College Physics A Strategic Approach (Volume 2) [2 ed.] 9780321598530

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Brief Contents Force and Motion CHAPTER 1

Representing Motion

2

CHAPTER

2

Motion in One Dimension

CHAPTER

3

Vectors and Motion in Two Dimensions

30

CHAPTER

4

Forces and Newton's Laws of Motion

CHAPTER

5

Applying Newton's Laws

CHAPTER

6

Circular Motion, Orbits, and Gravity

CHAPTER

7

Rotational Motion

CHAPTER

8 Equilibrium and Elasticity

67

102

131 166

200 232

It4;i'" Conservation Laws CHAPTER CHAPTER CHAPTER

9

Momentum

260

10 Energy and Wotk 289 11 Using Energy 322

Properties of Matter CHAPTER

12 Thermal Properties of Matter

CHAPTER

13 Fluids

362

405

1!Xi;i"'. Oscillations and Waves CHAPTER CHAPTER CHAPTER

14 Oscillations 444 15 Traveling Waves and Sound 477 16 Superposition and Standing Waves

507

1!Xi;i.. Optics CHAPTER

17

CHAPTER

18 Ray Optics 574 19 Optical Instruments

CHAPTER

Wave Optics

544

609

Electricity and Magnetism CHAPTER

20 Electric Fields and Forces

CHAPTER

21

CHAPTER

22 Current and Resistance

Electric Potential

642

675 7 12

23 Circuits 739 CHAPTER 24 Magnetic Fields and Forces CHAPTER

CHAPTER

25

CHAPTER

26 AC Electricity

776

Electromagnetic Induction and Electromagnetic Waves

816

852

'itd;i"... ' Modern Physics CHAPTER CHAPTER CHAPTER CHAPTER

27 28 29 30

Relativity

886

Quantum Physics

922

Atoms and Molecules Nuclear Physics

991

954

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c COLLEGE

...... ., . RANDALL D . KNIGHT California Polytechnic State Uni ve rsity. San Luis Obispo

'I)

BRIAN JONES Colorado State Un i vers ity

STUART FIELD Col o rado State University

Addison·Wesley Boston Colum bu s Indianapo lis New Yo rk San Francisco Upper Sad dle Rive r Amste rda m Cape Town Dubai London Mad rid Mi lan Munich Paris Montreal Toronto

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Real-World Applications Applications of biological or medical interest are marked BID in the li st below. MeAT-style Passage Problems are marked BID below. Other end-of-chapter problems of biological or medical interest are marked BlO in the chapter. "Try It Yourself' experiments are marked TIY .

Chapter 1 Deprh gauges 6 BID Scales of nerve cells vs galaxies 11

Accuracy oflongjumps 12 Tty How raU are you reaUy! 12 Mars Climate Orbiter: unit error

15 Navigating geese 23

Chapter 2 BID Tree rings 34

Rocker propulsion 122- 123 A mountain railway 123

Chapter 5 TIY Physics scudents can't

jump 141 Weighrless astronauts 142 Anri -lock brakes 146 Skydiver rerminal speed 150 BID Tracrion 154-155 Sropping disrances 157

Crash cushions 42

Solar sails 45 BID Swan's rakeoff 47 BID Chameleon rongues 48 Runway design 51 Braking distance 52 TIY A reaccion rime challenge 53

BID A springbok's pronk 55 BID Cheerah vs. gazelle 57

Chapter 3 BID Fish shape for lungingvs. veering 72 Designing speed-ski slopes 80 Oprimizingjavelin throws 82 TIY A game of catch in a moving

vehicle 86 Hollywood srunrs 86-87 Physics of fielding 89 Designing roller coasters 92 BID Record-breaking frogjumps 93

Chapter 4 Voyager and Newton's first

law 103 Tty Gerring rhe kerch up our 104 Searbelrs and Newton's first

law 104 Racing bike drag 111 Tty Feel rhe difference (ine"ia) 115 Race~car driver mass 116

Chapter 6 Clockwise clocks 170 Rotation of a compact disc 172 Scottish heavy hammer throw 173 Car cornering speed 176-177 Wings on Indy racers 177 Banked racetrack turns 177 BKl Maximum walking speed 178 BID How you sense "up" 180 Fast-spinning plancts 180 BID Cenrrifuges 181 TIV Human centrifuge 182 Rotating space stations 184 Variable graviry 187 Walking on rhe moon 188 Huming wirh a sling 191-192

Chapter 7 Srarring a bike 206 Designing wheelchair hand-rims

206 Turning a capstan 208 Camera stabilizers 211 TIV Hammering home inertia 214 Golf putter moment of inertia 217

Rolling vs. sliding: ancient movers

Chapter 8 BID Muscle forces 233 BID Finding rhe body's cenrer of graviry 236 Rollover safery for cars 237 Tty Balancing soda can 238 BID Human srabiliry 239 Tty Impossible balance 239 Elasriciry of a golf ball 240 BID Spider silk 244 BID Bone strength 244-245

Chapter 9 BID Oprimizing frogjumps 263 Tty Warer balloon carch 265

BID Ram skull adaprarions 265 BID Hedgehog spines 265 BID Squid propulsion 273 Ice-skating spins 278 Hurricanes 279 Aerial 6refighring 280

Chapter 10 Flywheel energy storage on the ISS 300 Why racing bike wheels are lighr 301 BID Energy srorage in rhe Achilles rendon 304 TIV Agitating Jroms 305

BID Jumping locusrs 307 Crash helmers 312 Runaway-truck ramps 314

Chapter 11 BID Energy in rhe body: inpurs 326 BID Calorie conrenr of foods 327 BID Energy in the body: omputs 327

BID Daily energy use for mammals

221 and rep riles 329 Bullers and Newron'srhird law 122 Spinning a gyroscope 222 BID Energy and locomorion 331 Get complete eBook Order by email at [email protected]

xvi

Get complete eBook Order by email at [email protected] Real-World Applications

Optical molasses 333 Temperature in space 334 TI Y Energetic cooking 335 Refrigeracors 341 Revecsible heat pumps 342 TIY Typing Shakespeare 345 BlO En[[opy in biological syscems 347 Efficiency of an automobile 348

Chapter 12 ~O Infrared images 362,393,394

Frost on Mars 366 ~O Swim-bladder damage co caugh~ fish 368 ~O Diffusion in the lungs 370 Chinook winds 377 Thermal expansion joints 379 TIY Thermal expansion to the rescue 380

BID Survival of aquatic life in winter 381 Hurricane season 382 Tin pest 383

BID Frogs that survive freezing 383 ~O Keeping cool 384 Gasoline engines 390 Carpet vs. tile for comfort 392 ~O

Penguin feathers 393 Heat transfer on earth 394 ~O Breathing in cold air 395 Ocean temperature 403

Chapter 13 Submarine windows 409 Pressure zones on weacher maps

411 Tire gauges 411 Barometers 412 BIO Measuring blood pressure 414 BIO Blood pressure in giraffes 414 BIO Body-fat measurements 416 Floating icebergs and boats 417 Hot-ait balloons 419 TIY Pressure forces 424

BIO BID BID BID BID BID

Airplane lift 424 Prairie dog burrows 424 Measuring arterial pressure 425 Cardiovascular disease 429 Intravenous transfusions 430

Chapter 14 BID Heart rhythms 445

BID TIY

BID

BIO TIY

BID

BID BID

Metronomes 451 Bitd wing speed 452 SHM in your microwave 454 Swaying buildings 455 Measuring mass in space 457 Weighing DNA 458 Car collision rimes 460 Pendulum prospecting 460 Animallocomotion 462 How do you hold your arms! 462 Gibbon brachiation 462 Shock absorbers 464 Tidal resonance 465 Musical glasses 466 Hearing (resonance) 467 Springboard diving 467 Spider-web oscillations 476

Chapter 15 BID Echolocation 477,488, 506 BID Ftog wave-sensors 480 BIO Spider vibration sense 481 TIY Distance co a lightning strike 483

Controlling exhaust noise 526

BKl The bat detector 528 Dogs' growls 529 Harmonics and harmony 537 Tsunamis 539

Chapter 17 TIY Observing interference 551

CD colors 556

BID Iridescent feathers 558 Antireflection coatings 558

Colors of soap bubbles and oil slicks 560 TIY Observing diffraction 565 Laser range finding 566 BID The Blue Morpho 573

Chapter 18 Shadow in a solar eclipse 578 Anti~gravity

mirrors 580

Optical image stabilization 583 Binoculars 584

Snell's window 585 Optical fibers 585 BKl Arthroscopic surgery 585 BIO Mirrored eyes (gigantocyptis) 594

BKl Range of hearing 488

Supermarket mirrors 596

Sonar imaging 488 BKl Ultrasound imaging 488 BID Owl ears 490 BID Blue whale vocalization 492

Optical fiber imaging 601 Mirages 608

Hearing in mice 493

BKl Hearing (cochlea hairs) 494 Solar surface waves 495 Red shifts in astronomy 497

BID Wildlife tracking with weathet radar 497 BID Doppler ultrasound imaging 498 Earthquake waves 499

Chapter 16 BID Shock wave lithotripsy 509 T I Y Through the glass darkly 512

The Tacoma bridge standing wave

515 Suing musical instruments 515 Microwave cold spots 516 BID Resonances of the ear canal 519 Wind musical instrumencs 520

BID Speech and hearing 521 Synthesizers 521 BID Vowels and formants 522

xvii

Chapter 19 BKl The Anableps "fout-eyed" fish 609 BKl A Naurilus eye 610 Cameras 610

BID The human eye 613 TlY Inverted vision 613

BID Seeing underwater 614 BKl Neat- and fatsightedness 615 Forced perspective in movies 6 17

BKl Microscopes 618-620,627 Telescopes above the atmosphere

621 Rainbows 623 BKl Absorption of chlorophyll 624 Fixing the HST 625

BID Optical and eieerron micrographs 627

BKl Visual acuity for a kestrel 629 BIO The blind spot 631 BID Surgical vision correction 635 BID Scanning confocal microscopy 637

Chapter 20

BID Saying "ah" 522 Blood pressure and flow 437 BKl Gel electrophoresis 642,664 Get complete by email at [email protected] Aerive noise reduerion 524 Scales of living creatutes 439 eBook Order TIY Charges on tape 645

xviii

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Real~World Applications

TlY Pulling water 648 ~O Bees picking up pollen 648 ~O Hydrogen bonds in DNA 651 ~O Separating sperm cells 652

Electrostatic precipitators 660 ~

Electric field of the heart 661 Static protection 663

Lightning rods 663 ~O Electrolocation 663 Cathode-ray tubes 665 ~O Flow cytometry 674

Chapter 21 BID Electropotencials around the

~O ~O ~O ~O

brain 675 Cause oflightning 679,711 Membrane potential 680 Medical linear accelerators 683 Shark elecrroreceptors 692 The electrocardiogram 694 Random-access memory 697

Camera flashes 700 00 Defibrillators 700 Fusion in the sun 702

Chapter 22 BID Percentage body-fat measurement

712,730-731 Monito ring corrosion in power

lines 714 TIY Listen to your potencial 718

The electtic torpedo tay 719 Fuel cells 719 Lightbulb filaments 722 Testing drinking water 723 00 Impedance tomography 723 Photoresistor night lights 725 Cooking hot dogs with electricity

728 Lightbulb failure 738

Chapter 23 00 Electric fish 739,775 Christmas- tree lights 745

H eadlight wiring 747 Transducers in measuring devices

749 Flashing bike light 755 Inrermittenr windshield wipers 757 BID Electricity in the nervous

system 757-764 BID Electrical nature of nerve and

muscle cells 758

Get complete BID Interpreting brain clecrrical potentials 762

Soil moisture measuremenr 765

The ground fault interrupter 871

BKl Cardiac defibrillators 774

Halogen bulbs 879 T he greenhouse effect 881

Chapter 24 BIO Magnetic resonance

imaging 776,789 TIY Buzzing magnets 780 Hard disk data storage 781 BKl Magnetocardiograms 787 The au rora 793

BID Mass spectrometers 794 BID Electromagnetic flowm eters 795 TlY Magnets and TV screens 795 Electric motors 802 BK) Magnetotactic bacteria 804 Loudspeaker cone function 805

The velocity selector 814 Ocean potentials 815

Chapter 25 BlO Color vision in animals 816,840 BID External pacemaker programming

817 BKl Shatk navigation 819 Generators 821 TlY Dynamo flashlights 821 Credit card readers 826

Magnetic braking 828 BID Transcranial magnetic stimulation

829 Radio transmission 830

The solar furnace 832 Polarizers 833 Polarization analysis 834

00 Honeybee navigation 834 TlY Unwanced tran smissions 837

Colors of glowing objects 838 BIO Infrared sensors in snakes 839 Astronomical images 841 Tethered satellite circuits 841

Metal detectors 851

Chapter 26 Charging electric tOothbrushes 856 Transformets 855-857 Power transmission 858

Household wiring 859 BlO Electrical safety 861 The lightning crouch 862 TIV TestingGFI circuits 863 Laptop trackpads 864

Chapter 27 Global Positioning Systems 886, 902, 913-914 The Stanford Linear Accelerator

905 Hyperspace in movies 910 Nuclear fission 913

00 Pion therapy 921

Chapter 28 00 Electron microscopy 922,936 00 X-ray imaging 923 BlO X-ray diffraction of DNA 925 BlO Biological effects of UV 928 00 Frequencies for photosymhesis 929 BIO Waves, photons and vision 931 TIV

High-energy moonlight 932 Photographing photOns 933 Scanning tunneling microscopy

943 BID Magnetic resonance imaging 945

Chapter 29 00 Spectroscopy 955 Colors of nebulae 956 Sodium filters for telesco pes 977 00 Fluorescence 979 BlO LASIK surgery 982 Compact fluorescent lighting 982

Light-emitting diodes 990

Chapter 30 00 Bone scans 991, 1010 Measuring past earth temperature

993 Nuclear fusion in the sun 996

Nuclear power 997 Plutonium "batteries" 1006

00 Radiocarbon dating 1007 BIO Radioactive isotopes for medicine

1002 BID Gamma-ray medical sterilization

1008 BIO Radiation dose from environmenral, medical sources 1009

00 Nuclea r medicine 1009 00 Nuclear imaging, PET scans 1010-1012

Under-pavement car detectors 866 Cerenkov radiation 1015 Cleaning upby computer 867 eBook Order email power at [email protected] Nuclear fission 1022 BIO Nuclear magnetic resonance 870

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Detailed Contents

c Preface to the Instructor

IV

Studying fot and Taking the MCAT Exam

x

P reface to the Student

XIV

Real-World Applications

XVI

'4#·'40'"

3.1 Using Veccots 3.2 Using Veccots on Motion Diagrams 3.3 Coordinate Systems and Vector

II14;i.1 Force and Motion OVERVIEW

Components

3.4 Motion on a Ramp 3.5 Relative Motion 3.6 Motion in Two Dimensions: Projectile

1

W hy Things Change

Vectors and Motion in Two Dimensions

Motion

ft

0

3.7 Projectile Motion: Solving Problems 3.8 Motion in Two Dimensions: Circular Motion SUMMARY

. ~(

QUESTIONS AND PROBLEMS

·y:t·,;;:Ut' 14U·!;llill Representing Motion 1.1 Motion: A First Look 1.2 Position and Time: Putting Numbers on Nature

1.3 Velocity 1.4 A Sense of Scale: Significant Figures,

2 3 6 9

Scientific Notation, and Units

11

1.5 Vecmrs and Morion: A First Look 1.6 Where Do We Go From Here?

17 22 24 25

SUMMARY QUESTIONS AND PROBLEMS

·4n·';·4;iJ 2.1 2.2 2.3 2.4 2.5 2.6

Motion in One Dimension Describing Motion Uniform Motion Instantaneous Velocity Acceleration Motion with Constanc Acceleration Solving One-Dimensional Motion

Problems 2.7 Free Fall SUMMARY QUESTIONS AND PROBLEMS

30 31 36 39 42 44 48 52 58 59

4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8

Forces and Newton's laws of Motion What Causes Motion? Force

A Short Catalog of Forces Identifying Forces What Do Fo rces Do? Newmn's Second Law Free-Body Diagrams Newton's Third Law SUMMARY QUESTIONS AND PROBLEMS

