Physics III
Homework VI
CJ
Chapter 22: 4, 11, 17, 25, 32, 42
22.4. Model: Two closely spaced slits produce a double-slit interference pattern.
Visualize: The interference pattern looks like the photograph of Figure 22.3(b).
Solve:
The fringe spacing is
y 
L
d
L
d
y

 589  10
9

m 150  10 2 m
4.0  10
3

m
 0.221 mm
22.11. Model: A diffraction grating produces an interference pattern.
Visualize: The interference pattern looks like the diagram of Figure 22.8.
Solve: (a) A grating diffracts light at angles sin m  m d . The distance between adjacent slits is d  1 mm 600 
1.667  106 m  1667 nm. The angle of the m  1 fringe is
 500 nm 
  17.46
 1667 nm 
1  sin1 
The distance from the central maximum to the m  1 bright fringe on a screen at distance L is
y1  L tan1   2 m tan17.46  0.629 m
(Note that the small angle approximation is not valid for the maxima of diffraction gratings, which almost always have angles
10.) There are two m  1 bright fringes, one on either side of the central maximum. The distance between them is
y  2y1  1.258 m .
(b) The maximum number of fringes is determined by the maximum value of m for which sinm does not exceed 1
because there are no physical angles for which sin   1. In this case,
sin m 
m m  500 nm 

d
1667 nm
We can see by inspection that m  1, m  2, and m  3 are acceptable, but m  4 would require a physically
impossible sin 4  1 . Thus, there are three bright fringes on either side of the central maximum plus the central maximum
itself for a total of 7 bright fringes.
22.17. Model: A narrow slit produces a single-slit diffraction pattern.
Visualize: The intensity pattern for single-slit diffraction will look like Figure 22.14.
Solve: Angle   0.70  0.0122 rad is a small angle (<< 1 rad). Thus we use Equation 22.20 to find the wavelength of
light. The angles of the minima of intensity are
p  p
 
a p
p


a
 0.10  10
p  1, 2, 3,
3

m  0.0122 rad 
2
 611 nm
22.25. Model: An interferometer produces a new maximum each time L2 increases by
difference r to increase by .
Visualize: Please refer to the interferometer in Figure 22.20.
Solve:
From Equation 22.33, the number of fringe shifts is
1
2
 causing the path-length
m 
2L2



2 1.00  10 2 m
656.45  10
9
  30,467
m
22.32. Model: Two closely spaced slits produce a double-slit interference pattern.
Solve:
The light intensity of a double-slit interference pattern at position y is
 d 
Idouble  4I1 cos2 
y
 L 
where I1 is the intensity due to each wave. The intensity of a bright fringe is 4I1.
For the central maximum, m  0. So, ycentral  0 m . The first minimum occurs at y1   L / 2d . Thus, the halfway position
between the maximum and the minimum is y   L 4d . Substituting into the intensity equation, the intensity at the halfway
position is
  d L 
2 
Idouble  4I1 cos2 
  4I1 cos  4   2I1

L
4
d


 
22.42. Model: A diffraction grating produces an interference pattern that is determined by both the slit spacing and the
wavelength used. The visible spectrum spans the wavelengths 400 nm to 700 nm.
Solve: According to Equation 22.16, the distance ym from the center to the mth maximum is ym  L tan m . The
angle of diffraction is determined by the constructive-interference condition d sin m  m , where m  0, 1, 2, 3, … The width
of the rainbow for a given fringe order is thus w  yred  yviolet. The slit spacing is
d
1 mm 1.0  103 m

 1.6667  106 m
600
600
For the red wavelength and for the m  1 order,
d sin1  1   1  sin 1

d
 sin 1
700  109 m
 24.83
1.6667  106 m
From the equation for the distance of the fringe,
yred  L tan1   2.0 m tan  24.83  92.56 cm
Likewise for the violet wavelength,
 400  10 9 m 
  13.88  yviolet   2.0 m  tan 13.88   49.42 cm
6
 1.6667  10 m 
1  sin 1 
The width of the rainbow is thus 92.56 cm  49.42 cm  43.1 cm.