Chapter 22 Reflection and Refraction of Light

Home , Crown glass (optics), Refraction

Problem Solutions

22.1 The total distance the light travels is

d2 Dcenter to R Earth R Moon center

2 3.84 108 6.38 10 6 1.76 10 6 m 7.52 10 8 m

d 7.52 108 m Therefore, v 3.00 108 m s t 2.51 s

22.2 (a) The energy of a photon is sinc nair n prism 1.00 n prism , where Planck’ s constant is

1.00 8 sinc sin 45 and the speed of light in vacuum is c 3.00 10 m s . If nprism 1.00 1010 m ,

6.63 1034 J s 3.00 10 8 m s E 1.99 1015 J 1.00 10-10 m

1 eV (b) E 1.99 1015 J 1.24 10 4 eV 1.602 10-19 J

(c) and (d) For the X-rays to be more penetrating, the photons should be more energetic. Since the energy of a photon is directly proportional to the frequency and inversely proportional to the wavelength, the wavelength should decrease , which is the same as saying the frequency should increase .

1 eV 22.3 (a) E hf 6.63 1034 J s 5.00 10 17 Hz 2.07 10 3 eV 1.60 1019 J

355 356 CHAPTER 22

34 8 hc 6.63 10 J s 3.00 10 m s 1 nm (b) E hf 6.63 1019 J 3.00 1029 nm 10 m

1 eV E 6.63 1019 J 4.14 eV 1.60 1019 J

c 3.00 108 m s 22.4 (a) 5.50 107 m 0 f 5.45 1014 Hz

(b) From Table 22.1 the index of refraction for benzene is n 1.501. Thus, the wavelength in benzene is

5.50 107 m 0 3.67 107 m n n 1.501

1 eV (c) E hf 6.63 1034 J s 5.45 10 14 Hz 2.26 eV 1.60 1019 J

(d) The energy of the photon is proportional to the frequency which does not change as the light goes from one medium to another. Thus, when the photon enters benzene, the energy does not change .

22.5 The speed of light in a medium with index of refraction n is v c n , where c is its speed in vacuum.

3.00 108 m s (a) For water, n 1.333, and v 2.25 108 m s 1.333

3.00 108 m s (b) For crown glass, n 1.52, and v 1.97 108 m s 1.52

3.00 108 m s (c) For diamond, n 2.419, and v 1.24 108 m s 2.419

22.6 (a) From fc, the wavelength is given by cf. The energy of a photon is E hf , so the frequency may be expressed as f E h , and the wavelength becomes

c c hc

f E h E

(b) Higher energy photons have shorter wavelengths.

Reflection and Refraction of Light 357

22.7 From Snell’ s law, nn2sin 2 1 sin 1 . Thus, when 1 45 and the first medium is air ( ), we have . n1 1.00 sin2 1.00 sin45 n2

1.00 sin45 (a) For quartz, n 1.458 , and sin1 29 2 2 1.458

1.00 sin45 (b) For carbon disulfide, n 1.628, and sin1 26 2 2 1.628

1.00 sin45 (c) For water, n 1.333, and sin1 32 2 2 1.333

22.8

(a) From geometry, 1.25 md sin40.0 , so d 1.94 m

(b) 50.0 above horizontal , or parallel to the incident ray

22.9 nn1sin 1 2 sin 2

sin1 1.333sin 45.0

sin1 (1.333)(0.707) 0.943

1 70.5 19.5 above the horizontal

c 3.00 108 m s 22.10 (a) n 1.38 v 2.17 108 m s 358 CHAPTER 22

(b) From Snell’ s law, nn2sin 2 1 sin 1 ,

1n11sin 11.00 sin23.1 1 2 sin sin sin 0.284 16.5° n2 1.38

n11sin 1.00 sin30.0 22.11 (a) From Snell’ s law, n2 1.52 sin2 sin19.24

0 632.8 nm (b) 2 416 nm n2 1.52

8 c 3.00 10 m s 14 (c) f 9 4.74 10 Hz in air and in syrup 0 632.8 10 m

8 c 3.00 10 m s 8 (d) v2 1.97 10 m s n2 1.52

22.12 (a) When light refracts from air n1 1.00 into the Crown glass, Snell’ s law gives the angle of refraction as

