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Physics 112
Homework 8 (solutions)
(2004 Fall)
Solutions to Homework Questions 8
Chapt22, Problem-6:
The two mirrors in Figure P22.6 meet
at a right angle. The beam of light in the vertical plane P strikes mirror 1 as
shown. (a) Determine the distance the reflected light beam travels before
striking mirror 2. (b) In what direction does the light beam travel after being
reflected from mirror 2?
Solution:
The first point to note here is that everything happens in a the
dashed Plane P (only specular reflection need be considered). Thus
it is possible (and useful) to redraw the geometry in only 2D
(ie the situation as ‘seen’ in plane P).
Once we have done that, the problem becomes one of ‘simple’
trigonometry plus the law of reflection (applied twice – once at
each mirror).
The 2D-view is shown right. So ….
(a) Where we are asked for the distance d.
Well we apply the law of reflection for mirror1, then given the
distance of that that reflection from mirror2, we can use simple
trig to calculate the distance requested
1.25 m = d sin 40.0° , so d = 1.94 m
Mirror 2
50°
i2 = 50°
40°
d
i1 = 40°
40°
50°
.(b) Where we are asked for the direction following the 2nd reflection
Here (again) using simple trig, we can calculate all the angles
1.25 m
associated with the “angle of incidence” of the 2nd reflection.
.Armed with this we simply apply the Law of Reflection again to obtain
50.0° above horizontal , or parallel to the incident ray.
50°
Mirror 1
Chapt22, Problem-7: An underwater scuba diver sees the Sun at an apparent angle of 45.0° from the vertical.
What is the actual direction of the Sun?
Solution:
Here one simply has to use Snell’s Law – obviously a sketch helps…
n1 sin !1 = n2 sin !2
sin !1 = 1.333 sin 45.0°
sin !1 = (1.333)(0.707) = 0.943
so !1 = 70.5° " 19.5° above the horizontal
!1
n1 = 1.00
!2
n2 = 1.333
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Physics 112
Homework 8 (solutions)
(2004 Fall)
Chapt22, Problem-13: A ray of light is incident on the surface of a block of clear ice at an angle of 40.0°
with the normal. Part of the light is reflected and part is refracted. Find the angle between the reflected and refracted light.
Solution:
Again, a straight application of Snell’s law:
40.0# $
#n sin !1 &
"1 # (1.00 ) sin 40.0° &
!2 = sin "1 % 1
( = sin %
( = 29.4° ,
1.309
$
'
$ n2 '
and the law of reflection gives ! = "1 = 40.0° .
air
"
ice
Hence, the angle between the reflected and refracted rays is
!2
! = 180° " #2 " $ = 180° " 29.4° " 40.0° = 111°
Chapt22, Problem-22: A submarine is 300 m horizontally out from the shore and 100 m beneath the surface
of the water. A laser beam is sent from the sub so that it strikes the surface of the water at a point 210 m from the shore. If
the beam just strikes the top of a building standing directly at the water’s edge, find the height of the building.
Solution:
The angle of incidence at the water
surface is
!2
air
n = 1.00
90.0
m
h
&
#
!2
!1 = tan "1 %
( = 42.0°
$ 100 m '
90.0 m
Then, Snell’s law gives the angle of
refraction as
#n
sin !1 &
"1# (1.333 ) sin 42.0° &
!2 = sin "1 % water
( = 63.1° ,
( = sin %
'
nair
$
1.00
$
'
210 m
210 m
so the height of the building is h =
=
= 107 m
tan!2 tan63.1°
!1
water
n = 1.333
!1
100 m
210 m
Submarine
Chapt22, Problem-24: A narrow beam of ultrasonic waves
reflects off the liver tumor in Figure P22.24. If the speed of the wave is
10.0% less in the liver than in the surrounding medium, determine
the depth of the tumor.
Solution:
" nmedium %
' sin 50.0° .
