelengths through dispersion.

207

208

Answers to Test Your Knowledge Questions

TYK.20.2 plain what you thin would happen if you changed the angle of the light approaching our homemade prisms. s we change the angle of incidence angle of approaching light the angle of deviation in the prism changes as well which means that the dispersed light will move along the paper.

at lens not having any impact on the image and the ‘cornea’ producing a magnied image. owever our homemade cornea would have had more power with the glycerine relative to the water. is is shown in quation .

Chapter 21

et’s say our ‘cornea’ has a radius of curvature of 1.m this actual value doesn’t matter so much as long as it’s constant across oth equations.

TYK.21.1 plain why the distance etween the hot spots is equal to half the wavelength of a microwave. s the amplitude increases upwards or downwards away from the midpoint the energy also increases meaning that oth the pea and the trough of the microwave prole will produce the highest amount of intensity heat. erefore the microwave will produce two hotspots – one at the pea and one at the trough. e distance etween these points is equal to half a wavelength. TYK.21.2 plain why we needed to tae the turntale out of the microwave for it to wor. e turntale allows the interference produced y the microwave to e evenly spread around the food as it rotates through the hot spots. or this eperiment to wor we need to mae sure the hot spots don’t move so the turntale needed to e removed.

Chapter 22 TYK.22.1 plain how the human eye can focus distant light onto the ac of the eye. e human eye can focus light ecause it has a conve front surface the cornea with a small radius of curvature and the cornea has a dierent refractive inde to the air .. TYK22.2 hy did we use water for the refractive inde dierence ater is clear which means the image will still e visile through the ‘lenses’ and it has a refractive inde very similar to that of the human eye water . cornea .. TYK.22.3 plain why the image is sometimes a little distorted near the edges of the ‘cornea’ in our demonstration see ig. . for an eample of this distortion. ese are eamples of wavefront aerrations that ecome increasingly pertinent as light travels through the peripheral edges of a lens. n a way this maes our homemade ‘cornea’ even more realistic as a real cornea also induces more aerrations as light moves away from the optical centre. TYK.22.4 f we had lled our lenses with glycerine a viscous clear liquid of refractive inde . do you thin the results would have een the same iscuss your thoughts. e eperiment would have een a lot sticier  ut the results would largely have een the same with the

 5 n9 – n  r

ith water  5 . – .  1. 5 1. ith glycerine  5 . – .  1. 5 1.

Chapter 23 TYK.23.1 plain how constructive interference is produced with two coherent light sources. onstructive interference is produced when multiple waves arrive in phase with n path dierence. is results in a larger amplitude of the resultant wave which produces right light or visile colour. TYK23.2 f two coherent waves arrive at a detector ‘in phase’ what would the path dierence e To arrive in phase with one another the waves either need a path dierence of  so they’ve travelled the same distance or one wave will need to have travelled a multiple of the wavelength n to arrive at the same point in the phase. TYK.23.3 plain why the interference pattern varies vertically in our eperiment. n this eperiment the lm was held vertically in order to ensure that the thicness would vary along the lm. is variation occurs ecause gravity pulls the lm downwards meaning that the lm ‘pools’ towards the ottom of the ringaperture therey maing the lm thicer towards the ottom. is change in thicness alters the path dierence etween the ray reected at the front surface and the ray reected at the ac surface. TYK.23.4 hy do you thin there’s no visile interference pattern towards the very top of the lm see ig. . fter a few seconds of eing held vertically the lm at the top ecomes too thin  nanometres to e precise so all the wavelengths of light produce destructive interference when reected ac to us at this point meaning we can ust see through the lm.

