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355

Chapter 22

Reflection and Refraction of Light

Problem Solutions

22.1 The total d istance the light travels is

center to Earth Mooncenter

8 6 6 8

2

2 3.84 10 6.38 10 1.76 10 m 7.52 10 m

d D R R

Therefore, 8

87.52 10 m3.00 10 m s

2.51 s

dv

t

22.2 (a) The energy of a photon is air prism prismsin 1.00c n n n , where Planck’ s constant is

prism

1.00sin sin 45c

n and the speed of light in vacuum is 83.00 10 m sc . If

101.00 10 m ,

34 8

15

-10

6.63 10 J s 3.00 10 m s1.99 10 J

1.00 10 mE

(b) 15 4

-19

1 eV1.99 10 J 1.24 10 eV

1.602 10 JE

(c) and (d) For the X-rays to be more penetrating, the photons should be more

energetic. Since the energy of a photon is d irectly proportional to the frequency and

inversely proportional to the wavelength, the wavelength should decrease , which is

the same as saying the frequency should increase .

22.3 (a) 34 17 3

19

1 eV6.63 10 J s 5.00 10 Hz 2.07 10 eV

1.60 10 JE hf

356 CHAPTER 22

(b)

34 8

19

2 9

6.63 10 J s 3.00 10 m s 1 nm6.63 10 J

3.00 10 nm 10 m

hcE hf

19

19

1 eV6.63 10 J 4.14 eV

1.60 10 JE

22.4 (a) 8

7

0 14

3.00 10 m s5.50 10 m

5.45 10 Hz

c

f

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

wavelength in benzene is

7

70 5.50 10 m3.67 10 m

1.501n

n

(c) 34 14

19

1 eV6.63 10 J s 5.45 10 Hz 2.26 eV

1.60 10 JE hf

(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.

(a) For water, 1.333n , and 8

83.00 10 m s2.25 10 m s

1.333v

(b) For crown glass, 1.52n , and 8

83.00 10 m s1.97 10 m s

1.52v

(c) For d iamond, 2.419n , and 8

83.00 10 m s1.24 10 m s

2.419v

22.6 (a) From f c , the wavelength is given by c f . 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, 2 2 1 1sin sinn n . Thus, when

1 45 and the first medium is air

(1 1.00n ), we have 2 2

sin 1.00 sin45 n .

(a) For quartz, 2 1.458n , and 1

2

1.00 sin45sin 29

1.458

(b) For carbon d isulfide, 2 1.628n , and 1

2

1.00 sin45sin 26

1.628

(c) For water, 2 1.333n , and 1

2

1.00 sin45sin 32

1.333

22.8

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

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

22.9 1 1 2 2sin sinn n

1sin 1.333sin 45.0

1sin (1.333)(0.707) 0.943

1 70.5 19.5 above the horizontal

22.10 (a) 8

8

3.00 10 m s1.38

2.17 10 m s

cn

v

358 CHAPTER 22

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

1 1 11 12

2

1.00 sin23.1sinsin sin sin 0.284 16.5°

1.38

n

n

22.11 (a) From Snell’ s law, 1 12

2

1.00 sin30.0sin1.52

sin sin19.24

nn

(b) 02

2

632.8 nm416 nm

1.52n

(c) 8

14

9

0

3.00 10 m s4.74 10 Hz

632.8 10 m

cf in air and in syrup

(d) 8

8

2

2

3.00 10 m s1.97 10 m s

1.52

cv

n

22.12 (a) When light refracts from air 1 1.00n into the

Crown glass, Snell’ s law gives the angle of

refraction as

1

2 Crown glasssin sin 25.0 n

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


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

wavelengths is:

Crown glass400 nm: 1.53n ;