·4n·';·441 5.1 5.2 5.3 5.4 5.5 5.6 5.7 5.8

Applying Newton's laws Equilibrium Dynamics and Newton's Second Law Mass and Weight Normal Forces Friction

Drag Interacting Objects

Ropes and Pulleys SUMMARY QUESTIONS AND PROBLEMS

67 68 71

74 79 82 84 86 89 94 95

102 103 104 107 III 113 115 118 120 124 125

131 132 135 138 142 143 148 150 153 158 159

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xx

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Conservation Laws OVERVIEW

.41t,'1·4;11

Circular Motion, Orbits, and Gravity

6.1 Uniform Circular Mo tion 6.2 Speed, Velocity. and Acceleration 6.3 6.4 6.5 6.6 6.7

in U niform Circular Motion Dynamics of Unifo rm Circular Motion Apparenr Forces in Circular Motion Circular Orbits and Weighrlessness Newton's Law of Gravity Gravity and Orbits SUMMARY QUESTIONS AND PROBLEMS

19U"114;D

Rotational Motion 7.1 The Rotation of a Rigid Body 7.2 Torque 7.3 Gravitational Torque and the Cenrer of Gravity 7.4 Rotational Dynamics and Moment of Inertia 7.5 Using Newton's Second Law for Rotation 7.6 Rolling Motion SUMMARV QUESTIONS AND PROBLEMS

19161114;·:' 8.1 8.2 8.3 8.4

Equilibrium and Elasticity Torque and Static Equilibrium

Stability and Balance Springs and Hooke's Law Stretching and Compressing Materials SUMMARY QUESTIONS AND PROBLEMS

PART I SUMMARY ONE STEP BEYOND

PART I PROBLEMS

Force and Motion Dark Matter and the Structure of the Universe

Why Some Things Stay the Same

259

166 167 171 173 179 182 185 189 193 194

200 201 204 209 213 217 220 224 225

232 233 237 239 242 247 248 254 255 256

·91t·'1·400'

Momentum 9.1 Impulse 9.2 Momenmm and the Impulse. Momentum Theorem

260 26 1 262

9.3 Solving Impulse and Momentum Problems

9.4 Conservation of Momenrum 9.5 Inelastic Collisions 9.6 Momentum and Collisions in Two Dimensions

9.7 Angular Momentum SUMMARY QUESTIONS AND PROBLEMS

191!"1141111Energy and Work 10.1 10.2 10.3 10.4 10.5 10.6

The Basic Energy Model Work Kinetic Energy Potential Energy

Thermal Enetgy

266 268 274 275 276 28 1 282

289 290 294 298 301 304

Using the Law of Conservation

of Energy 10.7 Energy in Collisions 10.8 Power SUMMARY QUESTIONS AND PROBLEMS

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306 309 312 315 316

Get complete eBook Order by email at [email protected] Deta il ed Con te nts

'A#,'I·§;II'

Co

11 .1 11.2 11 .3 11.4 11.5 11.6 11.7 11 .8 11.9

Using Energy Transforming Energy Energy in ,he Body: Energy Inpurs Energy in ,he Body: Energy Ourpurs Thermal Energy and Temperature Heat and the First Law of Thermodynamics Heat Engines Heat Pumps Enrropy and ,he Second Law of Thermodynamics Sysrems, Energy, and Enrropy SUMMARY QUE STIONS AND PROBLEMS

PART II SUMMARY ONE STEP BEYOND

Conservation Laws Order Out of C haos

PART II PROBLEMS

322 323 326 327 331

334 338 341

'AiOI·§;I@1 13.1 13.2 13.3 13.4 13.5 13.6 13.7

Fluids

405

Fluids and Densi,y Pressure Measuring and Using Pressure Buoyancy Fluids in Motion Fluid Dynamics Viscosity and Poiseuille's Equation

406 407 411 415 419 422 427 431 432

SUMMA RY QUESTIONS AND PROBLEMS

343 346 349 350 356 357 358

xxi

PART HI SUMMARY ONE STEP BEYOND

Properties of Matter Size and Life

PART III PROBLEMS

438 439 440

Oscillations and Waves OVERVIEW

MOtion That Repeats Again and Again

443

Properties of Matter OVERVIEW

Beyond , he Particle Model

36]

'AiOI.§;1 tI

Oscillations

14.1 Equilibrium and Oscillation 14.2 Linear Restoring Forces 14.3 14.4 14.5 14.6 14.7

and Simple Harmonic Motion D escribing Simple Harmonic Motion Energy in Simple Harmonic Motion Pendulum Motion Damped Oscillations Driven Oscillations and Resonance SUMMARY QUESTIONS AND PROBLEMS

'A#·'I·§jlfA 12.1 12.2 12.3 12.4 12.5

Thermal Properties of Matter

T he Aromic Model of Marrer The Aromic Model of an Ideal Gas Ideal~ Gases Processes Thermal Expansion Specific Heat and Heat ofTransformJtion 12.6 Calorimetry 12.7 Thermal Properties of Gases 12.8 Heat Transfer

362 363 365 371 378

'A#·'I'§jl,., Traveling Waves and Sound 15.1 The Wave Model 15.2 Traveling Waves 15.3 Graphical and Mathematical D escrip tions of Waves

15.4 Sound and Ligh, Waves 381 15.5 Energy and Inrensiry 385 15.6 Loudness of Sound 387 390 15.7 The Doppler Effecr and Shock Waves SUMMARY 396 SUMMARY QUESTIONS AND PROBLEMS QUESTIONS AND PROBLEMS 397 Get complete eBook Order by email at [email protected]

444 445 447 449 455 460 463 465 469 470 477 478 479

483 487 490 492 495 500 501

xxii

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18.6 Image Formation with Spherical

Iln'!il;ll~ Superposition and Standing 507 508 509 511 516 520

Waves 16.1 The Principle of Superposition 16.2 Standing Waves 16.3 Standing Waves on a String 16.4 Standing Sound Waves 16.5 Speech and Hearing 16.6 The Interference of Waves from

SUMMARY QUESTI ONS AND PR O BLEM S

PART IV SUMMARY

Oscillations and Waves Waves in rhe Earth and the Ocean

PART IV PROBLEMS

Light is a Wave

i4#·!i·§jiQI

Optic al Instruments 19.1 The Camera 19.2 The Human Eye 19.3 The Magnifier 19.4 The Microscope 19.5 The Telescope 19.6 Color and Dispersion 19.7 Resolution of Optical Instruments

7

538

SUMMARY QUESTIONS AND PROBLEMS

539 540 PART

11,*1•• Optics OVERVIEW

SUMMARY QUESTIONS AND PROBLEMS

523 527 530 531

Two Sources

16.7 Beats

ONE STEP BEYOND

Mirro rs

18.7 The Thin-Lens Equation

v SUMMARY

ONE STEP BEVOND

Optics Scanning Confocal Microscopy

PART V PROBLEMS

593 597 602 603

609 610 613 616 618 620 622 624 630 631 636 637 638

543

lectricity and Magnetism OVERVIEW

Charges, Currents. and Fields

Electric Fields and Forces 20.1 20.2 20.3 20.4 20.5 20.6 20.7

Charges and Forces Charges, Atoms, and Molecules

Coulomb's Law The Concept of the Electric Field Applications of the Electric Field Conduccors and Electric Fields

Wave Optics 17.1 What Is Light! 17.2 The Interference of Light 17.3 The Diffraction Grating 17.4 Thin-Film Interference 17.5 Single-Slit Difftaction 17.6 Circular-Aperm re Diffraction SUMMARY QUESTIONS AND PROBLEMS

&#·'4·5;11:1 Ray Optics 18.1 18.2 18.3 18.4 18.5

544 545 548 553 556 560 564 567 568

642 643 649 651 655 658 662

Forces and Torques in Eleccric

Fields

·;U4414;IU

641

SUMMARY QUESTIONS AND PROBLEMS

Electric Potential

663 667 668

675

21 .1 Electric Potemial Energy and Electric Potemial

21 .2 Sou rces of Eleccric Potemial 21 .3 Electric Potemial and Conservation of Energy 21 .4 Calculating The Electric Pocemial 21 .5 Connecting Potemial and Field 21 .6 The Electrocardiogram 21 .7 Capacitance and Capacicors 21 .8 Dielectrics and Capacitors 21.9 Energy and Capacitors

574 The Ray Model of Light 575 Reflection 578 Refraction 581 Image Formation by Refraction 586 SUMMARY Thin Lenses: Ray Tracing 587 QUESTION S AND PROBLEMS Get complete eBook Order by email at [email protected]

676 678 680 684 691 694 695 698 699 703 704

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24.3 Electric Currents Also Create Magnecic Fields 24.4 Calculating the Magnetic Field Due to a Current

xxiii

782 785

24.5 Magnetic Fields Exert Forces on Moving Charges

789

24.6 Magnetic Fields Exert Forces on Currents

795

24.7 Magnetic Fields Exert Torques on Dipoles

799

24.8 Magnets and Magnetic Materials SUMMARY QUESTIONS AND PROBLEMS

i4:h';:'if f1 i4:£·'1·8#jA

Current and Resistance 22.1 A Model of Current 22.2 Defining and Describing Current

22.3 Batteries and emf 22.4 Connecting Potential and Current 22 .5 Ohm's Law and Resistor Circuits 22.6 Energy and Power SUMMARY QUESTIONS ANO PROBLEMS

712

713 715 717 720 724 727 732 733

Electromagnetic Ind u ction and Electromagnetic Waves

25.1 25.2 25.3 25.4 25.5

Induced Currents Motional emf Magnetic Flux Faraday's Law Induced Fields and Electromagnetic Waves 25.6 Properties of Electromagnetic Waves

23.1 23.2 23 .3 23 .4 23 .5 23.6

Circuits

739

Circuit Elements and Diagrams

740 741 743 748 750

Kitchhoffs Laws

Series and Parallel Circuits Measuring Voltage and Current More Complex Circuits CapacitOrs in Parallel and Series 23 .7 Circuits 23 .8 Electricity in the Nervous System SUMMARY QUESTIONS AND PROBLEMS

752 755 757 766 767

Electromagnetic Waves

25.8 The Electromagnetic Specrrum SUMMARY

i4#·"·8if .lld

AC Electricity

817 818 821 825 829 831 835 836 843 844 852

26.1 Alternating Current 26.2 AC Electricity and

853

Transformers 26.3 Household Electricity

855 859

26.4 Biological Effects and Electrical Safety 26.5 Capacitor Circuits 26.6 Inductors and Inductor

861 863

Circuits

26.7 Oscillation Circuits SUMMARY QUESTIONS AND PROBLEMS

i4:£·'1·8if 4 Magnetic Fields and ' Forces

816

25.7 The Photon Model of

QUESTIONS AND PROBLEMS

i4:hl·8#"

803 806 807

PART VI SUMMARY

Electricity and Magnetism

776 ONE STEP BEYOND The Greenhouse Effect and Global Warming 24.1 Magnetism 777 VI PROBLEMS 24.2 The Magnetic FieldeBook Order by email 778 at PART Get complete [email protected]

865 867 873 874 880 881 882

xxiv

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'40,'4'4i*;:1 Atoms and Molecules

Modern Physics OVERVIEW

New Ways of Looking ar rhe World

29.1 Spectroscopy 29 .2 A toms 29.3 Bohr's Model of Aromic

885

Quantization

29.4 The Bohr Hydrogen Arom 29.5 The Quantum -Mechanical Hydrogen Arom 29.6 Multielectron Atoms 29.7 Excited States and Spectra 29.8 Molecu les 29.9 Stimulated Emission and Lasers SUMM ARY QUESTIONS AND PROBLEMS

14#·"·ii;'14 27 .1 27 .2 27 .3 27 .4 27 .5 27 .6 27 .7 27.8

Relativity

886

Relariviry: Wh ats Ir All About!

887 887 891 894 897 899 904

Galilean Relativity Eins tein's Principle of Relariviry Events and Measurements

The Relariviry of Simulraneity Time Dilation Length Contraction

Velocities of Objects in Special Relarivity

27 .9 Relativistic Momentum 27 .10 Relativistic Energy SUMMARY QUESTIONS AND PROBLE MS

906 908 910 915 916

1·'t.!414;iUI 30.1 30.2 30.3 30.4 30.5 30.6

28.1 28.2 28.3 28.4 28.5 28.6 28.7 28.8

X Rays and X-Ray Diffracrion The Phococleccric Effecc Phocons Maner Waves

Energy Is Quantized Energy Levels and Quantum Jumps The Uncertainty Principle Applicarions and Implicacions

of Quanrum Theory SUMMARY QUESTIONS AND PROBLEMS

922 923 925 931 933 936 939 940 943 946 947

955 957 960 963 969 971 974 978 980 984 985

Nuclear Physics

991

Nuclear Strucrure

992 994 997 999 1003

Nuclear Srabiliry Forces and Energy in the N ucleus Radiati on and Radioac tivity

Nuclear Decay and H alf-Lives Medical Applicarions of Nuclear Physics 30.7 T he Ulrimare Building Blocks of Mactcr SUMMARY

QUESTIONS AND PROB LEM S

PART VII SUMMARY ONE STEP BEYOND

'AiBA,Sifi :1 Quantum Physics

954

Modern PhysiCS The Physics of Very Cold Aro ms

PART VII PROBLEMS

1007 1011 1016 1017 1023 1024 1025

Appendix A Machemacics Rev iew

A-1

Appendix B Periodi c Table of che Elements Appendix C AccivPhysics OnLine Accivicies

A-3

Appendix D Acomic and Nuclear D aca

A-5

Answers ro Odd-Numbered Problems Credits

A-9

Index

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A-4

C-1

I-I

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These images of animal eyes and light-sensing organs reveal the wide range of structures that can produce a visual sense. There is a certain similarity among eyes as well; did you spot the false eyes, patterns that are designed to mimic the appearance of an eye?

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Light Is a Wave Tsaac Newton is best know n for hi s studi es of mec hani cs and th e three laws th at bear his name, but he also did important earl y work on opti cs. He was the first person to carefull y study how a pri sm breaks white light into colors. New to n was a stro ng proponent of th e "corpu scle" theory of li ght, arguing that li ght co nsists of a stream of particles. In fac t, Ne wt on wasn' t quite ri ght. As you will see, th e beautiful colors of a peacock 's feath ers and the shimmery rainbow of a soap bubble both depend on the

fact that li ght is a wave, not a particle. In particular, li ght is an electromag neti c wave, although these chapters depe nd on nothing morc than the "waviness" of light waves for your unde rstanding. The wave theory we developed in Pm1 TV will be put to good use in Part V as we begin our in vesti gati on of li ght and optics with an analys is of the wave model of light.

The Ray Model Yet Newton was correc t in hi s observation th at li ght see ms to trav el in strai ght lines, so m e thin g we wo uldn ' t ex pect a wave to do . C on sequ e ntl y, o ur investi gati ons of how light works will be greatl y aided by another model of light, the ray model, in whi ch li ght tra ve ls in strai ght lines, bounces from mirrors, and is bent by lenses. The ray model will be an excell e nt tool for analy zin g man y of th e practi cal appli cati ons of optics. Whe n you look in a mirror, you see an image of yourself th at appears to be behind the mirror. We will use the ray model of light to dete m1ine just how it is that mirrors and lenses form images. At the sam e time, we will need to reconcile the wave and ray modeJs, learning how they are related to eac h other and whe n it is appropri ate to use each.

Working with Light The nature of light is quite subtl e and e lusive. In Parts VI and vn, we will turn to the qu es ti o n of just what li g ht is . As we w ill see, li ght has both wave- like and particl e- like aspects. For now, however, we will set thi s questi on as ide and work with the wave a nd ray models to develop a practi cal understandin g of light. This will lead us, in Chapter 19, to an anal ys is of some common optical instrum ents. We will ex pl ore how a camera captures images and how te lescopes and mi c ro scopes work. Ultimately, th e fact that you are readin g thi s book is due to th e optics of the first optical instrument you ever llsed, your eye! We will in vesti gate the optics of the eye, learn how the cornea and lens be nd li ght to c reate an image on your retina, and see how g lasses or contact lenses can be used to correct the image should it be out of foc lls.

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LOOKING AHEAD .. Th e go al of Ch apte r 17 is t o understand and use t he w ave m odel of light.

