Homework Set #10 Solutions Chapter 24
24.8
In a double-slit interference pattern the distance from the central maximum to the
position of the m th order bright fringe is given by
 L 
ym  m 

 d 
where d is the distance between the splits and L is the distance to the screen. Thus, the
spacing between the first- and second-order bright fringes is
9
  L    600  10 m   2.50 m  
y  y2  y1  [2  1] 
  0.0300 m  3.00 cm
  1
0.050  103 m
 d  

24.10
The angular deviation from the line of the central maximum is given by
 y
 
 1.80 cm 


  tan 1    tan 1 
  0.737
L
140 cm
(a) The path difference is then
  d sin   0.150 mm sin  0.737  1.93 103 mm  1.93 m
(b)



  3.00 
9
 643 10 m 
  1.93 106 m  
(c) Since the path difference for this position is a whole number of wavelengths, the
waves interfere constructively and produce a maximum at this spot.
24.11
As shown in the figure at the right, the path
difference in the waves reaching the telescope is
  d 2  d1  d 2 1  sin   . If the first minimum
   2 
occurs when   25.0 , then
  180    90.0     40.0 , and
d2 

1  sin 

 250 m 2 
1  sin 40.0
 350 m
Thus, h  d2 sin 25.0  148 m
24.14
With nglass  nair and nliquid  nglass , light reflecting from the air-glass boundary experiences
a 180° phase shift, but light reflecting from the glass-liquid boundary experiences no
shift. Thus, the condition for destructive interference in the two reflected waves is
2nglass t  m
where
m  0, 1, 2,
For minimum (non-zero) thickness, m  1 giving
24.22
t

2nglass

580 nm
 193 nm
2 1.50 
With a phase reversal due to reflection at each surface of the magnesium fluoride layer,
there is zero net phase difference caused by reflections. The condition for destructive
interference is then
1
1 


2 t   m   n   m  
, where m  0, 1, 2,
2
2  n film


For minimum thickness, m  0 , and the thickness is
t   2m  1

4 n film
550  10
 1
9
4 1.38 
m
  9.96  10
8
m  99.6 nm
24.23
There is a phase reversal upon reflection at each surface of the film and hence zero net
phase difference due to reflections. The requirement for constructive interference in the
reflected light is then
2 t  mn  m

n film
, where m  1, 2, 3, 
With t  1.00 105 cm  100 nm , and n film  1.38 , the wavelengths intensified in the
reflected light are

2 n film t
m

2 1.38 100 nm 
, with m  1, 2, 3, 
m
Thus,   276 nm, 138 nm, 92.0 nm 
and none of these wavelengths are in the visible spectrum
24.45
From Malus’s law, the intensity of the light transmitted by the first polarizer is
I1  I i cos2 1 . The plane of polarization of this light is parallel to the axis of the first plate
and is incident on the second plate. Malus’s law gives the intensity transmitted by the
second plate as I 2  I1 cos2 2  1   I i cos2 1 cos2 2  1  . This light is polarized parallel
to the axis of the second plate and is incident upon the third plate. A final application of
Malus’s law gives the transmitted intensity as
I f  I 2 cos2 3   2   I i cos2 1 cos2  2  1  cos2 3   2 
With 1  20.0, 2  40.0, and 3  60.0 , this result yields
I f  10.0 units  cos2  20.0 cos2 20.0 cos2 20.0  6.89 units
24.47
(a) If light has wavelength  in vacuum, its wavelength in a medium of refractive
index n is n   n . Thus, the wavelengths of the two components in the specimen
are
n 
1
and

n1

546.1 nm
 413.7 nm
1.320
n 
2

n2

546.1 nm
 409.7 nm
1.333
(b) The number of cycles of vibration each component completes while passing
through the specimen are
N1 
t
n
1
and

1.000  10-6 m
 2.417
413.7  10-9 m
N2 
t
n
2

1.000  10-6 m
 2.441
409.7  10-9 m
Thus, when they emerge, the two components are out of phase by
N 2  N1  0.024 cycles . Since each cycle represents a phase angle of 360°, they
emerge with a phase difference of
 =  0.024 cycles  360 cycle   8.6
24.52
Assuming the glass plates have refractive indices greater than that of both air and water,
there will be a phase reversal at the reflection from the lower surface of the film but no
reversal from reflection at the top of the film. Therefore, the condition for a dark fringe is
2 t  mn  m   n film  for m  0, 1, 2,
If the highest order dark band observed is m  84 (a total of 85 dark bands counting the
m  0 order at the edge of contact), the maximum thickness of the wedge is
tmax 
mmax    84   
 42 


2  n film  2  1.00 
When the film consists of water, the highest order dark fringe appearing will be
n 
 1.333 
mmax  2 tmax  film   2  42   
  112

  


Counting the zeroth order, a total of 113 dark fringes are now observed.