# n liver &
v
= liver = 0.900 ,
v medium
From Snell’s law, sin ! = $
But,
so,
n medium c vmedium
=
n liver
c v liver
12.0 cm
50.0#
d
! = sin "1[(0.900 )sin 50.0° ] = 43.6° .
From the law of reflection,
d=
12.0 cm
= 6.00 cm , and
2
h=
d
6.00 cm
=
= 6.30 cm
tan! tan ( 43.6°)
2
h
nmedium
!
!
Tumor
nliver
Physics 112
Homework 8 (solutions)
(2004 Fall)
Chapt22, Problem-30: A certain kind of glass has an index of refraction of 1.650 for blue light of
wavelength 430 nm and an index of 1.615 for red light of wavelength 680 nm. If a beam containing these two colors is
incident at an angle of 30.00° on a piece of this glass, what is the angle between the two beams inside the glass?
Solution:
The angles of refraction for the two wavelengths are
# n sin !i &
"1 # ( 1.000 ) sin 30.00° &
!red = sin "1% air
( = 18.04°
( = sin %
'
$
1.615
$ nred '
# n sin !i &
"1# (1.000 ) sin 30.00° &
and !blue = sin "1 % air
( = sin %
( = 17.64°
$
1.650
'
$ n blue '
Thus, the angle between the two refracted rays is !" = "red # "blue = 18.04° #17.64° = 0.40°
Chapt22, Problem-32: The index of refraction for
violet light in silica flint glass is 1.66 and that for red light is
1.62. What is the angular dispersion of visible light passing
through an equilateral prism of apex angle 60.0° if the angle
of incidence is 50.0°? (See Fig. P22.32.)
Solution:
For the violet light, n glass = 1.66 , and
# n sin ! &
1i
!1r = sin "1 % air
(
n
$
glass
'
# 1.00 sin 50.0° &
= sin "1 %
( = 27.5°
'
$
1.66
! = 90° " #1r = 62.5°, $ = 180.0° " 60.0° " ! = 57.5°,
!2i = 90.0° " # = 32.5° . The final angle of refraction
and
of the violet light is
nair = 1.00
60.0#
"
!1i = 50.0#
!1r !2i
# n glass sin !2i &
"1 # 1.66 sin 32.5° &
!2r = sin "1 %
( = 63.2°
( = sin %$
'
n air
1.00
$
'
(
%
!2r
nglass
)
Following the same steps for the red light n glass = 1.62 gives
!1r = 28.2°, " = 61.8°, # = 58.2°, ! 2i = 31.8°, and !2r = 58.6°
Thus, the angular dispersion of the emerging light is Dispersion = !2r
violet
" !2r
red
= 63.2° " 58.6° = 4.6°
Chapt22, Problem-36: A beam of light is incident from air on the surface of a liquid. If the angle of
incidence is 30.0° and the angle of refraction is 22.0°, find the critical angle for the liquid when surrounded by air.
Solution:
Using Snell’s law, the index of refraction of the liquid is found to be
n air sin !i (1.00 ) sin 30.0°
=
= 1.33
sin22.0°
sin !r
# n &
# 1.00 &
Thus, !c = sin "1 % air ( = sin "1 %
( = 48.5°
$ 1.33 '
$ n liquid '
n liquid =
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Physics 112
Homework 8 (solutions)
Chapt22, Problem-40: A jewel thief hides a diamond by
(2004 Fall)
placing it on the bottom of a public swimming pool. He places a circular
raft on the surface of the water directly above and centered on the diamond,
as shown in Figure P22.40. If the surface of the water is calm and the pool
is 2.00 m deep, find the minimum diameter of the raft that would prevent the
diamond from being seen.
Solution:
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 (with apex at the
diamond) whose half angle is q, where q is greater than
or equal to the critical angle.
rmin
h
The critical angle at the water-air boundary is
!
!
!
!