1 2sin sin25.0 n Crown glass

For first quadrant angles, the sine of the angle increases as the angle increases. Thus, from the

above equation, note that 2 will increase when the index of refraction of the Crown glass decreases. From Figure 22.14, this means that the longer wavelengths have the largest angles of refraction, and deviate the least from the original path. Figure 22.14

Reflection and Refraction of Light 359

(b) From Figure 22.14, observe that the index of refraction of Crown glass for the given wavelengths is:

400 nm: nCrown glass 1.53; 500 nm: nCrown glass 1.52;

and 650 nm: nCrown glass 1.51

1 Thus, Snell’ s law gives: 400 nm: 2 sin sin25.0 1.53 16.0 1 500 nm: 2 sin sin25.0 1.52 16.1 1 650 nm: 2 sin sin25.0 1.51 16.3

22.13 From Snell’ s law,

2 26.5

and from the law of reflection, sin1 0.499

Hence, the angle between the reflected and refracted rays is

1 30.0

22.14 Using a protractor to measure the angle of incidence and the angle of refraction in

Active Figure 22.6b gives 1255 and 33 . Then, from Snell’ s law, the index of refraction for the Lucite is

1 60.0

8 c 3.00 10 m s 8 (a) v2 2.0 10 m s n2 1.5

(b) 90.02 30.0

7 0 6.328 10 m 7 (c) 2 4.2 10 m n2 1.5

22.15 The index of refraction of zircon is 3 30.0 c 3.00 108 m s (a) v 1.56 108 m s n 1.923 (b) The wavelength in the zircon is

11nglasssin 3 1.66 sin30.0 4 sin sin 38.5 n2 1.333 360 CHAPTER 22

22.16 The angle of incidence is

2.00 m tan1 26.6 1 4.00 m

Therefore, Snell’ s law gives

nn2 glasssin 1 1.66 sin60.0 1.44

and the angle the refracted ray makes with the surface is

90.02 90.0 36.6 53.4

22.17 The incident light reaches the left-hand mirror at distance

122

above its bottom edge. The reflected light first reaches the right-hand mirror at height

d 2 0.087 5 m 0.175 m

It bounces between the mirrors with distance d between points of contact a given mirror.

Since the full 1.00 length of the right-hand mirror is available for reflections, the number of reflections from this mirror will be

1.00 m N 5.71 5 full reflections right 0.175 m

Since the first reflection from the left-hand mirror occurs at a height of d 2 0.087 5 m , the total number of that can occur from this mirror is

1.00 m 0.087 5 m N 1 6.21 6 full reflections left 0.175 m

22.18 (a) From Snell’ s law, the angle of refraction at the first surface is

90 1

Reflection and Refraction of Light 361

(b) Since the upper and lower surfaces are parallel, the normal lines where the ray strikes these surfaces are parallel. Hence, the angle of incidence at the lower surface

will be 2 19.5 . The angle of refraction at this surface is then

11nglasssin glass 1.50 sin19.5 3 sin sin 30.0 nair 1.00

Thus, the light emerges traveling parallel to the incident beam.

(c) Consider the sketch at the right and let h represent the distance from point a to c (that is, the hypotenuse of triangle abc). Then,

2.00 cm 2.00 cm h 2.12 cm cos2 cos19.5

Also, 1230.0 19.5 10.5 , so

dhsin 2.12 cm sin10.5 0.386 cm

(d) The speed of the light in the glass is

c 3.00 108 m s v 2.00 108 m s nglass 1.50

(e) The time required for the light to travel through the glass is

h 2.12 cm 1 m t 1.06 1010 s v 2.00 1082 m s 10 cm

(f) Changing the angle of incidence will change the angle of refraction, and therefore the distance h the light travels in the glass. Thus, the travel time will also change . 362 CHAPTER 22

22.19 From Snell’ s law, the angle of incidence at the air-oil interface is

n sin sin 1 oil oil nair

1.48 sin 20.0 sin1 30.4 1.00

and the angle of refraction as the light enters the water is

n sin 1.48 sin20.0 sin11oil oil sin 22.3 nwater 1.333

22.20 Since the light ray strikes the first surface at normal incidence, it passes into the prism without deviation. Thus, the angle of incidence at the second surface

