Chapter 17 continued
r i
42°
r, mirror 90° r
90° 42°
48°
17.2 Curved Mirrors
page 480
Level 1
61. A concave mirror has a focal length of
10.0 cm. What is its radius of curvature?
r 2f 2(10.0 cm) 20.0 cm
62. An object located 18 cm from a convex
mirror produces a virtual image 9 cm from
the mirror. What is the magnification of the
image?
di
C
F
■
Figure 17-21
real; inverted; larger
Level 2
65. Star Image Light from a star is collected by
a concave mirror. How far from the mirror
is the image of the star if the radius of
curvature is 150 cm?
Stars are far enough away that the
light coming into the mirror can be
considered to be parallel and parallel
light will converge at the focal point.
0.5
Since r 2f,
m do
(9 cm)
18 cm
r
2
150 cm
2
f 75 cm
66. Find the image position and height for the
object shown in Figure 17-22.
1
3
mirror is , what is the boy’s height?
hi
m ho
hi
ho 3.8 cm
m
F
16 cm
0.60 m
13
31 cm
1.8 m
64. Describe the image produced by the object
in Figure 17-21 as real or virtual, inverted
or upright, and smaller or larger than the
object.
■
Figure 17-22
1
1
1
f
do
di
d f
do f
di o
(31 cm)(16 cm)
31 cm 16 cm
366
Solutions Manual
Physics: Principles and Problems
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
63. Fun House A boy is standing near a
convex mirror in a fun house at a fair. He
notices that his image appears to be
0.60 m tall. If the magnification of the
Chapter 17 continued
33 cm
what is the magnification of the image?
hi
di
ho
do
m diho
hi do
(33 cm)(3.8 cm)
31 cm
4.1 cm
67. Rearview Mirror How far does the image
of a car appear behind a convex mirror,
with a focal length of 6.0 m, when the car
is 10.0 m from the mirror?
1
1
1
do
di
f
d f
do f
di o
(10.0 m)(6.0 m)
10.0 m (6.0 m)
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
3.8 m
68. An object is 30.0 cm from a concave mirror
of 15.0 cm focal length. The object is
1.8 cm tall. Use the mirror equation to
find the image position. What is the image
height?
1
1
1
do
di
f
d f
do f
di o
(30.0 cm)(15.0 cm)
30.0 cm 15.0 cm
30.0 cm
hi
di
m ho
do
diho
hi do
(30.0 cm)(1.8 cm)
(30.0 cm)
1.8 cm
69. Dental Mirror A dentist uses a small mirror with a radius of 40 mm to locate a
cavity in a patient’s tooth. If the mirror is
concave and is held 16 mm from the tooth,
Physics: Principles and Problems
(40 mm)
2
r
2
f 20 mm
1
1
1
do
di
f
d f
do f
(16 mm)(20 mm)
16 mm 20 mm
di o 80 mm
di
(80 mm)
16 mm
m 5
do
70. A 3.0-cm-tall object is 22.4 cm from a
concave mirror. If the mirror has a radius of
curvature of 34.0 cm, what are the image
position and height?
r
2
f 34.0 cm
2
17.0 cm
1
1
1
do
di
f
d f
do f
di o
(22.4 cm)(17.0 cm)
22.4 cm 17.0 cm
70.5 cm
hi
di
ho
do
m diho
hi do
(70.5 cm)(3.0 cm)
22.4 cm
9.4 cm
Level 3
71. Jeweler’s Mirror A jeweler inspects a
watch with a diameter of 3.0 cm by placing
it 8.0 cm in front of a concave mirror of
12.0-cm focal length.
a. Where will the image of the watch
appear?
1
1
1
do
di
f
Solutions Manual
367
Chapter 17 continued
d f
do f
(8.0 cm)(12.0 cm)
8.0 cm 12.0 cm
di o 24 cm
1
1
1
do
di
f
fdo
b. What will be the diameter of the image?
hi
di
ho
do
di do f
(10.0 cm)(150 cm)
150 cm (10.0 cm)
9.4 cm
diho
(24 cm)(3.0 cm)
8.0 cm
hi do
di
(9.4 cm)
150 cm
m 0.063
do
hi mho (0.063)(12 cm) 0.75 cm
9.0 cm
72. Sunlight falls on a concave mirror and
forms an image that is 3.0 cm from the
mirror. An object that is 24 mm tall is
placed 12.0 cm from the mirror.
a. Sketch the ray diagram to show the
location of the image.
