4.
REASONING The drawing at
the right shows the laser beam
after reflecting from the plane
mirror. The angle of reflection
is , and it is equal to the angle
of incidence, which is 33.0.
Note that the angles labeled  in
the drawing are also 33.0,
since they are angles formed by
a line that intersects two
parallel lines. Knowing these
angles, we can use trigonometry
to determine the distance dDC,
which locates the spot where
the beam strikes the floor.
L

1.10 m
33.0

A
B
1.80 m

D
C
Floor
SOLUTION Applying trigonometry to triangle DBC, we see that
tan  
d BC
or
d DC
d DC 
d BC
tan 
(1)
The distance dBC can be determined from d BC  1.80 m  d AB , which can be substituted into
Equation (1) to show that
d DC 
d BC
tan 

1.80 m  d AB
tan 
(2)
The distance dAB can be found by applying trigonometry to triangle LAB, which shows that
d
tan   AB
or
dAB  1.10 m  tan 
1.10 m
Substituting this result into Equation (2), gives
dDC 
1.80 m  dAB 1.80 m  1.10 m  tan  1.80 m  1.10 m  tan 33.0


 1.67 m
tan 
tan 
tan 33.0
13- REASONING AND SOLUTION The ray diagram is shown below. (Note: f  –50.0 cm
and do  25.0 cm )
Scale
Reflected ray
10.0 cm
Virtual Image
Object
10.0 cm
25.0 cm
6.67 cm
16.7 cm
F
a. The ray diagram indicates that the image is 16.7 cm behind the mirror
C
.
b. The ray diagram indicates that the image height is 6.67 cm .
23. REASONING When paraxial light rays that are parallel to the principal axis strike a
convex mirror, the reflected rays diverge after being reflected, and appear to originate from the
focal point F behind the mirror (see Figure 25.16). We can treat the sun as being infinitely far
from the mirror, so it is reasonable to treat the incident rays as paraxial rays that are parallel to
the principal axis.
SOLUTION
a. Since the sun is infinitely far from the mirror and its image is a virtual image that lies
behind the mirror, we can conclude that the mirror is a convex mirror .
b. With d i = –12.0 cm and d o   , the mirror equation (Equation 25.3) gives
1
1
1
1 1
1





f
do di
 di
di
Therefore, the focal length f lies 12.0 cm behind the mirror (this is consistent with the
reasoning above that states that, after being reflected, the rays appear to originate from the
focal point behind the mirror). In other words, f  –12.0 cm . Then, according to
Equation 25.2, f  – 21 R , and the radius of curvature is
R  –2 f  –2(–12.0 cm) = 24.0 cm
30. REASONING We will start by drawing the two situations in which the object is 25 cm and
5 cm from the mirror, making sure that all distances (including the radius of curvature of the
mirror) and heights are to scale. For each location of the object, we will draw several rays to
locate the image (see the Reasoning Strategy for convex mirrors in Section 25.5). Once the
images have been located, we can readily answer the questions regarding their positions and
heights.
SOLUTION The following two ray diagrams illustrate the situations where the objects are
at different distances from the convex mirror.
Image
1
3
Object
1
1
1
1
25 cm
F
C
Image
1
1
1
3
1
Object 1
F
C
5 cm
a. As the object moves closer to the mirror, it can be seen that the magnitude of the image
distance becomes smaller .
b. As the object moves closer to the mirror, the magnitude of the image height becomes
larger .
c. By measuring the image heights, we find that the ratio of the image height when the
object distance is 5 cm to that when the object distance is 25 cm is 3 .
34. REASONING The mirror equation relates the object and image distances to the focal
length. Thus, we can apply the mirror equation once with the given object and image
distances to determine the focal length. Then, we can use the mirror equation again with the
focal length and the second object distance to determine the unknown image distance.
SOLUTION According to the mirror equation, we have
1
1
1


d
d
f
o1i1

and
First position of object
1
1
1


d
d
f
o2i2

Second position of object
Since the focal length is the same in both cases, it follows that
1
1
1
1



d o1 d i1 d o2 d i2
1
1
1
1
1
1
1






 0.071 cm 1
d i2 d o1 d i1 d o2 25 cm 17 cm 19 cm
b
g
d i2  14 cm
The negative value for di2 indicates that the image is located 14 cm behind the mirror.
43. CONCEPT QUESTIONS a. According to the discussion about relative velocity in
Section 3.4, it follows that vIY = vIM + vMY.
b. As you walk perpendicularly toward the stationary mirror, you perceive the mirror
moving toward you in the opposite direction. Thus, vMY = –vYM.
c. The velocities vYM and vIM have the same magnitude. This is because the image in a
plane mirror is always just as far behind the mirror as the object is in front of it. For
instance, if you move 1 meter perpendicularly toward the mirror in 1 second, the magnitude
of your velocity relative to the mirror is 1 m/s. But your image also moves 1 meter toward
the mirror in the same time interval, so that the magnitude of its velocity relative to the
mirror is also 1 m/s. The two velocities, however, have opposite directions.
SOLUTION According to the discussion in Section 3.4, we have
vIY = vIM + vMY = vIM – vYM
(1)
Remembering that the magnitudes of both velocities vYM and vIM are the same and that the
direction in which you walk is positive, we have
vYM = +0.90 m/s
and
vIM = –0.90 m/s
The velocity vIM is negative, because its direction is opposite to the direction in which you
walk. Substituting these values into Equation (1), we obtain
vIY = (–0.90 m/s) – (+0.90 m/s) = –1.80 m/s
Chapter 26
3.REASONING Since the light will travel in glass at a constant speed v, the time it takes to pass
perpendicularly through the glass is given by t  d / v , where d is the thickness of the glass. The
speed v is related to the vacuum value c by Equation 26.1: n  c / v.
SOLUTION Substituting for v from Equation 26.1 and substituting values, we obtain
t
d nd 1.5  4.0 10 –3 m 