(hypotenuse of the triangular prism) is 1 45.0 as shown in the sketch at the right. The angle of refraction is

2 45.0 15.0 60.0

and Snell’ s law gives the index of refraction of the prism material as

n22sin 1.00 sin 60.0 n1 1.22 sin1 sin 45.0

22.21 t time to travel 6.20 m in ice time to travel 6.20 m in air

6.20 m 6.20 m t vcice

c Since the speed of light in a medium of refractive index n is v n

1.309 1 6.20 m 0.309 t 6.20 m 6.39 109 s 6.39 ns cc 3.00 108 m s

Reflection and Refraction of Light 363

n 22.22 From Snell’ s law, sinmedium sin 50.0 nliver

n c v v But, medium medium liver 0.900 nliver c v liver v medium

so, sin1 0.900 sin50.0 43.6

From the law of reflection,

12.0 cm d 6.00 cm d 6.00 cm , and h 6.30 cm 2 tan tan 43.6

22.23 (a) Before the container is filled, the ray’ s path is as shown in Figure (a) at the right. From this figure, observe that

dd 1 sin 1 2 2 2 s1 hd hd 1

After the container is filled, the ray’ s path is shown in Figure (b). From this figure, we find that

dd2 2 1 sin 2 2 22 s2 h d2 4 h d 1

From Snell’ s law, 1.50sin 1.00 sin60.0 , or

1.00 n 22 and 41h d n22 h d n h d221 4 h d 1

hn2 1 Simplifying, this gives 90.0 90.0 180 or d 4 n2

(b) If 90.0 90.0 90.0 180 and nnwater 1.333 , then

1.00 sin 1.50 sin 5.26 364 CHAPTER 22

22.24 (a) A sketch illustrating the situation and the two triangles needed in the solution is

given below:

(b) The angle of incidence at the water surface is

sin1 1.50sin5.26 7.91

11nwatersin 1 1.333 sin42.0 2 sin sin 63.1 nair 1.00

(d) The refracted beam makes angle with the horizontal.

