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```1010
CHAPTER 37
DIFFRACTION
CHAPTER 37
1.
2.
3.
4.
5.
6.
(a) expand; (b) expand
(a) second side maximum; (b) 2:5
(a) red; (b) violet
diminish
(a) increase; (b) same
(a) left; (b) less
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
(a) contract; (b) contract
(a) the m = 5 minimum; (b) (approximately) the maximum between the m = 4 and
m = 5 minima
with megaphone (larger opening, less diraction)
(a) a and c tie, then b and d tie; (b) a and b tie , then c and d tie
four
red
(a) larger; (b) red
(a) less; (b) greater; (c) greater
(a) decrease; (b) same; (c) in place
(a) decrease; (b) decrease; (c) to the right
(a) A; (b) left; (c) left; (d) right
CHAPTER 37
12.
DIFFRACTION
1011
(a) increase; (b) rst order
Solutions to Exercises & Problems
1E
The condition for a minimum in a single-slit diraction pattern is given by Eq. 37-3:
a sin = m, where a is the slit width, is the wavelength, and m is an integer. Solve for
: = a sin =m = (0:022 mm)(sin 1:8 )=1 = 6:9 10 3 mm = 690 nm:
2E
(a) = sin 1 (1:50 cm=2:00 m) = 0:430 :
(b) For the mth diraction minimum a sin = m: Solve for a:
m = 2(441 nm) = 0:118 mm :
a = sin
sin 0:430
3E
Use a sin = m. The angle is measured from the forward direction, so for the situation
described in the problem it is 0:60 , for m = 1. Thus
m = 633 10 9 m = 6:04 10 5 m :
a = sin
sin 0:60
4E
(a) Use a sin = m. For = a and m = 1 the angle is the same as for = b and
m = 2. Thus a = 2b .
(b) Let ma be the integer associated with a minimum in the pattern produced by light with
wavelength a and let mb be the integer associated with a minimum in the pattern produced
by light with wavelength b . A minimum in one pattern coincides with a minimum in the
other if they occur at the same angle. This means ma a = mb b . Since a = 2b , the
minima coincide if 2ma = mb . Thus every other minimum of the b pattern coincides with
a minimum of the a pattern.
5E
(a) Use Eq. 37-3 to calculate the separation between the rst (m1 = 1) and fth (m2 = 5)
minima:
m
D (m m ) :
y = D sin = D a = D
m
=
a
a 2 1
1012
CHAPTER 37
DIFFRACTION
solve for a:
10 6 mm)(5 1) = 2:5 mm :
a = D(m2 y m1 ) = (400 mm)(5500:
35 mm
(b) For m = 1
(1)(550 10 6 mm) = 2:2 10 4 :
=
sin = m
a
2:5 mm
The angle is = sin 1 (2:2 10 4 ) = 2:2 10 4 rad:
6E
From Eq. 37-3
a = m = 1 = 1:41 :
sin sin 45:0
7E
(a) A plane wave is incident on the lens so it is brought to focus in the focal plane of the
lens, a distance of 70 cm from the lens.
(b) Waves leaving the lens at an angle to the forward direction interfere to produce an
intensity minimum if a sin = m, where a is the slit width, is the wavelength, and m
is an integer. The distance on the screen from the center of the pattern to the minimum
is given by y = D tan , where D is the distance from the lens to the screen. For the
conditions of this problem
(1)(590 10 9 m) = 1:475 10 3 :
=
sin = m
a
0:40 10 3 m
This means = 1:47510 3 rad and y = (7010 2 m) tan(1:47510 3 rad) = 1:010 3 m.
8P
The condition for a minimum of intensity in a single-slit diraction pattern is a sin = m,
where a is the slit width, is the wavelength, and m is an integer. To nd the angular
position of the rst minimum to one side of the central maximum set m = 1:
1 = sin
1
= sin
a
1
589 10 9 m = 5:89 10 4 rad :
1:00 10 3 m
If D is the distance from the slit to the screen, the distance on the screen from the center
of the pattern to the minimum is y1 = D tan 1 = (3:00 m) tan(5:89 10 4 rad) = 1:767 10 3 m.
CHAPTER 37
1013
DIFFRACTION
To nd the second minimum set m = 2:
2 = sin
1
2(589 10 9 m) = 1:178 10 3 rad :
1:00 10 3 m
The distance from the pattern center to the minimum is y2 = D tan 2 = (3:00 m) tan(1:178
10 3 rad) = 3:534 10 3 m. The separation of the two minima is y = y2 y1 =
3:534 mm 1:767 mm = 1:77 mm.