What Is Light?

Interference of Light

You'lIlcarn that li ght has aspects of waves, rays, and particles. We ' ll develop mode ls for eac h of these in thi s and coming chapters; this chapter w ill concen trate on the wave model of ligh t.

Like all waves, li ght waves of the same freq uency can interfere constructi ve ly or destruct ive ly.

Double·slit interference When li ght shines o n two narrow, cl osely spaced slits, interference frin ges are seen on a screen beh ind the slits. Interference is a clear indi cation of the wave nature of li ght.

Fringes seen with green lig ht.

The diffraction grating The colors of soap bubbles can be understood using th e

w a ve m od e l of light.

To understand the bending of light by a co ntact lens, the ray model (C hapter 18) is appropriate.

Solar cells generate electri city from sunlight. We'll use the phot on mo del (C hapter 28) t o understand how.

Many closely spaced slits or grooves form a ditTraction gratin g, capab le of break ing whi te li ght into its component co lors. The m icroscop ic grooves in a CD act as a diffraction grating, leading to its colo rfu l appeara nce.

Looking Back .. 15.4 Ught and electrom agnetic waves

Thin·film interference

Diffraction One of the most bas ic aspects of waves is that they can be nd . o r diffract. around the edges of objec ts. The diffracti on of light, although difficult to obse rve because of li gh t' s small wavelength, is an ind icat ion that li ght is a wave. A careful exami nation of the sha dow of this razor blade Get complete shows fringe s due to diffraction of light.

Interference is al so poss ible between waves refl ecting o ff the front and back surfaces of a lhin transparent film.

Looking Back .. 15.5 Circular and plane waves 16.3 Reflection s 16.6 Interference

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The colors in an oi l slick result f rom th in·film interference. [email protected]

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17.1

What Is Light?

545

17.1 What Is Light? What is light? The first Greek scientists and philosophers did not make a distinction between li ght and vision. Li ght, to them, was not so mething that ex isted apart from seeing. But gradually there arose a view that tight actually "ex ists," that Light is some so rt of physical entity that is present regardl ess of whether or not someo ne is looking. Newton, in addition to doing pioneering work in mathematics and mechanics in the l660s, was also one of the early investigators of light. He believed that li ght consists of very small , light, fast particles, which he called corpuscles, traveling in straight lines. Newton was vigorously opposed by Robert Hooke (of Hooke's law) and the Dutch scienti st Christiaan Huygens, who argued that light was some sort of wave. Although th e debate was Li ve ly, and sometim es acrimonious, Newton eventually prevailed. The be lief that light co nsists of corpuscles was not seri ously questioned for more than a hundred years after Newton's death . The situation changed dramatically in l80 I) when the English scientist Thomas Youn g ann ounced that he had produced interference betwee n two waves of light. Youn g's ex periment, which we will analyze in thi s chapter, was a painstakingly difficult ex perim ent with the technology of his era. Nonetheless, Young's ex pe rime nt appeared to settle the debate in favor of a wave theory of li ght because interference is a di stinctl y wave-like phenome non. But if light is a wave, why does it sometimes see m to travel in straight lines? And just what kind of wave is it?

Models of Light As Newton, Youn g, and othe rs found, the nature of light is e lusive. Light is the chameleon of the physical world. Under some circumstances, light acts like particles traveling in straight lines. But chan ge the circumstances, and li ght shows the sa me kind s of wave-like behavior as so und waves or water waves. Change the circumstances yet again , and light exhibits behavior that is neithe r wave-like nor particlelike but has characteristics of both. Rathe r than an all-encompassing "th eory of light," it will be bette r to develop three models of light. Each model successfuU y explains the behavior of li ght within a certain domain-that is, within a certain range of physical situations. Our task will be twofold:

J. To develop clear and distinct models of light. 2. To learn the conditions and circum stances for which each model is valid. The seco nd task is especially important. We' ll begin with a brief summary of all three models, so that yo u will have a road map of where we're headed. Each of these model s wiU be developed in the comin g chapters. All three models of light are needed to The wave model: The wave model of light is the most widely applicable model, explain the operation of a CD player and responsible for the widely known "fact" that li ght is a wave. It is certainl y true that, the rainbow colo rs of the CD itself. under man y circumstances, light ex hibits th e same behavior as so und or water waves. Lasers and electro-optical devices, critical technologies of the 2 1st century, are best understood in te rm s of the wave model of li ght. Some aspects of the wave model of li ght were introduced in Chapters 15 and 16, and the wave model is the primary focu s of thi s chapter. The study of light as a wave is called wave optics. The ray model: An equally well-known "fact" is that li ght trave ls in a straight line. These strai ght-line paths are called light rays. In Newton's view, light rays are the traj ectori es of particle-like corpuscles of light. The properties of prisms, mirrors, lenses, and optical instrume nts suc h as te lescopes and microscopes are best understood in term s of light rays. Unfortunately, it' s difficult to reconcile the stateme nt " light travels in a strai ght line" with the stateme nt " light is a wave." For th e most part, waves and rays are mutually exclusive models of li ght. One of our mostimportant tasks will be to lea rn when each mode l is appropriate. The ray model of li ght, the Get complete by email at [email protected] basis of ray optics, is the subj ect ofeBook the nextOrder chapter.

546

Get complete eBook Order by email at [email protected] CHAPTER 17

Wave Optics

FIGURE 17 . 1 Light passing through an opening makes a sharp-edged shadow.

The photon model: Modern technology is increasingl y reliant on quantum physics. In the quantum world, light behaves like neither a wave nor a particle. Instead, Jj ght consists of photolls that have both wave-like and particle-like properties. Photons are the qU(lJIta of light. Much of the quantum theory of light is beyond the sco pe of this textbook, but we will take a peek at th e important ideas in Chapters 25 and 28 of this text.

The Propagation of Light Waves shows an everyday observation about light. The sunli ght passing through the door makes a sharp-ed ged shadow as it falls upon the floor. Thi s behavior is exactly what yo u would ex pect if light consisted of little particles traveling outward from the source in straight lines. Indeed, it was the observation of sharp-edged shadows that led Newton to propose the idea of corpuscles of light. Is this behavior, where light seems to tra vel in st raight lines, consistent with th e motion of a wave? Li ght waves oscillate so rapidly-at so me IOt4 Hz-that we cannot hope to directly observe the crests and troughs of its electric or magnetic field s. Instead, let's use ordinary water waves to illu st rate so me basic properti es of wave motion common to all waves, including light. FIGURE 17 .2 shows a water wave, as seen looking strai ght down on the water surface. The wave e nters from the top of the picture, then passes though a window-like opening in a barrier. After passing through the open ing, th e wave spreads oul to till the space behind the opening. This spreading of a wave is the phenomenon called diffraction. Diffraction is a sure sign that whatever is passing through the opening is a wave. The straight-line travel of light appears to be incompatible with this spreading of a wave. Notice, however, that the width of the opening in Figure 17.2 is only slightl y larger than the wavelength of the water wave. As FIGURE 17 .3 shows, so mething quite differe nt occurs when we make the opening much wider. Rather than spreading out, now the wave co ntinues to move straight forward, with a well-defined boundary between where the wave is moving and its "shadow," where there is no wave. This is just th e behavior observed for light in Figure 17. 1. Whether a wave spreads out or travels straight ahead, with sharp shadows on both sides, ev identl y depends on th e size of the objects with which th e wave interacts. The spreading of diffraction becomes noticeable only when an opening or object is " narrow," comparable in size to the wavelength of the wave. Thus we wou ld expect a light wave to spread out in this way only when it passes objects that are comparable in size to the wavelength of light. But the wavelength of light is extremely short. Later in thi s chapter we'll see how its wavelength can be meas ured, and we'll find that a typical wavelength of li ght is only about 0.5 .um. When light waves interact with everyday-sized objects, such as the opening in Figure L7.1, the situation is like that of the water wave in Figure 17. 3. The wave travel s straight ahead, and we'll be able to use the ray model of light. Only when the size of an object or an opening approaches the wavelength of light does the wave-like spreading become important. For example, FIGURE 17 .4 on the next page shows th e shadow of a 0.5-mm-diam eter pin. For an object this small , th e shadow is !lot sharp, but shows a series of bright and dark fringes as we move away from the center of th e shadow. One of the goals of thi s chapter is to unde rstand the origin of such fringes, a goal for which we' ll need the wave model of light. FIGURE 17 . 1

FIGURE 17 .2 A water wave passing through a narrow opening in a barrier.

A wave approm.: hes

f

a narrow openi ng

'"

from th is side.

~

T he wave spre ads out behind the opening.

FIGURE 17 .3 A water wave passing through an opening that's many wavelengths wide.

.:---.

\~~: - -~ ,

"S hadow"

I

Light Is an Electromagnetic Wave "Shadow"

If light is a wave, what is it that is waving? This was the question posed at the start of the 19th century by Young's demonstration of the interference of light. As we brietly noted in Chapter IS , it was ultimately established, through theoretical and experimental efforts by numero us scie nti sts, that light is an electromagnetic wave, an oscilof electric We will examine the nature of electromagnetic Get completelation eBook Orderand by magnetic email atfields. [email protected]

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17.1

waves in more detail in Part Vl after we introduce th e ideas of electric and magnetic fields. For now we can say that light waves are a "self-sustaining oscillation of the electromagnetic field." Being se lf-sustaining means that electromagnetic waves require 110 material medium in order to travel; hence electromagnetic waves are not mechanical waves. Fortunately, we can learn about the wave properties of light without having to understand electromagnetic fields. In fact, the discovery that light propagates as a wave was made 60 years before it was realized that light is an electromagnetic wave. We, too, will be able to learn much about the wave nature of light without having to know just what it is that is waving. Recall fhatall electromagnetic waves, including light waves, travel in a vacuum at the same speed, called the speed of light. Its value is Vlight

=

C

What Is Light?

547

FIGURE 17.4 Diffraction of light by the point of a pin.

8

= 3.00 X 10 mls

where the symbol c is used to designate the speed of light. Recall also that the wavelengths of light are extremely small, ranging from about 400 nm for violet light to 700 nm for red light. Electromagnetic waves with wavelengths outside this range are not visible to the human eye. A prism is able to spread the different wavelengths apart, from which we learn that "white light" is aLI the colors, or wavelengths, combined. The spread of colors seen with a pri sm, or seen in a rainbow, is called the visible spectrum. If the wavelengths of light are incredibly small, the oscillation frequencies are unbelievably high. The frequency for a 600 nm wavelength of light is v

f= - = A

3.00 X LOs m/s 600XI0

9

m

14

=5.00X 10 Hz

The frequencies of light waves are roughly a factor of a trillion (10 12 ) higher than sound frequencies.

The Index of Refraction Light waves travel with speed c in a vacuum, but they slow down as they pass through transparent materials such as water or glass or even, to a very slight extent, ai.r. The slowdown is a consequence of interactions between the electromagnetic field of the wave and the electrons in the material. The speed of light in a material is characterized by the material's index of refraction It, defined as 11=

speed of light in a vacuum speed of ljght in the material

c

v

( 17.1 )

where v is the speed of light in the materiaL The index of refraction of a material is always greater than 1 because v is always less than c. A vacuum has II = I exactly. Table 17.1 lists the indices of refraction for several material s. Liquids and solids have higher indices of refraction than gases, simpl y because they have a much higher density of atoms for th e light to interact with.

ImJDI .... An accurate value for the index of refraction of air is relevant only in very precise measurements. We will assume /lair = 1.00 in this text. ...

White light passing through a prism is spread out into a band of colors called the visible spectrum.

If th e speed of a light wave changes as it enters into a transparent material , such as glass, what happen s to the light's frequency and wavelength? Because v = Af, TABLE 17.1 Typical indices of refraction either A or f or both have to change when v changes. Material Index of refraction As an analogy, think of a sound wave in the air as it impinges on the surface of a I exactly pool of water. As the ai.r oscillates back and forth , it periodjcally pushes on the sur- Vacuum l.OOO3 face of the water. These pushes generate the compressions of the so und wave that Air continues on into the water. Because each push of th e air causes one compression of Water 1.33 the water, the frequen cy of the sound wave in the water must be exactly the same as Glas!> 1.50 the frequency of the sound wave in the air. In other words, the frequency of a wave Diamond 2.42 does not change as thecomplete wave moves from one medium to another. Get eBook Order by email at [email protected]

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CHAPTER 17

Wave Optics

FIGURE 17.5 Light passing through a

transparent material with index of refraction II. A Iran"pare11l material in wh ich li ghttra \'e l ~ slowe r. at speed I' = d n

Vacuumll = I

-

\.

Index

II

/I

= I

The same is true for electromagnetic waves, although the pushes are a bit more complex as the electric and magnetic fi elds of the wave interact with th e atoms at th e surface of the material. Nonetheless, th e frequency does not change as the wave moves from one material to another. FIGURE 17 .5 shows a light wave passing through a transparent material with index of refraction n. As the wave travels through a vacuum it has wavelength Avac and frequ ency f w.tc such that AVlJcf vac = c. In th e material , Amat};nat = v = cill. The freq uency does not change as th e wave enters (fm;]! = f~'ac), so the wavelength must change. The wavelength in the material is A m~!

=-

II

fmal

\

C

C

"'-vat·

= -- = -- = -n/mal l~fvac Il

(17.2)

The wavelength in the transparent material is shorter than the wavelength in a vacuum. This makes sense. Suppose a marching band is marching at one step per second at a speed of I m/s. Suddenly th ey slow their speed to ~ mls but maintain their march at one step per second. The only way to go slower while marching at the same pacejs to take smaller sfeps. When a Light wave enters a material, the only way it can go slower while oscillating at th e same freq uenc y is to have a shorter wave-

T he wavelength inside the mate rial decrea."es. but the rreque ncy doesn ' t c hange .

le/lgth.

Analyzing light traveling through glass

EXAMPLE 17 1

Orange light with a wavelength o f 600 nm is incide nt on a I.OO-mm-thick glass microscope s lide . a. What is the li ght speed in the glass? b. How many wavelengths of the li ght are ins ide the s lide? SOLVE

a. From Table 17.1 we see that the index of refraction of glass is II glass = 1.50. Thus the speed of li ght in glass is

c Vg lass

= --

n glass

=

3.00 X lOll

1.50

m/s

2.00 X 10' rnls

A ljght wave travels through three transparent materials of equal thickness. Rank in order, from the highest to lowest, the indices of refraction Ill' 1l 2' and 11 3 '

b. Because Ilai r = 1.00, the wavelength of the light is the same in air and vacu um: A v~e = A uir = 600 nm. Thus the wavelength in side the glass is

,\

glass

A vae 600 om 7 = - - = - - - = 400 nm = 400 X 10- In 1.50 . II glass

N wavelength s span a distance d = NA, so the number of wavelengths in d = 1.00 mm is

d 1.00 X 10- 3 In N=- = A 4.00 X 10 7 m

2500

ASSESS The fact that 2500 wavelengths fil wi thin I mm shows how smaU the wavelengths of li ght are.

-

17.2 The Interference of Light Consider the situation shown in FIGURE 17 .6 . Light passes through a narrow openinga slit-that is only 0.1 mm wide, about twice the width of a human hair. This situation is similar to Figure 17.2, where water waves spread out after passing through a narrow opening. When light waves pass through a narrow slit, they too spread out Get completebehind eBook at wave [email protected] theOrder slit, justby as email the water did behind the opening in the barrier. The li ght

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17.2 The Interference of Light

is exhibiting diffractioll, the sure sign of waviness. We will look at diffraction in more detail later in the chapter. For now, we merely need the observation that li ght does, indeed, spread out behind an opening that is sufficiently small.

549

FIGURE 17.6 Light, just like a water wave, sp reads out behind a slit in a screen if the slit is sufficient ly small.

The

' p,:::: e, we can substitute (Jill from EquationI7.? for tan(Jmin Equation 17.4 to find that the mth bright fringe occurs at position

111

=

0, 1, 2, 3, ...

(l7.8)

Positions of bright fringes for double-sli t interference at screen di stance L

The interference pattern is sy mmetrical , so there is an mth bright fringe at the same di stance on both sides of the center. You can see this in Figure 17.?b.

