Download Chapter 35: Interference 35-55 THINK The index of refraction of oil is

Chapter 35: Interference
35-55
THINK The index of refraction of oil is greater than that of the air, but smaller than that
of the water.
EXPRESS Let the indices of refraction of the air, oil, and water be n1, n2, and n3,
respectively. Since n1  n2 and n2  n3 , there is a phase change of  rad from both
surfaces. Since the second wave travels an additional distance of 2L, the phase difference
is
2

(2 L)
2
where 2   / n2 is the wavelength in the oil. The condition for constructive interference
is
2
(2 L)  2m ,
2
or

2 L  m , m  0, 1, 2,...
n2
ANALYZE (a) For m  1, 2,..., maximum reflection occurs for wavelengths

2n2 L 2 1.20  460 nm 

 1104 nm , 552 nm, 368nm...
m
m
We note that only the 552 nm wavelength falls within the visible light range.
(b) Maximum transmission into the water occurs for wavelengths for which reflection is a
minimum. The condition for such destructive interference is given by
1 

2L   m  
2  n2



4n2 L
2m 1
which yields  = 2208 nm, 736 nm, 442 nm … for the different values of m. We note that
only the 442 nm wavelength (blue) is in the visible range, though we might expect some
red contribution since the 736 nm is very close to the visible range.
LEARN A light ray reflected by a material changes phase by  rad (or 180°) if the
refractive index of the material is greater than that of the medium in which the light is
traveling. Otherwise, there is no phase change. Note that refraction at an interface does
not cause a phase shift.
35-81
THINK There is a small difference between the wavelength in air and the wavelength in
vacuum.
EXPRESS Let 1 be the phase difference of the waves in the two arms when the tube has
air in it, and let 2 be the phase difference when the tube is evacuated. If  is the
wavelength in vacuum, then the wavelength in air is /n, where n is the index of
refraction of air. This means
4 n  1 L
2n 2
1   2  2 L





L
M
N
b g
O
P
Q
where L is the length of the tube. The factor 2 arises because the light traverses the tube
twice, once on the way to a mirror and once after reflection from the mirror. Each shift by
one fringe corresponds to a change in phase of 2 rad, so if the interference pattern shifts
by N fringes as the tube is evacuated, then
b g
4 n  1 L
 2 N .

ANALYZE Solving for n, we obtain
c
c
h
h
60 500  109 m
N
n  1
 1
 100030
.
.
2L
2 5.0  102 m
LEARN The interferometer provides an accurate way to measure the refractive index of
the air (and other gases as well).