9P
Let the rst mimimum be a distance y from the central axis which is perpendicular to the
speaker. Then sin = y=(D2 + y2 )1=2 = m=a = =a (for m = 1). Slove for y:
y = p D2 = p D 2
(a=) 1
(af=vs ) 1
100 m
= 41:2 m :
=p
[(0:300 m)(3000 Hz)=(343 m/s)]2 1
10P
From y = mD=a we get
D m = (632:8 nm)(2:60) [10 ( 10)] = 24:0 mm :
y = mD
=
a
a
1:37 mm
11E
From Eq. 37-4
= 2 (x sin ) = 5892nm
0:10 mm (sin 30 ) = 267 rad :
2
This is equivalent to 267 rad 84 = 2:79 rad = 160 :
12E
(a) = sin 1 (1:1 cm=3:5 m) = 0:18 :
(b) Use Eq. 37-6:
(0:025 mm)(sin 0:18 ) = 0:46 rad :
= a
sin
=
538 nm
1014
CHAPTER 37
DIFFRACTION
13P
If you divide the original slit into N strips and represent the light from each strip, when it
reaches the screen, by a phasor, then at the central maximum in the diraction pattern you
add N phasors, all in the same direction and each with the same amplitude. The intensity
there is proportional to N 2 . If you double the slit width you need 2N phasors if they are
each to have the amplitude of the phasors you used for the narrow slit. The intensity at
the central maximum is proportional to (2N )2 and is therefore 4 times the intensity for
the narrow slit. The energy reaching the screen per unit time, however, is only twice the
energy reaching it per unit time when the narrow slit is in place. The energy is simply
redistributed. For example, the central peak is now half as wide and the integral of the
intensity over the peak is only twice the analogous integral for the narrow slit.
14P
Think of the Huygens' explanation of diraction phenomenon. When A is in place only
the Huygens' wavelets that pass through the hole get to point P . Suppose they produce a
resultant electric eld EA . When B is in place the light that was blocked by A gets to P
and the light that passed through the hole in A is blocked. Suppose the electric eld at P
is now EB . The sum EA + EB is the resultant of all waves that get to P when neither A
nor B are present. Since P is in the geometric shadow this is zero. Thus EA = EB and
since the intensity is proportional to the square of the electric eld, the intensity at P is
the same when A is present as when B is present.
15P
(a) The intensity for a single-slit diraction pattern is given by
2
sin
I = Im 2 ;
where = (a=) sin , a is the slit width and is the wavelength. The angle is measured
from the forward direction. You want I = Im =2, so
sin2 = 21 2 :
(b) 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 3 signicant digits,
(c) Since = (a=) sin ,
1
= sin a :
Now = = 1:39= = 0:442, so
= sin
1
0:442 :
a
CHAPTER 37
1015
DIFFRACTION
The angular separation of the two points of half intensity, one on either side of the center
of the diraction pattern, is
= 2 = 2 sin
(d) For a= = 1:0,
for a= = 5:0,
and for a= = 10,
1
0:442 :
a
= 2 sin 1 (0:442=1:0) = 0:916 rad = 53 ;
= 2 sin 1 (0:442=5:0) = 0:177 rad = 10 ;
= 2 sin 1 (0:442=10) = 0:0884 rad = 5:1 :
16P
(a) The intensity for a single-slit diraction pattern is given by
2
I = Im sin2 ;
where = (a=) sin . Here a is the slit width and is the wavelength. To nd the
maxima and minima, set the derivative of I with respect to equal to zero and solve for
. The derivative is
dI = 2I sin ( cos sin ) :
d
m
3
The derivative vanishes if 6= 0 but sin = 0. This yields = m, where m is an integer.
Except for m = 0 these are the intensity minima: I = 0 for = m.
The derivative also vanishes for cos sin = 0. This condition can be written tan = .
These are the maxima.
(b) The values of that satisfy tan = can be found by trial and error on a pocket
calculator or computer. Each of them is
slightly less than one of the values (m +
2
few are 0, 4:4934, 7:7252, 10:9041, 14:0662,
and 17:2207. They can also be found graphically. As in the diagram to the right, plot
y = tan and y = on the same graph.
The intersections of the line with the tan curves are the solutions. The rst two solutions listed above are shown on the diagram.
y
y=tan α
y=α
0
π /2
π
3π /2
1016
CHAPTER 37
DIFFRACTION
(c) Write = (m + 12 ) for the maxima. For the central maximum, = 0 and m =
For the next, = 4:4934 and m = 0:930. For the next = 7:7252 and m = 1:959.
1.
2
17P
Since the slit width is much less than the wavelength of the light, the central peak of the
single-slit diraction pattern is spread across the screen and the diraction envelope can
be ignored. Consider 3 waves, one from each slit. Since the slits are evenly spaced the
phase dierence for waves from the rst and second slits is the same as the phase dierence
for waves from the second and third slits. The electric elds of the waves at the screen
can be written E1 = E0 sin(!t), E2 = E0 sin(!t + ), and E3 = E0 sin(!t + 2), where
= (2d=) sin . Here d is the separation of adjacent slits and is the wavelength.
