Reflection and Refraction of Light CHAPTER OUTLINE

25.1 The Nature of Light 25.2 The Ray Model in Geometric Optics ANSWERS TO QUESTIONS 25.3 The Wave Under Reflection 25.4 The Wave Under Q25.1 The ray approximation, predicting sharp shadows, is valid for λ << d. Refraction For λ ~ d diffraction effects become important, and the light waves 25.5 Dispersion and Prisms 25.6 Huygens’s Principle will spread out noticeably beyond the slit. 25.7 Total Internal Reflection 25.8 Context ConnectionOptical Fibers

Q25.2 With a vertical shop window, streetlights and his own reflection can impede the window shopper’s clear view of the display. The tilted shop window can put these reflections out of the way. Windows of airport control towers are also tilted like this, as are automobile windshields.

FIG. Q25.2

Q25.3 The right-hand fish image is light from the right side of the actual fish, refracted toward the observer, and the second image is light from the left side of the fish refracted toward the observer.

Q25.4 An echo is an example of the reflection of sound. Hearing the noise of a distant highway on a cold morning, when you cannot hear it after the ground warms up, is an example of acoustical refraction. You can use a rubber inner tube inflated with helium as an acoustical lens to concentrate sound in the way a lens can focus light. At your next party, see if you can experimentally find the approximate focal point!

Q25.5 Suppose the light moves into a medium of higher refractive index. Then its wavelength decreases. The frequency remains constant. The speed diminishes by a factor equal to the index of refraction.

687 688 Reflection and Refraction of Light Q25.6 If a laser beam enters a sugar solution with a concentration gradient (density and index of refraction increasing with depth) then the laser beam will be progressively bent downward (toward the normal) as it passes into regions of greater index of refraction.

Q25.7 Diamond has higher index of refraction than glass and consequently a smaller critical angle for total internal reflection. A brilliant-cut diamond is shaped to admit light from above, reflect it totally at the converging facets on the underside of the jewel, and let the light escape only at the top. Glass will have less light internally reflected.

Q25.8 The index of refraction of diamond varies with the frequency of the light. Different color- components of the white light are refracted off in different directions by the jewel. The diamond disperses light to form a spectrum, as any prism does.

Q25.9 A faceted diamond or a stone of cubic zirconia sparkles because the light entering the stone from above is totally internally reflected and the stone is cut so the light can only escape back out the top. If the diamond or the cubic zirconia is immersed in a high index of refraction liquid, then the total internal reflection is thwarted and the diamond loses its “sparkle”. For an exact match of index of refraction between cubic zirconia and corn syrup, the cubic zirconia stone would be invisible.

Q25.10 Take a half-circular disk of plastic. Center it on a piece of polar-coordinate paper on a horizontal corkboard. Slowly move a pin around the curved side while you look for it, gazing at the center of the flat wall. When you can barely see the pin as your line of sight grazes the flat side of the block, θ the light from the pin is reaching the origin at the critical angle c . You can conclude that the index 1 of refraction of the plastic is . θ sin c

Q25.11 The light with the greater change in speed will have the larger deviation. If the glass has a higher index than the surrounding medium, X travels slower in the glass.

Q25.12 Total internal reflection occurs only when light moving originally in a medium of high index of refraction falls on an interface with a medium of lower index of refraction. Thus, light moving from air (n = 1) to water (n = 133. ) cannot undergo total internal reflection.

Q25.13 Highly silvered mirrors reflect about 98% of the incident light. With a 2-mirror periscope, that results in approximately a 4% decrease in intensity of light as the light passes through the periscope. This may not seem like much, but in low-light conditions, that lost light may mean the difference between being able to distinguish an enemy armada or an iceberg from the sky beyond. Using prisms results in total internal reflection, meaning that 100% of the incident light is reflected by each prism. That is the “total” in total internal reflection. At the surfaces of entry into and exit from the prisms, antireflective coatings can minimize light loss.

Q25.14 Light from the lamps along the edges of the sheet enters the plastic. Then it is totally internally reflected by the front and back faces of the plastic, wherever the plastic has an interface with air. If the refractive index of the grease is intermediate between 1.55 and 1.00, some of this light can leave the plastic into the grease and leave the grease into the air. The surface of the grease is rough, so the grease can send out light in all directions. The customer sees the grease shining against a black background. The spotlight method of producing the same effect is much less efficient. With it, much of the light from the spotlight is absorbed by the blackboard. The refractive index of the grease must be less than 1.55. Perhaps the best choice would be 155..×= 100 124 ..

Chapter 25 689 Q25.15 At the altitude of the plane the surface of the Earth need not block off the lower half of the rainbow. Thus, the full circle can be seen. You can see such a rainbow by climbing on a stepladder above a garden sprinkler in the middle of a sunny day. Set the sprinkler for fine mist. Do not let the slippery children fall from the ladder.

Q25.16 A mirage occurs when light changes direction as it moves between batches of air having different indices of refraction because they have different densities at different temperatures. When the sun makes a blacktop road hot, an apparent wet spot is bright due to refraction of light from the bright sky. The light, originally headed a little below the horizontal, always bends up as it first enters and then leaves sequentially hotter, lower-density, lower-index layers of air closer to the road surface.