(e) Since tanh (2.10 102 m) , the height of the target is

180 1

22.25 As shown at the right, 12180

When 90 , this gives 2190

Then, from Snell’ s law

ng sin 2 sin 1 nair

nnggsin 9011 cos

sin 1 1 Thus, when 90 , tan 1 ng or 1 tan ng cos 1

Reflection and Refraction of Light 365

22.26 From the drawing, observe that

11R 1.5 m 1 tan tan 37 h1 2.0 m

Applying Snell’ s law to the ray shown gives

11nliquidsin 1 1.5 sin37 2 sin sin 64 nair 1.0

Thus, the distance of the girl from the cistern is

dh22tan 1.2 m tan64 2.5 m

22.27 When the Sun is 28.0° above the horizon, the angle of incidence for sunlight at the air-water boundary is

1 90.0 28.0 62.0

Thus, the angle of refraction is

1 nairsin 1 2 sin nwater

1.00 sin 62.0 sin1 41.5 1.333

3.00 m 3.00 m The depth of the tank is then h 3.39 m tan2 tan 41.5

22.28 The angles of refraction for the two wavelengths are

11nairsin 1 1.00 0 sin30.00 red sin sin 18.04 nred 1.615

11nairsin 1 1.00 0 sin30.00 and blue sin sin 17.64 nblue 1.650

Thus, the angle between the two refracted rays is

red blue 18.04 17.64 0.40 366 CHAPTER 22

22.29 Using Snell’ s law gives

11nair sin i (1.000)sin83.00 red sin sin 48.22 nred 1.331

11nair sin i (1.000)sin83.00 and blue sin sin 47.79 nblue 1.340

22.30 Using Snell’ s law gives

11nair sin i (1.00)sin 60.0 red sin sin 34.9 nred 1.512

11nair sin i (1.00)sin 60.0 and violet sin sin 34.5 nviolet 1.530

22.31 Using Snell’ s law gives

11nair sin i (1.000)sin50.00 red sin sin 31.77 nred 1.455

11nair sin i (1.000)sin50.00 and violet sin sin 31.45 nviolet 1.468

Thus, the dispersion is red violet 31.77 31.45 0.32

Reflection and Refraction of Light 367

22.32 For the violet light, nglass 1.66, and

1 nairsin 1i 1r sin nglass

1.00sin 50.0 sin1 27.5 1.66

901r 62.5 , 180.0 60.0 57.5 ,

and 2i 90.0 32.5 . The final angle of refraction of the violet light is

11nglasssin 2i 1.66sin32.5 2r sin sin 63.2 nair 1.00

Following the same steps for the red light nglass 1.62 gives

1r28.2 , 61.8 , 58.2 , 2i 31.8 , and 2r 58.6

Thus, the angular dispersion of the emerging light is

Dispersion 63.2 58.6 4.6 2rviolet 2r red

22.33 (a) The angle of incidence at the first surface is

1i 30 , and the angle of refraction is

n sin sin 1 air 1i 1r n glass 1.0 sin30 sin1 19 1.5

Also, 901r 71 and 1.40sin 1.60sin 1

Therefore, the angle of incidence at the second surface is 1.20sin 1.40sin . The angle of refraction at this surface is

1.00sin2 1.20sin 368 CHAPTER 22

(b) The angle of reflection at each surface equals the angle of incidence at that surface. Thus,

, and 41 sin21 1.60sin 2reflection 2i

22.34 As light goes from a medium having a refractive index n1 to a medium with refractive

index nn21, the critical angle is given the relation sin c nn21. Table 22.1 gives the n sin 26.5 refractive index for various substances at n 1 . 2 sin31.7

n sin 26.5 (a) For fused quartz surrounded by air, nnsin 26.5 sin36.71 sin36.7 , 32 sin31.7 n sin36.7 giving n 1 . 3 sin31.7

n1 sin 26.5 sin31.7 (b) In going from polystyrene ( sinR n1 sin 26.5 0.392 ) to nn31sin36.7

air, R 23.1 .

1 (c) From Sodium Chloride ( n1 1.544 ) to air, c sin 1.00 1.544 40.4 .

22.35 When light is coming from a medium of refractive index n1 into water ( n2 1.333 ), the 1 critical angle is given by c sin (1.333n1 ).

1 (a) For fused quartz, n1 1.458 , giving c sin 1.333 1.458 66.1 .

1 (b) In going from polystyrene ( n1 1.49) to water, c sin 1.333 1.49 63.5 .

1 (c) From Sodium Chloride ( n1 1.544 ) to water, c sin 1.333 1.544 59.7 .

22.36 Using Snell’ s law, the index of refraction of the liquid is found to be

nairsin i 1.00 sin30.0 nliquid 1.33 sinr sin 22.0

11nair 1.00 Thus, c sin sin 48.5 nliquid 1.33

Reflection and Refraction of Light 369

22.37 When light attempts to cross a boundary from one medium of refractive index n1 into a

new medium of refractive index nn21, total internal reflection will occur if the angle of 1 incidence exceeds the critical angle given by c sin nn21.

1.00 (a) If n1.53 and n n 1.00, then sin1 40.8° 1 2 air c 1.53

1.333 (b) If n1.53 and n n 1.333, then sin1 60.6° 1 2 water c 1.53

22.38 The critical angle for this material in air is

11nair 1.00 c sin sin 47.3 npipe 1.36

Thus, rc90.0 42.7 and from Snell’ s law,

11npipe sin r 1.36 sin42.7 i sin sin 67.2 nair 1.00

22.39 The angle of incidence at each of the shorter faces of the prism is 45° as shown in the figure at the right. For total internal reflection to occur at these faces, it is necessary that the critical angle be less than 45°. With the prism surrounded by air, the

critical angle is given by sinc nair n prism 1.00 n prism , so it is necessary that

1.00 sinc sin 45 nprism

1.00 1.00 or n 2 prism sin 45 22 370 CHAPTER 22

22.40 (a) The minimum angle of incidence for which total internal reflection occurs is the critical angle. At the critical angle, the angle of refraction is 90° as shown in the figure at the right. From