The phasor diagram is shown to the right. It yields
E3
E = E0 cos + E0 + E0 cos = E0 (1 + 2 cos )
for the amplitude of the resultant wave. Since the intensity of a wave is proportional to the square of the
electric eld, we may write I = AE02 (1+2 cos )2 , where
A is a constant of proportionality. If Im is the intensity at the center of the pattern, for which = 0, then
Im = 9AE02 . Take A to be Im =9E02 and obtain
φ φ
E2
E
I = I9m (1 + 2 cos )2 = I9m 1 + 4 cos + 4 cos2 :
φ
φ
ωt
E1
18E
Use Eq. 37-12: sin = 1:22=d. In our case = 2:5 =2 = 1:25 , so
:22 = 1:22(550 nm) = 31 m :
d = 1sin
sin 1:25
19E
(a) Use the Rayleigh criteria. To resolve two point sources the central maximum of the
diraction pattern of one must lie at or beyond the rst minimum of the diraction pattern
of the other. This means the angular separation of the sources must be at least R =
1:22=d, where is the wavelength and d is the diameter of the aperture. For the headlights
of this problem
10 9 m) = 1:34 10 4 rad :
R = 1:22(550
5:0 10 3 m
(b) If L is the distance from the headlights to the eye when the headlights are just resolvable
and D is the separation of the headlights, then D = L tan R LR , where the small angle
CHAPTER 37
1017
DIFFRACTION
approximation tan R R was made. This is valid if R is measured in radians. Thus
L = D=R = (1:4 m)=(1:34 10 4 rad) = 1:0 104 m = 10 km.
20E
(a) Use Eq. 37-14:
10 6 mm) = 1:3 10 4 rad :
R = 1:22 d = (1:22)(540
5:0 mm
(b) The linear separation is D = LR = (160 103 m)(1:3 10 4 rad) = 21 m:
21E
The minimum separation is
8
10 9 m) = 50 m :
Dmin = dem R = dem 1:22 d = (1:22)(3:82 105:1m)(550
m
22E
3 m)(4:0 10 3 m)
Lmax = D = 1:22D=d = (5:0 1:10
= 30 m :
22(550 10 9 m)
R
23E
3 m)
1
:
22
(1
:
22)(250
mm)(500
10
Dmin = LR = L d =
= 30:5 m :
5:00 mm
24E
(a) Use Rayleigh's criterion: two objects can be resolved if their angular separation at the
observer is greater than R = 1:22=d, where is the wavelength of the light and d is
the diameter of the aperture (the eye or mirror). 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 tan R LR , where R is measured in radians. The small angle approximation
tan R R was made. Thus
10
10 9 m) = 1:1 107 m = 1:1 104 km :
D = 1:22dL = 1:22(8:0 510:0 m)(550
10 3 m
This distance is greater than the diameter of Mars. One part of the planet's surface cannot
be resolved from another part.
1018
CHAPTER 37
DIFFRACTION
(b) Now d = 5:1 m and
10
10 9 m) = 1:1 104 m = 11 km :
D = 1:22(8:0 10 5:m)(550
1m
25E
2 m)(4:0 10 3 m)
D
D
(5
:
0
10
6 m = 1600 km :
Lmax = = 1:22=d =
=
1
:
6
10
9
1:22(0:10 10 m)
R
26E
= (6:2 103 m)(1:22)(1:6 10 2 m) = 53 m :
Dmin = LR = L 1:22
d
2:3 m
27P
(a) The diameter is
= (2000 103 m)(1:22)(1:40 10 9 m) = 17:1 m :
D = LR = L 1:22
d
0:200 10 3 m
(b) I=Im = (d=D)2 = (0:200 10 3 m=1:71 m)2 = 1:37 10
10 :
28P
(a)
6 m)(1:5 10 3 m)
D
2(50
10
= 0:19 m :
Lmax = 1:22=d =
1:22(650 10 9 m)
(b) The wavelength of the blue light is shorter so Lmax / 1 will be larger.
29P
According to Rayleigh's criterion (Eq. 37-14) R = 1:22=d. In our case R D=L, where
D = 60 m is the size of the object your eyes must resolve, and L is the limiting viewing
distance in question. Also d = 3:00 mm is the diameter of your pupil. Solve for L:
(60 m)(3:00 mm) = 27 cm :
=
L = 1Dd
:22
1:22(550 nm)
CHAPTER 37
1019
DIFFRACTION
30P
(a) Use Eq. 37-12:
= sin
= sin
1
1
1:22 = sin
d
1
1:22(vs =f )
d
(1:22)(1450 m/s) = 6:8 :
(25 103 Hz)(0:60 m)
(b) Now f = 1:0 103 Hz so
1:22 =
d
(1:22)(1450 m/s) = 2:9 > 1 :
(1:0 103 Hz)(0:60 m)
Since sin cannot exceed 1 there is no minimum.
31P
Use 2 = 1:22=d = D=L:
m/mi)(500 10 9 m) = 4:7 cm :
d = 1:22DL = (1:22)(220 mi)(1610
(30 ft)(0:305 m/ft)
32P
From R = 1:22=d = D=L we get
9
103 m) = 0:36 m :
d = 1:22DL = (1:22)(550 100:30m)(160
m
33P
(a) The rst minimum in the diraction pattern is at an angular position , measured
from the center of the pattern, such that sin = 1:22=d, where is the wavelength and
d is the diameter of the antenna. If f is the frequency then the wavelength is = c=f =
(3:00 108 m/s)=(220 109 Hz) = 1:36 10 3 m. Thus
= sin
1
1:22 = sin
d
1
1:22(1:36 10 3 m) = 3:02 10 3 rad :
55:0 10 2 m
The angular width of the central maximum is twice this, or 6:04 10 3 rad (0:346 ).
