355 Chapter 22 Reflection and Refraction of Light Problem Solutions 22.1 The total distance the light travels is center to Earth Moon center 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 8 7.52 10 m 3.00 10 ms 2.51 s d v t 22.2 (a) The energy of a photon is air prism prism sin 1.00 c n n n , where Planck’ s constant is prism 1.00 sin sin 45 c n and the speed of light in vacuum is 8 3.00 10 ms c . If 10 1.00 10 m , 34 8 15 -10 6.63 10 Js 3.00 10 ms 1.99 10 J 1.00 10 m E (b) 15 4 -19 1 eV 1.99 10 J 1.24 10 eV 1.602 10 J E (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 . 22.3 (a) 34 17 3 19 1 eV 6.63 10 J s 5.00 10 Hz 2.07 10 eV 1.60 10 J E hf
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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
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 (