SOLUTIONS TO PROBLEMS

Section 25.1 The Nature of Light

No problems in this section

Section 25.2 The Ray Model in Geometric Optics

Section 25.3 The Wave Under Reflection

Section 25.4 The Wave Under Refraction

P25.1 (a) From geometry, 1.sin. 25m =°d 40 0 Mirror 2

Light 50° so d = 194. m. Mirror 2 beam i = 50° 40.0° 2 ° (b) 50. 0 above the horizontal 40° d P 1.25 m i1= 40° 40° Mirror 1 50° or parallel to the incident ray. 50° 1.25 m Mirror 1

FIG. P25.1

P25.2 (a) Method One:

αθ=°− The incident ray makes angle 90 1

with the first mirror. In the picture, the law of reflection implies that θθ= ′ 11. FIG. P25.2

βθ=°−′ =−= θα Then 9011 90 .

In the triangle made by the mirrors and the ray passing between them,

β +°+=90γ 180 °

γβ=°−90 . continued on next page 690 Reflection and Refraction of Light Further, δ =°−==90γ β α

and ∈=δα = .

Thus the final ray makes the same angle with the first mirror as did the incident ray. Its direction is opposite to the incident ray.

Method Two:

The vector velocity of the incident light has a component vy perpendicular to the first mirror

and a component vx perpendicular to the second. The vy component is reversed upon the

first reflection, which leaves vx unchanged. The second reflection reverses vx and leaves vy unchanged. The doubly reflected ray then has velocity opposite to the incident ray.

++ (b) The incident ray has velocity vvvxyzijk$$$. Each reflection reverses one component and leaves the other two unchanged. After all the reflections, the light has velocity −− − vvvxyzijk$$$, opposite to the incident ray.

P25.3 The incident light reaches the left-hand mirror at distance Mirror Mirror

af100.tan.. m 500°= 00875 m reflected beam above its bottom edge. The reflected light first reaches the 1.00 m right-hand mirror at height

2bg 0.. 087 5 m= 0 175 m. 5.00°

It bounces between the mirrors with this distance between 1.00 m points of contact with either. FIG. P25.3 100. m Since = 572. 0.175 m

the light reflects five times from the right-hand mirror and six times from the left .

θθ= n1 P25.4 Using Snell’s law, sin2 sin 1 n2

θ =° 2 25. 5

λ λ ==1 2 442 nm . FIG. P25.4 n1

Chapter 25 691 θ = θ *P25.5 The law of refraction nn1122sin sin can be put into the more general form

c θθ= c sin1 sin 2 v1 v2 θθ sin1 = sin 2 vv1 2

In this form it applies to all kinds of waves that move through space.

° θ sin35 . = sin 2 343 ms 1 493 ms θ = sin2 0 . 266 θ =° 2 15. 4

The wave keeps constant frequency in

vv f ==1 2 λλ 1 2

λ λ ==v21 1 493 msaf 0. 589 m = 2 256. m v1 343 ms

× 8 ==c 300. 10 ms=×14 P25.6 (a) f − 474. 10 Hz λ 6. 328× 10 7 m

λ 632. 8 nm (b) λ ==air =422 nm glass n 1.50

c 300. × 108 ms (c) v ==air =×200. 108 ms = 200 Mms glass n 150.

θ = θ P25.7 nn1122sin sin

θ =° sin1 1333 . sin 45 θ == sin1 afaf1 . 33 0 . 707 0 . 943 θ =°→° 1 70.. 5 19 5 above the horizon

FIG. P25.7

*P25.8 We find the angle of incidence:

θ = θ nn1122sin sin θ =° 1. 333 sin1 1 . 52 sin 19 . 6 θ =° 1 22. 5

The angle of reflection of the beam in water is then also 22. 5° .

692 Reflection and Refraction of Light *P25.9 (a) As measured from the diagram, the incidence angle is 60°, and the refraction angle is 35°. sinθ v sin35° v From Snell’s law, 2 = 2 , then = 2 and the speed of light in the block is θ ° sin 1 v1 sin60 c 20. × 108 ms .

(b) The frequency of the light does not change upon refraction. Knowing the wavelength in a vacuum, we can use the speed of light in a vacuum to determine the frequency: c = fλ , thus − 300..×= 1089f ej 6328 × 10 , so the frequency is 474. 1 THz.

(c) To find the wavelength of light in the block, we use the same wave speed relation, v = fλ , so ×=814 × λ λ = 2.. 0 10ej 4 741 10 , so glass 420 nm .

θθ= P25.10 (a) nn1122sin sin

1.sin. 00 30 0°=n sin. 19 24 °

n = 152.

× 8 ==c 300. 10 ms=×14 (c) f − 474. 10 Hz in air and in syrup. λ 6. 328× 10 7 m

c 300. × 108 ms (d) v == =×198. 108 ms = 198 Mms n 152.

v 198. × 108 ms (b) λ == = 417 nm f 474. × 1014 s

c 300. × 108 ms P25.11 (a) Flint Glass: v == =×1. 81 108 ms = 181 Mms n 166.

c 300. × 108 ms (b) Water: v == =×225. 108 ms = 225 Mms n 1333.

c 300. × 108 ms (c) Cubic Zirconia: v == =×136. 108 ms = 136 Mms n 220.

θθ= °= ° P25.12 nn1122sin sin : 1. 333 sin 37 . 0n2 sin 25 . 0

c c n ==190.: v ==×158. 108 ms = 158 Mms 2 v 190.