O1 Ray 1
Ray 2
C F
Horizontal scale:
1 block 1.0 cm
Vertical scale:
1 block 4 mm
I1
b. Use the mirror equation to calculate the
image position.
fd
do f
pages 480–481
Level 1
74. A light ray strikes a plane mirror at an angle
of 28° to the normal. If the light source is
moved so that the angle of incidence
increases by 34°, what is the new angle of
reflection?
i i, initial 34°
28° 34°
62°
r i
62°
75. Copy Figure 17-23 on a sheet of paper.
Draw rays on the diagram to determine the
height and location of the image.
(3.0 cm)(12.0 cm)
12.0 cm 3.0 cm
o
di 4.0 cm
c. How tall is the image?
di
4.0 cm
m 0.33
do
12.0 cm
3.0 cm
8.0 cm
F
4.0 cm
hi mho (0.33)(24 mm)
8.0 mm
73. Shiny spheres that are placed on pedestals
on a lawn are convex mirrors. One such
sphere has a diameter of 40.0 cm.
A 12-cm-tall robin sits in a tree that is
1.5 m from the sphere. Where is the image
of the robin and how tall is the image?
■
Figure 17-23
r 20.0 cm, f 10.0 cm
368
Solutions Manual
Physics: Principles and Problems
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
1
1
1
do
di
f
Mixed Review
Chapter 17 continued
b. What is the image height?
O1
I1
Horizontal scale:
1 block 1.0 cm
Vertical scale:
2 blocks 1.0 cm
hi 1.0 cm
di 2.7 cm
hi
m ho
F
diho
hi do
(22.9 cm)(2.4 cm)
30.0 cm
The image height is 1.0 cm, and its
location is 2.7 cm from the mirror.
Level 2
76. An object is located 4.4 cm in front of a
concave mirror with a 24.0-cm radius.
Locate the image using the mirror equation.
78. What is the radius of curvature of a concave
mirror that magnifies an object by a factor
of 3.2 when the object is placed 20.0 cm
from the mirror?
hi
m r
f 2
ho
24.0 cm
2
12.0 cm
1
1
1
do
di
f
d f
do f
di o
(4.4 cm)(12.0 cm)
4.4 cm 12.0 cm
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
1.8 cm
6.9 cm
77. A concave mirror has a radius of curvature
of 26.0 cm. An object that is 2.4 cm tall is
placed 30.0 cm from the mirror.
a. Where is the image position?
r
2
f 26.0 cm
2
13.0 cm
1
1
1
do
di
f
d f
di o
do f
(30.0 cm)(13.0 cm)
30.0 cm 13.0 cm
22.9 cm
di mdo
(3.2)(20.0 cm)
64 cm
1
1
1
do
di
f
d d
do di
i
f o
(20.0 cm)(64 cm)
20.0 cm (64 cm)
29 cm
r 2f
(2)(29 cm)
58 cm
79. A convex mirror is needed to produce an
image one-half the size of an object and
located 36 cm behind the mirror. What
focal length should the mirror have?
hi
di
ho
do
m diho
do hi
(36 cm)h
o
ho
2
72 cm
1
1
1
do
di
f
Physics: Principles and Problems
Solutions Manual
369
Chapter 17 continued
dodi
a. What kind of mirror would do this job?
f do di
An enlarged, upright image results
only from a concave mirror, with the
object inside the focal length.
(72 cm)(36 cm)
72 cm (36 cm)
b. What is its radius of curvature?
72 cm
di
m 80. Surveillance Mirror A convenience store
uses a surveillance mirror to monitor the
store’s aisles. Each mirror has a radius of
curvature of 3.8 m.
do
di mdo (7.5)(14.0 mm)
105 mm
a. What is the image position of a customer
who stands 6.5 m in front of the mirror?