Index age number folloe by ‘b’ inicates boes, ‘f ’ inicates gures an ‘t’ inicates tables 3D lms, polarisation, 137 A Aberrations, 206 angular frequency, 149–10 blurring, 149 cromatic, 79–1, 0f istorting, 149 in uman eye, 10–12 measurement of, 12 remoal of, 13–14 in lenses, 10 limiting sperical, 21, 22f types of, 149–10 aefront, 149 ernie polynomials, 149–10, 11f Aberrometer, 12 lenslets onto sensor, 13f aefront, 12 Absorption,  Aaptie optics, 13 imaging system, 14f Aitie colours, 6 Airy isc, iraction, 104 Amplitue, aelengt vs., 97 Angiograpy, optical coerence tomograpy, 19 Angle ‘AD’, 17, 1b, 1f Angle of eiation, 3, 196, 19 minimum, 6–7, 6t Angle of iraction, 101 Angle of incience, 1, 16f Angle of refraction, 1, 1f, 16f, 19 Angle, polarisation, 134 Angular frequency, 149–10 Angular magnication, telescopes, 127–12 Anterior angle, gonioscopy, 142, 142f Anterior eye, gonioscopy, 142f Antisolar point, 7–79, 79f Aperture, iraction, 100 Aperture stop cameras, 123 telescopes, 12 Apical angle, 79, 79f, 3 Applanation tonometry, 147, 147f, 14f Aqueous rainage, gonioscopy, 142 Asperical cure surfaces, reection at, 64, 64f Asperical mirrors, 64 Astigmatism, 2, 10 Atmosperic refraction, 26, 26f Ais meriian, 2, 3 B ac erte istance, 0, 1f, 2 ase notation, prism, 91–92 icone tin lens, 2b inoculars, 123

irefringence, 134, 13f lurring, aberrations, 149 rester angle, 134, 134f C amera, 123–12 aperture stop, 123 ept of el, 12, 12f el of ie, 124 focal lengts f , 123–124, 12f magnication m, 124 obscuras, 123, 124b pinole, 124f singlelens, 124f amera obscura, 123, 124b eperiment on, 163–166, 164f, 16f troublesooting, 166t translucent, 163 upsieon image, 16–166, 16f, 166f anelas c, 116 See also uminous intensity anelas per square metre cm2, 11 See also uminance ataracts, 129–130 entre, lens ticness, 39 romatic aberration, 79–1, 0f, 204 of cron glass lens, 197 of int glass lens, 197 in uman eye, 1–2, 2f ircular apertures, 104 ircular iraction, 103f, 104 ircular polarisation, 132 olours, 6–7, 6f olour temperature, 120, 121f onstructie interference, 101, 10–106, 17, 20 ontact tonometry See Applanation tonometry onergence, 12 one cornea, 179 poer, 19 ornea cone, cure sape, 179 curature, 11 ioptric poer, 179 refraction, 179 refractie ine, obect troug, 11f, 12f osine square la of illumination, 117 ritical angle, 20 prism, 7– rosscyliner tecnique, f rosse polarisers, 133, 133f scans See n face faceup scans urature, 11 yclic quarilaterals, 3–4, 4–b yliner ais, 2 ylinrical error, 2– crosscyliner tecnique, 4–, f ylinrical poer, focimeter, 110–111, 111f 209

210

Index

D Deformable mirror, 13, 13f Dept of el cameras, 12, 12f lo ision ais, 129 Destructie interference, 101, 10–106 Deiation of ligt, prism, 3– Diraction airy isc, 104 angle of iraction, 101 aperture, 100 circular, 103f, 104 enition, 100 farel, 104 raunofer, 104 resnel, 104, 10f grating, 103–104 uygen’s principle, 100 interference patterns, 103 maima, 101, 102f, 103f minima, 101, 103f multiple slit, 103–104 nearel, 104 out of pase, 101 pattern, 101, 102f, 103 in pase, 101 principles of, 97 rectilinear propagation of ligt, 101f single slit, 101–103, 102f slit it, 100–101, 101f, 103–104 aefronts, 100, 100f aelet, 100 Diractionlimit, optical systems, 104 Dioptres, 13, 27 prism, 90 Dioptric poer, 179 Diplopia, 91, 91f, 92f Direct gonioscopy, 143, 143f Direct optalmoscopy, 140–141, 140f, 141f Dispersion enition, 77 prism, 171 rainbos, 77–79, 7f, 7t, 171 reection, 172, 172f refraction, 172 Distorting, aberrations, 149 Diergence, 12, 12f calculations, 13–14 types, 12 Double ision See Diplopia Dsape mirror, 143 Duocrome, 1–2 E ge ticness, lens, 39 lectric el electric ector, 131 oscillating, 131 polarisation by reection, 133, 134f lectromagnetic raiation, interference properties of, 17 lectromagnetic spectrum , 3 energy, 176f frequency, 176f aelengt, 176f mmetropic eye, 4 n face faceup scans, 19 ntrance pupil, telescopes, 12