Chapter 25 693 θ θθ= θ = −1 F n11sin I P25.13 nn1122sin sin : 2 sin G J H n2 K

° θ = −1 R100.sin 30 U =° 2 sin S V 19. 5 T 150. W

θ θ 2 and 3 are alternate interior angles formed by the ray cutting parallel normals. FIG. P25.13 θθ== ° So, 3219. 5

θ = θ 150.sin34 100 .sin θ =° 4 30. 0

θθ= θ P25.14 sin12nw sin 1 = 62° air n = 1.00 θθ==1 1 °−°= 28° sin 21sin sina90 . 0 28 . 0f 0 . 662 1333. 1. 333 water − θ ==°sin1 0 . 662 41 . 5 θ n = 1.333 2 a f 2 d 300. m h == = 339. m h θ ° tan 2 tan41.5

3.0 m

FIG. P25.14

P25.15 For αβ+=90 °

θαβθ′ +++ = ° with 12180

θθ′ +=° we have 1290 .

θθ′ = Also, 11

θθ= and 1sin12n sin . FIG. P25.15

θθθ=−= Then, sin111nn sinbg90 cos

θ sin 1 == θ θ = −1 θ n tan 1 1 tan n . cos 1

694 Reflection and Refraction of Light F n I 12.0 cm *P25.16 From Snell’s law, sinθ = G medium J sin50 . 0° 50.0° H nliver K d nmedium n nmedium cvmedium vliver liver But, ===0900., h θ n cv v liver liver medium θ

− so, θ =°=°sin1 af0900 . sin 500 . 436 . . Tumor

From the law of reflection, FIG. P25.16

12. 0 cm d 600. cm d ==600. cm, and h == = 630. cm 2 tanθ tana43 . 6°f

θθ= P25.17 At entry, nn1122sin sin

°= θ or 1.sin. 00 30 0 1 .sin 50 2

θ =° 2 19.. 5

The distance h the light travels in the medium is given by

200. cm cosθ = 2 h

200. cm FIG. P25.17 or h = = 212. cm . cos19.5°

αθ=−= θ °−°=° The angle of deviation upon entry is 1230.. 0 19 5 10 .. 5

d The offset distance comes from sinα = : d =°=af221.sin.. cm 105 0388 cm . h

200. cm P25.18 The distance, h, traveled by the light is h = = 212.. cm cos19.5°

c 300. × 108 ms The speed of light in the material is v == =×200.. 108 ms n 150.

−2 h 212. × 10 m − Therefore, t == =×106.. 1010 s = 106 ps v 2.00× 108 m s

Chapter 25 695 P25.19 Applying Snell’s law at the air-oil interface,

θ =° nnairsin oil sin20 . 0

yields θ =°30.. 4

Applying Snell’s law at the oil-water interface

θ ′ =° nnw sinoil sin20 . 0

yields θ ′ =°22.. 3

FIG. P25.19

*P25.20 From the figure we have wba=+2

wa− 700 µ m− 1 µ m b = = = 349. 5 µ m 2 2

b 349. 5 µ m tanθ == =0.. 291θ = 16 2 ° 22t 1 200 µ m

For refraction at entry,

θ = θ nn1122sin sin θ ° θ ==−−1 n22sin 11155.sin. 162 ==° − 1 sin sin sin0 . 433 25 . 7 n1 100.

P25.21 Consider glass with an index of refraction of 1.5, which is 3 mm thick. The speed of light in the glass is

310× 8 ms =×2108 ms. 15.

−−3 3 310× m 310× m − The extra travel time is − ~.10 11 s 210× 8 ms 310× 8 ms

600 nm For light of wavelength 600 nm in vacuum and wavelength = 400 nm in glass, 1.5

× −3 × −3 310 m − 310 m 3 the extra optical path, in wavelengths, is − − ~.10 wavelengths 410m× 7 610× 7 m

696 Reflection and Refraction of Light P25.22 See the sketch showing the path of the light ray. α and γ are angles of incidence at mirrors 1 and 2.

For triangle abca,

22αγβ++=° 180

or βαγ=°−+180 2bg. (1)

Now for triangle bcdb,

a900..°−αγθf +bg 90 0 °− + = 180 ° FIG. P25.22

or θ =+α γ . (2)

Substituting Equation (2) into Equation (1) gives βθ=°−180 2 .

Note: From Equation (2), γ =−θ α . Thus, the ray will follow a path like that shown only if α < θ . For α > θ , γ is negative and multiple reflections from each mirror will occur before the incident and reflected rays intersect.

Section 25.5 Dispersion and Prisms

P25.23 From the dispersion curve for fused quartz in the chapter (that is, the graph of its refractive index versus wavelength) we have

= = nv 1. 470 at 400 nm and nr 1458. at 700 nm.

θθ= θθ= Then 1.sin. 00 1 470 sinv and 100.sin. 1 458 sinr

θθ δδθθ−=−=−−11F sin I − F sin I rvrvsin G J sin G J H 1458. K H 1470. K

° ° −−11F sin30 . 0 I F sin30 . 0 I ∆δ = sin G J − sin G J =°0171. H 1. 458 K H 1. 470 K

θ θ = θ θ = −1 F n11sin I P25.24 nn1122sin sin : 2 sin G J H n2 K

° θ = −1 F 100.sin. 300 I =° 2 sin G J 19. 5 H 150. K

The surface of entry, the surface of exit, and the ray within the prism form a triangle. Inside the triangle the angles must add up according to FIG. P25.24

°−θ + °+ °−θ = ° 90... 023 60 0 90 0 180

θ = °−°+°−°+°=° 3 di(.90 0 19 .) 5 60 . 0 180 90 . 0 40 . 5

θ ° θ = θ θ = −−1 F n33sin I = 1 F 150.sin. 405 I =° nn3344sin sin : 4 sin G J sin G J 77. 1 H n4 K H 100. K Chapter 25 697

θθ= P25.25 At the first refraction, 1.sin 00 12n sin .