Crown glass500 nm: 1.52n ;

and Crown glass650 nm: 1.51n

Thus, Snell’ s law gives: 1

2400 nm: sin sin25.0 1.53 16.0

1

2500 nm: sin sin25.0 1.52 16.1

1

2650 nm: sin sin25.0 1.51 16.3

22.13 From Snell’ s law,

2 26.5

and from the law of reflection, 1sin 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 1 255 and 33 . Then, from Snell’ s law, the index of

refraction for the Lucite is

1 60.0

(a) 8

8

2

2

3.00 10 m s2.0 10 m s

1.5

cv

n

(b) 290.0 30.0

(c) 7

702

2

6.328 10 m4.2 10 m

1.5n

22.15 The index of refraction of zircon is 3 30.0

(a) 8

83.00 10 m s1.56 10 m s

1.923

cv

n

(b) The wavelength in the zircon is

glass 31 1

4

2

sin 1.66 sin30.0sin sin 38.5

1.333

n

n

360 CHAPTER 22

(c) The frequency is 1c

22.16 The angle of incidence is

1

1

2.00 mtan 26.6

4.00 m

Therefore, Snell’ s law gives

2 glass 1sin 1.66 sin60.0 1.44n n

and the angle the refracted ray makes with the surface is

290.0 90.0 36.6 53.4

22.17 The incident light reaches the left-hand mirror at

d istance

1 22

above its bottom edge. The reflected light first

reaches the right-hand mirror at height

2 0.087 5 m 0.175 md

It bounces between the mirrors with d istance 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

right

1.00 m5.71 5 full reflections

0.175 mN

Since the first reflection from the left-hand mirror occurs at a height of 2 0.087 5 md ,

the total number of that can occur from this mirror is

left

1.00 m 0.087 5 m1 6.21 6 full reflections

0.175 mN

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

190


(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

glass glass1 1

3

air


1.00

n

n

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

(c) Consider the sketch at the right and let h represent the d istance from point a to c

(that is, the hypotenuse of triangle abc). Then,

2

2.00 cm 2.00 cm2.12 cm

cos cos19.5h

Also, 1 2 30.0 19.5 10.5 , so

sin 2.12 cm sin10.5 0.386 cmd h

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

8

8

glass

3.00 10 m s2.00 10 m s

1.50

cv

n

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

10

8 2

2.12 cm 1 m1.06 10 s

2.00 10 m s 10 cm

ht

v

(f) Changing the angle of incidence will change the angle of refraction, and therefore

the d istance 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

1 oil oil

air

1

sinsin

1.48 sin 20.0sin 30.4

1.00

n

n

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

1 1oil oil

water

1.48 sin 20.0sinsin sin 22.3

1.333

n

n

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

2 21

1

1.00 sin 60.0sin1.22

sin sin 45.0

nn

22.21 time to travel 6.20 m in ice time to travel 6.20 m in airt

ice

6.20 m 6.20 mt

v c

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

vn

9

8

6.20 m 0.3091.309 16.20 m 6.39 10 s 6.39 ns

3.00 10 m st

c c


22.22 From Snell’ s law, medium

liver

sin sin 50.0n

n

But, medium medium liver

liver liver medium

0.900n c v v

n c v v

so, 1sin 0.900 sin50.0 43.6

From the law of reflection,

12.0 cm

6.00 cm2

d , and 6.00 cm

6.30 cmtan tan 43.6

dh

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

1

2 2 21

1sin

1

d d

s h d h d

After the container is filled , the ray’ s path is

shown in Figure (b). From this figure, we find

that

2

2 222

2 2 1sin

2 4 1

d d

s h d h d

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

2 2

1.00

1 4 1

n

h d h d

and 2 22 24 1h d n h d n

Simplifying, this gives 90.0 90.0 180 or 2

2

1

4

h n

d n

(b) If 90.0 90.0 90.0 180 and water 1.333n n , 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