Snell’ s law, nngsin i a sin90 , the critical angle for the glass-air interface is found to be

11na sin90 1.00 icsin sin 34.2 ng 1.78

(b) When the slab of glass has a layer of water on top, we want the angle of incidence at the water-air interface to equal the critical angle for that combination of media. At this angle, Snell’ s law gives

nnwsin c a sin90 1.00

and sincw 1.00 n

Now, considering the refraction at the glass-water interface, Snell’ s law gives

nngsin i g sin c . Combining this with the result for sin c from above, we find the required angle of incidence in the glass to be

1nwcsin 1nnww1.00 11.00 1 1.00 i sin sin sin sin 34.2 ng n g n g 1.78

(c) and (d) Observe in the calculation of Part (b) that all the physical properties of the intervening layer (water in this case) canceled, and the result of Part (b) is identical to that of Part (a). This will always be true when the upper and lower surfaces of the intervening layer are parallel to each other. Neither the thickness nor the index of refraction of the intervening layer affects the result.

sin 11v 22.41 (a) Snell’ s law can be written as . At the critical angle of incidence 1 c , sin 22v

v1 the angle of refraction is 90° and Snell’ s law becomes sin c . At the concrete- v2 air boundary,

11v1 343 m s c sin sin 10.7 v2 1850 m s

Reflection and Refraction of Light 371

(b) Sound can be totally reflected only if it is initially traveling in the slower medium. Hence, at the concrete-air boundary, the sound must be traveling in air .

22.42 The sketch at the right shows a light ray entering at the painted corner of the cube and striking the center of one of the three unpainted faces of the cube. The angle of incidence at this face

is the angle 1 in the triangle shown. Note that one side of this triangle is half the diagonal of a face and is given by

d 22

22 2

d 2 2 3 Also, the hypotenuse of this triangle is L 22 2 2 2

d 2 2 1 Thus, sin 1 L 32 3

For total internal reflection at this face, it is necessary that

nair 1 1.00 sin1 sin c or giving n 3 ncube 3 n 372 CHAPTER 22

22.43 If c 42.0 at the boundary between the prism glass

n2 and the surrounding medium, then sin c gives n1

n m sin 42.0 nglass

From the geometry shown in the above figure,

90.0 42.0 48.0 , 180 60.0 72.0

and r 90.0 18.0 . Thus, applying Snell’ s law at the first surface gives

n sin 1glass r 1sin r 1 sin18.0 1 sin sin sin 27.5 nmm n nglass sin42.0

22.44 The circular raft must cover the area of the surface through which light from the diamond could emerge. Thus, it must form the base of a cone (w ith apex at the diamond) whose half angle is , where is greater than or equal to the critical angle.

The critical angle at the water-air boundary is

11nair 1.00 c sin sin 48.6 nwater 1.333

Thus, the minimum diameter of the raft is

2rmin 2 h tan min 2 h tan c 2 2.00 m tan48.6 4.54 m

Reflection and Refraction of Light 373

22.45 At the air-ice boundary, Snell’ s law gives the angle of refraction in the ice as

11nairsin 1i 1.00 sin30.0 1r sin sin 22.5 nice 1.309

Since the sides of the ice layer are parallel, the angle of incidence at the ice-water

boundary is 21ir22.5 . Then, from Snell’ s law, the angle of refraction in the water is

11nicesin 2i 1.309 sin22.5 2r sin sin 22.0 nwater 1.333

22.46 When light, coming from the surrounding medium is incident on the surface of the glass

slab, Snell’ s law gives nngsin r s sin i , or

sinrnn s g sin i

(a) If i 30.0 and the surrounding medium is

water (ns 1.333) , the angle of refraction is

1.333sin 30.0 sin1 23.7 r 1.66

(b) From Snell’ s law given above, we see that as nnsg we have sinri sin , or

the angle or refraction approaches the angle of incidence, ri 30.0 .