1020
CHAPTER 37
DIFFRACTION
(b) Now = 1:6 cm and d = 2:3 m, so
2 m) 1
:
22(1
:
6
10
1
= sin
=
8
:
5
10
2:3 m
The angular width of the central maximum is 1:7 10 2 rad (0:97 ).
34P
(a) The angular separation is
= (1:22)(550 10 9 m) = 8:8 10 7 rad = 0:1800 :
= R = 1:22
d
0:76 m
(b) The distance is
1012 km/ly)(0:18) = 8:4 107 km :
D = LR = (10 ly)(9:46(3600)(180)
(c) The diameter is
:18)()(14 m) = 2:5 10 5 m = 0:025 mm :
d = 2R L = 2(0(3600)(180)
35P
(a) Since = 1:22=d, the larger the wavelength the larger the radius of the rst minimum
(and second maximum, ect). Therefore the white pattern is outlined by red lights (with
longer wavelength than blue lights).
(b)
1:22(7 10 7 m) = 1:3 10 4 m :
d = 1:22
1:5(0:50 )(=180 )=2
36P
The energy of the beam of light which is projected onto the moon is concentrated is a
circular spot of diameter d1 , where d1 =dem = 2 = 2(1:22=d0 ), with d0 the diameter of
the mirror on Earth and dem the Earth-Moon separation. The fraction of energy picked
up by the reector of diameter d2 on the Moon is then 0 = (d2 =d1 )2 . This reected light,
upon reaching the Earth, has a circular cross section of diameter d3 satisfying d3 =dem =
2 = 2(1:22=d2 ). The fraction of the reected energy that is picked up by the telescope is
then 00 = (d0 =d3 )2 : Thus the fraction of the original energy picked up by the detector is
2 2
d
d
0
2
0
00
= =
=
d3
d1
2
d0 d2
d0 d2
=
(2:44dem =d0 )(2:44dem =d2 )
2:44dem
4
(2
:
6
m)(0
:
10
m)
= 2:44(0:69 10 6 m)(3:82 108 m) 4 10
13 :
4
CHAPTER 37
1021
DIFFRACTION
37E
Bright interference fringes occur at angles given by d sin = m, where d is the slit
separation, is the wavelength, and m is an integer. For the slits of this problem d = 11a=2,
so a sin = 2m=11. The rst minimum of the diraction pattern occurs at the angle 1
given by a sin 1 = and the second occurs at the angle 2 given by a sin 2 = 2, where a
is the slit width. You want to count the values of m for which 1 < < 2 , or what is the
same, the values of m for which sin 1 < sin < sin 2 . This means 1 < (2m=11) < 2. The
values are m = 6, 7, 8, 9, and 10. There are ve bright fringes in all.
38E
The number is 2(d=a) 1 = 2(2a=a) 1 = 3:
39E
It is clear from Eq. 37-5 that for a single slit of width 2a the diraction pattern is given by
I = Im
sin(2) ;
2
where = a sin =. Now, if we put d = a in Eq. 37-19, then = = a sin =, and
Eq. 37-19 reduces to
I = Im
(cos )2
sin 2
= Im
2 sin cos 2
2
= Im
sin(2)
2
2
;
where the trigonometric identity sin(2) = 2 sin cos was used. Thus Eq. 37-19 indeed
reduces to the diraction pattern for a single slit of width 2a.
40P
(a) Let the location of the fourth bright fringe coincide with the rst minimum of diraction
pattern: sin = 4=d = =a, or d = 4a.
(b) Any bright fringe which happens to be at the same location with a diraction minimum
will vanish. So let sin = m1 =d = m2 =a = m1 =4a = m2 =a, or m1 = 4m2 where
m2 = 1; 2; 3; : The fringes missing are thus the 4th, the 8th, the 12th, , i.e., every
fourth fringe is missing.
41P
The angular location of the mth bright fringe is given by d sin = m so the linear separation between two adjacent fringe is
D :
y = (D sin ) = Ddm = D
m
=
d
d
1022
CHAPTER 37
DIFFRACTION
42P
(a) 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 rst diraction
minimum occurs at the angle 1 given by a sin 1 = , where a is the slit width. The
diraction peak extends from 1 to +1 , so you want to count the number of values of
m for which 1 < < +1 , or what is the same, the number of values of m for which
sin 1 < sin < + sin 1 . This 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 nine fringes.
(b) The intensity at the screen is given by
I = Im
cos2 sin 2
;
where = (a=) sin , = (d=) sin , and Im is the intensity at the center of the pattern.