quialent lens, 34–37 eton’s formulae, 3–36, 36b magnication using, 37 principal planes, 34–3 ray tracing, 69–70 rules for, 70 it pupil, telescopes, 12 yepiece lens, telescopes, 12, 127–12 F arel iraction, 104 ar point, 0–1 ibrebase ourieromain  D, 17–1, 17f ibrebase timeomain  D, 16–17, 17f ept plane, 17f of eye, 1 faceon plane, 17f locoerence, nearinfrare ligt source, 16–17 measurement beam, 16–17 moeable mirror, 16–17 reference beam, 16–17 reectance, 16–17 superluminescent ioe, 16–17 iel of ie cameras, 124 lo ision ais, 129 telescopes, 12 iel stop, telescopes, 12 ocal lengt bac erte, 32, 32f cameras, 123–124, 12f cone sperical mirror, 196 front erte, 32, 32f poer of tin lens, 2–29 ray tracing, 6–66 reection, 61–62, 62f sperical cure surfaces, 21 tic lenses, 42–43 tin lens primary, 29 seconary, 2–29 ocal plane, 144–14 ocal point ray tracing, 6–66 reuce eye, seconary, 4 reection, 61–62 sperical cure surfaces, 21, 21f tin lens primary, 29 seconary, 2–29 ocimeter, , 20 ais eel, 111 cylinrical poer, 110–111, 111f escription, 109 focusing system of, 109 graticule of, 109 oriontal prism, 112 lens frame rest, 109 negatie sperecyliner form, 111 obseration system, 109 optical centre, 109 parts of, 109, 110f poer eel, 110 set up, 110f sperical poer, 110 target of, 109, 110f toric, 110–111 types of, 109 ertical prismatic ierences, 111–112

Index

oucault nife test, 144–14, 144f, 14f raunofer iraction, 104 requency, 3–4 electromagnetic spectrum, 176f aelengt vs., 97 resnel iraction, 104, 10f resnel lenses, 4, 4f G alilean telescopes, 126, 126f, 200 eometric optics, principles of, 97 lare, 134, 13f oniolens, 143, 143f onioscopy, 142–143 anterior angle, 142, 142f anterior eye, 142f aqueous rainage, 142 irect, 143, 143f goniolens, 143 inirect, 143f intraocular pressure , 142 total internal reection, 142 H uman eye aberrations in, 10–12 measurement of, 12 remoal of, 13–14 cromatic aberration in, 1–2, 2f conergence to incoming ligt, 47 cornea, 4f feature of, 47 focus ligt, 20 reuce eye, 4–49 uygen’s principle, 100, 20 yperopia, 4, 0–2 refractie error, 146–147 I lluminance cosine square la of, 117 inerse square la of, 117, 119 las of, 117f ligt sources an, 119, 119f, 120f potometry, 117–11, 117f mage iagram of, 66, 67f istance, ergence to tin lens, 30–31, 30f formation, reection, –60, 62–63 prism, –90, f, 9f ergence, tic lenses, 43–4 mbertic principle, 147 ncient ligt rays, 1 nirect gonioscopy, 143f nirect optalmoscopy, 141–142, 141f nnity, 12 nterference, 13 constructie, 10–106, 17, 20 enition, 13 estructie interference, 10–106 to measure istances, 99–100 out of pase, 10–106 patterns, 103 pattern it eperiment, 13–1, 1f in pase, 10–106 principles of, 97 properties of electromagnetic raiation, 17 tin lm, 104–106, 106f, 13 type of, 13

nterpupillary istance D, 112 ntraocular pressure  gonioscopy, 142 measurement see Applanation tonometry nerse square la of illumination, 117 ris, 123 J acson crosscyliner  See rosscyliner tecnique K elins , 120, 121f See also olour temperature eplerian epler telescopes, 126, 127f, 200 L ateral isplacement, 17–19, 17f aterally inerse reection,  as of reection, 202 engt of arc, 114 ens aberrations in, 10 istortion, 11f aapte, 129–130 enition, 27 iagram of, 66 equialent, 34–37 eton’s formulae, 3–36, 36b principal planes, 34–3 forms, 27, 2f front an bac surface, 2f optical centre of, 10 planocone, 194 planocylinrical, 2, 3f positie an negatie, 2f, 29f prism, 91–92 frame rest, focimeter, 109 sperocylinrical, 2, 3f ticness, 39–40 See also ¢ic lens centre, 39 ege ticness, 39 planocone lens, 39f, 40f sagitta sag, 39 tin, 27 See also ¢in lens enslets, 12 igt absence of, 7–9, f absorption,  colours, 6–7, 6f electromagnetic spectrum, 3 frequency, 3–4 interaction it obects, –6 natural, 132f particle vs. ae, 4b potometry, 113–116, 113f colour temperature, 120, 121f illuminance, 117–11, 117f luminance, 11–119, 11b, 11f luminous u, 116 luminous intensity, 116–117 measurements of, 116–119 potons, 3 reection,  spee of, b transmission,  ergence of, 179 isible, 3 aelengt, 3 aeparticle uality, 3