θ = The critical angle at the second surface is given by n sin3 100 . :

θ = −1 F 100. I =° or 3 sin G J 41.. 8 H 150. K

θθ=°− FIG. P25.25 But, 2360.. 0

θ <° Thus, to avoid total internal reflection at the second surface (i.e., have 3 41.) 8

θ >° it is necessary that 2 18.. 2

θθ= θ >°= Since sin12n sin , this becomes sin1 150 . sin 182 . 0468 .

θ >° or 1 27.. 9

θθ= P25.26 At the first refraction, 1.sin 00 12n sin . Φ The critical angle at the second surface is given by θ 1 θ 2 θ = θ = −1 F 100. I θ n sin3 100 . , or 3 sin G J . 3 H n K

°−θθ + °− +Φ = ° But bgbg 90.. 023 90 0 180 FIG. P25.26

θθ=−Φ which gives 23.

θ < −1 F 100. I Thus, to have 3 sin G J and avoid total internal reflection at the second surface, H n K

θ >−Φ −1 F 100. I it is necessary that 2 sin G J . H n K

θθ= θ >−LΦ −1 F 100. IO Since sin12n sin , this requirement becomes sin1 n sin sin G J NM H n KQP

θ >−−−11F LΦ F 100. IOI or 1 sinGn sin sin G J J . H NM H n KQPK

θ >−−−12F ΦΦI Through the application of trigonometric identities, 1 sinH n 1 sin cos K .

θ Φ θ As n increases, 1 increases. As increases, both of the terms inside the parentheses increase, so 1 increases. It becomes easier to produce total internal reflection. If Φ decreases below − −12F − 12I θ tan Hejn 1 K it becomes necessary for 1 to be negative for total internal reflection to occur. That is, the incident ray must be on the other side of the normal to the first surface, but the equation is still true. If the quantity n2 −−1 sinΦΦ cos is greater than 1, total internal reflection happens for θ all values of 1 . However, if this quantity is greater than n, the ray internal to the prism never reaches the second surface.

698 Reflection and Refraction of Light sinθ P25.27 For the incoming ray, sinθ = 1 . 2 n

° θ = −1 F sin50 . 0 I =° Using the figure to the right, bg2 sin G J 27. 48 violet H 166. K

° θ = −1 F sin50 . 0 I =° bg2 sin G J 28.. 22 red H 162. K FIG. P25.27

θθ=°− For the outgoing ray, 3260. 0

− and sinθθ= n sin : θ =°=°sin1 166 . sin 3252 . 6317 . 43 bg4 violet

θ =°=°−1 bg4 red sin162 . sin 3178 . 5856 . .

∆θθ=−=°−°=° θ The angular dispersion is the difference 44bgviolet bg 4 red 63.. 17 58 56 4 .. 61

Section 25.6 Huygens’s Principle

P25.28 (a) For the diagrams of contour lines and wave fronts and rays, see Figures (a) and (b) below. As the waves move to shallower water, the wave fronts bend to become more nearly parallel to the contour lines.

(b) For the diagrams of contour lines and wave fronts and rays, see Figures (c) and (d) below. We suppose that the headlands are steep underwater, as they are above water. The rays are everywhere perpendicular to the wave fronts of the incoming refracting waves. As shown, the rays bend toward the headlands and deliver more energy per length at the headlands.

(a) Contour lines (b) Wave fronts (c) Contour lines (d) Wave fronts and rays and rays

FIG. P25.28

Chapter 25 699 Section 25.7 Total Internal Reflection

P25.29 n sinθ = 1 . From the table of refractive indices

−1 F 1 I (a) θ = sin G J =°24. 4 H 2419. K

−1 F 1 I (b) θ = sin G J =°37. 0 H 166. K

−1 F 1 I (c) θ = sin G J =°49. 8 H 1309. K

sinθ v P25.30 2 = 2 (a) θ sin 1 v1

θ =° and 2 90. 0 at the critical angle

sin90 . 0° 1850 ms = θ sin c 343 ms

θ ==°−1 so c sinaf0185 . 107 . .

(b) Sound can be totally reflected if it is traveling in the medium where it travels slower: air .

θ = n2 P25.31 sin c n1

=°== nn21sin88 . 8bgbg 1 . 000 3 0 . 999 8 1 . 000 08 FIG. P25.31

700 Reflection and Refraction of Light

P25.32 For plastic with index of refraction n ≥ 142. surrounded by air, the critical angle for total internal reflection is given by

θ = −−11F 11I ≤ F I =° c sinG J sin G J 44. 8 . H nK H 142. K

In the gasoline gauge, skylight from above travels down the plastic. The rays close to the vertical are totally reflected from the sides of the slab and from both facets at the lower end of the plastic, where it is not immersed in gasoline. This light returns up inside the plastic and makes it look bright. Where the plastic is immersed in gasoline, with index of refraction about 1.50, total internal reflection should not happen. The light passes out of the lower end of the plastic with little reflected, making this part of the gauge look dark. To frustrate total internal reflection in the gasoline, the index of refraction of the plastic should be n < 212.