1sin 1.50sin5.26 7.91

(c) Snell’ s law gives the angle of refraction as

1 1water 12

air

1.333 sin42.0sinsin sin 63.1

1.00

n

n

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

(e) Since 2tan (2.10 10 m)h , the height of the target is

1180

22.25 As shown at the right, 1 2 180

When 90 , this gives 2 190

Then, from Snell’ s law

2

1

air

1 1

sinsin

sin 90 cos

g

g g

n

n

n n

Thus, when 90 , 11

1

sintan

cosgn or

1

1 tan gn


22.26 From the drawing, observe that

1 1

1

1

1.5 mtan tan 37

2.0 m

R

h

Applying Snell’ s law to the ray shown gives

liquid 11 1

2

air

sin 1.5 sin 37sin sin 64

1.0

n

n

Thus, the d istance of the girl from the cistern is

2 2tan 1.2 m tan64 2.5 md h

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 air 12

water

1

sinsin

1.00 sin 62.0sin 41.5

1.333

n

n

The depth of the tank is then 2

3.00 m 3.00 m3.39 m

tan tan 41.5h

22.28 The angles of refraction for the two wavelengths are

1 1air 1

red

red

sin 1.00 0 sin 30.00sin sin 18.04

1.615

n

n

and 1 1air 1

blue

blue

sin 1.00 0 sin30.00sin sin 17.64

1.650

n

n

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

1 1air

red

red

sin (1.000)sin83.00sin sin 48.22

1.331

in

n

and 1 1air

blue

blue

sin (1.000)sin83.00sin sin 47.79

1.340

in

n


1 1air

red

red

sin (1.00)sin 60.0sin sin 34.9

1.512

in

n

and 1 1air

violet

violet

sin (1.00)sin 60.0sin sin 34.5

1.530

in

n


1 1air

red

red

sin (1.000)sin 50.00sin sin 31.77

1.455

in

n

and 1 1air

violet

violet

sin (1.000)sin 50.00sin sin 31.45

1.468

in

n

Thus, the d ispersion is red violet 31.77 31.45 0.32


22.32 For the violet light, glass 1.66n , and

1 air 1i1r

glass

1

sinsin

1.00sin 50.0sin 27.5

1.66

n

n

1r90 62.5 , 180.0 60.0 57.5 ,

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

glass 2i1 1

2r

air

sin 1.66sin 32.5sin sin 63.2

1.00

n

n

Following the same steps for the red light glass 1.62n gives

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

Thus, the angular d ispersion of the emerging light is

2r 2rviolet red63.2 58.6 4.6Dispersion

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

1 30i, and the angle of refraction is

1 air 11

glass

1

sinsin

1.0 sin30sin 19

1.5

ir

n

n

Also, 1r90 71 and

11.40sin 1.60sin

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

angle of refraction at this surface is

21.00sin 1.20sin

368 CHAPTER 22

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

Thus,

2 1sin 1.60sin , and 2 2ireflection

41

22.34 As light goes from a medium having a refractive index 1n to a medium with refractive

index 2 1n n , the critical angle is given the relation

2 1sin c n n . Table 22.1 gives the

refractive index for various substances at 12

sin 26.5

sin31.7

nn .

(a) For fused quartz surrounded by air, 13 2

sin 26.5sin 26.5 sin36.7 sin36.7

sin31.7

nn n ,

giving 13

sin36.7

sin31.7

nn .

(b) In going from polystyrene ( 11

3 1

sin 26.5 sin 31.7sin sin 26.5 0.392

sin 36.7R

nn

n n) to

air, 23.1R .

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

.

22.35 When light is coming from a medium of refractive index 1n into water (

2 1.333n ), the

critical angle is given by 1

1sin (1.333 )c n .

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

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

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

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

liquid

1.00 sin 30.0sin1.33

sin sin 22.0

air i

r

nn

Thus, 1 1air

liquid

1.00sin sin 48.5

1.33c

n

n


22.37 When light attempts to cross a boundary from one medium of refractive index 1n into a

new medium of refractive index 2 1n n , total internal reflection will occur if the angle of

incidence exceeds the critical angle given by 1

2 1sinc n n .