211

212

Index

igt rays, 201 incient, 1 normal, 12f, 16b inear magnication eton’s formulae using, 37 reection, 63–64 ergence to tin lens, 29–30, 30f magnie, 30 minie, 30 paraial rays, 29 inear polarisation, 132 onger aelengt, polarisation eperiment, 16 ongsigteness See yperopia o ision ais, 123, 12–130 aapte lenses, 129–130 assistie tecnology, 130 ept of el, 129 el of ie, 129 anel an atform magniers, 12–129 magnication eices, 129t magniers, 12–129 telescopic monocular, 12–129 isual angle, 12, 129f uminance, 11–119, 11b, 11f uminous u, 116 uminous intensity, 116–117 u l, 117, 117f See also lluminance M agnetic el, oscillating, 131 agnication cameras, 124 telescopes, 127–12 agniers, 12–129 aima iraction, 101, 102f, 103f multiple ae, 99 icelson interferometer, 99–100, 100f, 1–16 icroaes, 17 aelengt of, 17 amplitue prole an, 176f eperiment, 17–177, 176t inima iraction, 101, 103f multiple ae, 99 inuscyliner form, 2–3 irrors asperical, 64 multiple plane, 9–60, 9f, 60f paraboloial, 64 sperical, 71f onoculars, 123, 129 ultiple plane mirrors, reection, 9–60, 9f, 60f ultiple slit iraction, 103–104 ultiple tin lens, 31–34 in contact, 31, 31f separate by istance, 32–34 stepalong meto, 33–34 erte poer, 32–33 ultiple ae, 9–99, 9f, 99f coerent, 9 complete constructie interference, 99 complete estructie interference, 99 ientical aes, 99f interference, 99 maima, 99 minima, 99 out of pase, 9, 9f

ultiple ae (Continued) pat ierence, 99 in pase, 9, 9f pase ierence, 9 pase of, 9f resultant ae, 9 yopia, 4, 49–0 refractie error, 146–147 N earel iraction, 104 eutralisation retinoscopy, 144–14 eton’s formulae, 3–36, 36b magnication using, 37 O bect, 12–13 iagram of, 66, 67f bectie lens telescopes, 12, 127–12  See ptical coerence tomograpy   angiograpy A, 19 ptalmoscopy irect, 140–141, 140f, 141f inirect, 141–142, 141f pposite angles, 4–b ptical ais, poer, 19 ptical coerence tomograpy  angiograpy, 19 clinical applications, 19 coerent, 1 constructie interference, 1, 16f conentional vs. en face, 19 escription, 16 estructie interference, 1, 16f brebase ourieromain, 17–1, 17f brebase timeomain, 16–17, 17f inpase, 1 interferometry an, 16–19 ligt reision, 1 icelson interferometer, 1–16 outofpase, 1 principle of, 99–100 resultant ae, 1 of retinal layers, 16f septsource, 17f ae, 1 ptical instrument, 123, 206 ptically isotropic, 134 ptical symmetry, 134 ptical systems, 202 P araboloial mirrors, 64 arallel, 12 araial rays, 21, 29, 6 article vs. ae, 4b eascrests, ae, 97 enumbra, 7– otograpy, polarisation, 136 otometer, 20 angles, 113–116, 114f escription, 113 ligts, 113–116, 113f colour temperature, 120, 121f illuminance, 117–11, 117f luminance, 11–119, 11b, 11f