1
1
1
do
di
f
A mirror that is used for surveillance
is a convex mirror. So the focal
length is the negative of half the
radius of curvature.
dodi
di do
16 mm
r 2f (2)(16mm)
32 mm
r
2
f 3.8 m
2
1.9 m
1
1
1
do
di
f
(14.0 mm)(105 mm)
14.0 mm (105 mm)
f 82. The object in Figure 17-24 moves from
position 1 to position 2. Copy the diagram
onto a sheet of paper. Draw rays showing
how the image changes.
d f
do f
di o
1
C
2
F
1.5 m
1.0 m
b. What is the image height of a customer
who is 1.7 m tall?
hi
di
ho
do
1.5 m
2.0 m
2.5 m
m ■
diho
hi do
O1
(1.5 m)(1.7 m)
6.5 m
Level 3
81. Inspection Mirror A production-line
inspector wants a mirror that produces an
image that is upright with a magnification
of 7.5 when it is located 14.0 mm from a
machine part.
370
Solutions Manual
O2
1
Ray 1
Ray 2
Ray 2
0.38 m
Ray 1
Figure 17-24
2
C
F
I1
I2
Horizontal scale:
1 block 10 cm
83. A ball is positioned 22 cm in front of a spherical mirror and forms a virtual image. If the
spherical mirror is replaced with a plane mirror, the image appears 12 cm closer to the mirror. What kind of spherical mirror was used?
Physics: Principles and Problems
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
(6.5 m)(1.9 m)
6.5 m (1.9 m)
Chapter 17 continued
The object position for both mirrors is
22 cm. So, the image position for the
plane mirror is 22 cm.
Because the spherical mirror forms a
virtual image, the image is located
behind the mirror. Thus, the image position for the spherical mirror is negative.
34 cm
1
1
1
do
di
f
d d
do di
i
f o
(22 cm)(34 cm)
22 cm (34 cm)
hi 2.4 m
di 14 m
C
F
I1
The image is 2.4 m tall, and it is 14 m
from the mirror.
86. A 4.0-cm-tall object is placed 12.0 cm from a
convex mirror. If the image of the object is
2.0 cm tall, and the image is located at
6.0 cm, what is the focal length of the mirror? Draw a ray diagram to answer the question. Use the mirror equation and the magnification equation to verify your answer.
O1 Ray 1
62 cm
I1
Ray 2
The focal length is positive, so the
spherical mirror is a concave mirror.
hi
di
ho
do
d d
do di
i
f o
hido
di ho
(12.0 cm)(6.0 cm)
12.0 cm (6.0 cm)
(0.28 m)(3.2 m)
1.6 m
1
1
1
do
di
f
d d
do di
i
f o
(3.2 m)(0.56 m)
3.2 m (0.56 m)
f 12 cm
1
1
1
do
di
f
m 0.56 m
F
Horizontal scale:
1 block 1.0 cm
Vertical scale:
3 blocks 2.0 cm
84. A 1.6-m-tall girl stands 3.2 m from a convex
mirror. What is the focal length of the mirror
if her image appears to be 0.28 m tall?
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
Horizontal scale:
1 block 1.0 m
Vertical scale:
2 blocks 1.0 m
Ray 2
di di, plane 12 cm
22 cm 12 cm
Ray 1
O1
12 cm
Thinking Critically
pages 481–482
87. Apply Concepts The ball in Figure 17-25
slowly rolls toward the concave mirror on the
right. Describe how the size of the ball’s image
changes as it rolls along.
0.68 m
85. Magic Trick A magician uses a concave
mirror with a focal length of 8.0 m to make a
3.0-m-tall hidden object, located 18.0 m from
the mirror, appear as a real image that is seen
by his audience. Draw a scale ray diagram to
find the height and location of the image.