For the third bright interference fringe d sin = 3, so = 3 rad and cos2 = 1. Similarly,
= 3a=d = 3=5:00 = 0:600 rad and (sin )2 =2 = (sin 0:600)2 =(0:600)2 = 0:255.
The intensity ratio is I=Im = 0:255.
43P
(a) The rst minimum of the diraction pattern is at 5:00 so a = = sin = 0:440 m=
sin 5:00 = 5:05 m.
(b) Since the fourth bright fringe is missing d = 4a = 4(5:05 m) = 20:2 m:
(c) For the m = 1 bright fringe
= (5:05 m) sin 1:25 = 0:787 rad ;
= a sin
0:440 m
so the intensity of the m = 1 fringe is
I = Im
sin 2
= (7:0 mW/cm2 )
0:787
2
= 5:7 mW/cm2 ;
which agrees with what Fig. 37-39 indicates. Similarly for m = 2 I = 2:9 mW/cm2 , also
in agreement with Fig. 37-39.
44P
As the phase dierence is varied from zero to , both the intensity prole of the diraction and the location of the interference maximum change. At = , the original central
diraction envelop is now a minimum, and the two maxima of the diraction intensity
prole are now centered where the rst minima were. The locations of the intensity maxima/minima due to interference exchange, with the original locations of the maxima now
those of minima, and vice versa.
CHAPTER 37
1023
DIFFRACTION
As is further varied from to 2, the intensity pattern is gradually changed back,
resuming the original pattern at = 2.
45E
(a) d = 20:0 mm=6000 = 0:00333 mm = 3:33 m:
(b) Let d sin = m (m = 0; 1; 2; ), we nd = 0 for m = 0, = sin 1 (=d) =
sin 1 (0:589 m= 3:30 m) = 10:2 for m = 1, and similarly 20:7 for m = 2; 32:2
for m = 3; 45 for m = 4, and 62:2 for m = 5. Since jmj=d > 1 for jmj 6 these
are all the maxima.
46E
(a) Let d sin = m and solve for :
= (1:0 mm=200)(sin 30 ) = 2500 nm ;
= d sin
m
m
m
where m = 1; 2; 3 . In the visible light range m can assume the following values:
m1 = 4; m2 = 5 and m3 = 6. The corresponding wavelengths are 1 = 2500 nm=4 =
625 nm; 2 = 2500 nm=5 = 500 nm, and 3 = 2500 nm=6 = 416 nm.
(b) The colors are orange (for 1 = 625 nm), blue-green (for 2 = 500 nm), and violet (for
3 = 416 nm).
47E
The angular location of the mth order diraction maximum is given by m = d sin . To
be able to observe the fth-order one we must let sin jm=5 = 5=d < 1, or
=315 = 635 nm :
< d5 = 1:00 nm
5
So all wavelengths shorter than 635 nm can be used.
48E
The ruling separation is d = 1=(400 mm 1 ) = 2:5 10 3 mm. Diraction lines occur at
angles such that d sin = m, where is the wavelength and m is an integer. Notice that
for a given order the line associated with a long wavelength is produced at a greater angle
than the line associated with a shorter wavelength. Take to be the longest wavelength
in the visible spectrum (700 nm) and nd the greatest integer value of m such that is
less than 90 . That is, nd the greatest integer value of m for which m < d. Since
d= = (2:5 10 6 m)=(700 10 9 m) = 3:57 that value is m = 3. There are 3 complete
orders on each side of the m = 0 order. The second and third orders overlap (see 59P).
1024
CHAPTER 37
DIFFRACTION
49E
Let the total number of lines on the grating be N , then d = L=N where L = 3:00 cm. For
the second order diraction maximum d sin = (L=N ) sin = m = 2, so
= (3:00 10 2 m)(sin 33 ) = 13; 600 :
N = L 2sin
2(600 10 9 m)
50E
Use Eq. 37-25 for diraction maxima: d sin = m. In our case since the angle between the
m = 1 and m = 1 maxima is 26 the angle corresponding to m = 1 is = 26 =2 = 13 .
Solve for d:
m = (1)(550 nm) = 2:4 m :
d = sin
sin 13
51P
Let d sin = (L=N ) sin = m, we get
7 nm)(sin 30 )
(1
:
0
10
(
L=N
)
sin
=
= 500 nm :
=
m
(1)(10; 000)
52P
(a) Maxima of a two-slit interference pattern occur at angles given by d sin = m,
where d is the slit separation, is the wavelength, and m is an integer. The two lines
are adjacent so their order numbers dier 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 and 0:3d = (m + 1). Subtract the rst equation from the second to obtain
0:1d = , or d = =0:1 = (600 10 9 m)=0:1 = 6:0 10 6 m.
(b) Minima of the single-slit diraction 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 set = 90 and nd the largest value of m for which m < d sin . This is the
highest order that is diracted 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.