(a) If 1

1 2 air

1.001.53 and 1.00, then sin 40.8°

1.53cn n n

(b) If 1

1 2 water

1.3331.53 and 1.333, then sin 60.6°

1.53cn n n

22.38 The critical angle for this material in air is

1 1air

pipe

1.00sin sin 47.3

1.36c

n

n

Thus, 90.0 42.7r c and from Snell’ s law,

pipe1 1

air


1.00

r

i

n

n

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 air prism prismsin 1.00c n n n , so it is

necessary that

prism

1.00sin sin 45c

n

or prism

1.00 1.002

sin 45 2 2n

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, sin sin90g i an n , the

critical angle for the glass-air interface is found to be

1 1sin90 1.00sin sin 34.2

1.78

ai c

g

n

n

(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

sin sin90 1.00w c an n

and sin 1.00c wn

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

sin sing i g cn n . Combining this with the result for sin c from above, we find the

required angle of incidence in the glass to be

1 1 1 11.00sin 1.00 1.00

sin sin sin sin 34.21.78

w ww ci

g g g

n nn

n n n

(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.

22.41 (a) Snell’ s law can be written as 1 1

2 2

sin

sin

v

v. At the critical angle of incidence 1 c ,

the angle of refraction is 90° and Snell’ s law becomes 1

2

sin c

v

v. At the concrete-

air boundary,

1 11

2

343 m ssin sin 10.7

1850 m sc

v

v


(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 .

(c) Sound in air falling on the wall from most directions is 100% reflected , so the wall is a

good mirror.

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 d iagonal of a face and is given by

2 2

2 2 2

d

Also, the hypotenuse of this triangle is

2 22 2 3

2 2 2

dL

Thus, 1

2 2 1sin

33 2

d

L

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

air1

cube

sin sin c

n

n or

1 1.00

3 n giving 3n

372 CHAPTER 22

22.43 If 42.0c at the boundary between the prism glass

and the surrounding medium, then 2

1

sin c

n

n gives

glass

sin 42.0mn

n

From the geometry shown in the above figure,

90.0 42.0 48.0 , 180 60.0 72.0

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

glass1 1 1

1

glass

sin sin sin18.0sin sin sin 27.5

sin 42.0

r r

m m

n

n n n

22.44 The circular raft must cover the area of the surface

through which light from the d iamond could emerge.

Thus, it must form the base of a cone (w ith apex at the

d iamond) whose half angle is , where is greater than

or equal to the critical angle.

The critical angle at the water-air boundary is

1 1air

water

1.00sin sin 48.6

1.333c

n

n

Thus, the minimum diameter of the raft is

2 2 tan 2 tan 2 2.00 m tan48.6 4.54 mmin min cr h h


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

1 1air 11

ice


1.309

ir

n

n

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

boundary is 2 1 22.5i r

. Then, from Snell’ s law, the angle of refraction in the water

is

1 1ice 22

water


1.333

ir

n

n

22.46 When light, coming from the surrounding

medium is incident on the surface of the glass

slab, Snell’ s law gives sin sing r s in n , or

sin sinr s g in n

(a) If 30.0i and the surrounding medium is

water ( 1.333)sn , the angle of refraction is

11.333sin 30.0

sin 23.71.66

r

(b) From Snell’ s law given above, we see that as s gn n we have sin sinr i, or

the angle or refraction approaches the angle of incidence, 30.0r i .

(c) If s gn n , then sin ( )sin sinr s g i in n , or r i .

22.47 From Snell’ s law, sin sing r s in n , where gn is the refractive index of the glass and sn

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

19.6r, the angle of incidence must have been

1 1sin 1.52 sin19.6

sin sin 22.51.333

g r

i

s

n

n

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 22.5i .

374 CHAPTER 22

22.48 (a) For polystyrene surrounded by air, total internal

reflection at the left vertical face requires that

1 1air3

1.00sin sin 42.2

1.49c

p

n

n

From the geometry shown in the figure at the right,

2 390.0 90.0 42.2 47.8

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

2

1

air

sin 1.49 sin 47.8sin 1.10

1.00

pn

n

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 , 1sin 0.499 , and 1 30.0 .