θ = −1 F 150. I =° since c sin G J 45. 0 . H 212. K

Section 25.8 Context ConnectionOptical Fibers

θ ===nair 100. θ =° P25.33 sin c 0735. c 47. 3 npipe 136.

Geometry shows that the angle of refraction at the end is

φθ=°−=°−°=° 90.... 0c 90 0 47 3 42 7 .

Then, Snell’s law at the end, 1.sin.sin. 00θ =° 1 36 42 7 FIG. P25.33

gives θ =°67.. 2

The 2-µm diameter is unnecessary information.

θ =° P25.34 For total internal reflection, nn112sin sin90 . 0

θ = θ =° 150.sin.(.)1 133100 or 1 62. 4

P25.35 As the beam enters the slab,

°= θ 100.sin. 500 148 .sin2

θ =° giving 2 31.. 2 FIG. P25.35

155. mm The beam then strikes the top of the slab at x = from the left end. Thereafter, the beam 1 tan31 . 2° strikes a face each time it has traveled a distance of 2x1 along the length of the slab. Since the slab is − 420 mm long, the beam has an additional 420 mm x1 to travel after the first reflection. The number of additional reflections is continued on next page Chapter 25 701 420 mm − x 420 mm−°155.tan. mm 312 1 = = 81. 5 or 81 reflections ° 2x1 310.tan. mm 312

since the answer must be an integer. The total number of reflections made in the slab is then 82 .

P25.36 (a) A ray along the inner edge will escape if any ray escapes. Its angle of Rd− incidence is described by sinθ = and by n sinθ >°190. sin Then R

nRa − df nd > 1 nR−> nd R nR−> R nd R > . R n − 1

(b) As d → 0, R → 0 . This is reasonable. min FIG. P25.36

As n increases, Rmin decreases. This is reasonable.

As n decreases toward 1, Rmin increases. This is reasonable.

× −6 1. 40ej 100 10 m − (c) R = =×350 10 6 m min 040.

Additional Problems

*P25.37 For sheets 1 and 2 as described,

°= ° nn12sin26 . 5 sin 31 . 7 = 0849. nn12

For the trial with sheets 3 and 2,

°= ° nn32sin26 . 5 sin 36 . 7 = 0747. nn32

Now

= 0.. 747nn31 0 849 = nn31114.

For the third trial,

°=θ = θ nn13313sin26 . 5 sin 1 . 14 n sin θ =° 3 23. 1

702 Reflection and Refraction of Light *P25.38 Scattered light leaves the center of the photograph (a) in all θ =° ° horizontal directions between 1 0 and 90 from the θ 1 max normal. When it immediately enters the water (b), it is (a) ° θ gathered into a fan between 0 and 2max given by

θ = θ nn1122sin sin = θ 1.sin 00 90 1 . 333 sin2 max θ =° θ 2 max 48. 6 2 max

The light leaves the cylinder without deviation, so the (b) viewer only receives light from the center of the photograph when he has turned by an angle less than 48.6°. When the paperweight is turned farther, light at the back surface undergoes total internal reflection (c). The viewer sees things outside the globe on the far side. (c)

FIG. P25.38

P25.39 Let nxaf be the index of refraction at distance x below the top of the atmosphere and nxaf== h n be its value at the planet surface.

F n − 1. 000I Then, nxa f =+1. 000 G J x . H h K

(a) The total time interval required to traverse the atmosphere is

hhdx nxaf 1 h L F n − 1000. I O ∆t ==z z dx : ∆t =+zM1000. G J xdxP 00v c c 0 N H h K Q

h afn − 1. 000 F hh2 I F n + 1000. I ∆t =+ = G J . c ch HG 2 KJ c H 2 K

h (b) The travel time in the absence of an atmosphere would be . c

F n + 1000. I Thus, the time in the presence of an atmosphere is G J times larger . H 2 K

θθ′ == ° θθ= P25.40 (a) 1130. 0 nn1122sin sin

°= θ 100.sin. 300 155 .sin2 θ =° 2 18. 8

− F n sinθ I FIG. P25.40 (b) θθ′ ==30. 0 °θ = sin 1 11 11 2 G n J H 2 K ° −1 F 155.sin. 300 I = sin G J =°50. 8 H 1 K continued on next page Chapter 25 703 (c), (d) The other entries are computed similarly, and are shown in the table below.

incidence reflection refraction incidence reflection refraction 0 0 0 0 0 0 10.0 10.0 6.43 10.0 10.0 15.6 20.0 20.0 12.7 20.0 20.0 32.0 30.0 30.0 18.8 30.0 30.0 50.8 40.0 40.0 24.5 40.0 40.0 85.1 50.0 50.0 29.6 50.0 50.0 none* 60.0 60.0 34.0 60.0 60.0 none* 70.0 70.0 37.3 70.0 70.0 none* 80.0 80.0 39.4 80.0 80.0 none* 90.0 90.0 40.2 90.0 90.0 none*

*total internal reflection

1 3 P25.41 For water, sinθ ==. c 43 4

θ ==°−1 Thus c sinaf0 . 750 48 . 6

and d = 2100.tan m θ afc

d =°=af200.tan.. m 486 227 m. FIG. P25.41

θ P25.42 Call 1 the angle of incidence and of reflection on the θ α left face and 2 those angles on the right face. Let θ β represent the complement of 1 and be the θ α = γ βδ= complement of 2 . Now and because they are pairs of alternate interior angles. We have

A =+=+γδαβ

and BA=++=++=αβαβ AA2.

