Homework-12
Chapter 26
26. REASONING The light ray traveling in the oil can only penetrate into the water if it does
not undergo total internal reflection at the boundary between the oil and the water. Total
internal reflection will occur if the angle of incidence θ = 71.4° is greater than the critical
angle θc for these two media. The critical angle is found from
sin c 
n2
n1
(26.4)
where n2 = 1.333 is the index of refraction of water (see Table 26.1), and n1 = 1.47 is the
index of refraction of the oil.
SOLUTION Solving Equation 26.4 for θc, we obtain
 n2 
1  1.333 
  sin 
  65.1
n
1.47


 1
c  sin 1 
Comparing this result to θ = 71.4°, we see that the angle of incidence is greater than the
critical angle (θ > θc). Therefore, the ray of light will not enter the water ; it will instead
undergo total internal reflection within the oil.
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31. REASONING AND SOLUTION
a. Using Equation 26.4 and the refractive index for crown glass given in Table 26.1, we find
that the critical angle for a crown glass-air interface is
 1.00 
  41.0
 1.523 
c  sin 1 
The light will be totally reflected at point A since the incident angle of 60.0° is greater than
c. The incident angle at point B, however, is 30.0° and smaller than c. Thus, the light will
exit first at point B .
b. The critical angle for a crown glass-water interface is
 1.333 
  61.1
 1.523 
c  sin 1 
The incident angle at point A is less than this, so the light will first exit at point A .
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36. REASONING Using the value given for the critical angle in Equation 26.4 (sin c = n2/n1),
we can obtain the ratio of the refractive indices. Then, using this ratio in Equation 26.5
(Brewster’s law), we can obtain Brewster’s angle B.
SOLUTION From Equation 26.4, with n2 = nair = 1 and n2 = nliquid, we have
sin  c  sin 39 
1
(1)
nliquid
According to Brewster’s law,
tan  B 
n2
1

n1 nliquid
(2)
Substituting Equation (2) into Equation (1), we find
tan  B 
1
nliquid
 sin 39  0.63
 B  tan –1  0.63  32
or
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38. REASONING Given the height h of
the laser above the glass pane, the
distance d between the edge of the pane
and the point where the laser reflects
depends upon the angle θ between the
incident ray and the vertical (see the
drawing). The height h is adjacent to
the angle θ, and the distance d is
d
opposite, so from tan  
(Equation
h
1.3) we have that
Laser
θ
h
d  h tan 
θ
d
Glass pane
(1)
The glass pane is horizontal, so the normal to the surface is vertical, and we see that θ is also
the angle of incidence when the incident ray strikes the glass pane. The reflected beam is
100% polarized, so we conclude that the angle θ of incidence is equal to Brewster’s angle θB
n
for the glass-air interface: θ = θB. Brewster’s angle is given by tan  B  2 (Equation 26.5),
n1
where n2 = 1.523 and n1 = 1.000 are the indices of refraction for crown glass and air,
respectively (see Table 26.1). Therefore,
tan   tan  B 
n2
n1
(2)
SOLUTION Substituting Equation (2) into Equation (1), we obtain
n 
 1.523 
d  h tan   h  2    0.476 m  
  0.725 m
 1.000 
 n1 
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43. REASONING The angle of each refracted ray in the crown glass can be obtained from
Snell’s law (Equation 26.2) as ndiamond sin 1 = ncrown glass sin 2, where 1 is the angle of
incidence and 2 is the angle of refraction.
SOLUTION The angles of refraction for the red and blue rays are:
Blue ray
Red ray
n
sin 1 
  2.444  sin 35.00 
  sin 1 
  66.29
 ncrown glass 
1.531




 2  sin 1 
diamond
n
sin 1 
 2.410 sin 35.00 
  sin 1 
  65.43
 ncrown glass 
1.520



 2  sin 1 
diamond
The angle between the blue and red rays is
 blue   red  66.29  65.43  0.86
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50. REASONING
a. Given the focal length (f = −0.300 m) of the lens and the image distance di, we will
1 1 1
 
employ the thin-lens equation
(Equation 26.6) to determine how far from the
d o di f
van the person is actually standing, which is the object distance do. The image and the
person are both behind the van, so the image is virtual, and the image distance is negative:
di = −0.240 m.
b. Once we have determined the object distance do, we will use the magnification equation
h
d
m  i   i (Equation 26.7) to calculate the true height ho of the person from the height
ho
do
hi = 0.34 m of the image.
SOLUTION
a. Solving Equation 26.6 for do, we obtain
1
1 1
 
d o f di
or
do 
1
1

 1.2 m
1 1
1
1


f di 0.300 m  0.240 m 
b. Taking the reciprocal of both sides of
hi
ho

di
do
(Equation 26.7) and solving for ho
yields
d 
 1.2 m 
ho  hi  o     0.34 m  
  1.7 m
hi
di
d

0.240
m


 i
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ho

do
or
52. REASONING The height of the mountain’s image is given by the magnification equation
as hi = –hodi/do. To use this expression, however, we will need to know the image distance
di, which can be determined using the thin-lens equation. Knowing the image distance, we
can apply the expression for the image height directly to calculate the desired ratio.
SOLUTION According to the thin-lens equation, we have
1
1
1


di do
f
(1)
For both pictures, the object distance do is very large compared to the focal length f.
Therefore, 1/do is negligible compared to 1/f, and the thin-lens equation indicates that di = f.
As a result, the magnification equation indicates that the image height is given by
hi  –
ho di
do
–
ho f
do
(2)
Applying Equation (2) for the two pictures and noting that in each case the object height ho
and the focal length f are the same, we find
 hi 5 km
 hi 14 km
 ho f 
 –

 do 5 km  do 14 km 14 km



 2.8
5.0 km
 ho f 
do 

5 km
 –

d
o 

14 km
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