INTERFERENCE, 14TH EDITION
35
35.5. IDENTIFY: If the path difference between the two waves is equal to a whole number of wavelengths,
constructive interference occurs, but if it is an odd number of half-wavelengths, destructive interference occurs.
SET UP: We calculate the distance traveled by both waves and subtract them to find the path difference.
EXECUTE: Call P1 the distance from the right speaker to the observer and P2 the distance from the left
speaker to the observer.
(a) P1 = 8.0 m and P2 = (6.0 m)2 + (8.0 m)2 = 10.0 m. The path distance is
DP = P2 - P1 = 10.0 m – 8.0 m = 2.0 m.
(b) The path distance is one wavelength, so constructive interference occurs.
(c) P1 = 17.0 m and P2 = (6.0 m)2 + (17.0 m)2 = 18.0 m. The path difference is 18.0 m –17.0 m = 1.0 m,
which is one-half wavelength, so destructive interference occurs.
EVALUATE: Constructive interference also occurs if the path difference 2l , 3l , 4l , etc., and destructive
interference occurs if it is l /2, 3l /2, 5l /2, etc.
ml
35.7. IDENTIFY: The value of y20 is much smaller than R and the approximate expression ym = R
is accurate.
d
y20 = 10.6 ´10-3 m.
SET UP:
EXECUTE: d =
20Rl (20)(1.20 m)(502 ´10-9 m)
=
= 1.14 ´10-3 m = 1.14 mm.
y20
10.6 ´10-3 m
EVALUATE: tan q 20 =
y20
so q 20 = 0.51° and the approximation sinq 20 » tanq 20 is very accurate.
R
35.13. IDENTIFY and SET UP: The dark lines are located by d sinq = (m + 1 )l . The distance of each line from
2
the center of the screen is given by y = Rtanq .
EXECUTE: First dark line is for m = 0 and d sinq1 = l /2.
sin q1 =
sinq 2 =
l
2d
=
550 ´10-9 m
2(1.80 ´10-6 m)
= 0.1528 and q1 = 8.789°. Second dark line is for m = 1 and d sinq 2 = 3l /2.
æ 550 ´10-9 m ö
3l
= 3ç
÷ = 0.4583 and q 2 = 27.28°.
-6
2d
è 2(1.80 ´10 m) ø
y1 = Rtanq1 = (0.350 m)tan8.789° = 0.0541 m.
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35-1
35-2
Chapter 35
y2 = Rtanq 2 = (0.350 m)tan27.28° = 0.1805 m.
The distance between the lines is Dy = y2 - y1 = 0.1805 m - 0.0541 m = 0.126 m = 12.6 cm.
EVALUATE: sinq1 = 0.1528 and tanq1 = 0.1546. sinq 2 = 0.4583 and tanq 2 = 0.5157. As the angle
increases, sinq » tanq becomes a poorer approximation.
ml
35.14. IDENTIFY: For small angles: ym = R
.
d
SET UP: First-order means m = 1.
EXECUTE: The distance between corresponding bright fringes is
Dy =
Rm
(4.00 m)(1)
Dl =
(660 - 470) ´ (10-9 m) = 2.53´10-3 m = 2.53 mm.
-3
d
(0.300 ´10 m)
EVALUATE: The separation between these fringes for different wavelengths increases when the slit
separation decreases.
35.18. IDENTIFY:
f path difference
relates the path difference to the phase difference f .
=
2p
l
SET UP: The sources and point P are shown in Figure 35.18.
æ 524 cm - 486 cm ö
EXECUTE: f = 2p ç
÷ø = 119 radians.
2 cm
è
EVALUATE: The distances from B to P and A to P aren’t important, only the difference in these distances.
Figure 35.18
35.21. IDENTIFY: The phase difference f and the path difference r1 - r2 are related by f =
2p
l
(r1 - r2 ). The
æfö
intensity is given by I = I0 cos 2 ç ÷ .
è 2ø
c 3.00 ´108 m/s
=
= 2.50 m. When the receiver measures intensity I0 , f = 0.
f 1.20 ´108 Hz
2p
2p
EXECUTE: (a) f =
(r - r ) =
(1.8 m) = 4.52 rad.
l 1 2 2.50 m
SET UP: l =
æfö
æ 4.52 rad ö
(b) I = I0 cos 2 ç ÷ = I0 cos2 ç
= 0.404I0 .
2 ÷ø
è 2ø
è
EVALUATE: (r1 - r2 ) is greater than l /2, so one minimum has been passed as the receiver is moved.
35.24. IDENTIFY: Require destructive interference for light reflected at the front and rear surfaces of the film.
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Interference
35-3
SET UP: At the front surface of the film, light in air (n = 1.00) reflects from the film (n = 2.62) and there
is a 180° phase shift due to the reflection. At the back surface of the film, light in the film (n = 2.62)
reflects from glass (n = 1.62) and there is no phase shift due to reflection. Therefore, there is a net 180°
phase difference produced by the reflections. The path difference for these two rays is 2t, where t is the
505 nm
thickness of the film. The wavelength in the film is l =
.
2.62
EXECUTE: (a) Since the reflection produces a net 180° phase difference, destructive interference of the
æ 505 nm ö
reflected light occurs when 2t = ml. t = m ç
= (96.4 nm)m. The minimum thickness is 96.4 nm.
è 2[2.62] ÷ø
(b) The next three thicknesses are for m = 2, 3 and 4: 192 nm, 289 nm, and 386 nm.
EVALUATE: The minimum thickness is for t = l0 /2n. Compare this to Problem 35.23, where the
minimum thickness for destructive interference is t = l0 /4n.
35.31. IDENTIFY: Require destructive interference between light reflected from the two points on the disc.
SET UP: Both reflections occur for waves in the plastic substrate reflecting from the reflective coating, so
they both have the same phase shift upon reflection and the condition for destructive interference
(cancellation) is 2t = (m + 1 )l , where t is the depth of the pit. l =
2
EXECUTE: 2t =
l
. t=
l0
n
. The minimum pit depth is for m = 0.
l l0 790 nm
=
=
= 110 nm = 0.11mm.
4 4n 4(1.8)
2
EVALUATE: The path difference occurs in the plastic substrate and we must compare the wavelength in
the substrate to the path difference.
© Copyright 2016 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.