FIG. P25.42

P25.43 (a) We see the Sun moving from east to west across the sky. Its angular speed is

∆θ 2π rad − ω == =727.. × 10 5 rad s ∆t 86 400 s

The direction of sunlight crossing the cell from the window changes at this rate, moving on the opposite wall at speed

−− vr==ω af2.. 37 mej 7 27 × 1054 rads = 1 . 72 × 10 ms = 0 . 172 mms. continued on next page 704 Reflection and Refraction of Light (b) The mirror folds into the cell the motion that would occur in a room twice as wide:

vr==ω 20174bg.. mm s = 0345 mm s .

(c), (d) As the Sun moves southward and upward at 50.0°, we may regard the corner of the window as fixed, and both patches of light move northward and downward at 50.0° .

P25.44 (a) 45. 0° as shown in the first figure to the right.

(b) Yes

If grazing angle is halved, the number of reflections from the side faces is doubled. FIG. P25.44

P25.45 Horizontal light rays from the setting Sun pass above the hiker. The light rays are twice refracted and once reflected, as in Figure (b). The most intense light reaching the hiker, (the light that represents the visible rainbow,) is located between angles of 40° and 42° from the hiker’s shadow.

The hiker sees a greater percentage of the violet Figure (a) inner edge, so we consider the red outer edge. The

radius R of the circle of droplets is

R =°=af800.sin.. km 420 535 km .

Then the angle φ, between the vertical and the radius where the bow touches the ground, is given by

200.. km 200 km cosφ ===0. 374 R 5.35 km

φ =° or 68. 1 . Figure (b)

The angle filled by the visible bow is FIG. P25.45 360°−af 2 × 68. 1 ° = 224 °

224° so the visible bow is = 62. 2% of a circle . 360°

This striking view motivated Charles Wilson’s 1906 invention of the cloud chamber, a standard tool of nuclear physics. The effect is mentioned in the Bible Ezekiel 1:29.

Chapter 25 705 P25.46 Light passing the top of the pole makes an angle of incidence φθ=°− 1 90. 0 . It falls on the water surface at distance from the pole

Ld− s = 1 tanθ

φ φφ= and has an angle of refraction 2 from 1.sin 00 12n sin .

= φ Then sd22tan

and the whole shadow length is

− φ +=Ld+ F −1 F sin 1 II ss12 d tan sin G J tanθ G H n KJ H K

− θ +=Ld+ F −1 F cos II ss12 d tanG sin G JJ FIG. P25.46 tanθ H H n KK

° 200. m F −1 F cos40 . 0 II = + af200.tansin mG G JJ = 379. m tan40 . 0° H H 133. KK

2 S′ Lnn− O 152..− 100 2 P25.47 (a) 1 = 21= L O = 00426. M + P M + P S1 Nnn21Q N152.. 100Q

2 S′ Lnn− O 100..− 152 2 (b) If medium 1 is glass and medium 2 is air, 1 = 21= L O = 0.. 042 6 M + P M + P S1 Nnn21Q N100.. 152Q

There is no difference .

= P25.48 (a) With n1 1

= and nn2

′ − 2 S1 = F n 1I the reflected fractional intensity is G + J . S1 H n 1K

The remaining intensity must be transmitted:

22 S n − 1 2 afafnn+−−11nn22++−+−21 nn 21 4n 2 =−1 F I = = = . G + J + 2 + 22+ S1 H n 1K afn 1 afn 1 afn 1

S n − 1 2 42419af. (b) At entry, 2 =−1 F I = = 0828. . G + J + 2 S1 H n 1K a2419. 1f

S At exit, 3 = 0.. 828 S2

S F S IF S I 2 Overall, 3 = G 3 JG 2 J ==af0828.. 0685 S1 H S2 KH S1 K

or 68.. 5% 706 Reflection and Refraction of Light

4n P25.49 Define T = as the transmission coefficient for one an + 1f2 encounter with an interface. For diamond and air, it is 0.828, as in Problem P25.48.

As shown in the figure, the total amount transmitted is

TT22+−+−+−af111 T2 T2 af T4 T2 af T6

2 2n ++KKTTa1 −f +

We have 11−=−T 0. 828 = 0 . 172 so the total transmission is

2246 a0.... 828f 1++++a 0 172f a 0 172f a 0 172f K .

246 FIG. P25.49 To sum this series, define F =++++1afafaf 0... 172 0 172 0 172 K.

2246 Note that afafafaf0172....F =+++ 0 172 0 172 0 172 K , and 2246 1+=++++=af 0.... 172FF 1 afafaf 0 172 0 172 0 172 K .

1 Then, 10=−FFaf. 172 2 or F = . 1− a 0. 172f2

a0828. f2 The overall transmission is then = 0706. or 70.. 6% 1− af 0. 172 2

P25.50 (a) As the mirror turns through angle θ, the angle of incidence increases by θ and so does the angle of reflection. The incident ray is stationary, so the reflected ray turns through angle 2θ . The ω angular speed of the reflected ray is 2 m . The speed of the dot of ω light on the circular wall is 2 m R .