 2.0 10 –11 s
8
v
c
3.00 10 m/s
8- SOLUTION In the ice of thickness di, the speed of light is vi, and the travel time is ti = di/vi.
Similarly, the travel time in the quartz sheet is tq = dq/vq. Therefore, the desired distance ct is
 d dq 
c
c
ct  c ti  tq  c  i    di  d q
 vi vq 
vi
vq




Since Equation 26.1 gives the refractive index as n = c/v and since Table 26.1 gives the
indices of refraction for ice and quartz as ni = 1.309 and nq = 1.544, the result just obtained can
be written as follows:
ct  di
c
c
 dq  di ni  dq nq   2.0 cm 1.309   1.1 cm 1.544   4.3 cm
vi
vq
19. The drawing at the right shows the
geometry of the situation using the same
notation as that in Figure 26.6. In addition to
the text's notation, let t represent the thickness
of the pane, let L represent the length of the
ray in the pane, let x (shown twice in the
figure) equal the displacement of the ray, and
let the difference in angles 1 – 2 be given by
x
3
Air ( n 3 = n 1)
x
L
. We wish to find the amount x by which the
emergent ray is displaced relative to the
incident ray. This can be done by applying
Snell's law at each interface, and then making
use of the geometric and trigonometric
relations in the drawing.
2
2
t

Glass (Glass
n2) ( n2 )
Air ( n1)
1
SOLUTION If we apply Snell's law (see Equation 26.2) to the bottom interface we obtain
n1 sin 1  n2 sin  2 . Similarly, if we apply Snell's law at the top interface where the ray
emerges, we have n2 sin  2  n3 sin  3  n1 sin  3 . Comparing this with Snell's law at the bottom
face, we see that n1 sin 1  n1 sin  3 , from which we can conclude that 3 = 1. Therefore, the
emerging ray is parallel to the incident ray.
From the geometry of the ray and thickness of the pane, we see that L cos  2  t , from
which it follows that L  t /cos  2 . Furthermore, we see that x  L sin   L sin 1 –  2  .
Substituting for L, we find
x  L sin(1 –  2 ) 
t sin(1 –  2 )
cos  2
Before we can use this expression to determine a numerical value for x, we must find the
value of 2. Solving the expression for Snell's law at the bottom interface for 2, we have
sin  2 
n1 sin 1 (1.000) (sin 30.0)

 0.329
n2
1.52
or
 2  sin –1 0.329  19.2
Therefore, the amount by which the emergent ray is displaced relative to the incident ray
is
x
t sin (1 –  2 ) (6.00 mm) sin (30.0–19.2)

 1.19 mm
cos  2
cos 19.2
30. REASONING Total internal reflection
will occur at point P provided that the angle 
in the drawing at the right exceeds the critical
angle. This angle is determined by the angle 2
P

at which the light rays enter the quartz slab.
We can determine 2 by using Snell’s law of
refraction and the incident angle, which is
given as 1 = 34°.


SOLUTION Using n for the refractive index of the fluid that surrounds the crystalline quartz
slab and nq for the refractive index of quartz and applying Snell’s law give
n sin 1  nq sin  2
sin  2 
or
n
sin 1
nq
(1)
But when  equals the critical angle, we have from Equation 26.4 that
sin   sin  c 
n
nq
(2)
According to the geometry in the drawing above,  = 90° – 2. As a result, Equation (2)
becomes
n
sin  90 –  2   cos  2 
(3)
nq
Squaring Equation (3), using the fact that sin22 + cos22 = 1, and substituting from
Equation (1), we obtain
cos2  2  1– sin 2  2  1–
n2
n2
2
sin


1
nq2
nq2
Solving Equation (4) for n and using the value given in Table 26.1 for the refractive
index of crystalline quartz, we find
n
nq
1+sin 1
2