53P
(a) For the maximum with the greatest value of m (= M ) we have M = a sin < d,
so M < d= = 900 nm=600 nm = 1:5, or M = 1. Thus three maxima can be seen, with
m = 0; 1:
CHAPTER 37
1025
DIFFRACTION
(b) From Eq. 37-28
d sin = tan = 1 sin
hw = Nd cos = Nd
cos N
N
1 tan sin 1 600 nm = 0:051 :
= 1000
900 nm
1
d
54P
The angular positions of the rst-order diraction lines are given by d sin = , where d is
the slit separation and is the wavelength. Let 1 be the shorter wavelength (430 nm) and
be the angular position of the line associated with it. Let 2 be the longer wavelength
(680 nm) and let + be the angular position of the line associated with it. Here
= 20 . Then d sin = 1 and d sin( + ) = 2 . Use a trigonometric identity to
replace sin( +) with sin cos +cos
p sin , then use the equation for the rst line to
replace sin p
with 1 =d and cos with 1 21 =d2 . Afterpmultiplying by d you should obtain
1 cos + d2 21 sin = 2 . Rearrange to get d2 21 sin = 2 1 cos .
Square both sides and solve for d. You should get
r
2
2
d = (2 1 cos 2) + (1 sin )
sin s
[(680 nm) (430 nm) cos 20 ]2 + [(430 nm) sin 20 ]2
sin2 20
= 914 nm = 9:14 10 4 mm :
=
There are 1=d = 1=(9:14 10 4 mm) = 1090 rulings per mm.
55P
Use Eq. 37-25: m = d sin . For m = 1
= (1:73m) sin(17:6 ) = 523 nm ;
= d sin
m
1
and for m = 2
37:3 ) = 524 nm :
= (1:73m)sin(
2
Similarly we may compute the values of corresponding to the angles for m = 3 . The
average value of these 0 s is 523 nm.
1026
DIFFRACTION
CHAPTER 37
56P
1
The dierence in path lengths between the two adjacent light rays shown to the right is x = jAB j +
jBC j = d sin + d sin . The condition for bright
fringes to occur is thus
x = d (sin + sin ) = m ;
where m = 0; 1; 2; :
ψθ
2
d
A
ψ
C
θ
B
57P
From the gure to the right we see that the angular
deviation of the rst-order maximum from the incident direction is = + 1 . Here sin 1 = =d
sin = (600 nm=1:50 m) sin = 0:400 sin :
Thus
sin
d
1
= + sin (0:400 sin ) :
= + 1 = + sin
1
incident wave
diffracted wave
ψ
ψ
The plot is as follows. The angles are given in radians.
1
0.8
0.6
0.4
0
0.2
0.4
0.6
0.8
1
1.2
58P
From Eq. 37-25 we get
m ;
(sin ) = m
=
d
d
1.4
1.6
θ1
CHAPTER 37
1027
DIFFRACTION
but for small ; sin (d sin =d) = cos so
=
dmcos
p
m
d 1 sin2 =
m
p
=
:
d 1 (m=d)2
(d=m)2 2
p
59P
From d sin = m we see that if two spectral lines (labeled 1 and 2, respectively) overlap,
meaning they share the same value of , then m1 1 = m2 2 . For m1 = 2 and m2 = 3 this
becomes
1 = m2 = 3 :
2 m1 2
Since 400 nm < 2 < 1 < 700 nm we can always nd suitable values of 1 and 2 which
satisfy this condition. For example 1 = 600 nm, and 2 = 400 nm, or 1 = 660 nm and
2 = 440 nm, etc. So these two spectra always overlap, regardless of the value of d.
60P
In this case a = d=2, and the formula for the locations of the mth diraction minimum,
m = a sin = (d=2) sin , may be re-written as (2m) = d sin , which we recognize as
the formula for the location of the (2m)-th maximum of interference. Thus all the (2m)-th
(i.e., even) orders of maxima will be eliminated (exept m = 0).
61P
At the locationpof the hole sin 50 mm=30 cm = 0:164, and from m = d sin we nd
sin 5:0 cm= (30 cm)2 + (5:0 cm)2 = 0:164; so
= (1:00 106 nm=350)(0:164) = 470 nm :
m = d sin
Since for white light > 400 nm the only integer m allowed here is m = 1. At one edge of
the hole = 477 nm, while at the other edge
6
= d sin 0 = 1:00 10 nm
350
"
#
50 mm + 10 mm
p
= 560 nm :
(30 cm)2 + (6:0 cm)2
So the range of wavelength is from 470 to 560 nm.
62P
The derivation is similar to that used to obtain Eq. 37-27. At the rst minimum beyond
the mth principal maximum two waves from adjacent slits have a phase dierence of =
2m + (2=N ), where N is the number of slits. This implies a dierence in path length
1028
CHAPTER 37
DIFFRACTION
of L = (=2) = m + (=N ). If m is the angular position of the mth maximum
then the dierence in path length is also given by L = d sin(m + ). Thus d sin(m +
hw ) = m + (=N ). Use the trigonometric identity sin(m + hw ) = sin m cos hw +
cos m sin hw . Since hw is small, we may approximate sin hw by hw in radians
and cos hw by unity. Thus d sin m + d hw cos m = m + (=N ). Use the condition
d sin m = m to obtain d hw cos m = =N and
:
hw = Nd cos
m
63E
Let R = = = Nm and solve for N :
:6 nm + 589:0 nm)=2 = 491 :
N = m = (589
2(589:6 nm 589:0 nm)
64E
(a) Solve from R = = = Nm:
500 nm
=
= Nm
(600= mm)(5:0 mm)(3) = 0:056 nm56 pm :
(b) Since sin = mmax =d < 1,
1
mmax < d = (600= mm)(500
10 6 mm) = 3:3 ;
thus mmax = 3. No higher orders of maxima can be seen.
