Chapter 36: Diffraction
36-15
THINK The relative intensity in a single-slit diffraction depends on the ratio a/, where
a is the slit width and  is the wavelength.
EXPRESS The intensity for a single-slit diffraction pattern is given by
I  Im
sin 2 
2
where Im is the maximum intensity and  = (a/)sin. The angle  is measured from the
forward direction.
ANALYZE (a) We require I = Im/2, so
1
sin 2    2 .
2
(b) We evaluate sin2  and  2 2 for  = 1.39 rad and compare the results. To be sure
that 1.39 rad is closer to the correct value for  than any other value with three significant
digits, we could also try 1.385 rad and 1.395 rad.
(c) Since  = (a/)sin ,
  sin 1
Now / = 1.39/ = 0.442, so
  sin 1
 I
F
G
Ha J
K.
0.442 I
F
G
Ha J
K.
The angular separation of the two points of half intensity, one on either side of the center
of the diffraction pattern, is
0.442
  2  2 sin 1
.
a
(d) For a/ = 1.0,
  2sin 1  0.442 1.0  0.916rad  52.5 .
F
IJ
G
H K
(e) For a/ = 5.0,
(f) For a/ = 10,
  2sin 1  0.442 5.0  0.177 rad  10.1 .
b
g
  2 sin 1 0.442 10  0.0884 rad  5.06 .
LEARN As shown in Fig. 36-8, the wider the slit is (relative to the wavelength), the
narrower is the central diffraction maximum.
36-21
THINK We apply the Rayleigh criterion to estimate the linear separation between the
two objects.
EXPRESS If L is the distance from the observer to the objects, then the smallest
separation D they can have and still be resolvable is D = LR, where R is measured in
radians.
ANALYZE (a) With small-angle approximation, R  1.22 / d , where  is the
wavelength and d is the diameter of the aperture. Thus,
c
hc
h
. 8.0  1010 m 550  109 m
122
. L 122
D

 11
.  107 m = 1.1  104 km .
3
d
5.0  10 m
This distance is greater than the diameter of Mars; therefore, one part of the planet’s
surface cannot be resolved from another part.
(b) Now d = 5.1 m and D 
1.22  8.0 1010 m  550 109 m 
5.1m
 1.1104 m  11 km .
LEARN By the Rayleigh criterion for resolvability, two objects can be resolved only if
their angular separation at the observer is greater than  R  1.22 / d .
36-43
THINK For relatively wide slits, the interference of light from two slits produces bright
fringes that do not all have the same intensity; instead, the intensities are modified by
diffraction of light passing through each slit.
EXPRESS The angular positions  of the bright interference fringes are given by d sin 
= m, where d is the slit separation,  is the wavelength, and m is an integer. The first
diffraction minimum occurs at the angle 1 given by a sin 1 = , where a is the slit
width. The diffraction peak extends from –1 to +1, so we should count the number of
values of m for which –1 <  < +1, or, equivalently, the number of values of m for
which
– sin 1 < sin  < + sin 1.
The intensity at the screen is given by
sin  I
G
c hF
H J
K
2
I  I m cos2 
where  = (a/)sin ,  = (d/)sin , and Im is the intensity at the center of the pattern.
ANALYZE (a) The condition above means – 1/a < m/d < 1/a, or –d/a < m < +d/a. Now
d/a = (0.150  10–3 m)/(30.0  10–6 m) = 5.00,
so the values of m are m = –4, –3, –2, –1, 0, +1, +2, +3, and +4. There are 9 fringes.
(b) For the third bright interference fringe, d sin  = 3, so  = 3 rad and cos2  = 1.
Similarly,  = 3a/d = 3/5.00 = 0.600 rad and
sin  I F
sin 0.600 I
F
G
G
J
H K H0.600 J
K 0.255 .
2
2
The intensity ratio is I/Im = 0.255.
LEARN The expression for intensity contains two factors: (1) the interference factor
cos 2  due to the interference between two slits with separation d, and (2) the diffraction
factor [(sin  ) /  ]2 which arises due to diffraction by a single slit of width a. In the limit
a  0, (sin  ) /   1, and we recover Eq. 35-22 for the interference between two slits
of vanishingly narrow slits separated by d. Similarly, setting d = 0 or equivalently,  = 0,
we recover Eq. 36-5 for the diffraction of a single slit of width a. A plot of the relative
intensity is given below.
36-49
THINK Maxima of a diffraction grating pattern occur at angles  given by d sin  = m,
where d is the slit separation,  is the wavelength, and m is an integer.
EXPRESS If two lines are adjacent, then their order numbers differ by unity. Let m be
the order number for the line with sin  = 0.2 and m + 1 be the order number for the line
with sin  = 0.3. Then,
0.2d = m, 0.3d = (m + 1).
ANALYZE (a) We subtract the first equation from the second to obtain 0.1d = , or
d = /0.1 = (600  10–9m)/0.1 = 6.0  10–6 m.
(b) Minima of the single-slit diffraction pattern occur at angles  given by a sin  = m,
where a is the slit width. Since the fourth-order interference maximum is missing, it must
fall at one of these angles. If a is the smallest slit width for which this order is missing,
the angle must be given by a sin  = . It is also given by d sin  = 4, so
a = d/4 = (6.0  10–6 m)/4 = 1.5  10–6 m.
(c) First, we set  = 90° and find the largest value of m for which m < d sin. This is the
highest order that is diffracted toward the screen. The condition is the same as m < d/
and since
d/ = (6.0  10–6 m)/(600  10–9 m) = 10.0,
the highest order seen is the m = 9 order. The fourth and eighth orders are missing, so the
observable orders are m = 0, 1, 2, 3, 5, 6, 7, and 9. Thus, the largest value of the order
number is m = 9.
(d) Using the result obtained in (c), the second largest value of the order number is m = 7.
(e) Similarly, the third largest value of the order number is m = 6.
LEARN Interference maxima occur when d sin  = m, while the condition for
diffraction minima is a sin  = m. Thus, a particular interference maximum with order
m may coincide with the diffraction minimum of order m. The value of m is given by
d sin  m
d 

 m    m.
a sin  m
a
Since m = 4 when m  1, we conclude that d/a = 4. Thus, m = 8 would correspond to the
second diffraction minimum ( m  2 ).