Physics: Principles and Problems
C
■
F
Figure 17-25
Solutions Manual
371
Chapter 17 continued
Beyond C, the image is smaller than the
ball. As the ball rolls toward the mirror,
the image size increases. The image is
the same size as the ball when the ball
is at C. The image size continues to
increase until there is no image when
the ball is at F. Past F, the size of the
image decreases until it equals the
ball’s size when the ball touches the
mirror.
mirror, what is the focal length of the concave mirror?
di, initial do, initial
6.0 cm
di di, initial (8.0 cm)
6.0 cm (8.0 cm)
14.0 cm
1
1
1
do
di
f
88. Analyze and Conclude The object in
Figure 17-26 is located 22 cm from a
concave mirror. What is the focal length
of the mirror?
d d
do di
i
f o
(6.0 cm)(14.0 cm)
6.0 cm (14.0 cm)
f f 1.0101 cm
22 cm
■
91. Analyze and Conclude The layout of the
two-mirror system shown in Figure 17-11 is
that of a Gregorian telescope. For this
question, the larger concave mirror has a
radius of curvature of 1.0 m, and the
smaller mirror is located 0.75 m away. Why
is the secondary mirror concave?
r
2
f do
2
22 cm
2
11 cm
89. Use Equations Show that as the radius of
curvature of a concave mirror increases to
infinity, the mirror equation reduces to the
relationship between the object position
and the image position for a plane mirror.
92. Analyze and Conclude An optical arrangement used in some telescopes is the
Cassegrain focus, shown in Figure 17-27.
This telescope uses a convex secondary
mirror that is positioned between the
primary mirror and the focal point of
the primary mirror.
Convex
secondary mirror
As f → , 1/f → 0. The mirror equation
then becomes 1/do 1/di, or do di.
90. Analyze and Conclude An object is located
6.0 cm from a plane mirror. If the plane mirror is replaced with a concave mirror, the
resulting image is 8.0 cm farther behind the
mirror. Assuming that the object is located
between the focal point and the concave
372
Solutions Manual
Concave
primary mirror
F
Telescope tube
Eyepiece
■
Figure 17-27
a. A single convex mirror produces only
virtual images. Explain how the convex
Physics: Principles and Problems
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
The smaller mirror is concave to produce
a real image at the eyepiece that is
upright. The light rays are inverted by the
first concave mirror and then inverted
again by the secondary concave mirror.
Figure 17-26
Chapter 18 continued
83. If an object is 10.0 cm from a converging
lens that has a focal length of 5.00 cm, how
far from the lens will the image be?
n
ng
A
c sin1 sin1 1.00
1.52
1
1
1
f
do
di
41.1°
When the light ray in the glass strikes
the surface at a 62° angle, total internal
reflection occurs.
d f
do f
o
di (10.0 cm)(5.00 cm)
10.0 cm 5.00 cm
10.0 cm
62°
28°
62°
28°
45°
45°
18.2 Convex and Concave Lenses
page 510
Level 1
81. The focal length of a convex lens is 17 cm.
A candle is placed 34 cm in front of the
lens. Make a ray diagram to locate the
image.
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
di 34 cm
O1
F
Horizontal scale:
1 block 2 cm
di 34 cm
F
I1
82. A converging lens has a focal length of
25.5 cm. If it is placed 72.5 cm from an
object, at what distance from the lens will
the image be?
1
1
1
f
do
di
d f
do f
o
di (72.5 cm)(25.5 cm)
72.5 cm 25.5 cm
39.3 cm
The image is 39.3 cm from the lens.
Physics: Principles and Problems
d
do
h
ho
m i i
d
m
do i
(24 cm)
0.75
32 cm
1
1
1
f
do
di
Ray 1
Ray 2
Level 2
84. A convex lens is needed to produce an
image that is 0.75 times the size of the
object and located 24 cm from the lens on
the other side. What focal length should be
specified?
d d
do di
o i
f (32 cm)(24 cm)
32 cm 24 cm
14 cm
85. An object is located 14.0 cm from a convex
lens that has a focal length of 6.0 cm. The
object is 2.4 cm high.
a. Draw a ray diagram to determine the location, size, and orientation of the image.
O1
Ray 1
hi 1.8 cm
di 10.5 cm
Ray 2
F
Horizontal scale:
1 block 1.0 cm
Vertical scale:
1 block 0.4 cm
F
I1
Solutions Manual
389
Chapter 18 continued
b. Solve the problem mathematically.
