Chapter 37
Questions: 1, 3, 4, 5, 6, 8, 9, 10, 11, 13, 14, 16
Problems: 1, 3, 9, 11, 15, 17, 25, 31, 36, 41, 49, 51, 55, 57, 61
Questions
1. The maximum intensity in a di¤raction grating occurs at d sin = m ;
where is the wavelength. As increases must also increase. Thus a
di¤raction grating bends red light more than blue.
3. Yes, as long as the di¤erence in length was less than the coherence
length of the light.
4. It becomes too thin for any constructive interference, see example 37-4.
5. Eyeglasses are too thick. This means that the order of interference is
so large that all interference e¤ects are smeared out.
6. If your …ngers are close enough together it is possible to secondary
di¤raction peaks typical of single-slit di¤raction.
8. The wavelength of light is much shorter than the wavelength of sound.
Hence di¤ractions e¤ects allow us to hear around corners, but not see around
them.
9. The EM …elds from the two slits of the monochromatic light wave that
add. It is the square sum of these …elds that gives the intensity and the cross
term that results from this sum gives rise to interference.
10. It is the …elds that add not the intensity. Thus the intensity is
proportional to
I / (E1 + E2 )2 = E12 + 2E1 E2 + E22 ;
I = I1 + I2 + 2 hE1 E2 i :
For incoherent sources the time average of E1 E2 vanishes. Thus we can
simply add the intensities.
11. The widths of the peaks due to constructive interference are much
narrower for a grating versus two slits. This increase in resolution enables a
di¤raction grating to resolve interference peaks between two slightly di¤erent
wavelenghts.
13. For a broad slit, the intensity pattern is tightly centered about the
centerline. As the slit narrows this intensity pattern broadens and it is possible to observe small secondary peaks in the intensity on the wings of this
pattern. Continuing to narrow the slit results in a very broad intensity pattern and the secondary peaks occur at greater separation..
1
14. In the center both wavelengths of light form a constructive intensity
peak. For higher order peaks, the red light occurs at larger angles than the
blue, see …gure 37-7.
16. You need several di¤erent orientations in order to see the lattice
spacings in di¤erent directions through the crystal (Not all crystals are simple
cubic.)
Problems
1. The constructive interference pattern is determined from the criterion
d sin = m :
10 2 ; we …nd
For m = 1; d = 15 m, and sin = :071=2:2 = 3:227
= d sin = 15
10
6
3:227
10
2
= 484nm.
3. (a) The constructive interference pattern is determined from the criterion
d sin = m :
For adjacent …gures at small angles this expression reduces to
= 5mm=L; = 633 10 6 mm, and d = :12mm, we have
633
10
9
= 12
10
55
10
L
3
!L=
60 10
633 10
=d
: For
8
9
L = 95cm
(b) The fringe spacing, `; is proportional to the wavelength, hence
`=5
480
= 3:8mm
633
9. The constructive interference pattern is determined from the criterion
d sin = m :
If instead the path di¤erence is 1=4 of a wavelength versus a full wavelength
then
sin
1
1 600 10 6
=
= 6 10 4 ;
4d
4
:25
180
' 6 10 4 rad =
6 10 4 = :034 :
=
2
11. The constructive interference pattern is determined from the criterion
d sin = m :
(a) The height of the m fringe is (L is the distance to the screen)
ym = L tan
ym = L q
d2
m
=L
sin
cos
m
m
m
sin
m
sin2
1
m
= Lq
m =d
1
(m )2 =d2
(m )2
For the case d = 6:5 m,
fringe is
y2 = 1:7 q
= Lp
= :633 m, and L = 1:7m the height to the m = 2
2
(6:5
633
10 6 )2
9
10
(2
= 33:76cm
633
10 9 )2
The height to the m = 1 fringe is
y1 = 1:7 q
(6:5
633
10
10
6 )2
9
= 16:63cm
(633
The separation between these two fringes,
y = 33:76
10
9 )2
y; is
16:63 = 17:13cm
(b) The height to the m = 4 fringe is
y4 = 1:7 q
4
(6:5
633
10 6 )2
9
10
(4
= 71:90cm
633
10 9 )2
The height to the m = 3 fringe is
y4 = 1:7 q
3
(6:5
10
633
6 )2
9
10
(3
= 51:93cm
633
The separation between these two fringes,
y = 71:09
10
9 )2
y; is
51:93 = 19:97cm
3
15. (a) The maxima occur at the same locations as that for two-slit
di¤raction. Hence
sin
1
sin
2
:633 10 6
= :0844 ! 1 = 4:8
d
7:5 10 6
2
2 :633 10 6
=
=
= :169 ! 2 = 9:7
d
7:5 10 6
=
=
(b) The minima occur when
d sin =
m
:
N
For a grating with 5 slits (N = 5) the third minima occurs when m = 3:
The wavelength is = 633nm and the slit spacing is d = 7:5 m: Hence the
angular location for the third minima is
sin
=
3
3
= 5:064
5d
10
2
!