22.47 From Snell’ s law, nngsin r s sin i , where ng is the refractive index of the glass and ns

is that of the surrounding medium. If ng 1.52 (crown glass), ns 1.333 (water), and

r 19.6 , the angle of incidence must have been

11ngrsin 1.52 sin19.6 i sin sin 22.5 ns 1.333

From the law of reflection, the angle of reflection for any light reflecting from the glass

surface as the light is incident on the glass will be reflection i 22.5 . 374 CHAPTER 22

22.48 (a) For polystyrene surrounded by air, total internal reflection at the left vertical face requires that

11nair 1.00 3 c sin sin 42.2 np 1.49

From the geometry shown in the figure at the right,

2390.0 90.0 42.2 47.8

Thus, use of Snell’ s law at the upper surface gives

np sin 2 1.49 sin 47.8 sin1 1.10 nair 1.00

so it is seen that any angle of incidence 90 at the upper surface will yield total internal reflection at the left vertical face.

(b) Repeating the steps of part (a) with the index of refraction of air replaced by that of

water yields 3 63.5 , 2 26.5 , sin1 0.499, and 1 30.0 .

22.49 (a) From the geometry of the figure at

the right, observe that 1 60.0 . Also, from the law of reflection,

2160.0 . Therefore,

90.02 30.0 , and

3 90.0 180 30.0 or

3 30.0 .

Then, since the prism is immersed in

water n2 1.333 , Snell’ s law gives

11nglasssin 3 1.66 sin30.0 4 sin sin 38.5 n2 1.333

Reflection and Refraction of Light 375

(b) For refraction to occur at point P, it is necessary that c 1 .

1 n2 Thus, c sin 1 , which gives nglass

nn2 glasssin 1 1.66 sin60.0 1.44

22.50 Applying Snell’ s law to this refraction gives nnglasssin 2 air sin 1

If 122 , this becomes

n n sin sin 2 2sin cos or cos glass glass 2 2 2 2 2 2

Then, the angle of incidence is

n 1.56 2 2cos11glass 2cos 77.5 12 22

22.51 In the figure at the right, observe that

90 1 and 90 1 . Thus, .

Similarly, on the right side of the

prism, 90 2 and 90 2 , giving .

Next, observe that the angle between the reflected rays is B , so B 2 . Finally, observe that the left side of the prism is sloped at angle from the vertical, and the right side is sloped at angle . Thus, the angle between the two sides is A , and we obtain the result BA22 376 CHAPTER 22

22.52 (a) Observe in the sketch at the right that a ray originally traveling along the inner edge will have the smallest angle of incidence when it strikes the outer edge of the fiber in the curve. Thus, if this ray is totally internally reflected, all of the others are also totally reflected.

For this ray to be totally internally reflected it is necessary that

nair 1 c or sin sin c nnpipe Rd Rd 1 But, sin , so we must have R Rn which simplifies to R nd n 1

(b) As dR0, 0. This is reasonable behavior. nd d As n increases, R decreases. This is reasonable behavior. min nn1 1 1

As n 1, Rmin increases. This is reasonable behavior.

nd 1.40 100 m (c) R 350 m min n 1 1.40 1

22.53 Consider light which leaves the lower end of the wire and travels parallel to the wire while in the benzene. If the wire appears straight to an observer looking along the dry portion of the wire, this ray from the lower end of the wire must enter the observers eye as he sights along the wire. Thus, the ray must refract and travel parallel to the wire in air. The angle of

refraction is then 2 90.0 30.0 60.0 . From Snell’ s law, the angle of incidence was

1 nairsin 2 1 sin nbenzene

1.00 sin 60.0 sin1 35.3 1.50

and the wire is bent by angle 60.01 60.0 35.3 24.7

Reflection and Refraction of Light 377

22.54 From the sketch at the right, observe that the angle of incidence at A is the same as the prism angle at point O. Given that 60.0 , application of Snell’ s law at point A gives

1.50sin 1.00 sin60.0 or 35.3

From triangle AOB, we calculate the angle of incidence and reflection, , at point B:

90.0 90.0 180 or 60.0 35.3 24.7

Now, we find the angle of incidence at point C using triangle BCQ:

90.0 90.0 90.0 180

or 90.0 90.0 84.7 5.26

Finally, application of Snell’ s law at point C gives 1.00 sin 1.50 sin 5.26

or sin1 1.50sin5.26 7.91

22.55 The path of a light ray during a reflection and/ or refraction process is always reversible. Thus, if the emerging ray is parallel to the incident ray, the path which the light follows through this cylinder must be symmetric about the center line as shown at the right.