1.544
1+sin 34
2
 1.35
(4)
41. REASONING Because the refractive index of the glass depends on the wavelength (i.e., the
color) of the light, the rays corresponding to different colors are bent by different amounts in the
glass. We can use Snell’s law (Equation 26.2: n1 sin 1  n2 sin  2 ) to find the angle of refraction
for the violet ray and the red ray. The angle between these rays can be found by the subtraction
of the two angles of refraction.
SOLUTION In Table 26.2 the index of refraction for violet light in crown glass is 1.538,
while that for red light is 1.520. According to Snell's law, then, the sine of the angle of refraction
for the violet ray in the glass is sin  2  (1.000 /1.538) sin 45.00  0.4598 , so that
 2  sin 1(0.4598)  27.37
Similarly, for the red ray, sin  2  (1.000 /1.520) sin 45.00  0.4652 , from which it
follows that
 2  sin 1 (0.4652)  27.72
Therefore, the angle between the violet ray and the red ray in the glass is
27.72 – 27.37  0.35
Chapter 27
5.
SSM REASONING The slit separation d is given by Equation 27.1 with m  1; namely
d   / sin . As shown in Example 1 in the text, the angle  is given by   tan 1 ( y / L ) .
SOLUTION The angle  is
  tan 1
0.037 m I
F
G
H4.5 m J
K 0.47 
Therefore, the slit separation d is
d 

sin 

490  10 –9 m
 6.0  10 –5 m
sin 0.47 
15. REASONING To solve this problem, we must express the condition for constructive
interference in terms of the film thickness t and the wavelength  film of the light in the soap
film. We must also take into account any phase changes that occur upon reflection.
SOLUTION For the reflection at the top film surface, the light travels from air, where the
refractive index is smaller (n = 1.00), toward the film, where the refractive index is larger
(n = 1.33). Associated with this reflection there is a phase change that is equivalent to onehalf of a wavelength. For the reflection at the bottom film surface, the light travels from the
film, where the refractive index is larger (n = 1.33), toward air, where the refractive index is
smaller (n = 1.00). Associated with this reflection, there is no phase change. As a result of
these two reflections, there is a net phase change that is equivalent to one-half of a
wavelength. To obtain the condition for constructive interference, this net phase change
must be added to the phase change that arises because of the film thickness t, which is
traversed twice by the light that penetrates it. For constructive interference we find that
2 t  21  film   film , 2  film , 3 film , ...
or
c h
2 t  m  21  film ,
where m  0, 1, 2, ...
Equation 27.3 indicates that  film   vacuum / n . Using this expression and the fact that m = 0
for the minimum thickness t, we find that the condition for constructive interference
becomes
c h

IJ
c hF
G
Hn K
2 t  m  21  film  0  21
or
t 
 vacuum
4n

vacuum
611 nm
 115 nm
4 1.33
b g
19. REASONING This problem can be solved by using Equation 27.4 for the value of the
angle  when m  1 (first dark fringe).
SOLUTION
a. When the slit width is W = 1.8  10 –4 m and   675 nm  675  10 –9 m , we find,
according to Equation 27.4,
L(1) 675  10
I
F
 sin M
G
J
HW K N 1.8  10
  sin 1 m
1
O
P
Q
–9
m
 0.21
m
–4
b. Similarly, when the slit width is W = 1.8  10 –6 m and   675  10 –9 m , we find
L
675  10
M
N 1.8  10
  sin 1 (1)
–9
–6
O
P
Q
m
 22 
m
25. REASONING AND SOLUTION Using Equation 27.4 for the first-order dark fringes
(m = 1) and referring to Figure 27.24 in the text, we see that
sin   ( 1)

W

y
L2  y 2
Since the distance L between the slit and the screen equals the width 2y of the central bright
fringe, this equation becomes

W

y
bg
2y
2
 y2

1
5

0.447
31. REASONING AND SOLUTION According to Rayleigh's criterion (Equation 27.6), the
minimum angular separation of the two objects is
 min ( in rad)  1.22

D
 1.22
F
565  10 m I
G
H 1.02 m J
K 6.76  10
–9
–7
rad
Therefore, from the figure above, the separation y of the two objects is
y  L min  ( 3.75  10 4 m)(6.76  10 –7 rad ) = 0.0254 m
37. REASONING AND SOLUTION The geometry of the situation is shown below.

Fi rst d ark
fri nge
y

d
Midp oint of
centra l bri ght
fri nge
L
From the geometry, we have
tan  
y 0.60 mm

 0.20
L 3.0 mm
or
 = 11.3
Then, solving Equation 27.7 with m  1 for the separation d between the slits, we have
c
h
(1) 780  10 –9 m
m
d 


sin 
sin 11.3
4.0  10 –6 m
47. REASONING AND SOLUTION
To decide whether constructive or destructive
interference occurs, we need to determine the wavelength of the wave. For electromagnetic
waves, Equation 16.1 can be written f   c , so that

c 3.00  10 8 m

 5.60  10 2 m
3
f 536  10 Hz
Since the two wave sources are in phase, constructive interference occurs when the path
difference is an integer number of wavelengths, and destructive interference occurs when
the path difference is an odd number of half wavelengths. We find that the path difference is
8.12 km – 7.00 km = 1.12 km. The number of wavelengths in this path difference is
1.12  103 m / 5.60  102 m   2.00 . Therefore,
constructive interference occurs .