ImlD .... Equations 17.7 and 17.8 do 1101 apply to th e inte rference of sound waves from two loudspeakers. The approximations we've used (small angles, L » d) are usuall y not valid for the much longer wavelengths of so und waves . ..... EX AMPLE 17 2

Observing interference It 's acrually not that hard to observe doub le-s lit interference. Place a piece of alu minum foil on a hard sunace and cut two parallel slits, about I mm apart. using a razor blade. Now hold the slits up to your eye and look at a di stant small bright light , such as a streetlight at night. Because of diffract ion , you' ll see the li ght source spread out in a direct ion perpel/dicular to the long direction of the slits. Superimposed on the diffraction pattern is a fin e pattern of interference max ima and mini ma, as seen in the photo above of Chri stmas tree lights taken through two slits in foil.

How far do the waves travel?

Li ght from a helium -neon laser (A = 633 nm) illuminates two slits spaced DAD mm apart. A view ing screen is 2.0 m behind the slits. A bright frin ge is observed at a point 9.5 mm fro m the cen ter of the screen. What is the fringe number m, and how much farther does the wave from one slit travel to thi s po int than the wave from the other slit ?

SOLVE

Solving Equation 17.8 fo r m g ives

Ymd (9.5 X 10- 3 m)(OAO X 10- 3 m) m= = =3 ilL (633 X 10 9 m)(2.0 m) Then the ex tra di stance traveled by o ne wave co mpared to the other is

ar = lilA = 3(633 X 10- 9 m) = 1.9 X A bright frin ge is observed when one wave has traveled an integer number o f wavele ngth s farth er lhan the other. Thus we know that ~r must be iliA , where 111 is an integer. We can find m from Equation 17.8.

PREPARE

ASSESS The path-l ength diffe rences in two -slit interference are generally very small , j ust a few wavelengths of li ght. Here, 61' is o nl y about o ne part in a million of the 2 m di stance tra veled by the waves!

Equation 17.8 predi cts that the interference pattern is a series of equally spaced brigbt lines on the screen, exactly as shown in Figure 17.7h. How do we kn ow the fringes are equally spaced? The fringe spacing between fring e m and frin ge m + lis 1\.)'-)' -)'", -",+ 1

(m

+ d

I )AL

mAL AL - -d- -- - d

10- 6 m

( 17.9)

Because ..1.y is independent of 111 , allY two bright fringes have the sam e spacing. The dark fringes in the photograph are bands of destructive interference. You Get eBook Ordernce byoccurs emailatatposition [email protected] learned in Chapter 16complete that destru ctive interfere s where the

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17

CHAPTER

Wave Optics

FIGURE 17 .9 Symbols used to describe two-slit interference.

y

Path-len gth differe nce Ar 10 a bri ght or durk frin ge

Ym is rhe the

position ?f the

IIIth

\..

.;. 4A

Y4

TnU XTnlWll.

,

7A/2 3A 5A/2

Y3

- Y,

.. .......• Y~

r' is tht ' p~s ilion of the

Y2 Y; Y,

mth m ini mum .

111

=

I

( 17.10)

We can use Equation 17.6 for dr and the small-angle approximation to find that the dark frin ges are located at positions

Y'm =

3A(l

A

(Ill + ~) 1IL

111

d

2

=

0, I, 2, 3, .

(17. 11 )

Positi o ns of dark frin ges fo r do ubl e-s li t in terfe rence

A/2

m=O

OA

Centmll1. . . .

We have used Y;I1 ' with a prime, to disti.nguish the location of the mth m.inimum from the mth maximum at YI11' You can see from Equation 17.11 that the dark fringes are located exactly halfway between the bright fringes. FIGURE 17.9 summarizes the symbols we use to describe two-slit inte rference. FIGURE 17 .10 is a graph of the double-slit intensi ty versus y. Notice the unusual orientation of the graph , with the intensity increasing toward the le ft so that the y-ax is can match the experimental layout. You can see that the intensity oscil.lates between dark fringes, where the intensity is zero, and equally spaced bright fringes of maxi mum i.nte nsity. The maxima occur at positions where YIIl = mAL/d.

muximum

FIGURE 17.10 Intensity of the interference fringes in the double-slit experiment.

The frin ge intensity

y

decreuses becuusc thc fig lH intcns ity from clu.: h slit by itself is 110 t un ifonn . ".

m

EXAMPLE 17 3

Measuring the wavelength of light

A double-slit interference pattern is observed on a sc reen 1.0 m behind two slits spaced 0.30 mm apart From the center of one parti cu lar frin ge to thecenter of the ninth bright fringe from this one is 1.6 e m. What is the wavelength of the light? PREPARE It is not always obvious which fringe is the cen tral maximum . Slight imper-

fections in the slits can make the interference fringe pattern less than idea l. However, you do not need to identify the m = 0 fringe because you can make use of the fact that the frin ge spac in g ~ y is uniform. The interference patte rn looks like the photograph of Figure 17.7h.

Fringe

spacing Ay

m = 0, 1, 2, 3, ...

2A

111=2

Y~

o

path-le ngth difference of the waves is a whole number of wavelengths plus half a wavelength:

'----I

SOLVE

USlits

The fringe spac ing is

Ccntral ------

1.6cm

maximum

~y = -- = 1 8X I0 3 1n

9

Thc bright frin ges ~ are equall y spaced . .....•

.

Using this frin ge spac ing in Equation 17.9, we find Ihalthe wavelength is

d

A = -~ )'

L

lncreas ing - - - - - - - - j o light intensity

.

3.0 X 10 4 m =-'-'-"'----'" ( l. 8 X 10- 3 In) 1.0 m

= 5.4 X 10- 7 III = 540 nm

It is customary to express the wavelengths of light in nanometers. Be sure to do this as you solve problems. ASSESS You learned in Chapter 15 that visible light spans the wavelength range 400-700 nm, so finding a wavelength in this ran ge is reasonable. In fact, it's because of experiments like the double-slit experiment that we're able to measure the wavelengths of li ght.

STOP TO TH I NK 17 3 Light of wavelength A, illuminates a double slit, and interfere nce fringes are observed on a screen behind the slits. Whe n the wavelength is changed to Az• the fringes get closer together. Is Az larger or smaller than AI ?

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17.3 The Diffraction Grating

553

17.3 The Diffraction Grating Suppose we were to replace th e doubl e slit with an opaque screen that has N closely spaced slits. When illuminated from one side, eac h of these slits becomes the source of a Li ght wave that diffracts, or spreads Ollt, behind the slit. Such a multi-slit device is called a diffraction grating . The Li ght inte nsity pattern on a screen be hind a diffraction grating is du e to the interference of N overlapped waves. FIGURE 17. '1 shows a diffraction gratin g in which N slits are equally spaced a distance d apart. This is a top view of the grating, as we look down on the experiment, and the slits extend above and be low the page. Onl y 10 sLits are shown here, but a practical grating will have hundreds or even thousands of slits. Suppose a plane wave of wavelength A approaches from the left. The crest of a plane wave arrives simultaneously at each of the sLits, causing the wave emergi ng from each slit to be in plwse with the wave emerging from every other slit- that is, all the emerging waves crest and trough si multan eo Lisly. Each of th ese eme rging waves spreads aLit, just like th e li ght wave in Figure 17.6, and after a short distance they aU overlap with each other and interfere. We want to know how the interfere nce pattern will appear on a screen behind the gratin g. The Light wave at the sc reen is the superposition of N waves, from N slits, as they spread and overlap. As we did with the double sLit, we'll assume that the di stan ce L to the sc reen is very large in co mpariso n with the sl it spaci ng d; hence the path followed by the light fro m one slit to a point on the screen is very Ilearly parallel to the path followed by the light from nei ghborin g slits. The paths ca nnot be perfectly parallel, of cou rse, or they wou ld never meet to interfere, but the slight deviation from perfec t parallelism is too small to noti ce. You can see in Fjgure 17.J 1 that th e wave from one slit travel s di stance D.r = dsin8 farther than the wave from the slit above it and D.r = dsinB less than the wave below it. This is th e same reasoning we used in Figure 17.8 to analyze the doubl e- slit ex periment.

FIGURE 17.11 Top view of a diffracti on grating w ith N = 10 slits.

N sli ls with

A

_--

....

~~ ~

Plane wave II approaching from lefl

FIGURE 17.12 Interference for a grating with fi ve slits.

Grating wi lh five slits

The path· lcngth di ffcrencc. Il r. betwccn these paths from fi ve adjaccnl sli ts is exaclly I A, s~ thai'" = I ,

/

Screen

• ~--------------

0,

0,

Y 111 = 2

Y,

III

o

", = 0

-Y ,

11/ = 1

-Y2_

l' Ilr between these paths is exactly 2A (III = 2) .

= I

Y,

m= 2

Appearance of screen

L

Figure 17. 11 was a magnifi ed view of the slits. FIGURE 17 .12 steps back to where we can see the viewing screen, for a gratin g with five slits. If th e angle B is such that D.r = dsin 8 = mA, where 111 is an integer, then the light wave arriving at the sc ree n from one slit wi.ll tra vel exactly m wavelengths more or less than light from the two slits next to it, so these waves will be exactly ill phase with each othe r. But eac h of those waves is in phase with waves from the sLits next to t.hem, and so on until we reach the end o f the gratin g. In other words, N light waves, from N different slits, will all be in phase with each other when they arrive at a point on the screen at angle Om such that

d sinB,., = mA

111

= 0, I , 2, 3, .

(17.12)

A ngles orbrig hL rringes d ue Lo a

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d iffraction g raling wi th s lits di stance d apart

Spreadi ng ci rcular wuves from euch slit overlap und interfere .

'. "The wave from each sli t travel s Il r = d sinO farther than Ihe wave from Ihe slilabove it.

554

CHAPTER

17

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16.4,16.5

Actj'v

physcs

The screen will have bright constructive-interference fringes at the values of 8m given by Equation 17.12. When this happens, we say that the light is "diffracted at angle 8m ." Because it's usually easier to measure distances rath er than angles, the position Ym of the I11th maximum is

(17. 13)

Ym = Lta n8m Pos ilions of brig hi frin ges due to a dirrraction grating di stance L rrom screen

The integer m is called the order of the diffraction. Practical gratings, with very small values for d, display only a few orders. Because d is llsuall y very small , it is customary to characterize a grating by the number of lilles per millimeter. Here " Line" is sy nonymous with "sLit," so the number of Lines per milLimeter is simply the inverse of the slit spacing d in miU.imeters.

Imm .. The condition for constructive interference in a grating of N sLits is identical to Equation 17.6 for just two slits. Equation 17.12 is simply the requirement that the path-length difference between adjacent slits, be they two or N, is m)... But unlike the angles in double-slit interference, the angles of constructive interference from a diffraction grating are generally not small angles. The reason is that the slit spacing d in a diffraction grating is usually so smaLl that AId is not a small number. Thus you C(IIl/wt use the small-angle approximation to si mpLify Equations 17.12and 17.13 . .. Although the condition for constructive interference is the same for a diffraction grating as it was for th e double sLit, the intensit y of the fringes for a grating differs from that for a double slit i.n two important ways. First, the bright frin ges of a diffraction grating are lIarrower than those of a double slit with the sa me slit spacing d. Second, the fringes are brighter. For a grating with N slits, it can be shown that th e maximum inten sity of a bright fringe is

(17. 14) Max imum intensity of a bri ght frin ge for a d irrraction grating Wilh N slits

QUADRATK

where II is the intensity of the wave from a single slit. For a double slit, with N = 2, th e intensity of the central maximum is 4 times that of eac h sLit alone. For a practical diffraction grating, with hundreds of slits, the bri ght fringe s are enormously brighter. Not only do the fringes get brighte r as N increases, they also get narrower. If you double the number of slits, for instance, twice as much light will reach the screen. But according to Equation 17.14, the maximum intensity of a bri ght fringe wou ld increase by a factor of 4. Since the re's only twice as much light to work with, the fringes must be narrower in order to concentrate the li ght in each fringe enough to make it four times brighter. To demonstrate th e sig nificance of the number of slits, FIGURE 17 .13 on the next page shows the interference pattern due to three diffraction gratings with the sa me slit spacing d but increasing numbers of slits. The case with N = 2 is that of doubl eslit interfe rence. As the number of slits is increased, the bright fringes become mu ch narrower. For a realistic diffraction grating, with N > 1000, the interference pattern Get completeconsists eBookofOrder emailofatvery [email protected] a smallby number bright and very narrow fringes while most of the scree n remains dark.

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17.3 The Diffraction Grating

555

FIGURE 17.13 The intensity on the screen due to three diffraction gratings. Notice that the

intensity axes have different scales.

N=2

N ~

10

i, \

As the num ber

0

-• •• •-

N ~

50

I 2500

I

0

slit s in the grating increases. the rringes get narrower and bri gh ter.

FIGURE 11.14 A diffraction grating can be

used to measure wave lengths of light.

Spectroscopy As we'll see in Chapter 29, eac h atomic element in the periodic tabl e, if appropriately excited by light, elec tricity, or collisions with other atoms, e mits light at only certain well-defined wavelengths. By accurately measuring these wavelengths, we can deduce the various elements in a sampl e of unknown composition. Molecules al so emjt light that is characteristic of their composition. The scie nce of measuring the wavelengths of atomjc and molecular emissions is called spectroscopy. Because their bright fringes are so di stinct, diffraction grating s are an ideal tool for spectroscopy. Suppose the Light inc ident on a grating consists of two slightly different wavelengths. According to Equation 17.l2, each wavelength will diffract at a slightly different angle and, if N is sufficiently large, we'll see two di stinct fringes on th e scree n. FIGURE 17 .14 illustrates this idea. By contrast, th e fringes in a doubl e-slit experiment are so broad that it would not be possible to di stingui sh the fringes of one wavelength from those of the other. EXAMPLE 17 4

~ ~

Blue light has :.I IOllger wavelength than violet, and thus d irfracts more .

\.~===11

l Grating

'· > ~~==I~ Al l wav~l cngths overlap .11 } = 0

~~I Light ------1 intensity

o

Measuring wavelengths emitted by sodium atoms

Light from a sodium lamp passes through a diffraction grat ing hav in g 1000 s lits per millimeter. The inte rference pattern is viewed on a sc reen J .000 m behind the grat in g. Two bright ye llow fringes are visi ble 72.88 cm and 73.00 cm from the central maximum. What are the wavelengths of these two fringe s? PREPARE This situation is simjlar to that in Figure 17.14. The two

fringes are very close together, so we expect the wavelengths to be only slightly different. No other yellow fringes are men tioned, so we will assume these two fringes are the tirst-order diffraction (m = 1). SOLVE The distance Ym of a bri ght fringe from the central maximum is related to the diffraction ang le by Ym = Ltan8m . Thus the diffraction angles of these two frin ges are

_ _,(YJ- ) -_ { 36.08

8 J - tan

The two wavelengths have separated.

L

0

36. 13 0

fringe at 72.88 em fringe at 73.00 em

These angles must sati sfy the interferen ce condition dsin8 1 = A, so the wavelengths are

A = dsin8 1 What is d? If a I mm length of the grating has 1000 sl its , then the spac in g from one s lit to lhe nex t mu st be 1/1000 mm, or d = 1.00 X 10 6 m. Thu s the wavelengths creating the two bright fringes are

A = dsin8 1 =

589.0 nm { 589.6 nm

fringe at 72.88 em fring e at 73 .00 e m

ASSESS In Chapter 15 you lea rned that yellow li ght has a wave-

length of about 600 nm , so our answer is reasonab le.

Instrume nts that measure and anaJyze spectra, called spectropholOmeters, are widely llsed in chemi stry, biology, and medicin e. Because each mol ec ul e has a di stin ct spectrum-a "fin ge rprint"-spectroscopy is used to identify specific biomolecules in tissue, dmgs in urine, and chlorophyll in seawater.

Reflection Gratings We have analyzed what is called a transmission grafillg, with many parallel slits. It's Get acomplete eBook Order spaced by email difficult to make such grating with many closely sLits.at [email protected] practi ce, most diffraction gratings are manufactured as rejlection gratings. The simplest refl ecti on

556

CHAPTER

17

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Optics

FIGURE 17.15 A reflection grating. Incident light

Different wavelengths diffracted at different angles

~ ~:: Few 11m A refl ection gmt ing can be made by cllll ing

p 1, 2• The ret1ecled wave does not have a phase change.