65E
If a grating just resolves two wavelengths whose mean is and whose separation is then its resolving power is dened by R = =. The text shows this is Nm, where N is
the number of rulings in the grating and m is the order of the lines. Thus = = Nm
and
656:3 nm = 3650 rulings :
N = m = (1)(0
:18 nm)
66E
(a) From R = = = Nm we nd
:496 nm + 415:487 nm)=2 = 23; 100 :
N = m = (415
2(415:96 nm 415:487 nm)
CHAPTER 37
1029
DIFFRACTION
(b) The maxima are found at
= sin
1
m = sin
d
1
(2)(415:5 nm)
:
=
28
:
7
7
4:0 10 nm=23; 100
67E
(a) From d sin = m we nd
m = 3(589:3 nm) = 1:0 104 nm = 10 m :
d = sin
sin 10
(b) The total width of the ruling is
Nd
=
L = Nd = Rd
m m
(589
nm)(10 m) = 3:3 103 m = 3:3 mm :
= 3(589:59:3nm
589:00 nm)
68E
The dispersion of a grating is given by D = d=d, where is the angular position of a
line associated with wavelength . The angular position and wavelength are related by
d sin = m, where d is the slit separation and m is an integer. Dierentiate this with
respect to to obtain (d=d) d cos = m or
d = m :
D = d
d cos Now m = (d=) sin , so
d sin = tan :
D = d
cos The trigonometric identity tan = sin = cos was used.
69E
(a) For the rst order maxima = d sin , which gives
= sin
1
= sin
d
1
589 nm
;
=
18
6
76 10 nm=40; 000
so from 68E D = tan = = tan 18 =589 nm = 0:032 = nm: Similarly for m = 2 and m = 3
we have = 38 and 68 , and the corresponding values of dispersion are 0:076 =nm and
0:24 =nm, respectively.
1030
CHAPTER 37
DIFFRACTION
(b) R = mN = 40000 m = 40; 000 for (m = 1); 80; 000 (for m = 2); and 120; 000 (for
m = 3).
70P
(a) We require that sin = m1;2 =d sin 30 , where m = 1; 2 and 1 = 500 nm. This gives
nm) = 2400 nm :
d sin230s = 2(600
sin 30
For a grating of given totla width L we have N = L=d / d 1 , so we need to minimize d to
maximize R = mN / d 1 . Thus we choose d = 2400 nm.
(b) Let the third-order maximum for 2 = 600 nm be the rst minimum for the single-slit
diraction prole. This requires that d sin = 32 = a sin , or a = d=3 = 2400 nm=3 =
800 nm:
(c) Let sin = mmax 2 =d 1 to obtain
nm = 3 :
mmax d = 2400
800 nm
2
Since the third order is missing the only maxima present are the ones with m = 0; 1 and
2.
71P
(a) From the expression for the half-width hw (given by Eq. 37-28) and that for the
resolving power R (given by Eq. 37-32), we nd the product of hw and R to be
d sin = tan ;
hw R = Nd cos Nm = d m
=
cos d cos where we used m = d sin (see Eq. 37-25).
(b) For rst order m = 1, so the corresponding angle 1 satises d sin 1 = m = . Thus
the product in question is given by
sin 1 = p sin 1 = p
1
tan 1 = cos
1
(1= sin 1 )2 1
1 sin2 1
1
=p 12
=p
= 0:89 :
(d=) 1
(900 nm=600 nm)2 1
72P
(a) Since the resolving power of a grating is given by R = = and by Nm, the range of
wavelengths that can just be resolved in order m is = =Nm. Here N is the number
CHAPTER 37
1031
DIFFRACTION
of rulings in the grating. The frequency f is related to the wavelength by f = c, where
c is the speed of light. This means f + f = 0, so = (=f )f = (2 =c)f ,
where f = c= was used. The negative sign means simply that an increase in frequency
corresponds to a decrease in wavelength. If we interpret f as the range of frequencies
that can be resolved we may take it to be positive. Then
2 f = c
Nm
and
c :
f = Nm
(b) The dierence in travel time for waves traveling along the two extreme rays is t =
L=c, where L is the dierence in path length. The waves originate at slits that are
separated by (N 1)d, where d is the slit separation and N is the number of slits, so the
path dierence is L = (N 1)d sin and the time dierence is
t = (N 1)c d sin :
If N is large this may be approximated by t = (Nd=c) sin . The lens does not aect the
travel time.