(b) The two angles marked θ in the figure to the right are equal because their sides are perpendicular, right side to right side and left side to left side.

d ds We have cosθ = = xd22+ dx FIG. P25.50

ds and =+2ω xd22. dt m

dx ds xd22+ xd 22+ So = = 2ω . dt dt d m d

Chapter 25 707

P25.51 Define n1 to be the index of refraction of the surrounding medium and n2 to be that for the prism material. We can use the critical angle of 42.0° to find the n ratio 2 : n1

°= ° nn21sin42 . 0 sin 90 . 0 .

n 1 So, 2 = = 149.. n sin42 . 0° 1

θ FIG. P25.51 Call the angle of refraction 2 at the surface 1. The ray inside the prism forms a triangle with surfaces 1 and 2, so the sum of the interior angles of this triangle must be 180°.

°−θ + °+ °− ° = ° Thus, bg 90.... 02 60 0a 90 0 42 0f 180 .

θ =° Therefore, 2 18.. 0

θ =° Applying Snell’s law at surface 1, nn112sin sin18 . 0

θθ= F n2 I =°θ =° sin1 G J sin2 149 . sin 180 . 1 27.. 5 H n1 K

*P25.52 The picture illustrates optical sunrise. At the center of the earth, θ1 δ 6 637. × 10 m θ cosφ = 2 ×+6 1 6.37 10 m 8614 2 φ φ =°298. θ =− °=° 2 90 298.. 870 FIG. P25.52 At the top of the atmosphere

θ = θ nn1122sin sin θ =° 1sin1 1 . 000 293 sin 87 . 0 θ =° 1 87. 4

Deviation upon entry is

δ =−θ θ 12 δ =°−°=°87... 364 87 022 0 342

F 86 400 sI Sunrise of the optical day is before geometric sunrise by 0342..°G J = 82 2 s. Optical sunset H 360° K occurs later too, so the optical day is longer by 164 s .

708 Reflection and Refraction of Light P25.53 (a) For polystyrene surrounded by air, internal reflection requires

θ = −1 F 100. I =° 3 sin G J 42.. 2 H 149. K

θθ=°−=° Then from geometry, 2390... 0 47 8

θ =°= From Snell’s law, sin1 149 . sin 478 . 110 . .

This has no solution.

Therefore, total internal reflection always happens . FIG. P25.53

θ = −1 F 133. I =° (b) For polystyrene surrounded by water, 3 sin G J 63. 2 H 149. K

θ =° and 2 26.. 8

θ =° From Snell’s law, 1 30.. 3

since the beam is initially traveling in a medium of lower index of refraction.

δθ=−= θ ° θθ= P25.54 1210. 0 and nn1122sin sin

4 with n = 1 , n = . 1 2 3

θθ==−°−−1 1 θ Thus, 1 sinbgnn22 sin sin21 sin b10 . 0 g.

(You can use a calculator to home in on an approximate solution to this equation, testing different θ θ =° θ values of 1 until you find that 1 36. 5 . Alternatively, you can solve for 1 exactly, as shown below.)

θθ=−°4 We are given that sin11 sinbg10 . 0 . 3

3 This is the sine of a difference, so sinθθ=°−° sin cos10 . 0 cos θ sin 10 . 0 . 4 11 1

°=θθF °−3I Rearranging, sin10 . 0 cos11G cos 10 . 0 J sin H 4K

sin10 . 0° − = tanθ and θ ==°tan1 af0740 . 365 . . cos10 . 0°− 0 . 750 1 1

Chapter 25 709 400. cm P25.55 tanθ = 1 h

200. cm and tanθ = 2 h

2 θθ==2 2 θ tan12bg200 . tan 400 . tan 2

sin2 θ F sin2 θ I 1 = 400. 2 . (1) − 2 θ G − 2 θ J 1 sin 1 H 1 sin 2 K

θθ= Snell’s law in this case is: nn1122sin sin

θθ= sin121 . 333 sin .

2 θθ= 2 FIG. P25.55 Squaring both sides, sin1 1777 . sin 2 . (2)

1.sin 777 2 θ F sin2 θ I Substituting (2) into (1), 2 = 400. 2 . − 2 θ G − 2 θ J 11777.sin2 H 1 sin 2 K

2 0. 444 1 Defining x = sin θ , = . 11777− . xx1 −

Solving for x, 0.. 444−=− 0 444xx 1 1 . 777 and x = 0.. 417

θ θ ==°−1 From x we can solve for 2 : 2 sin0417 . 402 . .

200. cm 200. cm Thus, the height is h == = 236. cm . θ ° tan 2 tan40 . 2

P25.56 Observe in the sketch that the angle of incidence at point P is γ, and using triangle OPQ:

L sinγ = . R

RL22− Also, cosγγ=−1 sin2 = . R

γ = φ Applying Snell’s law at point P, 1.sin 00 n sin.

sinγ L Thus, sinφ == FIG. P25.56 n nR

nR22− L 2 and cosφφ=−1 sin2 = . nR

From triangle OPS, φα++a 90.. 0 °+°−=f bg 90 0 γ 180 ° or the angle of incidence at point S is α =−γ φ . Then, applying Snell’s law at point S

continued on next page 710 Reflection and Refraction of Light gives 100.sinθα==−nn sin sinbg γφ

LF L I nR22− L 2RL 2− 2F L IO or sinθγφγφ=−=nn sin cos cos sin MG J − G JP NMH RK nR R H nRKQP