3
= 2:9 :
There are only 4 minima between maxima. Hence the 6th minima occurs
when m = N + 2;
sin
6
=
1+
2
5
d
= :1182 !
6
= 6:79
17. The criterion for constructive interference is the same for a di¤raction
grating as it is for twin slits. For 3000 lines/cm, the slit spacing is d =
10 2 m=3000 = 1=3 10 5 m. Thus (a)
sin
1
1
520 10 9
= :1560,
d
10 5
= :1566rad = 8:975 :
=
=
3
(b) For the …fth order fringe
sin
5
5
520 10
d
10 5
= :8947rad = 51:3 :
=
=
5
3
9
= :78;
25. The angular location for the 6th order 484:3nm is found from
6
5
7
sin = 484:3nm = 581:2nm = 415:1nm:
d
d
d
4
Hence it could be a 5th order 581nm line or a 7th order 415nm line.
31. The criterion for constructive interference in a thin …lm is
2d = (m + 1=2)
n;
where n is the wavelength of the light inside the …lm. Light with a wavelength of = 550nm will have a wavelength of
n
= =n = 550nm=1:33 = 413:5nm,
inside the soap …lm. The minimum thickness for constructive interference
occurs at m = 0; hence
dmin =
n =4
= 103:4nm:
36. For the oil …lm ‡oating on water the …rst interface (air to oil) there
is a phase shift during the re‡ection. At the second interface (oil to water)
that is an additional phase shift. These will cancel each other out. Hence
the condition for constructive interference is
2d = m
For green light,
= m =1:25:
oil
= 500nm; the minimum thickness is
d=
2:5
= 200nm:
For the soap …lm suspended in air there is only one
the condition for constructive interference is
2d = (m + 1=2)
soap
phase shift. Hence
= (m + 1=2) =1:333:
The minimum thickness (m = 0) is
d=
4
(4=3)
= 93:75nm
41. The criterion for constructive interference in a thin …lm is again
2d = (m + 1=2)
5
n;
only now n = 1: The maximum thickness is d = :065mm = 65
From the expression just above we have
m = 2d=
1=2 = 130
103 =550 = 236:4
103 nm.
:5 = 235:9:
So the largest m is m = 235: Counting the fringe for m = 0 results in 236
observable fringes.
49. The intensity pattern for di¤raction from a single slit of width a is
given by
2
sin (( a sin ) = )
:
S = So
( a sin ) =
The …rst minima occurs when the argument of the sine function is equal to
: Thus = a sin : When this occurs at =2, the sine function is unity, and
= a:
51. The wavelength for a 29M Hz radio wave is
= 10:34m. The
intensity pattern for di¤raction from a single slit of width a is given by
S = So
sin (( a sin ) = )
( a sin ) =
2
:
For a slit spacing of a = 45m, the …rst minima occurs at
( a sin ) =
=
sin
=
;
10:34
= :23;
a
45:
:232rad = 13:3 :
=
=
The full angular width is
= 26:6 :
55. For a circular aperture the Rayleigh criteria is
1:22
:
D
For 5cm resolution from a distance of 100km with a wavelength of
550nm, the diameter of the lens must be
sin
D = 1:22
550
min
=
10 9 = :05=105 = 1:22
110
10
2
=
= 1:342m.
57. If the radius of the spot is given by r; then the subtended angle
for a spot at the focal length is simply r=f: From the Rayleigh criteria the
minimum angle is
1:22
r
= :
min =
D
f
6
Since f =D = 1:4, and the wavelength of light is
r = 1:22
= 580nm, we have
1:4 = 990nm = :99 m.
Thus the diameter of the spot is d = 1:98 m.
61. For a circular aperture the Rayleigh criteria is
min
=
1:22
:
D
Now one arcsec is 1=3600 of a degree or 4:85 10 6 rad. The diameter of an
aperature for this minimum angle for = 550nm is
D=
1:22 550 10
4:85 10 6
9
= :138m = 13:8cm.
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