Index

otometer (Continued) luminous u, 116 luminous intensity, 116–117 measurements of, 116–119 otons, 3 ysical optics, principles of, 97 inole camera, 124b, 124f lanar angle, 199–200 lanar at angle, 113–114 lane of incience polarisation, 132, 133f by reection, 133 lane surfaces, reection, –60 lanocone lens, 39f, 40f, 194 lanocylinrical lens, 2, 3f olarisation angle, 134 applications of, 13–137 3D lms, 137 potograpy, 136 sunglasses, 136 circular, 132 crosse polarisers, 133, 133f enition, 131 eperiment on, 167–16, 16f, 169f troublesooting, 169t linear, 132 longer aelengt, 16 plane of incience, 132, 133f ppolarise, 132 reection, 131 refraction, 131, 134–13 scattering, 131, 13 by scattering, 167, 16 sorter aelengt, 16 spolarise, 132 transmission, 131, 132–133, 132f types of, 131, 132 olariser, 132–133 crosse, 133, 133f ositiecyliner form, 2–3, 3f oer bac erte, 192 concae sperical glass, 190 concae sperical mirror, 197 cone sperical glass, 190 cross, 2–3 front erte, 192 of prism, 90–91, 90f reection, 61 sperical cure surfaces, 19–21 centre of curature, 19 cone, 19 optical ais, 19 raius of curature, 19, 20b tic lenses, 40–43, 41f, 44f erte, 42 of tin lens, 27–29 focal lengt, 2–29 single, 27–2, 2f tin lenses, 193 oere lens, ray tracing negatiely, 69, 69f, 69t, 70f positiely, 67–6, 6f, 6t oer meriian, 2, 3 rescription, 1–2 rincipal planes, 34–3

rism apical angle, 79, 79f base notation, 91–92 critical angle, 7– enition, 3 eiation of ligt, 3– ioptres, 90 ispersion, 171 eects in sperical lenses, 92, 92f, 93f eects of, 10 equilateral triangular, 19, 199 eperiment, 171–172 troublesooting, 173t focimeter, oriontal, 112 images, –90, f, 9f lenses, 91–92 optical, 171 poer of, 90–91, 90f refractie ine outsie, 4f, 7f total internal reection, 7–, 7f types of, 4f R aians, 113–114 aius, 114 aius of curature, 19, 20b cone sperical mirror, 196 tic lenses, 41, 41f ainbos, ispersion, 77–79, 7f, 7t, 171 ay, 11, 12f ay iagram, 6 ayleig criterion, 104, 10f ay tracing iagrams scales, 72–74, 73f, 74f equialent lenses, 69–70 rules for, 70 negatiely poere lens, 69, 69f, 69t, 70f positiely poere lens, 67–6, 6f, 6t principles of, 6–69, 66f sperical mirrors, 70–72, 71f, 72t negatiely poere, 72 ectilinear propagation of ligt, 7, 101f euce eye, 4–49, 49b, 0f poer of, 202 eecting telescopes, 126–127, 127f eection at asperical cure surfaces, 64, 64f enition, 7 ispersion, 172, 172f rst la of, 7 focal lengt, 61–62, 62f focal points, 61–62 image formation, –60, 62–63 laterally inerse,  las of, 7 linear magnication, 63–64 multiple plane mirrors, 9–60, 9f, 60f plane surfaces, –60 polarisation, 131 poer, 61 reerse,  reersibility principle, 7 secon la of, 7 at sperical cure surfaces, 60–64, 61f ee, 14–146, 146f efraction, 179 atmosperic, 26, 26f

