Physics
Chapter 25: Interference and Diffraction
Chapter 25
INTERFERENCE AND DIFFRACTION
Problems
1. (a) Strategy The wavelength and frequency are related by  f  c.
Solution Compute the wavelength of a 60-kHz EM wave.

c 3.00 108 m s

 5.0 km
f
6.0 104 Hz
(b) Strategy Compute the path difference to determine the phase difference.
Solution The path difference is l  (19 km  12 km)  21 km  10 km, which is equal to two wavelengths.
So, the path difference results in constructive interference, but the reflection of the signal from the helicopter
results in a half-wavelength phase difference, which cause destructive interference. To summarize:
Destructive interference occurs, since the path difference is 10 km and there is a  2 phase shift.
7. Strategy Since the light from each lamp is incoherent, the intensity of the lamps together is just the sum of the
intensities.
Solution Find the combined intensity of the lamps.
I  I 0  4 I 0  5I 0
17. Strategy At the air-oil boundary, reflected light is 180° out of phase, since noil  nair . For transmitted light
reflected at the oil-water boundary, the reflective ray is not phase shifted since nwater  noil. The two reflected
rays are 180° out of phase, so constructive interference occurs when the relative path length difference, 2t, is an
odd multiple of one half the wavelength in the oil,   0 / noil .
Solution Find the wavelength.
1
1 


2t   m      m   0
2
2  noil



2tnoil
.
Solving for 0 yields 0 
m  12
Substituting t  0.40 m and noil  1.50 gives 0 
1200 nm
.
m  12
Evaluating this for m  0, 1, 2, and 3 gives 0  2400 nm, 800 nm, 480 nm, and 343 nm, respectively. Of these
values, only 480 nm is in the visible spectrum.
29. Strategy Sketch the diagram according to the instructions in the Problem statement. Compare the results to
Eq. (25-10).
Solution
1
Chapter 25: Interference and Diffraction
Physics
The m  1 maxima are 6.9 cm from the central maximum . As measured by a protractor, the angles between
the m  1 maxima and the central maximum are 30 . Solve Eq. (25-10) for  .
d sin   m, so   sin 1
m
 1.0 cm 
1
 sin 1  m
  sin (0.50m).
d
 2.0 cm 
So, 0  0 and 1  30 for m  0 and m  1, respectively. Yes, the angles agree .
39. Strategy Find the slit separation. Then solve Eq. (25-10) for m using   412 nm and   90.
Solution Compute the slit separation.
1
d
 2.0000  106 m
5000.0 slits/cm
Find the number of orders.
d sin   m , so m 
d sin 

(2.0000 106 m) sin 90

412 109 m
Since m must be an integer, four orders can be observed.
 4.85.
49. Strategy Call the width of the central maximum W. Referring to Figure 25.33, we have W  2x, and by
trigonometry, x  D tan  , so W  2x  2D tan  . Assume  is a small angle. Use Eq. (25-12).
Solution Since  is assumed to be small, we have W  2D tan   2D sin  .

2L
From Eq. (25-12), we have sin   since m  1, so W becomes W 
.
a
a
2L 1 2L 1
If a is replaced with 2a, the new width is Wnew 

 W.
2a
2 a
2
Thus, the new width is half the old width.
65. Strategy The first single-slit diffraction minimum is coincident with the fifth-order double-slit interference
maximum. Use the equations for the double-slit interference maxima and single-slit diffraction minima. The
distance to the first diffraction minimum is half the width of the central maximum.
Solution
2
Physics
Chapter 25: Interference and Diffraction
(a) Find the width of the slits.
a sin   m  (1)   , so a 

sin 
.
Since x  (2.40 cm) 2  0.0120 m  D  2.4 m, sin   tan  
a
D
x

(510 109 m)(2.4 m)
 0.10 mm .
0.0120 m
x
and
D
(b) Find the relationship between the slit width a and the distance between the slits d using the fifth-order doubleslit interference maximum and the first single-slit diffraction minimum.
d sin  d 5
d sin   5 and a sin    , so
 
 5 and d  5a  5(0.102 mm)  0.51 mm .
a sin  a 
3