32.1.IDENTIFY: Since the speed is constant, distance x  ct.
SET UP: The speed of light is c  3.00 108 m/s . 1 yr  3.156  107 s.
x 3.84 108 m

 1.28 s
c 3.00 108 m/s
(b) x  ct  (3.00 108 m/s)(8.61 yr)(3.156 107 s/yr)  8.15 1016 m  8.15 1013 km
EVALUATE: The speed of light is very great. The distance between stars is very large compared to terrestrial
distances.
EXECUTE: (a) t 
32.3.
IDENTIFY: Apply c  f .
c  3.00 108 m/s
c 3.0 108 m s
EXECUTE: (a) f  
 6.0 104 Hz.

5000 m
SET UP:
(b) f 
(c) f 
c


3.0 108 m s
 6.0 107 Hz.
5.0 m

3.0 108 m s
 6.0 1013 Hz.
5.0 106 m
c

3.0 108 m s
 6.0 1016 Hz.
 5.0 109 m
EVALUATE: f increases when  decreases.
(d) f 
32.5.
c

c  f  . Emax  cBmax . k  2 /  .   2 f .
IDENTIFY:
SET UP: Since the wave is traveling in empty space, its wave speed is c  3.00 108 m/s .
c 3.00 108 m/s
 6.94 1014 Hz
EXECUTE: (a) f  
 432 109 m
(b) Emax  cBmax  (3.00 108 m/s)(1.25 106 T)  375 V/m
2 rad
 1.45 107 rad/m .   (2 rad)(6.94  1014 Hz)  4.36 1015 rad/s .
 432 109 m
E  Emax cos(kx   t )  (375 V/m)cos([1.45 107 rad/m]x  [4.36 1015 rad/s]t )
(c) k 
2

B  Bmax cos(kx  t )  (1.25  106 T)cos([1.45  107 rad/m]x  [4.36  1015 rad/s]t )
32.7.
EVALUATE: The cos(kx  t ) factor is common to both the electric and magnetic field expressions, since these
two fields are in phase.
IDENTIFY and SET UP: The equations are of the form of Eqs.(32.17), with x replaced by z. B is along the yaxis; deduce the direction of E.
EXECUTE:   2 f  2 (6.10 1014 Hz)  3.83 1015 rad/s
2 f  3.83 1015 rad/s
 
 1.28 107 rad/m

c
c
3.00 108 m/s
Bmax  5.80 104 T
k
2

Emax  cBmax  (3.00 108 m/s)(5.80 104 T)  1.74 105 V/m
B is along the y-axis. E  B is in the direction of propagation (the +z-direction). From this we can deduce the
direction of E , as shown in Figure 32.7.
E is along the x-axis.
Figure 32.7
E  Emax iˆ cos(kz  t )  (1.74 105 V/m)iˆ cos[(1.28 107 rad/m)z  (3.831015 rad/s)t ]
B  B ˆj cos(kz  t )   5.80  104 T  ˆj cos[(1.28  107 rad/m)z  (3.83  1015 rad/s)t ]
max
EVALUATE:
32.9.
E and B are perpendicular and oscillate in phase.
IDENTIFY and SET UP: Compare the E ( y, t ) given in the problem to the general form given by Eq.(32.17).
Use the direction of propagation and of E to find the direction of B.
(a) EXECUTE: The equation for the electric field contains the factor sin(ky  t ) so the wave is traveling in the
+y-direction. The equation for E ( y, t ) is in terms of sin(ky  t ) rather than cos(ky  t ); the wave is shifted in
phase by 90 relative to one with a cos(ky  t ) factor.
(b) E ( y, t )  (3.10 105 V/m)kˆ sin[ky  (2.65 1012 rad/s)t ]
Comparing to Eq.(32.17) gives   2.65 1012 rad/s
2 c
2 c 2 (2.998 108 m/s)
  2 f 
so  

