1 The source being isotropic means Asphere = 4r2 is used in the intensity definition I =
P/A, which further implies
2
I2 P / 4r 22  r1 


.
I1 P / 4r 12  r2 
(a) With I1 = 9.60  10–4 W/m2, r1 = 6.10 m, and r2 = 30.0 m, we find
I2 = (9.60  10–4 W/m2)(6.10/30.0)2 = 3.97  10–5 W/m2.
(b) Using Eq. 17–27 with I1 = 9.60  10–4 W/m2,  = 2(2000 Hz), v = 343 m/s and  =
1.21 kg/m3, we obtain
2I
sm 
 1.71  107 m.
2
 v
(c) Eq. 17-15 gives the pressure amplitude:
pm   v sm  0.893 Pa.
2- We apply Eq. 33-40 (once) and Eq. 33-42 (twice) to obtain
1
I = 2 I0 cos2 2cos2 (90 – 2) .
Using trig identities, we rewrite this as
(a) Therefore we find 2 = 2 sin–1
1
I
1 2
=
I0
8 sin (22) .
0.40 = 19.6.
(b) Since the first expression we wrote is symmetric under the exchange: 2  90 – 2,
then we see that the angle's complement, 70.4, is also a solution.
3- (a) No refraction occurs at the surface ab, so the angle of incidence at surface ac is 90°
– . For total internal reflection at the second surface, ng sin (90° – ) must be greater
than na. Here ng is the index of refraction for the glass and na is the index of refraction for
air. Since sin (90° – ) = cos , we want the largest value of  for which ng cos   na.
Recall that cos  decreases as  increases from zero. When  has the largest value for
which total internal reflection occurs, then ng cos  = na, or
  cos1
Fn I  cos F
1 I
G
J 48.9 .
G
J
. K
Hn K H152
a
1
g
The index of refraction for air is taken to be unity.
4- (a) Our first step is to form the image from the first lens. With p1 = 3.00 cm and f1 =
+4.00 cm, Eq. 34-9 leads to
1 1 1
   i1  12.0cm.
p1 i1 f1
The corresponding magnification is m1 = –i1/p1 = 4. This image serves the role of
“object” for the second lens, with p2 = 8.00 + 12.0 = 20.0 cm, and f2 = –4.00 cm. Now,
Eq. 34-9 leads to
1 1 1
   i2  3.33 cm .
p2 i2 f 2
5- (a) We choose a horizontal x axis with its origin at the left edge of the plastic. Between
x = 0 and x = L2 the phase difference is that given by Eq. 35-11 (with L in that equation
replaced with L2). Between x = L2 and x = L1 the phase difference is given by an
expression similar to Eq. 35-11 but with L replaced with L1 – L2 and n2 replaced with 1
(since the top ray in Fig. 35-36 is now traveling through air, which has index of refraction
approximately equal to 1). Thus, combining these phase differences and letting all lengths
be in m (so  = 0.600), we have
L2
L  L2
350
.
4.00  350
.
n2  n1  1
1  n1 
160
.  140
. 
1  140
.  0.833.


0.600
0.600
b g
b g
b
g
b
g
(b) Since the answer in part (a) is closer to an integer than to a half-integer, the
interference is more nearly constructive than destructive.
6- Assume the wedge-shaped film is in air, so the wave reflected from one surface
undergoes a phase change of  rad while the wave reflected from the other surface does
not. At a place where the film thickness is L, the condition for fully constructive
interference is 2nL  m  21 where n is the index of refraction of the film,  is the
wavelength in vacuum, and m is an integer. The ends of the film are bright. Suppose the
end where the film is narrow has thickness L1 and the bright fringe there corresponds to m
= m1. Suppose the end where the film is thick has thickness L2 and the bright fringe there
corresponds to m = m2. Since there are ten bright fringes, m2 = m1 + 9. Subtract
2nL1  m1  21  from 2nL2  m1  9  21  to obtain 2n L = 9, where L = L2 – L1 is
the change in the film thickness over its length. Thus,
b g
b g
L 
c
b
g
h
9
9 9 630  10 m

 189
.  106 m.
n
2 150
.
bg
7- We will make use of arctangents and sines in our solution, even though they can be
“shortcut” somewhat since the angles are [almost] small enough to justify the use of the
small angle approximation.
(a) Given y/D = 70/400 (both expressed here in centimeters), then

tan1(y/D) = 0.173 rad.
With d and  in micrometers, Eq. 36-20 then gives
=
d

sin =
sin(0.173 rad21.66 rad .


Thus, use of Eq. 36-21 (with a = 12 µm and = 0.60 µm) leads to
=
a
sin  =10.83 rad .

Thus,
IP sin 2
2
Im =    (cos ) = 0.00743 .
(b) Consider Eq. 36-25 with “continuously variable” m (of course, m should be an integer
for interference maxima, but for the moment we will solve for it as if it could be any real
number):
d
m = sin 6.9

which suggests that the angle takes us to a point between the sixth minimum (which
would have m = 6.5) and the seventh maximum (which corresponds to m = 7).
8- Since the slit width is much less than the wavelength of the light, the central peak of
the single-slit diffraction pattern is spread across the screen and the diffraction envelope
can be ignored. Consider three waves, one from each slit. Since the slits are evenly
spaced, the phase difference for waves from the first and second slits is the same as the
phase difference for waves from the second and third slits. The electric fields of the
waves at the screen can be written as
E1 = E0 sin (t),
E2 = E0 sin (t + ),
E3 = E0 sin (t + 2),
where  = (2d/) sin . Here d is the separation of adjacent slits and  is the wavelength.
The phasor diagram is shown below.
It yields
E  E 0 (1  2 cos  )
for the amplitude of the resultant wave. Since the intensity of a wave is proportional to
2
the square of the electric field, we may write I  AE02 1  2 cos , where A is a constant
of proportionality. If Im is the intensity at the center of the pattern, for which  = 0, then
I m  9 AE02 . We take A to be I m / 9 E02 and obtain
b
I
9- b
10- a
11- b
12- c
13- c
14- d
15- a
16- d
17- d
18- c
b
Im
1  2 cos 
9
g
g I9 c1  4 cos  4 cos  h.
2
m
2