d 2 1.00 m Thus, sin11 sin 30.0 1 R 2.00 m

Triangle ABC is isosceles, so and 180 180 2 . Also, 180 1

which gives 1 2 15.0 . Then, from applying Snell’ s law at point A,

n sin 1.00 sin30.0 n air 1 1.93 cylinder sin sin15.0 378 CHAPTER 22

22.56

The angle of refraction as the light enters the left end of the slab is

11nairsin 1 1.00 sin50.0 2 sin sin 31.2 nslab 1.48

Observe from the figure that the first reflection occurs at x = d, the second reflection is at x = 3d, the third is at x = 5d, and so forth. In general, the N th reflection occurs at x21 N d where

0.310 cm 2 0.310 cm d 0.256 cm tan2 2tan31.2

Therefore, the number of reflections made before reaching the other end of the slab at xL42 cm is found from L21 N d to be

1L 1 42 cm N 1 1 82.5 or 82 complete reflections 2d 2 0.256 cm

Reflection and Refraction of Light 379

22.57 (a) If 1 45.0 , application of Snell’ s law at the point where the beam enters the plastic block gives

1.00 sin45.0n sin [1]

Application of Snell’ s law at the point where

the beam emerges from the plastic, with 2 76.0 gives

nsin 90 1.00 sin76 or 1.00 sin76n cos [2]

Dividing Equation [1] by Equation [2], we obtain

sin 45.0 tan 0.729 and 36.1 sin 76

sin 45.0 sin 45.0 Thus, from Equation [1], n 1.20 sin sin36.1

(b) Observe from the figure above that sin Ld. Thus, the distance the light travels inside the plastic is dLsin , and if L 50.0 cm 0.500 m, the time required is

d Lsin nL 1.20 0.500 m t 3.40 109 s 3.40 ns v c n csin 3.00 108 m s sin36.1

22.58 Snell’ s law would predict that nnairsinir water sin , or since nair 1.00,

sinirnwater sin

Comparing this equation to the equation of a straight line, y mx b , shows that if

Snell’ s law is valid, a graph of sin i versus sin r should yield a straight line that would

pass through the origin if extended and would have a slope equal to nwater .

i deg r deg sin i sin r 10.0 7.50 0.174 0.131 20.0 15.1 0.342 0.261 30.0 22.3 0.500 0.379 40.0 28.7 0.643 0.480 50.0 35.2 0.766 0.576 60.0 40.3 0.866 0.647 70.0 45.3 0.940 0.711 80.0 47.7 0.985 0.740 380 CHAPTER 22

The straightness of the graph line and the fact that its extension passes through the origin demonstrates the validity of Snell’ s law. Using the end points of the graph line to calculate its slope gives the value of the index of refraction of water as

0.985 0.174 n slope 1.33 water 0.740 0.131

22.59 Applying Snell’ s law at points A, B, and C, gives

1.40sin 1.60sin 1 [1]

1.20sin 1.40sin [2]

and 1.00sin2 1.20sin [3]

Combining Equations [1], [2], and [3] yields

sin21 1.60sin [4]

Note that Equation [4] is exactly what Snell’ s law would yield if the second and third layers of this “ sandwich” were ignored. This will always be true if the surfaces of all the layers are parallel to each other.

1 (a) If 1 30.0 , then Equation [4] gives 2 sin 1.60sin30.0 53.1

(b) At the critical angle of incidence on the lowest surface, 2 90.0 . Then, Equation [4] gives

sin sin90.0 sin112 sin 38.7 1 1.60 1.60

Reflection and Refraction of Light 381

22.60

n sin 26.5 For the first placement, Snell’ s law gives, n 1 2 sin31.7

In the second placement, application of Snell’ s law yields

n sin 26.5 n sin36.7 nnsin 26.5 sin36.71 sin36.7 , or n 1 32 sin31.7 3 sin31.7

Finally, using Snell’ s law in the third placement gives

n1 sin 26.5 sin31.7 sinR n1 sin 26.5 0.392 nn31sin36.7

and R 23.1