Answers to Homework #4 for Chapter 35 - Interference
1. A radio transmitting station operating at a frequency of 120 MHz has two identical
antennas, A and B, that radiate in phase. Antenna B is 9.00m to the right of antenna A.
Consider point P between the antenna and along the line connecting them, a horizontal
distance x to the right of the right of antenna A. For what values of x will constructive
interference occur at point P?
Answers: 0.75m, 2.00m, 3.25m, 4.50m, 5.75m, 7.00m, and 8.25m.
2. Coherent light from a sodium-vapor lamp is passed through a filter that blocks
everything except for light of a single wavelength. It then falls on two slits separated
by 0.460 mm. In the resulting interference pattern on a screen 2.20 m away, adjacent
bright fringes are separated by 2.82 mm.
What is the wavelength, ?
d = 0.460mm
R = 2.20m
y = 2.82mm
y = y1 – y0 = y2 – y1 = y3 – y2 = …..
y = Rm
d
R
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3. In a two slit interference pattern, the intensity at the peak of the central maximum is
I 0.
a) At a point in the pattern where the phase difference between the waves from the
two slits is 60.0, what is the intensity?
Answers:
a) 0.750I0
b) What is the path difference for 480nm light from the two slits at a point where the
phase is 60.0?
Answers:
b) 80.0nm
4. Light of 691 nm (in air) is incident roughly perpendicularly on a soap film of
refractive index, n = 1.33 and thickness, t, as shown below. What are the two smallest
film thicknesses (in nm) for which the reflected light undergoes “destructive”
interference?
t
5. Two rectangular pieces of plane glass are laid one upon the other on a table. A thin
strip of blue paper is placed between them at one edge so that a very thin wedge of
air is formed. The plates are illuminated at normal incidence by 546nm light from a
mercury-vapor lamp. Interference fringes are formed, with 15.0 fringes per
centimeter. Find the angle,  of the wedge.
(Hint: 15 bright fringes/cm means that x = x1 - x0 = (1/15)cm = 6.67 x 10-4m, with the
m = 0 bright fringe at x0, and the m=1 bright fringe at x1 from x = 0. Also, x = Lt/h
where h = height of the wedge, L = length of the glass, and  = tan-1(h/L )
glass
h
glass
L
Answer:
 = 0.0234
6. Two speakers A and B are 4.00m apart, and each one is emitting a frequency of 600
Hz. However, because of signal delays in the cables, speaker A is one-fourth of a
period “ahead” of speaker B. For points far from the speakers, find “all” the angles
relative to the centerline (see figure below) at which the sound from the speakers
cancels. Include angles on both sides (above and below) of the centerline. The speed
of sound is 340m/s
A

4.00m
centerline
B