1
1
1
f
do
di
fnew 2f
d f
do f
o
di 2(6.00 cm)
(14.0 cm)(6.0 cm)
14.0 cm 6.0 cm
10.5 cm
d
do
h
ho
b. If the original lens is replaced with a lens
having twice the focal length, what are
the image position, size, and orientation?
m i i
d h
do
i o
hi (10.5 cm)(2.4 cm)
14.0 cm
1.8 cm, so the image is
inverted
12.0 cm
1
1
1
f
do
di
d f
do fnew
o new
di, new (15.0 cm)(12.0 cm)
15.0 cm 12.0 cm
60.0 cm
d
do
h
ho
m i i
d
86. A 3.0-cm-tall object is placed 22 cm in front
of a converging lens. A real image is formed
11 cm from the lens. What is the size of the
image?
h
d
m i i
ho
do
d h
do
h
i, new o
hi, new do
(60.0 cm)(3.0 cm)
15 cm
12 cm
The image is inverted compared to
the object.
i o
hi 22 cm
1.5 cm
The image is 1.5 cm tall.
Level 3
87. A 3.0-cm-tall object is placed 15.0 cm in
front of a converging lens. A real image is
formed 10.0 cm from the lens.
a. What is the focal length of the lens?
a. What are the object position and object
height?
1
1
1
f
do
di
df
i
do d f
i
(5.0 cm)(15.0 cm)
5.0 cm (15.0 cm)
1
1
1
f
do
di
7.5 cm
d d
do di
m i i
o i
f (15.0 cm)(10.0 cm)
15.0 cm 10.0 cm
6.00 cm
d
do
h
ho
d h
di
o i
ho (7.5 cm)(2.0 cm)
5.0 cm
3.0 cm
390
Solutions Manual
Physics: Principles and Problems
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
(11 cm)(3.0 cm)
88. A diverging lens has a focal length of
15.0 cm. An object placed near it forms a
2.0-cm-high image at a distance of 5.0 cm
from the lens.
Chapter 18 continued
b. The diverging lens is now replaced by a
converging lens with the same focal
length. What are the image position,
height, and orientation? Is it a virtual
image or a real image?
fnew f
(15.0 cm)
1
1
1
f
di
do
d f
do f
o
So di (125 m)(1.0000 m)
15.0 cm
125 m 1.00 m
1
1
1
fnew
do
di, new
d f
do fnew
o new
di, new (7.5 cm)(15 cm)
7.5 cm 15 cm
15 cm
h
ho
b. A 1000.0-mm lens is focused on an
object 125 m away. What is the image
position?
d
do
m i i
1.01 m 1.01103 mm
90. Eyeglasses To clearly read a book 25 cm
away, a farsighted girl needs the image to be
45 cm from her eyes. What focal length is
needed for the lenses in her eyeglasses?
1
1
1
f
di
do
d d
do di
o i
So f d
h
i, new o
hi, new do
(15 cm)(3.0 cm)
(25 cm)(45 cm)
25 cm (45 cm)
56 cm
7.5 cm
6.0 cm
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
This is a virtual image that is upright
compared to the object.
18.3 Applications of Lenses
pages 510–511
Level 1
89. Camera Lenses Camera lenses are
described in terms of their focal length.
A 50.0-mm lens has a focal length of
50.0 mm.
a. A camera with a 50.0-mm lens is
focused on an object 3.0 m away. What
is the image position?
1
1
1
f
di
do
do f
So di do f
(3.0103
mm)(50.0 mm)
3
3.010 mm 50.0 mm
51 mm
Physics: Principles and Problems
Level 2
91. Copy Machine The convex lens of a copy
machine has a focal length of 25.0 cm. A
letter to be copied is placed 40.0 cm from
the lens.
a. How far from the lens is the copy
paper?
1
1
1
f
di
do
d f
do f
o
di (40.0 cm)(25.0 cm)
40.0 cm 25.0 cm
66.7 cm
b. How much larger will the copy be?
h
d
i i
ho
do
d h
do
(66.7 cm)(h )
40.0 cm
i o o
hi 1.67ho
The copy is enlarged and inverted.
Solutions Manual
391
Chapter 18 continued
92. Camera A camera lens with a focal length
of 35 mm is used to photograph a distant
object. How far from the lens is the real
image of the object? Explain.
35 mm; for a distant object, do can be
considered at , thus 1/do is zero.
According to the thin lens equation,
di f.
