Solutions to Homework #9
Chapter
28
4.
Picture the Problem
: The figure shows a car traveling at
17 m/s perpendicular to the line connecting two radio
transmitters. The car picks up a maximum signal when it is
at point A.
Strategy:
Set the wavelength equal to the difference in
distances between the two towers. Use the Pythagorean
theorem to calculate the distance between the car and each
tower as a function of time. Set the difference in distance
length equal to half a wavelength (for destructive
interference) and solve for the time.
Solution:
1. (a)
Find the
difference in path lengths:
450 m
150 m
300 m
λ
= ∆
=

=
l
2. (b)
Find the distance to the
far tower as a function of time:
(
29
2
2
2
1
450 m
v t
=
+
l
3.
Find the distance to the near
tower as a function of time:
(
29
2
2
2
2
150 m
v t
=
+
l
4.
Set the difference equal to
half a wavelength:
(
29
(
29
set
2
2
2
2
2
2
1
2
450 m
150 m
150 m
2
v t
v t
λ

=
+

+
=
=
l
l
5
. Rearrange the expression, square
both sides, and evaluate:
(
29
(
29
(
29
(
29
(
29
(
29
(
29
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
450 m
150 m
150 m
450 m
150 m
2
300 m
150 m
150 m
v t
v t
v t
v t
v t
�
�
�
�
+
=
+
+
�
�
�
�
�
�
�
�
�
�
+
=
+
+
�
�
�
�
+
+
�
�
6.
Rearrange the equation and
square both sides again:
(
29
(
29
(
29
2
2
2
2
2
2
2
4
2
2
2
6
2
450 m
2
150 m
150 m
300 m
2.25
10
m
2.756
10
m
v t
v t
�
�

�
�
+
=
�
�
�
�
�
�
�
�
�
�
+
=
7.
Solve for
t
:
6
2
4
2
1
2.756
10
m
2.25
10
m
30 s
17 m/s
t
=

=
Insight:
As the car moves farther away, the path difference
∆
l
decreases until it is nearly zero and the waves
constructively interfere.
14.
Picture the Problem
: Monochromatic red light produces a
twoslit interference pattern.
The color of the light is then
changed to blue.
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Strategy:
Recall the principles that explain the twoslit
interference pattern in order to answer the conceptual
question.
Solution:
1. (a)
Blue light has a smaller wavelength; therefore,
equations 28–1 and 28–2 show that the angles to the fringes
will be smaller. We conclude that the spacing between the
fringes will decrease if the color of the light is changed to
blue.
2. (b)
The best explanation is
II
. The fringe spacing decreases
because blue light has a smaller wavelength than red light.
Statement I erroneously assumes a greater frequency means a
longer wavelength, and statement III ignores the relationship
between wavelength and color.
Insight:
If the wavelength is increased by using infrared light instead of red light, the spacing between the fringes
would increase.
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 Spring '09
 KOROTKOVA
 Physics, Work, Diffraction, Wavelength, NM

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