(c) Total internal reflection is polystyrene carbon disulfidenot possible since n n

22.49 (a) From the geometry of the figure at

the right, observe that 1 60.0 .

Also, from the law of reflection,

2 1 60.0 . Therefore,

290.0 30.0 , and

3 90.0 180 30.0 or

3 30.0 .

Then, since the prism is immersed in

water 2 1.333n , Snell’ s law gives

glass 31 1

4

2


1.333

n

n


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

.

Thus, 1 21

glass

sinc

n

n, which gives

2 glass 1sin 1.66 sin60.0 1.44n n

22.50 Applying Snell’ s law to this refraction gives glass 2 air 1sin sinn n

If 1 22 , this becomes

glass 2 2 2 2sin sin 2 2sin cosn or glass

2cos2

n

Then, the angle of incidence is

glass1 1

1 2

1.562 2cos 2cos 77.5

2 2

n

22.51 In the figure at the right, observe that

190 and 190 . Thus,

.

Similarly, on the right side of the

prism, 290 and

290 ,

giving .

Next, observe that the angle between

the reflected rays is B ,

so 2B . 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

2 2B A

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

c or air

pipe

1sin sin c

n

n n

But, sinR d

R , so we must have

1R d

R n

which simplifies to 1R nd n

(b) As 0, 0d R . This is reasonable behavior.

As n increases, min1 1 1

nd dR

n n decreases. This is reasonable behavior.

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

(c) min

1.40 100 m350 m

1 1.40 1

ndR

n

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 air 21

benzene

1

sinsin

1.00 sin 60.0sin 35.3

1.50

n

n

and the wire is bent by angle 160.0 60.0 35.3 24.7


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 1sin 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.

Thus, 1 1

1

2 1.00 msin sin 30.0

2.00 m

d

R

Triangle ABC is isosceles, so and 180 180 2 . Also, 1180

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

air 1cylinder

1.00 sin30.0sin1.93

sin sin15.0

nn

378 CHAPTER 22

22.56

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

1 1air 12

slab


1.48

n

n

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

2 1x N d where

2

0.310 cm 2 0.310 cm0.256 cm

tan 2 tan31.2d

Therefore, the number of reflections made before reaching the other end of the slab at

42 cmx L is found from 2 1L N d to be

1 1 42 cm

1 1 82.52 2 0.256 cm

LN

d or 82 complete reflections


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 sin45.0 sinn [1]

Application of Snell’ s law at the point where

the beam emerges from the plastic, with 2 76.0

gives

sin 90 1.00 sin76n or 1.00 sin76 cosn [2]

Divid ing Equation [1] by Equation [2], we obtain

sin 45.0

tan 0.729sin 76

and 36.1

Thus, from Equation [1], sin 45.0 sin 45.0

1.20sin sin36.1

n

(b) Observe from the figure above that sin L d . Thus, the d istance the light travels

inside the plastic is sind L , and if 50.0 cm 0.500 mL , the time required is

9

8

1.20 0.500 msin3.40 10 s 3.40 ns

sin 3.00 10 m s sin36.1

d L nLt

v c n c

22.58 Snell’ s law would predict that air watersin sini rn n , or since

air 1.00n ,

watersin sini rn

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 watern .

degi degr

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 valid ity 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

water

0.985 0.1741.33

0.740 0.131n slope

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

11.40sin 1.60sin [1]

1.20sin 1.40sin [2]

and 21.00sin 1.20sin [3]

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

2 1sin 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.

(a) If 1 30.0 , then Equation [4] gives 1

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

1 12

1

sin sin90.0sin sin 38.7

1.60 1.60


22.60

For the first placement, Snell’ s law gives, 12

sin 26.5

sin31.7

nn

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

13 2

sin 26.5sin 26.5 sin36.7 sin36.7

sin31.7

nn n , or 1

3

sin36.7

sin31.7

nn

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

11

3 1

sin 26.5 sin 31.7sin sin 26.5 0.392

sin 36.7R

nn

n n

and 23.1R

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