Level 3
93. Microscope A slide of an onion cell is
placed 12 mm from the objective lens of
a microscope. The focal length of the
objective lens is 10.0 mm.
a. How far from the lens is the image
formed?
1
1
1
f
di
do
do f
So di do f
(12 mm)(10.0 mm)
12 mm 10.0 mm
6.0101 mm
b. What is the magnification of this image?
di
6.0101 mm
5.0
mo do
12 mm
1
1
1
f
di
do
d f
do f
o
di mm)(20.0 mm)
(10.0
10.0 mm 20.0 mm
20.0 mm, or 20.0 mm beneath
the eyepiece
d. What is the final magnification of this
compound system?
d
(20.0 mm)
i
me 2.00
do
10.0 mm
mtotal mome (5.0)(2.00)
1.0101
392
Solutions Manual
a. What are the image position, height,
and orientation as formed by the objective lens? Is this a real or virtual image?
1
1
1
f
do
di
d f
do f
o
di (425 cm)(20.0 cm)
425 cm 20.0 cm
21.0 cm
d
do
h
ho
m i i
d h
do
i o
hi (21.0 cm)(10.0 cm)
425 cm
0.494 cm
This is a real image that is inverted
compared to the object.
b. The objective lens image becomes the
object for the eyepiece lens. What are
the image position, height, and orientation that a person sees when looking
into the telescope? Is this a real or
virtual image?
do, new 25.0 cm di
25.0 cm 21.0 cm
4.0 cm
1
1
1
fnew
do, new
di, new
d
f
do, new fnew
o, new new
di, new (4.0 cm)(4.05 cm)
4.0 cm 4.05 cm
3.2102 cm
Physics: Principles and Problems
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
c. The real image formed is located
10.0 mm beneath the eyepiece lens.
If the focal length of the eyepiece is
20.0 mm, where does the final image
appear?
94. Telescope The optical system of a toy
refracting telescope consists of a converging
objective lens with a focal length of
20.0 cm, located 25.0 cm from a converging
eyepiece lens with a focal length of
4.05 cm. The telescope is used to view a
10.0-cm-high object, located 425 cm from
the objective lens.
Chapter 18 continued
ho, new hi
0.494 cm
d
do
h
ho
m i i
d
h
do, new
i, new o, new
hi, new (3.210 cm)(0.494 cm)
2
4.0 cm
4.0101 cm
This is a virtual image that is inverted compared to the object.
c. What is the magnification of the
telescope?
h
i, new
m ho
4.010 cm
1
10.0 cm
4.0
Mixed Review
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
pages 511—512
Level 1
95. A block of glass has a critical angle of 45.0°.
What is its index of refraction?
n
n1
2
sin c n
2
, n2 1.00 for air
n1 sin
c
1.00
n1 sin 45.0°
1.41
96. Find the speed of light in antimony trioxide
if it has an index of refraction of 2.35.
c
n v
c
v n
3.00108 m/s
2.35
1.28108 m/s
97. A 3.0-cm-tall object is placed 20 cm in front
of a converging lens. A real image is formed
10 cm from the lens. What is the focal
length of the lens?
1
1
1
do
f
di
d d
do di
o i
f (20 cm)(10 cm)
20 cm 10 cm
7 cm
Level 2
98. Derive n sin 1/sin 2 from the general
form of Snell’s law of refraction,
n1 sin 1 n2 sin 2. State any assumptions
and restrictions.
The angle of incidence must be in air. If
we let substance 1 be air, then n1 1.000. Let n2 n. Therefore,
n1 sin 1 n2 sin 2
sin 1 n sin 2
sin 1
n
sin 2
99. Astronomy How many more minutes
would it take light from the Sun to reach
Earth if the space between them were filled
with water rather than a vacuum? The Sun
is 1.5108 km from Earth.
Time through vacuum
8
(1.510 km)(1000 m/1 km)
d
t 8
c
3.0010 m/s
5.0102 s
Speed through water
3.00108 m/s
c
v n
1.33
2.26108
m/s
Time through water
8
(1.510 km)(1000 m/1 km)
d
t 8
v
2.2610 m/s
660 s
t 660 s 500 s 160 s
(160 s)(1 min/60 s) 2.7 min
Physics: Principles and Problems
Solutions Manual
393