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
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:
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
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
glass glass1 1
3
air
sin 1.50 sin19.5sin sin 30.0
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
Reflection and Refraction of Light 363
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
Reflection and Refraction of Light 365
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
22.30 Using Snell’ s law gives
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
22.31 Using Snell’ s law gives
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
Reflection and Refraction of Light 367
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
Reflection and Refraction of Light 369
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
sin 1.36 sin42.7sin sin 67.2
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
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 .
(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
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
1 1air 11
ice
1.00 sin30.0sinsin sin 22.5
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
1.309 sin22.5sinsin sin 22.0
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
sin 1.66 sin30.0sin sin 38.5
1.333
n
n
Reflection and Refraction of Light 375
(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
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 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
1.00 sin50.0sinsin sin 31.2
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
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 sin45.0 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
Reflection and Refraction of Light 381
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