II

2r =

557

(17.16)

Conditi on for deslrllctive interference with e ither 0 or 2 refl ecti ve phase changes Get complete eBook Order by email at [email protected] Conditi o n for constructi ve interference with o nl y I re fl ecti ve phase change

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CHAPTER

17

Wave Optics

mmlI ~ Equations 17.1 5 and 17.1 6 give th e film thicknesses th at yield co nstructive or destru cti ve interference. At oth er thicknesses, the waves will interfere neither fully constructive ly nor fully destructively, and the refl ected intensity will fall so mewhere between these two extremes . ..... These conditions are th e basis of a procedure to analyze thin -film interfe rence. TACTICS BO X 17 . 1

Analyzing thin-film interference

Follow the light wave as it passes through the film . The wave reflectin g from the second boundary travels an extra distance 21.

o

Note the indices of refracti on of the three media: the medium before th e film , the film itself, and the medium beyond the film. The first and third may be the same. There's a refl ecti ve phase change at any boundary where the index of refraction increases. fJ If !leither or both reflected waves undergo a phase change, the phase changes cancel and the effective path-length difference is !1d = 21. Use Equation 17. 15 for constructive interference and 17. 16 for destructive interference. @) [f ollly aile wave undergoes a phase change, the effective path-length difference is a d = 21 + ~ A. Use Equation 17. 15 for destru cti ve interference and 17 . 16 for constructive interference. Exercises 12, 13

..... Iridescent feathers BIO The gorgeous colors of the hummingb ird shown at the begin ning of [his chapter are due not 10 pigments bullo interference. Th is iridescence, presenl in so me bird feathers and insecl shell s. arises from biologica l structures whose size is similar to the wavelength of lighl. The sheen of an in sect, for in stance, is due to thin-film interference from muhiple thin layers in its shell. Peacock femhers are al so a layered structure, but each layer itse lf consists of nearly parallel rods of melanin, as shown in the micrograph. Ihat act as a diffraction grating. Thus a peacock feather comb ines thin-film interference and grating-like diffraction to produce its characteri stic multi-colored iridescent hues.

EXAMPLE 17 5

Designing an antireflection coating

To keep unwanted li ght from reflecting from the surface of eyeglasses or other lenses, a thin film of a material with an index of refraction 1/ = 1.38 is coated on to the pl astic len s (1/ = 1.55). It is desired to have destructi ve interference for A = 550 nm because that is the center of the visible spectrum . What is the thinnest film that will do thi s?

We follow the steps of Tactics Box 17.1. As the li g ht traverse s the film , it first reflects at the front surface of the coatin g. Here, the index of refraction increases from that of air (11 = 1.00) to thai of the film (II = 1.38), so there will be a retlect ive phase change. The light then reflects from the rear surface of the coati ng. The index again increases from that of the film (n = 1.38) to that of the plasti c (n = 1.55). With two phase chan ges, Tactics Box 17. 1 tell s us that we should use Equat ion 17. 16 fo r destructive interfe rence. PREPARE

SOLVE We can solve Equat ion 17. 16 for the thickness f that causes destructi ve interference:

1

=l::...(m +.!.)2 211

The thinnest film is the one for which m = 0, giving 550 nm

1=---

2( 1.38)

The glasses o n the top have a n an tirefle ction coating on Get eBook Order them . Tho se on th e bottom do complete not.

by

I x-= 100nm

2

ASSESS Interference e ffects occ ur when path -length d ifferences are on the order of a wavelength , so our answer of 100 nm seems email at [email protected] reasonable.

II

Get complete eBook Order by email at [email protected] 17.4

Thin Films of Air

Interference

559

FIGURE 17.20 Light and dark fringes

caused by thin -film interference du e to the air layer between two microscope slides.

A film need not be of a solid material. A thin layer of air sa ndwi ched between two glass surfaces also exhibits thin-film interference due to the waves that reflect off both inte rior air-glass boundaries. FIGURE 17 .20 shows two mi crosco pe slides being pressed toge the r. The light and dark " frin ges" occur because the slides are not exactly flat and they tou ch eac h other only at a few points. Everywhere else there is a thin layer of air between them. At so me points, the air layer's thi ckness is such as to give constructive interference (Light fringes), while at other places its thi c kness gives destructive interference (dark fringes). These fringes can be used to accurately measure the flatness of two glass plates, as Example 17.6 will show.

EXAMPLE 17 6

Thin ~ Film

Finding the fringe spacing from a wedge-shaped film of air

Two IS-em-long flal glass plates are separated by a lO-,um-thi ck spacer at one e nd, leaving a thin wedge of air between them , as shown in FIGURE 17.21 . The plates are illuminated by light from a sod ium lamp with wavelength A = 589 nm . Alternating bri ght and dark fringes are observed. What is the spacin g between two bright frin ges?

SOLVE Let x be the distance from the len e nd to a bright fringe.

From Figure 17.2 1, by s imilar triangles we have T x

or 1 = xTIL. From the co ndition for consuuctive interference, we then have

I)A

FIGURE 17 .21 Two glass plates w ith an air wedge between them. One wave reflects off this surface . .

Dlher retlects

\

xT ( 21: = III +"2 -;; There will be a bright frin ge for any integer value o f 111, and so the position of the mth frin ge, as measured from the left end, is

. . wh ile the

\

Xm ~ ~(m + !) 21lT 2

\ ~/s sucroce

~ Il

~-----------x ~----------~I

e xT =

L

l0/-tlll

We want to know the spac ing between two adjacent fringes, m and III + I , which is

~X = Xm+I- Xm= 21lT A~ (111+1 + ~) -~(m+ ~) =~ 2 211T 2 211T

L = 15 em

Evaluating, we Find

PREPARE The wave reflected from the lower pl ate has a reflective phase change, but the top reflection does no t because the index of refract ion decreases at the glass-air boundary. According to Tactics Box 17. 1, we should use Equati o n 17 .16 for con structive interfere nce:

21 ~ (m + !)~ 2 II

Thi s is a film of air, so here 1/ is th e index of refraction of air. Each integer value of 111 co rrespond s to a wedge thickness I for whi ch there is cons tru cti ve interference and thus a bright fringe.

The Colors of Soap Bubbles and Oil Slicks

~X

AL

= --- =

21lT

(5 .89 X 10- 7 m)(0. 15 m)

2(1.00)(10 X 10

6

m)

=

4.4 mm

ASSESS As the photo shows, if the two plates are very flat. the fringes will appear as strai ght lines perpendi cular to the direction of increasin g air thickness. However, if the plates are not quite flat, the fringes w ill appear curved. The amount of c urvature indicates the departure of the plates from perfect flatnes s.

TABLE 17.2 Wavelengths for

So far we ha ve co nsidered thin -film interference only for sing le wavelengths of constructive and destructive interference li ght. The bright colors of soa p bubbles and oil slicks on water are due to thin-film from a 470-nm -thick soap bubble. Visible wavelengths are shown in bold. interference of white light, which, as we've see n, is a mixture of illl wavelengths. A soap bubble is a very thin, spherical film of soapy water (1/ ~ l. 33) . Consider a soap Equation 1Il=1 m=2 111=3 film with thickness I = 470 nm. Light waves reflect from both surfaces of the film , and 2111 A eoll = - - - - I 833 nm 500 nm 357 11m these reflected waves interfere. The light reflecting from the front (air-water) surface has a 111 + "2 reflective phase change, but the back reflection does not. Thus Equation 17.16 describes 21/1 Get ce complete eBook Order by email at [email protected] constructi ve interferen and Equation 17 .15 is destructi ve interference. Table 17 .2 A"" ~ - 1250 nm 625 run 417 run 11/ shows wavelengths of constructi ve and destructive intetference for three values of fn.

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CHAPTER

FIGURE 17.22

17

Wave Optics

The colors of a soap bubble.

Light near the red e nd (625 nm) and violet e nd (4 17) of the s pec trum undergoes destructi ve interference; these colors are /1.ot re flected by the film. At the same time, Li g ht near 500 nm (green) interferes constructively and so is strongly reflected. Conseque ntly, a soap film of thi s thickn ess will appear green. Real soap bubbles and oil slicks have thicknesses that vary from point to point. At some thicknesses, green light is stro ngly reflected, while at others red or violet li g ht is. FIGURE 17.22 shows that the colors of a soa p bubble are predominately greens and

red/violets.

Colors in a vertical soap film

CONCEPTUAL EXAMPLE 17 7

shows a soap film in a wire loop. The loop is held venicaUy. Explain the colors see n in the film. FIGURE 17.23

FIGURE 17.23 A

soap film

in a wire loop.

Because of grav ity, the film is thi cker near the bottom and thinner at the top. It thus has a wedge shape, and the interferen ce pattern consists of lines of alternating construc ti ve and destructive interference, just as ror the air wedge of Example 17.6. Because

REASON

thi s soap rilm is iUuminated by white li ght, colors form as just discussed for any soap film. Notice that the very top or the film, which is extremely thin, appears black. This means that it is reflecting no light at all. When the film is very thin- much thinner than the wavelength of lightthere is almost no path-length difference between the two waves reflected off the front and the back of the fi lm. However, the wave reflected off the back undergoes a reflective phase change and is out of phase with the wave reflected off the front. The two waves thus always interfere destructively, no matter what their wavelength. Thi s simple ex perimen t shows directly that the two reflec ted waves have different reflective phase changes.

ASSESS

STOP TO TH INK 17 5 Re fl ection s from a thin layer of air between two glass plates cause constructive interference for a particular wavelength of li ght A. By how much mu st the thickness of thi s layer be increased for the interference to be destructive?

A. Al8

B. Al4

C. AI2

D. A

17.5 Single-Slit Diffraction

FIGURE 17.24 A single-slit diffraction

experiment. Secondary maxima

Viewi ng scrccn

We opened thi s chapter with a photograph of a water wave passing through a hole in a barrier, then spreading out on the other side. You then saw a photog raph showing that light, after passing a narrOw pin, also s preads out on the other side. This phenomenon is called diffraction. We ' re now ready to look at the detail s of diffraction. FIGURE 17.24 again shows the experimental arrangement for observing the diffrac tion of light through a narrow slit of width a. Diffraction through a tall , narrow slit of width a is known as single-slit diffractio n. A viewing screen is placed a di stance L behind the slit, and we wiLl assume that L » a. The light pattern on the viewing sc reen consists of a central maximum flanked by a seri es of weaker secondary maxima and dark fringes. Notice that the central maximum is significantly broader than the secondary maxima. It is also significantly brighter than the secondary max ima, although that is hard to tell here because thi s photograph has been overexposed to make the secondary maxima show up better.

Central maximum

Huygens' Principle

Dislance

Our analysis of the supe rposition of waves from di stinct so urces, such as two loudspeakers or the two slits in a doubl e-slit ex pe rime nt, has tacitly assumed that the sources are poillf sources, with no measurabl e extent. To understand diffraction, we need to think about th e propagation of an extended wave front. Thi s problem was completefirst eBook Order at [email protected] considered byby the email Dutch scientist Christiaan Huygens, a contemporary of Newton who argued that light is a wave.

Single slit of width {/

lncident light of wavelength A.

Get

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Single ~ Slit

561

Diffraction

Huyge ns Li ved before a mathemati cal theory of waves had been developed, so he developed a geo met ri cal model of wave propagati on. Hi s idea, which we now call Huygens' principle, has two steps:

1. Eac h point on a wave front is the source of a spherical wavelet that spreads out at the wave speed. 2. At a later time, the shape of the wave front is the curve that is tangent to all the wavelets. FIGUR E 17 .25 illustrates Huygens' principle for a plane wave and a spherical wave. As you can see, tl1 e cu rve tangent to the wavelets of a plane wave is a plane that has propagated to the ri ght. The curve tange nt to the wavelets of a spherical wave is a larger sphere. Huygens' principle is a visual device, not a theory of waves. No netheless, the full mathematical theory of waves, as it developed in the 19th centu ry, justifies Huygens' basic idea, although it is beyond the scope of this textboo k to prove it.

FIGURE 17.25 Huy g ens' principle applied to the propagation of plane waves and sph erica l waves.

(b) Spherical wave

(a) Pl ane wave

"I~ I 111(la

Ini tial wave front

wave frolll

., Each point o;~ the init ial wave from is the source of a spherica l wavelet.

.....

TI;~ wave front

)

at a Ialer ti me is tange nt to all the wavelet s.

The wave front at a later timc is tangent 10 all the wave lets.

Analyzing Single-Slit Diffraction shows a wave front passing through a narrow slit of width {/. According to Huygens' principle, each point on the wave front can be thought of as the source of a spherical wavelet. These wavelets overlap and interfere, producing the diffraction pattern seen on the viewi ng scree n. The full mathematical anal ysis, using every

FIGURE 17 .26a

ActjV phys os

16.6

FIGURE 17.26 Each point on th e wave front is a so urce of sp heri ca l wavelets . Th e superposition of th ese wavelets produces the di ffracti on pattern on the screen.

(a)

G reatly magnified view of slit

(c)

(b)

I

In itial wave front

3

5

a

Slit width (/

+---___

o

~

o

Each poin t on the wave front is paired with another poi nt distance a(2 away.

2 4 6

T

)

/

The wavelets going strai ght forw:lrd al l travel These w:lvelct:. all meet on the'sc reen The wavelets from each poi nt on the initial w.. ve front ove rlap and interfere. creati ng the same di stance to the screen. T hus they arri ve at 'll1gle fl. Wavelet 2 travels d istance Get complete eBook Order by email at [email protected] in phase and interfere constructively to produce ~r l:! = (ll(2)sin O farther than wavelet I. a diffraction pattem on the screen. the central max imum.

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CHAPTER 17

Wave

Optics

Water waves can be seen diffracting behind the "slit" between the two breakwaters. The wave pattern can be understood using Huygens' principle.

point on the wave front, is a fairly difficult problem in calculus. We' ll be sati sfied with a geometrical analysis based on just a few wavelets. FIGURE 17 .26b shows the paths of several wavelets as they travel strai ght ahead to the ce ntraJ point on the sc reen. (The screen is very far to the right in this magn ified view of the slit.) The paths to the screen are very nearly parallel to eac h other; thus all the wavelets travel the sa me di stance and arrive at the screen in phase with each other. The cOllstructive llItelterence between these wavelets produces the central maximum of the diffraction pattern at 0 = O. The situation is different at points away from the center of the sc reen. Wa ve lets I and 2 in FIGURE 17.26c start from points that are distance a12 apart. Suppose that dr 12, the extra di sta nce traveled by wavelet 2, happen s to be A12. In that case, wavelets 1 and 2 arrive out of phase and interfe re destructively. But if dr1 2 is A!2, th e n th e differe nce dr34 between paths 3 and 4 and the differe nce dr56 between paths 5 and 6 are also Al2. Those pairs of wavelets also interfere de structively. The superposition of all the wavelets produces perfect destructive interfere nce . Figure 17.26c happ ens to show six wavelets, but our conclusion is valid for any number of wavelets. The key idea is that every point on the wave front can be paired with another point that is distance a/2 away. [f the path-length difference is A!2, the wavelets that originate at these two points will arrive at the sc reen out of phase and interfere destructively. When we sum the displacements of all N wavelets, th ey will-pair by pair-add to zero. The viewing sc ree n at this position will be dark . This is the main idea of the analysis, one worth thinking about carefully. You can see from Figure [7.26c that L).r l2 = (a/2)sin8. This path-le ngth difference will be A!2, the condition for destructive interference, if

~.

(17.17)

or, equivalently, si n0 1 = Ala.

ImID .... Equation 17.17 cannot be sa tisfied if the slit width a is less than th e wavelength A. If a wave passes through an opening smaller than the wavelength, the central maximum of the diffraction pattern expands to completely fill the space behind th e opening. There are no minima or dark spots at any angle. This situation is uncommon for light waves, because A is so small , but quite common in the diffraction of sound and water waves . .... We can ex tend this idea to find other angles of perfect des tructive interference. Suppose eac h wavelet is paired with another wavelet from a point a/4 away. If dr betwee n these wavelets is AI2, then aJi N wavelets will again cancel in pairs to give complete destructive interfe rence. The angle O2 at which thi s occurs is found by replacing a!2 in Equation 17.17 with a/4, leading to the condition asin0 2 = 2A. This process can be continued, and we find that the general condition for complete destructive interfere nce is asinOp = pA

p = 1, 2,3,.