213

214

Index

efraction (Continued) enition, 14–16 ispersion, 172 ice in ater, 1b loa of ot air, 23–24, 24b polarisation, 131, 134–13 troug single material, 23–26 efractie error, 179 eye imaging, 139–140, 140f yperopia, 146–147 measurement of, 143–147 against moement, 144–14 it moement, 144–14 myopia, 146–147 obectie, 144 reuce eye, 4–49, 4f ceiner isc, 12f sperical, 49–2 subectie, 144 oring istance, 14–146 efractie ine cornea, 179 ierence, 11 eperiment, 179–12, 10f troublesooting, 12t obect troug cornea, 11f, 12f obect troug at lens, 11f, 12f outsie prism, 4f, 7f primary, 13, 13f of primary meium, 13 aelengt vs., 7t efractie telescopes, 126–127 esolution, optical systems, 104 etinal nere bre layer , 13 etinoscope, 14–146, 146f eerse reection,  eersibility principle, reection, 7 S accaes, 130 agitta sag, lens ticness, 39 cattering polarisation, 131, 13, 167, 16 ceiner isc, 12, 12f, 207 clieren imaging, 23–24, 2f aograp, 24–2b aos, 7 orter aelengt, polarisation eperiment, 16 ortsigteness See yopia inglelens cameras, 124f ingle slit iraction, 101–103, 102f ingle ae, 97–9 lit it iraction, 100–101, 101f, 103–104 nell’s la, 1–16 oli angles, 11, 11f, 116f pectral omain  D See ibrebase ourieromain  D pectral interference acquisition time, 17–1 measurement, 17–1, 17f spatial resolution, 17–1, 1b, 1f pee, b pee of ligt measurement, 176 troublesooting, eperiment, 176t perical cure surfaces, 19–21 caustic cure, 21 focal lengt, 21

perical cure surfaces (Continued) focal points, 21, 21f limiting sperical aberration, 21, 22f poer, 19–21 centre of curature, 19 cone, 19 optical ais, 19 raius of curature, 19, 20b reection at, 60–64, 61f perical lenses, prism, 92, 92f, 93f perical mirrors, ray tracing, 70–72, 71f, 72t negatiely poere, 72 perocylinrical lens, 2, 3f tepalong meto, 33–34 tic lenses, 44–4 teraians, 11 unglasses, 136 uperposition multiple ae, 9 principles of, 97 eptsource  , 1–19, 19f T elescopes, 123, 12–12 angular magnication, 127–12 aperture stop, 12 entrance pupil, 12 eit pupil, 12 eyepiece lens, 12, 127–12 el of ie, 12 el stop, 12 alilean, 126, 126f eplerian epler, 126, 127f magnication, 127–12 obectie lens, 12, 127–12 reecting, 126–127, 127f refractie, 126–127 terrestrial, 126 ¢ic lens, 194, 19 enition, 39, 202 focal lengt, 42–43 resnel equialent, 4, 4f image ergence, 43–4 poer, 40–43, 41f, 44f erte, 42 raius of curature, 41f stepalong meto, 44–4 irtual obect meto, 43–44 ¢in lm interference, 104–106, 106f, 13 eperiment, 13–1, 14f troublesooting, 1t ¢in lens biconcae, 190, 191 bicone, 2b enition, 27, 201 multiple, 31–34 in contact, 31, 31f separate by istance, 32–34 stepalong meto, 33–34 erte poer, 32–33 plus meniscus, 190 poer of, 27–29 focal lengt, 2–29 single, 27–2, 2f ergence to, 29–31 image istance, 30–31, 30f linear magnication, 29–30, 30f

Index

onometry, 147 applanation, 147, 147f otal internal reection gonioscopy, 142 prism, 7–, 7f ransmission,  polarisation, 131, 132–133, 132f U mbra, 7– npolarise ligt, 131, 206 V ergence, 11–14, 12f irection of ligt to alter, 1f far point, 49 image, 21–23 image, tic lenses, 43–4 of ligt, 179 obect, 21–23, 19 sperically cure surface, 20f to tin lens, 29–31 image istance, 30–31, 30f linear magnication, 29–30, 30f erte poer, 32–33 irtual obect meto, tic lenses, 43–44 isible, 3 isible ligt See also igt measurement of, 113 isual angle, lo ision ais, 12, 129f W ae, 11 energy of, 97 features of, 97–9, 97f

ae (Continued) multiple, 9–99, 9f, 99f coerent, 9 complete constructie interference, 99 complete estructie interference, 99 ientical aes, 99f interference, 99 maima, 99 minima, 99 out of pase, 9, 9f pat ierence, 99 in pase, 9, 9f pase ierence, 9 pase of, 9f resultant ae, 9 peascrests, 97 single, 97–9 trougs, 97 aefront, 11, 12f, 100, 100f aberrations, 149, 12 aelengt, 3, 201 vs. amplitue, 97 enition, 3 electromagnetic spectrum, 176f frequency, 3–4 vs. frequency, 97 ientication of, 4f measure of istance, 3 vs. refractie ine, 7t aelet, 100 aeparticle uality, 3, 11, 97 aes, polarisation, 131 Z ernie polynomials aberrations, 149–10, 11f, 207

215

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