 7.11104 m


(2.65 1012 rad/s)
(c)
E  B must be in the +ydirection (the direction in
which the wave is traveling).
When E is in the –z-direction
then B must be in the –xdirection, as shown in
Figure 32.9.
Figure 32.9
k
2
Emax

2.65 10 rad/s

 8.84 103 rad/m
c 2.998 108 m/s
 3.10 105 V/m


12
Emax 3.10 105 V/m

 1.03 103 T
c
2.998 108 m/s
Using Eq.(32.17) and the fact that B is in the iˆ direction when E is in the kˆ direction,
B  (1.03103 T)iˆ sin[(8.84 103 rad/m)y  (2.65 1012 rad/s)t ]
Then Bmax 
32.12.
EVALUATE: E and B are perpendicular and oscillate in phase.
IDENTIFY: Emax  cBmax .
SET UP: The magnetic field of the earth is about 104 T.
E 3.85 103 V/m
 1.28 1011 T.
EXECUTE: B  
c
3.00 108 m/s
EVALUATE: The field is much smaller than the earth's field.
33.25.
IDENTIFY: When unpolarized light passes through a polarizer the intensity is reduced by a factor of
1
2
and the
transmitted light is polarized along the axis of the polarizer. When polarized light of intensity I max is incident on
a polarizer, the transmitted intensity is I  I max cos 2  , where  is the angle between the polarization direction of
the incident light and the axis of the filter.
SET UP: For the second polarizer   60° . For the third polarizer,   90°  60°  30° .
EXECUTE: (a) At point A the intensity is I 0 / 2 and the light is polarized along the vertical direction. At point B
the intensity is ( I 0 / 2)(cos60°) 2  0.125I 0 , and the light is polarized along the axis of the second polarizer. At
point C the intensity is (0.125I 0 )(cos30°) 2  0.0938I 0 .
(b) Now for the last filter   90° and I  0 .
EVALUATE: Adding the middle filter increases the transmitted intensity.
33.27.
IDENTIFY and SET UP: Reflected beam completely linearly polarized implies that the angle of incidence equals
the polarizing angle, so  p  54.5. Use Eq.(33.8) to calculate the refractive index of the glass. Then use Snell’s
law to calculate the angle of refraction.
n
EXECUTE: (a) tan  p  b gives nglass  nair tan  p  (1.00) tan 54.5  1.40.
na
(b) na sin a  nb sinb
na sin  a (1.00)sin 54.5

 0.5815 and  b  35.5
nb
1.40
EVALUATE:
sin  b 
Note:   180.0   r  b and  r   a . Thus
  180.0  54.5  35.5  90.0; the
reflected ray and the refracted ray are
perpendicular to each other. This agrees
with Fig.33.28.
Figure 33.27
33.29.
IDENTIFY: From Malus’s law, the intensity of the emerging light is proportional to the square of the cosine of
the angle between the polarizing axes of the two filters.
SET UP: If the angle between the two axes is , the intensity of the emerging light is I = Imax cos2.
1
EXECUTE: At angle , I = Imax cos2, and at the new angle , I = Imax cos2. Taking the ratio of the
2
2
1
cos
I cos  2 I
 cos 
intensities gives max
, which gives us cos 
. Solving for  yields   arccos 

.
I max cos2 
I
2
 2 
EVALUATE: Careful! This result is not cos2.
33.31.
IDENTIFY: When unpolarized light of intensity I 0 is incident on a polarizing filter, the transmitted light has
intensity
1
2 0
I and is polarized along the filter axis. When polarized light of intensity I 0 is incident on a polarizing
filter the transmitted light has intensity I 0 cos 2  .
SET UP:
For the second filter,   62.0°  25.0°  37.0° .
EXECUTE: After the first filter the intensity is
1
2
I 0  10.0 W m 2 and the light is polarized along the axis of the
first filter. The intensity after the second filter is I  I 0cos 2 , where I 0  10.0 W m 2 and   37.0° . This
gives I  6.38 W m 2 .