(17.18)

When Op « I rad, which is almost always true for light waves, we can use the smaU-angl e approximation to write

A

8 = pP a

p = [. 2.3 •.

A ngles (in rad ians) or dark rri nges in sin gle·slit diffraction with slit width 1I

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(17.[9)

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563

17.5 Single-Slit Diffraction

Equation 17.19 gives the angles ill radiafls to the dark minima in the diffraction pattern of a single slit. Notice that p = 0 is explicitly excluded. p = 0 corresponds to th e straight-ahead position at 8 = 0, but you saw in Figures 17.6 and 17.26b that 0 = 0 is the central maximum, not a min.imum.

m:!IDI .... Equations 17.18 and 17.19 are mathematically the same as the condition for the filth maximum of the double-slit interference pattern. But the physical meaning here is quite diffe rent. Equation 17.19 locates the minima (dark fringes) of the s,ingle-slit diffraction pattern . ....

FtGURE 17.27 A graph of the intensity of a single-slit diffraction pattern.

y

Screen

Single slit

" = 2

°1

__ ----

---aJ --=== -\-l-- --- --CCll tral /

It is possible, although beyond the scope of thi s textbook, to calculate the entire light intensity pattern. The res ults of such a calculation are shown graphically in FIGURE 17 .27 . You can see the bright central maximum at 0 = 0, the weaker secondary maxima, and the dark points of de structive interference at the angles given by Equation 17.19. Compare thi s graph to the photograph of Figure 17.24 and make sure you see the agreement between the two.

maximum

,, =11 Width

I' ~

11

,, = 2 Light intensifY

o

L »a

The Width of a Single-Slit Diffraction Pattern We' ll find it useful , as we did for the double slit, to meaSure positions on the sc reen rather than angles. The position of the pth dark fringe, at aogle Op, is yp = LtanOp, where L is the di stance from the sLit to th e viewing screen. Using Equation 17.19 for Op and the small-angle approximation tanOp '::::: Op, we find that the dark fringes in the single-slit di.ffraction pattern are located at

pAL Y = P

p = 1, 2, 3, .

(I

( 17.20)

Position s of dark fringes for single-slit diffnlct ion with screen di stance L

Again, p = 0 is explicitly excluded because the midpoint on the viewing screen is the central max.imum, not a dark fringe. A diffraction pattern is dominated by the central maximum, which is much brighter than the secondary maxima. The width IV of the central maximum, shown in Figure 17.27, is defined as the di stance between the two p = Lminima on either side of the central maximum. Beca use the pattern is sy mmetrical , the width is simply IV = 2y,. This is

2AL

w = --

(\7.21)

(I

Wi dth of the central max imum for single-slit di ffrac ti on

The width of the central maximum is twice the spacing AVa between the dark fringes on either side. The farther away the sc reen (larger L), the wider the pattern of light on it becomes. In other words, the light waves are spreading out behind the slit, and they fill a wider and wider region as they travel farther. The central maximum of this single-slit An important implication of Equation t 7.21, one contrary to common sense, diffraction pattern appears white is that a narrower slit (smaller a) causes a wider diffraction patte rn. The smaller because it is overexposed. The width of the opening a wave squeezes through, the more it spreads out on the other the central maximum is clear. Get complete eBook Order by email at [email protected] side.

IV

564

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CHAPTER

17 Wave Optics

EXAMPLE 17 8

Finding the width of a slit

Li ght from a hel ium -neo n lase r (A = 633 nm) passes through a narrow sli t and is seen o n a screen 2.0 m behind the slil. The first minimum in the d iffractio n pattern is 1.2 cm from the middle of the cen tral max imum. How wide is the slit? Th e first minimum in a diffrac ti on pattern correspon ds to p = I. T he pos ition of thi s minimum is given as)'1 = 1.2 cm . We can then use Equ ati o n 17.20 to find the slit widlh Cl.

PREPARE

SOLVE

Eq uati on 17.20 g ives

pAL

a= - Yp

= 1.1

( I )(633 X 10- 9 m)(2.0 m)

X 10- ' m

0 .0 12 m

= 0.11

mm

Thi s value is typical of the slit widths used to observe sing le-sli t diffract ion. You can see that the small -angle approx imati on is well sati sfi ed.

ASSESS

STOP TO THINK 17 6 The fi gure shows two single-slit diffracti on pattern s. The di stance between the slit a nd the vi ewin g sc reen is the same in both cases. Which of th e foll owing could be tru e?

The slits are the same for both; A, > A,. The sLits are the same for both; A2 > AI. The wave le ngths are the same for both; a l > a2. The wavele ngths are th e same for both; (12 > al. The sLits and the wavelengths are the same for both ; PI F. The slits and the wavelengths are the sam e for both ; Ih

A. B. C. D. E.

> P2 . > P I.

17.6 Circular-Aperture Diffraction Diffraction occurs if a wave passes through an opening of any shape. Diffract.i on by a single slit establishes the basic ideas of diffracti on, but a co mm on situation of practi cal importance is diffracti on of a wave by a circular a pertu re . C ircul ar dLffracti on is mathemati calJ y more co mplex th an diffracti on from a slit, and we will prese nt results without deri vation . Consider some examp les . A loudspeaker cone ge nerates sound by the rapid osciJ lalion of a diaphragm, but the so und wave mllst pass through the circ ular aperture defined by the outer edge of th e speaker cone before it travel s into the room beyo nd. This is diffracti on by a circular aperture. Telescopes and mi croscopes are the reverse. Light wa ves from olltside need to e nter the instrum ent. To do so, they mu st pass through a circular lens. In fact, th e performance limit of optical in strume nts is determined by the diffraction of th e circular ope nings through whi ch the wa ves must pass. This is an issue we' ll loo k at more closely in Chapter 19. FIGURE 17 .28 shows a circular ape rture of diam eter D. Light waves passing through this aperture spread out to ge nerate a circular diffraction pattern . You should compare this to Figure 17.24 for a single slit to note the similarities and di fferences. The diffraction pattern still has a central maximum, now circular, and it is surrounded by a series of secondary bright frin ges. Most of the inte nsity is contained within the Get completecentral eBook Order max imum.by email at [email protected]

16.7

'v pAct, hyscs

Get complete eBook Order by email at [email protected] 17.6 FIGURE 17.28

565

The diffraction of light by a circular opening.

Circular apenure

c

Circular-Aperture Diffraction

P= 3 p ~ 2

0,

DiamClcrDJ

__ ---- p ~

_ ",,::::::::\I : :~- - ----Ccnlral / maximum

11

Width II' P= t

1

Lighl intensity

Angl e 0 1 locates th e first minimum in the intensity, where there is perfect destructive interference. A mathematicaJ analysis of circuJar diffraction finds that 1.22,\

e, = ~~

( 17.22)

D

where D is the diameter of the circuJar ope ning. This is very similar to th e result for a single slit, but not quite th e same. Equation 17.22 has assum ed the small-angl e appro ximation, whi c h is almost always vaJid for the diffraction of li ght but usuall y is 1101 valid for the diffraction of longer-wave length sound waves. Within the small-angle approximation, the widt h of the central maximum is

IV

= 2y , = 2Ltan8,

2.44,\L

'" ~D~-

( 17.23)

Width of central maximum for diffraction

-',,~'J

•

frol11 a ci rcu lar aperture of diameter D

Again, this is similar to, but not quite identical with, the width of the central max.imum in a single-slit diffraction pattern. The diam ete r of the diffraction pattern increases with di stance L, showing that light spreads out behind a circular aperture, but it decreases if the size D of the aperture is increased.

EXAMP LE 17 9

TRY IT YOURSELF Th is pi nhole aclS as a point source of li ght.

\

You are observing diffraction due 10 [hi s pil1l~ole.

Finding the right viewing distance

Li ght from a he lium -neon laser (A = 633 nm) passes through a O.SO-mm-diameter hole . How far away should a view ing screen be pl aced to observe a diffracti on pattern whose central maximum is 3.0 mm in di ame ter? SOLVE

Equation 17.23 gives us the appropriate scree n di stance:

wD

L ~ ~~ ~

2.44A

(3.0 X 10- 3 01)(5 .0 X 10-4 m) 2.44(633 X 10

9

m)

0.9701

Observing diffraction To observe diffraction from a circular aperture. use a pin to prick a very small hole in two pieces of aluAs we've see n, we need to use the wave model of li ght to understand the passage mi num foi l. Tape one pi ece up to a window of light through narrow apertures, where " narro w" means comparable in size to th e thaI faces the sun . Then, holding the other wavelength of the li ght. In the next two chapters, we' U co nsider the interaction of hole close to your eye, observe the sun Light with objects much larger than the wavelength. There, the ray model of light will through the pinhole on the window. You will see c lear diffraction frin ges around that pinbe more appropriate for desc ribing how light reflects from mirrors and refracts from hole. How do the frin ges vary if you change lenses. But the wave model will reappear in Chapter 19 when we study telesco pes lhe size of the pinho le near you r eye? (Don't and mi croscopes. We' ll find that the resolutioll of these instruments has a fundame n- look directly at the sun except throu gh the two pi nholes!) eBook Order by email at [email protected] tal limit set by theGet wavecomplete nature of light.

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CHAPTER

17 Wave Optics

Laser range finding

INTEGRATED EXAMPLE 17 10

Scientists use laser range finding to measure the di stance to the moon with great accuracy. A very brief (100 ps) laser pul se, with a wavelength of 532 nm, is fired at the moon , where it reflects off an array of 100 4.0-cm-diameter mirrors placed there by ApoUo 15 astron auts in 1971. The reflected laser light returns to earth , where it is collected by a telescope and detected. The average earth-moon distance is 384,000 km. The laser beam spreads o ut on it s way to the moon because of diffraction, reaching the mirrors with an intensity of 300 W/m 2 . The reflected beam sp reads out even more o n it s way back because of diffraction due to the circular aperture of the mirrors. a. What is the round-trip time for the laser pulse to travel to the moon and back ? b. 1f we want to measure the distance to the moon to an accuracy of 1.0 cm, how accurately mu st the arrival time of the returning pulse be measured? c. Because of the spread of the beam due to diffraction, the light arriving at earth from one of the mirrors will be spread over a circular spot. Estimate the diameter of thi s spot. d. What is the intensity of the laser beam when it arrives back at the earth? When the light reflects from one of the circuJar mirrors, diffraction causes it to spread alit as shown in FIGURE 17 .29 . The width of the central maximum for c ircular-aperture diffraction is given by Equation 17 .23.

a. The round-trip distance is 2L. Thus the round-trip travel time for the pulse, traveling at speed c, is

SOLVE

6x III ~ c

'0,,,1 angle ~8 ,.

'.

Mirror

C

3.00 X 1O~ mis

2.565

0.020 m

6.6 X 10 " ~ 66 ps

3.00 X lORmls

Thus the arrival time of the pul ses mu st be timed to an aCCllracy of about 70 ps. c. The li ght arriving at the moon reflects from c ircu lar mirrors of diameter D. Diffraction by these c ircu lar apertures causes the returning li g ht to spread out with angular width 28 1, where 8 1 is the angle of the first minimum on either s ide of the ce ntral maximum. Equation 17 .23 found th at the width of the central maximum- the diameter of the circ ular spot of light when it reaches the earth- is

2.44AU

2.44(532 X 10- 9 m)(3 .84 X 10' m)

w = --- ~

o

~

0.040 m

12,000 m

d. The li ght is reflected from c ircular mirrors of radius r = DI2 and area a = 1rr '2 = 1TD2/4. The power reflected from one mirror is, accordin g to Equation 15.11,

FIGURE 17 .29 The geometry of the returning laser beam.

When the beam reaches the earth. it has spread out into a ci rcle of diameter 11'.

2(3.84 X 10' m)

b. If we wish to meas ure the moon' s distance from the earth to an accuracy of ± 1.0 cm , then, because the laser beam travels both to and from the moon. we need to know the round-trip distance to an accuracy of ± 2.0 em. The lime it takes light to travel 6x = 2.0 cm is

PREPARE

Because of diffracti on. (he beam spreads by a

2L

IlI ~ - ~

p

~

la

~

0' , (0.040 m)' 1'If4 ~ (300 W/m -)'If 4

0.38 W

When the pulse returns to earth, it is now spread over the large area 7T \II 2/4. Thus the intens ity at the earth 's s urface from one mirror reflection is

on moon

~

P

0.38W

II = -,

L

(12,000 m)'

11' -

Reflected beam

returning to eart h

'If-

4

'If

3.3 X 10- 9 W/m'

4

There are 100 reflect in g mirrors in the array, so the total intens ity reaching earth is 1 = 1001 1 = 3.3 X 10- 7 W/m 2 • A large telescope is needed to detect thi s very smal.l intensity. Because we know the inten s ity of the laser beam as it strikes the mirrors, and we can eas ily find a mirror 's area, we can use Equation 15.11 to calculate the power of th e beam as it leaves a mirror. The inten s it y of the beam when it arrives back on earth will be much lower, and can also be found from Eq uation 15.11.

ASSESS A I.D-m-diameter telescope coUeels on ly a few phO/OIlS, or particles of light, in each returning pulse. Despite the difficult chall enge of detecting this very weak sig nal , the accuracy of these measurements is astounding: Th e latest ex perim ents can measure th e in stantaneous distance to the moon to ± I mm, a precis ion of3 parts in a trillion!

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567

SUMMARY The goal of Chapter 17 has been to understand and apply the wave model of light.

GENERAL PRINCIPLES

The Wave Model The wave model considers li ght to be a wave propagating throug h space. Interference and diffraction arc importa nl. The wave model is appropriate when light interacts w ith objects whose s ize is comparable to the wavelength of li ght or roughly less than about 0.1 mill.

Huygens' principle says that each point on a wave front is the source of a spheri cal wavelet. The wave front at a later lime is tangent to all the wavelets.

IMPORTANT CONCEPTS The index of refraction of a materiaJ determines the speed of light in that material: v = clll. The index of refraction of a material is always greater than 1, so that v is always less than c. The wavelength A in a material w ith index of re fract ion the wavelength "- vue in a vacuum: A = Avajn.

II

Diffraction is the spreading of a wave after it passes through an open ing.

;s shorter than

The!reqll eflCY of light does not change as it moves from one materia l to

anot her.

Constructive and destructive interference are due to the overl ap of two or more waves as they spread behind open ings.

APPLICATIONS Diffraction from a single slit

Interference from mUltiple slits

A s ing le slit of width a has a bright central maximum of width

Waves overlap as they spread out behind s ljts. Bright frin ges are seen on the viewing screen aI position s where tbe path-l ength difference tJ.r between successive slits is equal to mA, where m is an integer.

2AL a

w= - -

Double slit with separatio n d

that is fl an ked by weaker secondary maxima. Secondary maxima

Centml maximum

iliA

O"'= d Dark fringes

11111111111111

Eq uall y spaced bright fringe s are located at

' . is The f.rmge spacmg

mAL Y"'= - d-

m = 0, 1,2, ...

a )' = dAL

Dark fringes are located at angles such that asin0l' If Ala

«

= pA

p

= 1,2,3, ..

I, then from the small-angle approx imation,

er = -pAa

Diffraction grating wi th slit spac ing d

I

Very bright and narrow fringe s are located at angles and positions

Y,., = LtanO,.,

pAL a

V = --

•P

Circular aperture of diameter D

Thin-film interference

A bright central max imum of diameter

Interfere nce occurs between the waves reflected from the two surfaces of a thin film w ith index o f refraction II. A wave that reflects from a surface at whi ch the index of refract ion increases has a phase change.

2.44AL w= - - D is surrounded by circular secondary maxima. The first dark fringe is located at l.22A e, = D-

l. 22AL

Y'= - D -

complete eBook For an apermre ofGet any shape, a smaller ope nin gOrder ca uses aby greater spreading of the wave behind the openi ng.

Interference Constructive

oor 2 phase changes A

2r=m -

II

(III

Destructi ve 21 = + -'-)~ email at [email protected] 2 II

1 phase change

21 =

(III + -'-)~ 2

A

2/=111 -

II

II

568

CHAPTER

tMP)TM

!:!.!/

17

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O pti cs

For homework assigned on MasteringPhysics, go to

Problems labeled INT integrate significant material from earlier

www.masteringphysics.com

chapters;

BID are of biological or medical interest.

