Physics 112
Homework 11 (Ch24)
Due July 24
1. If a slit diffracts 650-nm light so that the first diffraction maximum is 4.0 cm wide on a
screen 1.50 m away, what will be the width of the diffraction maximum for light of
wavelength 420 nm?
Solution
The width of the first diffraction maximum is equal the distance between first minima:
y1  2 y1  4cm
For small angles

y1  x tan 1  x .
D
If we form the ratio of the expressions for the two wavelengths, we get

y1b y1b b
y1b  y1a b ;

 
y1a y1a a
a
y1b   4.0cm 
 420nm  , which gives
 650nm 
y1b  2.6cm.
2. How many lines per centimeter does a grating have if the third-order occurs at an 18.0° angle
for 630-nm light?
Solution
We find the slit separation from
d sin  m  m ;
d sin18.0   3  630  109 m  , which gives d  6.12  106 m  6.12  104 cm.
The number of lines/cm is
1
1

 1.64  103 lines cm.
d  6.12  104 cm 
3. What is the smallest thickness of a soap film (n  1.42) that would appear black if illuminated
with 480-nm light? Assume there is air on both sides of the soap film.
Solution
  480nm
n  1.42
t ?
We have 2t  min  m / n - destructive interference. For m=1:
  480 nm  
tmin  12 
 1  169 nm.
 1.42  
Physics 112
Homework 11 (Ch24)
Due July 24
4. How thick (minimum) should the air layer be between two flat glass surfaces if the glass is to
appear bright when 450-nm light is incident normally? What if the glass is to appear dark?
Solution
2t  m  12  - constructive interference. The minimum non-zero thickness is (m=0)
t min  14 450nm  113nm
2t  m - destructive interference. The minimum non-zero thickness is (m=1)
tmin  12  450nm1  225nm.
5. A thin film of alcohol (n  1.36) lies on a flat glass plate (n  1.51) . When monochromatic
light, whose wavelength can be changed, is incident normally, the reflected light is a
minimum for   512 nm and a maximum for   640 nm . What is the minimum thickness
of the film?
Solution
Constructive interference: a  640 nm
Destructive interference: b  512 nm
 a ,in   a / n1
b,in  b / n1
n1  1.36
The wave that reflects from the top surface of the alcohol has a phase change of
 , or 12  in terms of wave length.
The wave that reflects from the glass at the bottom surface of the alcohol has a phase change of
 , or 12  in terms of wave length.
for constructive interference 2t  ma a / n1
for destructive interference 2t  mb  12 b / n1
ma a / n1  mb  12 b / n1 ,
When we combine these two equations, we get:
2mb  1 a 640 5
or



2ma a
b 512 4
Thus we see that ma  mb  2 , and the thickness of the film is
t  ma a / 2n1   2  640nm / 2  1.36  471nm
Because of that:
Physics 112
Homework 11 (Ch24)
Due July 24
6. What is Brewster’s angle for a diamond submerged in water if the light is hitting the diamond
(n  2.42 ) while traveling in the water?
Solution
Because the light is coming from water to diamond, we find the angle from the vertical from
n
2.42
tan  p  diamond 
 1.82, which gives  p  61.2.
nwater
1.33
7. Light of wavelength 590 nm passes through two narrow slits 0.60 mm apart. The screen is
1.70 m away. A second source of unknown wavelength produces its second-order fringe 1.33
mm closer to the central maximum than the 590-nm light. What is the wavelength of the
unknown light?
Solution
For constructive interference, the path difference is a multiple of the wavelength:
d sin  m  m , m  0, 1, 2, 3, ... .
We find the location on the screen from: y  x tan  .
For small angles: sin  tan , which gives
ym  x
m
.
d
For the second-order fringes we have
2a
2
;
y2 b  x b .
d
d
When we subtract the two equations, we get
yd
 2x 
;
y  ya  yb     1  2   ;
y  1 
2x
 d 
 2 1.70m  
  590nm  2  , which gives 2  355nm.
1.33  103 m  
3
  0.60  10 m  
y2 a  x
8. What is the minimum (non-zero) thickness for the air layer between two flat glass surfaces if
the glass is to appear dark when 640-nm light is incident normally? What if the glass is to
appear bright?
The 12 - cycle phase change at the lower surface means the following
2t  m , m  0, 1, 2, ...
A) Destructive interference:
 640nm
We set m  1 to find the smallest nonzero value of t:
t 
 320nm.
2
2
2t   m  12   , m  0, 1, 2, ...
B) Constructive interference:
Solution:
We set m  0 to find the smallest value of t:
t

4

640nm
 160nm.
4