(c) Substitute the expressions you derived for t and f to obtain
c Nd sin = d sin = 1 :
f t = Nm
c
m
The condition d sin = m for a diraction line was used to obtain the last result.
73E
Bragg's law gives the condition for a diraction maximum:
2d sin = m ;
where d is the spacing of the crystal planes and is the wavelength. The angle is measured
from the normal to the planes. For a second order reection m = 2, so
2(0:12 10 9 m) = 2:6 10
d = 2 m
=
sin 2 sin 28
10 m = 0:26 nm :
74E
Use Eq. 37-34. For smallest value of let m = 1:
min = sin
1
m = sin
2d
1
(1)(30 pm)
2(0:30 103 pm) = 2:9 :
1032
CHAPTER 37
DIFFRACTION
75E
For rst order reection 2d sin 1 = and for the second order one 2d sin 2 = 2. Solve
for 2 :
2 = sin 1 (2 sin 1 ) = sin 1 (2 sin 3:4 ) = 6:8 :
76E
Use Eq. 37-34. From the peak on the left at angle 1 2d sin 1 = 1 , or 1 = 2d sin 1 =
2(0:94 nm) sin(0:75 ) = 0:025 nm = 25 pm. From the next peak 2 = 2d sin 2 = 2(0:94 nm)
sin(1:15 ) = 0:038 nm = 38 pm. You can check that the third peak from left is just the
second-order one for 1 .
77E
For the rst beam 2d sin 1 = A and for the second one 2d sin 2 = 3B . The values of d
and A can then be solved.
(a)
B = 3(97 pm) = 1:7 102 pm :
d = 2 3sin
2 sin 60
(b)
78E
2
A = 2d sin 1 = 2(1:7 102 pm)(sin 23 ) = 1:3 102 pm :
= 2d sin = 2(39:8 pm)(sin 30:0 ) = 39:8 pm :
79P
There are two unknowns, the X-ray wavelength and the plane separation d, so data for
scattering at two angles from the same planes should suce. The observations obey Bragg's
law, so
2d sin 1 = m1 and
2d sin 2 = m2 :
However, these cannot be solved for the unknowns. For example, use rst equation to
eliminate from the second. You obtain m2 sin 1 = m1 sin 2 , an equation that does not
contain either of the unknowns.
80P
The wavelengths satisfy m = 2d sin = 2(275 pm)(sin 45 ) = 389 pm. In the range of
wavelengths given, the allowed values of m are m = 3; 4, with corresponding wavelengths
being 389 pm=3 = 130 pm and 389 pm=4 = 97:2 pm, respectively.
CHAPTER 37
1033
DIFFRACTION
81P
The angle of incidence on the reection
planes is = 63:8 45:0 = 18:8 , and the
p
plane-plane separation is d = a0 = 2. Thus from 2d sin = we get
p
p
2 = p0:260 nm = 0:570 nm :
a0 = 2d = 2 sin
2 sin 18:8
82P
(a) The sets of planes with the next ve smaller
interplanar spacings (after a0 ) are shown in the diagram to the right. In terms of a0 the spacings are:
p
(i): a0 = 2 = 0:7071a0 ;
p
(ii): a0 = 5 = 0:4472a0 ;
p
(iii): a0 = 10 = 0:3162a0 ;
p
(iv): a0 = 13 = 0:2774a0 ;
p
(v): a0 = 17 = 0:2425a0 :
(b) Since any crystal plane passes through lattice
points its slope can be written as the ratio of
two integers. Consider a set planes with slope
m=n, as shown in the diagram to the right. The
rst and last planes shown pass through adjacent
lattice points along a horizontal line and there
are m 1 planes between. If h is the separation
of the rst and last planes, then the interplanar
spacing is d = h=m. If the planes make the angle
with the horizontal, then the normal to planes
(shown dotted) makes the angle = 90 .
The distance h is given by h = a0 cos and the
interplanar spacing is d = h=m = (a0 =m) cos .
Since
tan = n=m and cos =
p tan = m=n, p
2
1= 1 + tan = m= n2 + m2 . Thus
(i)
(iii )
(ii)
(iv)
(v)
ma0
= p a0 :
d = mh = a0 cos
m
n2 + m2
θ
na 0
φ
1034
CHAPTER 37
DIFFRACTION
83P
The angles of incidence which correspond to intensity maxima in reected beam of light
satisfy 2d sin = m, or
m(0:125 nm) = m :
sin = m
=
2d 2(0:252 nm) 4:032
Since j sin j < 1 the allowed values for m are m = 1; 2; 3; 4: Correspondingly the values of are = 14:4 ; 29:7 ; 48:1 ; and 82:8 . Therefore the crystal should be rotated counterclockwise by 48:1 45:0 = 3:1 or 82:8 45:0 = 37:8 , or clockwise by 45:0 14:4 = 30:6
or 45:0 29:7 = 15:3 :
84
= 0:143 rad, I=Im = 4:72 10 2 ;
= 0:247 rad, I=Im = 1:65 10 2 ;
= 0:353 rad, I=Im = 8:35 10 3
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