Problem difficulty is labeled as I (straightforward) to 11111 (challenging).

VIEW ALL SOLUTIONS QUESTIONS Conceptual Questions I. The freq uency of a li ght wave in air is 5.3 X 10 14 Hz. Is the fre-

2.

3.

4.

5.

6.

7.

quency of thi s wave hi gher, lower, or the same after the light e nters a piece of glass? Ra nk in order the following accord ing to the ir speeds, from slowest to fastest: (i) 425-n m-wavclength light th ro ugh a pane of glass, (ii) 500-n m-wavelength li ght through a ir, (iii ) 540-nm wave length light thro ugh water, (iv) 670- nm -wave1ength li ght through a di amond , and (v) 670-n m-wave length li ght through a vacuum. The wave le ngth of a light wave is 700 nm in a ir; thi s lig ht appears red. If thi s wave enters a pool of water, its wavelengt h becomes Aai/II = 530 nm. If you were sw immi ng underwate r, the li ght wo uld still appear red. Gi ven thi s, what property of a wave detennines its color? A do uble- slit interfe rence ex periment shows fr inges on a screen. The entire experimen t is the n immersed in water. Do the fri nges o n the scree n get cl oser together, farther apart , remain the same, or disappear en ti rely? Explain. Figure Q 17.5 shows the frin ges observed in a dou bl e-sli t interference experimen t whe n the two sli ts are ilJumin ated by FIGURE 017 .5 white light. The central max imum is whi te, but as we move away from the central max imum , the fr inges beco me less di st inct and more colorful. What is spec ial about the centra l max imum th at makes it white? Exp lain the presence of co lors in the oullying fr inges. In a double-sli t in terfe rence expe rimen t, interfere nce fringes are observed on a distan t screen. The width of both s!.i ts is then doubled without changing the distance between the ir ce nters. a. What happens to the spac ing of the frin ges? Ex plain. b. What happens to the in tensity of the bright fr inges? Expla in. Figure Q 17.7 shows th e viewABC D E ing scree n in a double-slit experime nt with mo nochromat ic li ght. Fringe C is the central max imum. a. What wi ll happen to th e frin ge spacing if the wave- FIGURE 017 .7 length of the li ght is decreased? b. What will happen [ 0 the fri nge spac ing if the spac ing between the slits is decreased? c. What will happen to the fri nge spac ing if the distance to the screen is decreased? d. Suppose the wavele ngth of the light is 500 nm. How much fart her is it from the dot on the screen in the cen ter of fringe Order by it is frocomplete m the dot to eBook the ri ght sl it? E to the le ft slit th an Get

8. Figure Q 17.7 is the inte rfere nce pattern seen on a view ing screen behind 2 slits. Suppose the 2 slits were repl aced by 20 sl its hav in g the same spacing d between adjacen t sli ts . a. Wo uld the num ber of fri nges o n the screen increase, decrease, or stay the same? b. Wo uld the frin ge spac in g in crease, decrease , or stay the same? c. Wo uld the widlh of eac h fr inge inc rease, decrease, o r Slay the same? d. Woul d the brighlness of eac h frin ge increase, decrease, or stay the same? 9. Figure Q 17.9 shows the light inte nsity o n a view in g screen behi nd a si ngle sli t of wi dth a. The light's wavelength is A. Is A < n, ,.\ = n, A > a, or is it not possible to tell? Ex pla in.

FIGURE 017 .9

10. Figure Q I7.I O shows the li ght intensity on a view ing scree n behi nd a c irc ul ar aperture . What happens to the widt h of the central max imum if a. The wavelength is increased? b. The diamete r of the apertu re is increased? c. How wi ll the screen appear if the aperture diameter is less FIGURE 017 .10 than the light wavelength ? 11. Why does li ght re flec ted fro m peacock feathe rs change color BID when you see the fea thers at a differe nt angle? 12. White li ght is inciden t on a diffract ion grating. What color is the central max imum of the interference pattern? 13. A soap bubble usually pops because some part of it becomes too thi n due to evaporati on or drainage of flu id. T he change in thi ckness also changes the color of li ght the bubb le refl ects. Wh y? 14 . An oil fil m on top of water has o ne patch th at is much thi nner than the wave length of visible li ght. The index of refractio n of the oil is less than that of water. Will the re fl ecti o n from th at extremely thin part of the film be bright or dark? Explain. 15. Shou ld the antireflect ion coat ing of a mi croscope objective lens designed for use with ul trav iolet light be thi nner, thicker, or the same thickness as the coating on a lens des igned for visible li ght? 16. If the thin wedge of air be tween the two pl ates of glass in Figu re 17.2 1 were replaced by water, would the distance between emailtheatfringes [email protected] inc rease, decrease, or remain the same? Explain .

Get complete eBook Order by email at [email protected] Problems 17. Example 17.5 showed that a thin film whose thi ckness is onequarter of the wavelength of light in the film serves as an antireIlec ti o n coating when coated o n g lass . In Example 17.5, " film < " glass ' If a quarter-wave thickness film with " film> " glass were used in stead, would the film still serve as an antirefl ection coat in g? Ex plain. 18. You are standing aga inst the wall near a corner of a large building. A friend is standing against the wall that is around the corner from you. You can' t see your fri en d. How is it that you can hear her when she talks to you?

Multiple-Choice Questions 19. I Li ght of wave length 500 nm in air e nters a glass block with index of refracti on n = 1.5. When the li g ht enters the block, which of the followin g properties of the light will not chan ge? A. The speed of the li ght B. The frequency of the li ght C. The wavelength of the light 20. j The freq uency of a light wave in air is 4.6 X 10 14 li z. Wh at is the wavelength of thi s wave after it enters a pool of water? A. 300 nm B. 490 nm C. 650 nm D . 870 nm 2 1. I Li ght passes throu gh a diffraction grating with a slit spacing of 0.001 mm. A viewing screen is 100 cm behind the grating. If the light is bille, with a wavelength of 450 nm, at about what di stance from the cenler of the interference pattern will the firslorder max imum appear? A. 5 cm B. 25 cm C. 50cm D. 100 cm

569

22. II Blue li ght of wavelength 450 nm passes through an interference gratin g with a slit spac in g of 0.001 mm and makes an interfere nce pattern o n the wall. How man y bright frin ges will be seen ? A. I B. 3 C. 5 D. 7 23. I Yellow light of wavelength 590 nm passes through a diffraction grat in g an d makes an int erference pattern on a scree n 80 em away. The fir st bright frin ges are 1.9 cm from the center of the pattern. How many lines per mm doe s thi s gratin g have? A. 20 B. 40 C. 80 D. 200 24. I Li ght passes through a ID-,um- wide slit and is viewed o n a sc ree n I m behind the sl it. If the width of the slit is narrowed, the band of li ght on the screen wiU A. Become narrower. B. Become wider. C. Stay about the same. 25. II Blue light of wavelength 450 nm passes thro ugh a 0.20-mmwide slit and illuminates a screen 1.2 m away. How wide is the centralmaximllm of the diffraction pattern ? A. 1.2 mm B. 2.0 mm D. 5.4 mm C. 2.7 mm 26. II A gree n laser beam of wavele ngth 540 nm passes throu gh a pinhole and illuminates a dartboard 3.0 m past the pinhole . The first minimum in the inten sity coincides with the ring surrounding the bull's-eye. 12 mm in diameter. What is the diameter of the pinhole? A. 0.14 mm B. 0.33 mm C. 0.59 mm D. 1.2 mm

VIEW ALL SOLUTIONS PROBLEMS Section 17.1 What Is Light?

Section 17.2 The Interrerence or Light

I. III a. How long does it take li ght to trave l throu gh a 3.0-mm7. II Two narrow slit s 50,um apart are illuminated with li ght of thick pi ece of window glass? wavelength 500 nm. What is the angle of the m = 2 bright b. Through what thickness of water co uld light travel in the frin ge in radians? In degrees? same amount of time? 8. III Li ght from a sodium lamp (A = 589 nm) illuminates two nar2. I a. How lon g (in ns) does it take li g ht to tra ve l 1.0 III in row slilS. The frin ge spacin g on a screen ISO cm behind the slits is 4.0 mm. What is the spacing (in mm) between the two slits? vacuum ? b. What di stance does li g ht travel in water, glass, and 9. II Two narrow slits are illuminated by light of wavelength A. The diamond during th e time that it travel s 1.0 m in a slits are spaced 20 wavelengths apart. What is the angle, in radivac uum ? ans, between the central maximum and the m = I bright fringe? 10. II A double- slit ex periment is perform ed with li ght of wave3. 1111 A 5.0-cm-thick layer of oil (n = 1.46) is sandwiched betwee n a 1.0-cm-thic k sheet of glass an d a 2.0-cm-thic k sheet of polylength 600 nm. The bright intelference frin ges are spaced 1.8 mm styrene plastic (II = 1.59). How long (in ns) does it take li ght apart on the vi ewing scree n. Wh at will the frin ge spacin g be if inci dent perpendicular to the glass to pass through thi s S.O-c mthe light is chan ged to a wavelength of 400 nm ? 11. 11111 Light fro lll a heliulll-neon laser (A = 633 nm) is used to illuthi ck sandwich? 4. II A li ght wave has a 670 nm wavelength in air. Its wavelength minate two narrow slits. The interference patte rn is observed on a sc reen 3.0 m behind the slilS. Twel ve bright frin ges are see n, in a transparent so lid is 420 nm. a. Wh at is the speed of li ght in thi s sol id ? spannin g a di stance of 52 mm . What is th e spacing (in mm) between the slits? b. What is the light's frequen cy in the sol id? 12. II Two narrow slits are 0.12 mm apart. Li g ht of wave length 5. II How much time does it take a pul se of li ght to travel through 550 nm illuminates the slits, causin g an interference pattern on 150 m of water? 6. II A helium-neon laser beam has a wavelength in air of633 nm . a screen 1.0 m away. Light from each slit travel s to the m = I It takes 1.38 ns for th e light to travel through 30.0 cm of an max imum on the right side of the central max imum. How much unknown liquid. What is the wave length of the laser beam in farth e r did the light from the le ft slit tra vel than th e li g ht from the liquid? the right slit ? Get complete eBook Order by email at [email protected]

570

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17

Wave Optics

13. III Consider a point P on the viewin g screen of a double-slit interference experiment. Thi s point is 75% of the way from the center of the 3rd bright fringe to the ce nter of the 4th bright fringe. If the wavelength of the light is 600 nm, what is the extra di stance that the wave from one s lit traveled compared to the wave from the other?

Section 17.3 The DitTraction Grating 14. III A diffraction grat in g with 750 s lits/mm is illuminated by light that gives a first-order diffraction angle of 34.0°. What is the wavelength o f the li ght ? 15. III! A I.O-cm-wide diffraction grating has 1000 slits. It is illuminated by li ght of wavelength 550 nm. What are the angles of the first two diffraction orders? 16. III Li ght of wavelength 600 nm illuminates a diffraction grat ing . The second-order maximum is at angle 39.5°. How many lines per millimeter does thi s gratin g have? 17. III A lab technician uses laser light with a wavelength of 670 nm to test a diffraction grating. When the grating is 40.0 cm from the scree n, the first-order maxima appear 6.00 cm from the ce nter of the pattern. How many lines per millimeter docs this grating have? 18. II The human eye can readily detect wavelengths from about 400 nm to 700 nm. If white light illuminates a diffraction gratin g havin g 750 lines/mm, over what range of angles does the vis ible 11/ = I spectrum extend? 19. III A diffraction grating with 600 lines/mm is illuminated w ith li ght of wavelength 500 nm. A very wide viewing screen is 2.0 m behind the grat in g. a. What is the distance between the two m = J fringes ? b. How many bright fringes can be seen on the sc reen ? 20. II A 500 line/mm diffraction grating is illuminated by li ght of wavelength 510 nm. How many diffraction orders are see n, and what is the angle of each?

Section 17.4 Thin-Film Interference

coating thickness that will minimi ze the reOection at the wavelength of 700 nm, where so lar ce lls are most effi c ien t? 26. III A thin film of MgF2 (II = 1.38) coats a pi ece of glass. Constructi ve interference is observed for the reflect ion of light with wavelengths of 500 nm and 625 nm. What is the thinnest film for which thi s can occur? 27. I Lookin g stra ight dow nward into a rain puddle whose surface is covered with a thin film of gasoline, you notice a sw irling pattern of colors caused by interference in side the gasol ine film . The point directly beneath YO ll is colored a beautiful iridescent g reen. You happen to remember that the index of refraction of gaso line is 1.38 and that the wavelength o f gree n li ght is about 540 nm. What is the minimum possible thickness o f the gasoline layer directly beneath you?

Section 17.5 Single-Slit Interference 28. II A helium-neo n laser (A = 633 nm) iLluminate s a s in gle sl it and is observed on a screen 1.50 m behind the slit. The distance between the first and second minima in the diffraction pattern is 4.75 mm. What is the width (in mm) of the sl it? 29. III For a demonstration , a professor uses a razor blade to cut a thin sl it in a piece of aluminum fo il. When she s hines a laser pointer (..\ = 680 nm) through the slit o nto a screen 5.5 m away, a diffraction pattern appears. The bright band in th e cen ter of the pattern is 8.0 cm wide. What is the width of the slit? 30. II A 0.50-mm-wide sl it is illuminated by light o f wavelength 500 nm. What is tbe width of the central maximum on a screen 2.0 m be hind the slit? 3 1. III The seco nd minimum in the diffraction pattern of a 0.10-mmwide sl it occurs at 0.700. What is the wavelength of the li ght ? 32. II What is th e width of a slit for which th e first minimum is at 45 ° when the slit is illuminated by a heli um -neon lase r (A ~ 633 nm)? Hint: The s mall-angle approximation is not valid at 45 °.

