Physics 295 Solutions to problems in Chapter 37
P37.2 ybright 
L
d
m

For m  1 ,
P37.5



3.40  103 m 5.00  104 m
yd

 515 nm
L
3.30 m
1
In the equation d sin    m   
2

The first minimum is described by
m0
and the tenth by m  9 :
sin  
Also,

1
9 
d
2
tan  
y
L
but for small ,
Thus, d 
d
P37.23


9.5 5 890  1010 m  2.00 m 
(a)
3
sin   tan 
9.5 9.5 L

sin 
y
 1.54  103 m  1.54 mm
7.26  10 m
The light reflected from the top of the oil film undergoes phase
reversal. Since 1.45  1.33 , the light reflected from the bottom
undergoes no reversal. For constructive interference of reflected
light, we then have
1

2nt   m   
2

or
m 
2  1.45 280 nm 
2nt

m   1 2
m   1 2
Substituting for m gives:
m 0,
0  1 620 nm (infrared)
m  1,
1  541 nm (green)
m 2,
2  325 nm (ultraviolet)
Both infrared and ultraviolet light are invisible to the human eye, so the
dominant color in reflected light is green .
(b)
The dominant wavelengths in the transmitted light are those that produce
destructive interference in the reflected light. The condition for destructive
interference upon reflection is
2nt  m
or
m 
2nt 812 nm

m
m
Substituting for m gives:
m  1,
1  812 nm (near infrared)
m 2,
2  406 nm (violet)
m 3,
3  271 nm (ultraviolet)
Of these, the only wavelength visible to the human eye (and hence the
dominant wavelength observed in the transmitted light) is 406 nm. Thus, the
dominant color in the transmitted light is violet .
P37.27
1

2nt   m    so
2

1 

t   m 
2  2n

Minimum
 1   500 nm 
t  
 96.2 nm
 2  2 1.30
P37.32 The condition for bright fringes is
2t 

2n
m

n
m  1, 2, 3,
.
From the sketch, observe that

 2  R  r 2 r 2
t  R  1  cos   R  1  1 
   
2  2  R
2R

The condition for
a bright fringe becomes
and ,
r2 
1 
 m 
R 
2 n
Thus, for fixed m
nr 2  constant
Therefore, nliquid r f2  nair ri2 and
nliquid  1.00
1.50 cm 2
 1.31
1.31 cm 2