# n &
# 1.00 &
!c = sin "1 % air ( = sin "1 %
( = 48.6°
$ 1.333 '
$ n water '
Thus, the minimum diameter of the raft is
2 rmin = 2 h tan !min = 2 h tan !c = 2 ( 2 .00 m ) tan 48.6° = 4.54 m
Chapt22, Problem-54: When you look through a window, by how much time is the light you see delayed
by having to go through glass instead of air? Make an order-of-magnitude estimate on the basis of data you specify. By how
many wavelengths is it delayed?
Solution:
Consider glass d = 4.0 mm thick, and with index of refraction n = 1.5 . The speed of light in the glass is
c 3.0 ! 108 m s
v= =
= 2 .0 ! 108 m s
n
1.5
The extra travel time is
!t =
d d
4.0 #10 "3 m
4.0 #10 "3 m
!11
" =
"
= 6.7 # 10"12 s ~10 s
v c 2.0 #108 m s 3.0 #108 m s
For light of wavelength !0 = 600 nm in vacuum, the wavelength in the glass is
!=
!0 600 nm
=
= 400 nm
1.5
n
and the extra path length as a number of wavelengths is
d d
4.0 $ 10#3 m 4.0 $10 #3 m
3
!d = #
=
#
= 3.3 $ 103 ~10
#9
#9
" "0 400 $10 m 600 $ 10 m
Chapt22, Conceptual-1: Under certain circumstances, sound can be heard from extremely far away. This
frequently happens over a body of water, where the air near the water surface is cooler than the air at higher altitudes. Explain
how the refraction of sound waves could increase the distance over which sound can be heard.
Solution:
Sound radiated upward at an acute angle to the horizontal is bent back towards the Earth by refraction.
This means that the sound can reach the listener by this path as well as by a direct path. Thus, the sound
is louder, allowing it to be heard from a greater distance.
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Physics 112
Homework 8 (solutions)
(2004 Fall)
Chapt22, Conceptual-2: What are some of the reasons that most ceilings are made of white, textured
material?
Solution:
Ceilings are generally painted white (or another light color) so they will reflect more light, making the room
brighter. Textured materials are often used on the ceiling to diffuse the reflected light and reduce ‘glare’
(specular reflections).
Chapt22, Conceptual-3: The color of an object is said to depend on wavelength. So, if you view colored
objects under water, in which the wavelength of light will be different, does the color change?
Solution:
The color will not change, for two reasons. First, despite the popular statement that the color depends on
wavelength, it actually depends on the frequency of the light, which does not change under water. Second,
when the light enters the eye, it travels through thhe fluid within the eye. Thus, even if color did depend
on wavelength, the important wavelength is that of the light in the ocular fluid, which does of course does
not change whether your head is under water or not.
Chapt22, Conceptual-5: Pass white light through a prism to produce a multicolored spectrum. Let that
spectrum fall on a piece of cardboard that has a slit in it such that only green light passes through. Nw let this green light
fall on a second prism. Describe the light that comes out of the second prism.
Solution:
A prism creates a multicolored spectrum when white light passes through it because white light contains
all visible wavelengths. Because the index of refraction varies wavelength, the different wavelengths
emerge from the prism traveling in slightly different directions. This produces the observed rainbow of
colors. If only one wavelength of light (or a very small range of wavelengths such as the small range known
as ‘green’) is allowed to enter the prism, as in the case of the second prism in the question, a single
wavelength (or small range - ‘green’) will emerge from it.
Chapt22, Conceptual-17: Why do astronomers looking at distant galaxies talk about looking back in
time?
Solution:
Light travels through the vacuum of space at a speed of 3x108 m s-1. Thus an image we see from a distant
star or galaxy must have been generated some time ago. For example, the star Altair is 16 lightyears away;
if we look at an image of Altair taken today, we are actually seeing what Altair looked like 16 years ago.
Hence we are effectively ‘looking back in time’
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