Section 17.6 Circular-Aperture Diffraction

33 . III A 0.50-mm-diameter ho le is illuminated by light o f wave21. II What is the thinnes t film of MgF2 (n = 1.38) on g lass that length 500 nm. Wh at is the width of the central maximum on a produces a strong re fl ecti o n for orange li ght with a wavelength sc ree n 2.0 m behind the slit? of 600 nm? II Light from a hel ium-neo n laser (A = 633 nm) passes through 34. 22. 1111 A very thin oil film (11 = 1.25) floats on water (11 = 1.33). a c ircular aperture and is observed on a screen 4.0 m behind the What is the thinnest film that produces a strong reflect ion for aperture. The w idth of the ce ntral maximum is 2.5 cm. What is green li ght with a wavelength of 500 nm? the diameter (in mm) of the hole? 23. II A film w ith 1/ = 1.60 is deposited o n g lass. What is the 35. III You want to photograph a c ircular diffraction pattern whose thinnest film that will produce co nstruct ive interference in the central maximum has a diameter of 1.0 cm. You have a heliumreflection of light with a wavelength of 550 nm? neon laser (..\ = 633 nm) and a 0.12-mm-diameter pinhole. How 24. II A ntireflec tion coat ings can be used on the inner surfaces of far behind the pinhole s ho uld you place the view ing screen? BID eyeglasses to reduce the reflection of stray light into the eye, II Infrared light of wavelength 2.5 JLm iUuminates a 0.20-mm36. thus reducing eyestrain . diameter hol e. What is the angle of the first dark fringe in rad ia. A 90-nm-thick coating is applied to the lens. What mu st ans? In degrees? be the coat in g's index of refract io n to be most effecti ve at 480 nm? Assume that the coating's index of refraction is less than that of the lens. b. If the index of refraction of the coating is 1.38, what thickGeneral Problems ness s hould the coat in g be so as to be most effect ive at 480 nm ? The thinnest possible coat ing is best. 37. III An advanced computer sends info rm ation to its various parts via infrared light pulses traveling through silicon fib ers 25. II SolarceUs are given antireflection coat in gs to maximi ze their efficiency. Conside r a silico n so lar ce ll (II = 3.50) coated with (1/ = 3.50). To acquire data from memory, the central processGet(11 complete eBook by emailingatunit [email protected] a layer of sili con dioxide se nd s a light-pulse request to the me mory unit. The = 1.45). W hat is the Order minimum

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38.

memory unit processes the req uest, then sends a data pulse back to the central processin g unit. The memory uni t takes 0.50 ns to process a request. If the information has to be obtained from me mo ry in 2.00 ns, what is the maximum distance the memory unit can be from the cen tral processi ng uni t? 1111 Figure PI7.38 shows the li ght Imensily (mW/m!) inte nsity o n a sc reen behind a double slit. The sl it spacing is 0.20 mm and the wavelength of the li ght is 600 nm. What is the distance from the slit s to the FtGURE P17 .38 screen? 11111 Figure P17.38 shows the li ght intens ity on a screen behi nd a doub le slit. The sl it spacing is 0.20 mm and the sc reen is 2.0 m behind the slits. What is the wave length of the light? II Your frie nd has been given a laser for her birthday. Unfonunate ly, she did not receive a manual with it and so she doesn' t know the wave le ngth that it e mits. You help her by performing a double-slit experiment, wi th sli ts separated by 0.36 mm . You fin d that the two brigh t fringes are 5.5 mm apart on a sc reen I .6 m from the slits. What is the wave length the laser emi ts? I A doub le sli t is illum in ated simultaneously with orange Ught of wave length 600 nm and light of an unk nown wavelength . The III = 4 bri ght fringe of the unknown wave lengt h overl aps the 11/ = 3 bright orange fringe. What is the unkn own wavelength? 11111 A laser beam, wit h a wavelengt h of 532 nm , is d irected exac tl y perpendic ular to a sc ree n hav in g two narrow slits spaced 0. 15 mm apart. Interfere nce fr in ges, in cluding a ce ntra l max imum , are observed on a sc ree n 1.0 m away. The direct ion of the beam is then slowly rOLaLed around an axis parallel to the sli ts to an ang le of 1.0°. By what distance does the cent ral maximum on the screen move? I A laser beam of wavelength 670 nm shines through a diffraction grat in g that has 750I ines/mm . Sketch the pallern that appears o n a sc reen 1.0 m behind the gratin g, not in g distances on your d rawi ng and expla inin g where these numbers co me from. IU The two most prom inen t wavele ngths in the light e milled by a hydrogen discharge lamp are 656 nm (red) and 486 nm (b lue). Li ght from a hydrogen lamp illumin ates a diffract ion grat in g wi th 500 lines/mOl, and the I.i ghl is observed on a sc ree n 1.50 m behind the grat ing. What is the distance between the firSl-order red and blue frin ges? 1111 A triple-s li t ex perimen t illuminates three equally spaced, narrow slits wi th light of wave len gth A. The intens ity of the wave fro m each slit is I I' Consider a point o n a di stant sc ree n at an ang le Stic h that the path-len gth differe nce between any two adjacent slits is AI2. Wh at is the intensity at thi s po in t? Give your answe r as a multiple of I I' II A diffrac ti on grating co nsists of 100 slits. If the number of sl its is increased to 200, wi th the same spac in g, by what fa cto r docs the maxi mum intensity of the bright frin ges on the sc reen increase? III A diffraction grating produces a first-order maximum at an angle of 20.0°. What is the angle of the second-order max imum? I A diffracti o n gratin g is illuminated simultaneously with red li ght of wave length 660 nm and li ght of an unknow n wavelength. The fifth-order maximum of the unk nown wave length exac tl y overlaps the third-order max imum of the red light. wavelength ?eBook Order by email at What is the unknown Get complete

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49. 1111 White li ght (400-700 nm) is inc ident on a 600 linelmm diffraction grating. What is the width of the fi rst-order rainbow on a screen 2.0 m behi nd the grating? 50. IIll1 For your sc ience fa ir project you need to des ign a diffraction gratin g that will di sperse the visib le spectrum (400-700 nm) over 30.0° in fi rst order. a. How many lines per millimeter docs your grati ng need? b. What is the first-order diffract io n angle of light fro m a sod ium lamp (A = 589 nm)? 5 1. III Figure P 17.5 1 shows the interference pattern on a screen 1.0 m beh ind an 800 li ne/m m diffract io n grating. What is the wavelength of the li ght?

89.7cmT\~ FIGURE P17 .S1

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52. III Figure PI 7.5 I shows the in terference pattern on a screen 1.0 m beh ind a diffraction g rat ing. T he wavele ngth of the li ght is 600 nm . How many lines per millimeter docs the grat ing have? 53. II Because sound is a wave, it is poss ible to make a diffraction INT grating for sound from a large board with several parallel slots for the sound to go through. When 10 kH z sound waves pass through such a grating, li steners 10m from the gratin g repon "lo ud spotstl 1.4 m on both sides of ce nter. Wh at is the spacin g between the s l o t ~? Use 340 m/s for the speed of sound. 54. 1111 The shiny surface of a CO is im printed with milli ons of tiny pits, arranged in a pattern of thousands of esse nti ally concentric c ircl es that act like a rellcction g ratin g whe n li ght shines on them. You decide to determi ne the distance between those c ircles by aiming a laser po inter (w ith A = 680 11m) perpendicul ar to the disk and measuring the diffraction pallern reflected on to a scree n 1.5 m from the d isk. T he ce ntral bri ght spot you expected to see is bl ocked by the laser pointe r itself. You do find two other bri ght spots separated by 1.4 m, o ne o n either side of the mi ssi ng central spot. T he rest of the panem is apparentl y diffracted at angles too great to show on your scree n. What is the distance between the c ircles o n the CO's surface? 55. III If sun ligh t shines stra ight onto a peacock feather, the feather BID appears bright blue when viewed from 15° o n e ither side of the inc ident bea m of sunl ight. The blue colo r is due to diffraction from the melan in bands in the fe ather barbu les, as was show n in the ph otograph on page 558 . Blue li ght with a wavelength of 470 nm is diffracted at 15° by these bands (thi s is the first-order diffraction) whil e ot he r wave le ngths in the sunli ght are diffra cted at different ang les. What is the spac ing o f the me lanin bands in the feather? 56. III The wings of some beetles Bk) have closely spaced paralle l li nes of melanin , causi ng the win g to act as a re nec ti on grati ng. Suppose sun light shines straigh t onto a beetle wing. If the me lanin lines on the wing are spaced 2.0 Mill apart , what is the fi rst-order diffraction angle for green li ght (A ~ 550 nm)? 57. 1111 A diffracti on gratin g hav in g 500 li nes/mm di ffracts visible light at 30°. What is the li ght's wave length ? [email protected]

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CHAPTER

17 Wave Optics

58. III Li ght emitted by Elemen t X passes through a diffraction grating hav ing 1200 lines/mm . The interference pattern is observed on a screen 75 .0 cm behind the grating. Bright fringes are seen on the scree n at di stances of 56.2 c m, 65.9 c m, and 93.5 cm from the central maximum. No other fringes are seen. a. What is the value of 111 fo r eac h of these diffracted wavelengths? Ex pl ain why only one value is poss ible. b. What are the wavelengths of light emitted by Element X? 59. II He lium atoms em it li ght at several wavelengt hs. Light from a helium la mp illuminates a diffraction grat ing and is observed on a scree n 50.00 cm behind the grat in g. The emi ss ion at wavele ngth 501.5 nm c reates a first-orde r bright fringe 2 1.90 cm from the centra l maximum. What is the wavelength of the bri ght frin ge that is 3 1.60 cm from the central maximum ? 60. II A sheet of glass is coated with a 500- nm-thick layer of o il (/I = 1.42). a . For what visible wavelengths of light do the reflected waves interfe re constructi vely? b. For what visible wavelength s of light do the reflected waves interfere destructively ? c . What is the colo r of reflected light? What is the color of tran smilled light? 61. A soap bubble is esse ntiall y a thin film of water surrounded by air. The colo rs you see in soap bubbles are prod uced by inte rference. What visible wavelength s of li g ht are strongly re nec ted fro m a 390-n m-thi ck soap bubble? What color wou ld such a soap bubbl e appear to be? 62. II In a single-sl it experime nt, the slit width is 200 times the wave length of the li g ht. What is the width of the ce ntral maximum on a screen 2.0 m be hind the slit? 63. III You need to use your cell phone, which broadcasts an 830 MHz sig nal , but you' re in an alley betwee n two mass ive, rad io-waveabsorbin g buildings that have onl y a 15 m space between them . What is th e angular width , in degrees , of th e electromagnetic wave after it emerges from between the buildings? 64. III Li ght from a sodium lamp (A = 589 nm) illumin ates a narrow sli t and is observed o n a screen 75 cm behind the slit. The di stance between the fiI st and third dark fringes is 7.5 mm. What is the width (in mm) of the sl it? 65. 111 1 The open ing to a cave is a tall , 3D-cm-wide crack. A bat that INT is preparin g to leave the cave emits a 30 kHz ultrasoni c chirp. How wide is the "sound beam" 100 m o utside the cave ope ning? Use v sound = 340 m/s. 66. I For what slit-width-to-wavelength rati o does the fir st minimum of a single-sl it diffraction pattern appear at (a) 300, (b) 60°, and (c) 90°?

Hint: The small-angle approx imat ion is not valid. 67. III Fi gure P1 7.67 shows the li ght intensity on a screen behind a sin gle slit. The wave length of t.he light is 500 nm and the screen is 1.0 m behind the slit. What is the width (i n mm) of the slit?

FIGURE P17 .67

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68. III Figure P 17.67 shows the li ght intensity on a scree n behind a singl e sli t. The wavelength of the light is 600 nm and the slit width is 0.15 mm. What is the di stance from the sl it to the Get complete eBook Order by screen?

69. II Figure P1 7.69 shows the light in tensity on a scree n 2.5 m be hind an aperture. The apertu re is illumin ated with li ght of wavelength 600 nm . a. Is the aperture a single slit or a double slit ? Explain. b. If the aperture is a sin gle slit , what is its width ? If it is a double sl it , what is the spac ing between the slits?

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70. II Figure P1 7.70 shows the l.ight intensity on a screen 2.5 m behind an aperture. The aperture is illuminated with li ght of wavelength 600 nm . a. Is the aperture a single slit or a double slit? Explain. b. If the aperture is a single slit, what is its width? If it is a double slit , what is the spaci ng between the slits? 7 1. III One day, aft er pulling down yo ur window shade, you not ice th at sunlight is pass in g through a pinhole in th e shade and making a small patc h of light on the far wall. Hav ing recently studi ed optic s in yo ur ph ys ics c lass, yo u' re not too surpri sed to see that the patch of light seems to be a c ircular diffraction pattern . It appears that the central max imum is about 3 cm across, and you e stimate th at the distance fro m the window shade to th e wall is about 3 Ill. Knowing th at the average wavelength of sunli ght is about 500 nm , estimate the diameter of the pi nhole. 72. 1111 A radar for tracking airc raft broadcasts a 12 GHz mi crowave INT beam from a 2.0- m-di ame ter c irc ular radar antenna. From a wave perspect ive , th e anten na is a circular aperture through which the microwaves diffract. a. What is the diameter of the radar beam at a distance of30 km ? b. If the ante nna emits 100 kW of power, what is the average microwave intensity at 30 km? 73. 11111 A helium-neon laser (A = 633 nm), shown in Figure PI 7 .73, is built with a glass lube of inside d iameter 1.0 mm. One mirror is partiaJl y transm itti ng to allow the laser beam out. An electri cal discharge in the tube causes it to glow like a neon li ght. From an opti cal perspective, the laser beam is a light wave th at dirfracts out through a I.O- mm-d iameter circular open ing. a. Ex pl ain why a laser beam can' t be pCIJeclly parall e l, with no sprcading. b. The angle 0 1 to the first minimum is call ed the divergence angle of a laser bea m. What is the di vergence angl c of thi s laser beam? c. What is the diameter (in mm) of the laser beam aft er it travels 3.0 m? d. What is the d iameter of the laser beam after it lravels 1.0 km?

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74. II In the laser range. findi ng experi ments of Examp le 17.10, the laser beam fired toward the moo n spreads Oll t as it trave ls beca use it d iffrac ts throug h a c ircu lar ex it as it leaves the laser. In order for the re fl ec ted li ght to be bri ght eno ugh to detect, the laser spot on the moon must be no more th an I km in d iameter. Stay in g w ithi n thi s

di ameter is acco mplished by us ing a spec ial Jarge-di ameter lase r. If A = 532 nm, what is the mini mum di a meter of the c irc ular opening from whi ch the laser beam emerges? The calth-moon distance is 384,000 km.

Passage Problems The Blue Mor pho Butterny

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The bri ll iant blue color o f a blue morpho butterfl y is, like the colors of peacock feat hers, due to inte rfere nce. Figure P 17.7Sa shows an easy way to demo nstrate thi s: If a drop of the clear so lvent acetone is placed o n th e wi ng of a blue morpho b utterfl y, the color changes fro m a brilli ant bl ue to an equally bri ll ian t green-returni ng to blue once the acetone evaporates. There wou ld be no change if the color were due to pi gmen t.

Light refl ections from layers have different pat h length s.

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A cross section of a scale from the w ing of a bl ue morpho b Uller~ fl y reveals the sou rce of the butterfly 'S co lor. As Figu re P1 7.75b shows, the sca les are covered wi th structures th at look like s mall C hristmas trees. Lig ht strikin g the w in gs reflects from d iffere nt lay~ ers of these structures, and the d iffer ing path le ngths cause the re fl ec ted li g ht to interfere construct ive ly or destruct ively, de p end~ ing o n the wave length. Fo r li ght at normal inc idence, b lue light exper iences co nstruc ti ve interfere nce whi le othe r colors undergo destructive interference a nd cancel. AcelO ne fill s the spaces in the scales w ith a fl ui d o f index of refract ion" = 1.38; thi s changes the cond itions for construct ive interference and results in a c hange in color. 75. I The coloring o f the blue mo rpho bu tterfl y is protecti ve. As the b utterfly flaps its w ings, the ang le at wh ich light strikes the w in gs cha nges. T hi s causes the b utter fl y's color to change and makes it d ifficult for a predator to fo ll ow. Th is color change is because: A. A d iffract ion pattern appears on ly at ce rtain angles. B. The index of refraction o f the w ing ti ss ues changes as the wi ng fl exes. e. The mot ion of the wings causes a Do ppler shift in the reflected light. D. As the a ngle changes. the d ifferences in path s amo ng li ght re fl ected fro m d iffere nt surfaces change, resulti ng in co n ~ stnlct ive interference fo r a d iffe rent color. 76. I T he change in colo r when aceto ne is p laced o n th e wing is d ue to the d iffere nce be twee n the in d ices of refrac tio n o f acetone and a ir. Co nsider li g ht of some part ic ul ar co lor. In acetone, A. The freq uency o flh e light is less than in air. B. The freq uency of the li ght is greater than in a ir. e. The wavelength of the light is less than in air. D. The wavelength of the li ght is greater than in air. 77. I The scales o n the butterfly wings are actuall y made o f a transparen t material with index of re[ract ion 1.56. Light refl ects from the surface of the scales because A. The scales' index of refrac ti o n is d ifferent fro m that of air. B. The scales' index of refrac ti on is simil ar to that of glass. e. The scales ' density is di ffe re nt fro m that o f air. D. Differen t colors of light have d ifferent wavelengths.

FIGURE P17 .7S

Stop to Think 17 .1: IlJ > " l > 11 2' A = AvaJI/ , so a shorter wave· length corresponds to a hi gher index of refract io n. Stop to Think 17 .2: B. W hen the screen is closer, you don't have (0 move as far fro m the center to reach the poin t where the path ~1eng th differe nce is one wavelength. Stop to Thin k 17.3: Sm a ller. The fr inge spacing portional to the wavelength A.

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Stop to T hink 17 .4: D. Longe r wavele ngths have larger d iffracti on angles. Red li g ht has a longer wave lengt h than vio let lig ht , so red li ght is d iffrac ted fill"lh er fro m the cen ter.

St op t o T hink 17.5: B . A n ex tra path d ifference of AI2 must be added to change fro m construc ti ve to destruct ive interference. In thin- fi lm in terference, one wave passes twice through the film . To inc rease the path length by Al2. the thi ckness needs to be in creased by only one~ h alf thi s, or Al4. St op to Thin k 17.6: B or C . The w idth of the centra l max imum , whi c h is propo rti o nal to Ala, has increased. T hi s could occur e ither because the wave length has increased or because the slit wid th has decreased.

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