L sinθ =−−−F nR22 L 2 R 2 L 2I R2 H K

− L and θ =−−−sin 1 L F nR22 L 2 R 2 L 2IO . NM R2 H KQP

P25.57 As shown in the sketch, the angle of incidence at point A is:

−−11F d 2I F 100. mI θ = sinG J = sin G J =°30. 0 . H R K H 2.00 mK

If the emerging ray is to be parallel to the incident ray, the path must be symmetric about the centerline CB of the cylinder. In the isosceles triangle ABC,

γ = α βθ=°− and 180 . FIG. P25.57 Therefore, αβγ++=180 °

becomes 2180α +°−=°θ 180

θ or α ==15.. 0 ° 2

Then, applying Snell’s law at point A,

n sinα = 100 . sinθ

sinθ sin30 . 0° or n == = 193. . sinα sin15 . 0°

P25.58 sinθ sinθθ sin 1 12θ sin 2 0.174 0.131 1.330 4 0.342 0.261 1.312 9 0.500 0.379 1.317 7 0.643 0.480 1.338 5 0.766 0.576 1.328 9 0.866 0.647 1.339 0 0.940 0.711 1.322 0 0.985 0.740 1.331 5

The straightness of the graph line demonstrates Snell’s proportionality. FIG. P25.58

The slope of the line is n =±1.. 327 6 0 01

and n =±1328... 08% Chapter 25 711

θ =° θ =° P25.59 (a) Given that 1 45. 0 and 2 76.. 0

Snell’s law at the first surface gives

n sinα =°1 . 00 sin 45 . 0 . (1)

Observe that the angle of incidence at the second surface is

β =°−90.. 0 α

Thus, Snell’s law at the second surface yields FIG. P25.59

nnsinβα=°−=° sinaf90 . 0 1 . 00 sin 76 . 0

or n cosα =° sin76 . 0 . (2)

sin45 . 0° Dividing Equation (1) by Equation (2), tanα = = 0729. sin76 . 0°

or α =°36.. 1

sin45 . 0° sin 45 . 0° Then, from Equation (1), n = = = 120.. sinα sin36 . 1°

L (b) From the sketch, observe that the distance the light travels in the plastic is d = . Also, sinα c the speed of light in the plastic is v = , so the time required to travel through the plastic is n

d nL 1200500..af m − ∆t == = =×340.. 109 s = 340 ns. v c sinα ej300.sin.×° 108 ms 361

*P25.60 Consider an insulated box with the imagined one-way mirror forming one face, installed so that 90% of the electromagnetic radiation incident from the outside is transmitted to the inside and only a lower percentage of the electromagnetic waves from the inside make it through to the outside. Suppose the interior and exterior of the box are originally at the same temperature. Objects within and without are radiating and absorbing electromagnetic waves. They would all maintain constant temperature if the box had an open window. With the glass letting more energy in than out, the interior of the box will rise in temperature. But this is impossible, according to Clausius’s statement of the second law. This reduction to a contradiction proves that it is impossible for the one-way mirror to exist.

ANSWERS TO EVEN PROBLEMS

P25.2 see the solution P25.8 22. 5°

P25.4 25. 5°; 442 nm P25.10 (a) 1.52; (b) 417 nm; (c) 474 THz; (d) 198 Mm/s P25.6 (a) 474 THz ; (b) 422 nm; (c) 200 Mm s 712 Reflection and Refraction of Light P25.12 158 Mm/s dn P25.36 (a) ; (b) yes, yes, yes; (c) 350 µ m n − 1 P25.14 339. m P25.38 Scattered light leaving the photograph in P25.16 6.30 cm all forward horizontal directions in air is gathered by refraction into a fan in the P25.18 106 ps water of half-angle 48.6°. At larger angles I see things on the other side of the globe, P25.20 25.7° reflected by total internal reflection at the back surface of the cylinder. P25.22 see the solution P25.40 (a) see the solution, the angles are θ ′= ° θ =° P25.24 30.0° and 19.5° at entry, 40.5° and 77.1° at 1 30. 0 and 2 18.; 8 exit (b) see the solution, the angles are θ ′= ° θ =° 1 30. 0 and 2 50.; 8 − P25.26 sin12F n −−1 sinΦΦ cos I (c) see the solution; (d) see the solution H K P25.42 see the solution P25.28 see the solution P25.44 (a) 45.0°; P25.30 (a) 10.7°; (b) air; (b) yes, such as 67.5°, see the solution (c) sound in air falling on the wall from most directions is 100% reflected P25.46 3.79 m

P25.32 Skylight incident from the above travels P25.48 (b) 68.5% down the plastic. If the index of refraction of the plastic is greater than 1.41, the rays 2ω xd22+ ω m ej close in direction to the vertical are totally P25.50 (a) 2R m ; (b) reflected from the side walls of the slab d and from both facets at the bottom of the plastic, where it is not immersed in P25.52 164 s

gasoline. This light returns up inside the ° plastic and makes it look bright. Where the P25.54 36. 5 plastic is immersed in gasoline, total − L internal reflection is frustrated and the P25.56 sin 1 L F nR22−− L 2 R 2 − L 2IO M 2 P downward-propagating light passes from N R H KQ the plastic out into the gasoline. Little light ==± is reflected up, and the gauge looks dark. P25.58 see the solution; n slope132808%..

P25.34 62.4° P25.60 see the solution