Chapter 33
Q 7: a, b, c
Q9: d, b, a, c
Problems
37. As the polarized beam of intensity I0 passes the first polarizer, its intensity is reduced
to I 0 cos 2  . After passing through the second polarizer which makes a 90 angle with the
first filter, the intensity is I  ( I 0 cos 2 ) sin 2   I 0 /10 which implies sin2  cos2  = 1/10,
or sin cos = sin2 /2 = 1/ 10 . This leads to  = 70° or 20°.
38. We note the points at which the curve is zero (2 = 0 and 90) in Fig. 33-44(b). We
infer that sheet 2 is perpendicular to one of the other sheets at 2 = 0, and that it is
perpendicular to the other of the other sheets when 2 = 90. Without loss of generality,
we choose 1 = 0, 3 = 90. Now, when 2 = 30, it will be  = 30 relative to sheet 1
and  = 60 relative to sheet 3. Therefore,
If 1
2
2
Ii = 2 cos () cos ( ) = 9.4% .
49. Consider a ray that grazes the top of the pole, as shown in the diagram that follows.
. m. The length of the shadow is x + L.
Here 1 = 90° –  = 35°, 1  0.50 m, and  2  150
x is given by
x  1 tan  1  (0.50 m) tan 35  0.35 m.
According to the law of refraction, n2 sin 2 = n1 sin 1. We take n1 = 1 and n2 = 1.33
(from Table 33-1). Then,
 2  sin 1
F
sin  I
Fsin 35.0IJ 2555
 sin G
G
J
K . .
.
Hn K H133
1
1
2
L is given by
L   2 tan  2  (150
. m) tan 25.55  0.72 m.
The length of the shadow is 0.35 m + 0.72 m = 1.07 m.
63. (a) A ray diagram is shown below.
Let 1 be the angle of incidence and 2 be the angle of refraction at the first surface. Let
3 be the angle of incidence at the second surface. The angle of refraction there is 4 =
90°. The law of refraction, applied to the second surface, yields n sin 3 = sin 4 = 1. As
shown in the diagram, the normals to the surfaces at P and Q are perpendicular to each
other. The interior angles of the triangle formed by the ray and the two normals must sum
to 180°, so 3 = 90° – 2 and
b
g
sin  3  sin 90 2  cos 2  1  sin 2  2 .
According to the law of refraction, applied at Q, n 1  sin 2  2  1. The law of refraction,
applied to point P, yields sin 1 = n sin 2, so sin 2 = (sin 1)/n and
sin 2  1
n 1
 1.
n2
Squaring both sides and solving for n, we get
n  1  sin 2  1 .
(b) The greatest possible value of sin2 1 is 1, so the greatest possible value of n is
nmax  2  141
. .
(c) For a given value of n, if the angle of incidence at the first surface is greater than 1,
the angle of refraction there is greater than 2 and the angle of incidence at the second
face is less than 3 (= 90° – 2). That is, it is less than the critical angle for total internal
reflection, so light leaves the second surface and emerges into the air.
(d) If the angle of incidence at the first surface is less than 1, the angle of refraction there
is less than 2 and the angle of incidence at the second surface is greater than 3. This is
greater than the critical angle for total internal reflection, so all the light is reflected at Q.
64. (a) We use Eq. 33-49:  B  tan 1nw  tan 1 (133
. )  531
. .
(b) Yes, since nw depends on the wavelength of the light.
Chapter 34
Q1: (a) a; (b) c
Q3: (a) from infinity to the focal point; (b) decrease continuously
Problems
3. The intensity of light from a point source varies as the inverse of the square of the
distance from the source. Before the mirror is in place, the intensity at the center of the
screen is given by IP = A/d 2, where A is a constant of proportionality. After the mirror is
in place, the light that goes directly to the screen contributes intensity IP, as before.
Reflected light also reaches the screen. This light appears to come from the image of the
source, a distance d behind the mirror and a distance 3d from the screen. Its contribution
to the intensity at the center of the screen is
Ir 
A
 3d 
2

A IP
 .
9d 2 9
The total intensity at the center of the screen is
I  IP  Ir  IP 
I P 10
 IP.
9 9
The ratio of the new intensity to the original intensity is I/IP = 10/9 = 1.11.
5. We apply the law of refraction, assuming all angles are in radians:
sin 
n
 w ,
sin   nair
which in our case reduces to '  /nw (since both  and ' are small, and nair  1). We
refer to our figure below.
The object O is a vertical distance d1 above the water, and the water surface is a vertical
distance d2 above the mirror. We are looking for a distance d (treated as a positive
number) below the mirror where the image I of the object is formed. In the triangle O AB
| AB | d1 tan  d1 ,
and in the triangle CBD
| BC | 2d 2 tan    2d 2  
2d 2
.
nw
Finally, in the triangle ACI, we have |AI| = d + d2. Therefore,
 d 2d   1
2d
| AC |
| AB |  | BC |
 d2 
 d 2   1  2   d 2  d1  2  d 2
tan 

nw  
nw

2  200cm 
 250cm 
 200cm  351cm.
1.33
d | AI | d 2 