37
Interference of Light Waves
CHAPTER OUTLINE
37.1
Conditions for Interference
37.2
Young’s DoubleSlit Experiment
37.3
Light Waves in Interference
37.4
Intensity Distribution of the
DoubleSlit Interference Pattern
37.5
Change of Phase Due to
ReF ection
37.6
Interference in Thin ±ilms
37.7
The Michelson Interferometer
ANSWERS TO QUESTIONS
Q37.1
(a)
Two waves interfere constructively if their
path difference is zero, or an integral multiple
of the wavelength, according to
δ
λ
=
m
, with
m
=
0123
, , , ,.
.. .
(b)
Two waves interfere destructively if their path
difference is a half wavelength, or an odd
multiple of
2
, described by
δλ
=+
⎛
⎝
⎞
⎠
m
1
2
, with
m
=
Q37.2
The light from the ﬂ
ashlights consists of many different
wavelengths (that’s why it’s white) with random time
differences between the light waves. There is no
coherence
between the two sources. The light from
the two ﬂ
ashlights does not maintain a constant phase
relationship over time. These three equivalent statements
mean no possibility of an interference pattern.
*Q37.3
(i) The angles in the interference pattern are controlled by
/
d
, which we estimate in each case:
(a) 0.45
µ
m/400
m
≈
1.1
×
10
−
3
(b) 0.7
m/400
m
≈
1.6
×
10
−
3
(c) and (d) 0.7
m/800
μ
m
≈
0.9
×
10
−3
. The ranking is
b
>
a
>
c
=
d.
(ii) Now we consider
L
/
d
:
(a) 4 m (0.45
m/400
m)
≈
4.4 mm
(b) 4 m (0.7
m/400
m)
≈
7 mm
(c) 4 m(0.7
m/800
m)
≈
3 mm
(d) 8 m(0.7
m/800
m)
≈
7 mm.
The ranking is
b
=
d
>
a
>
c.
*Q37.4
Yes.
A single beam of laser light going into the slits divides up into several fuzzyedged beams
diverging from the point halfway between the slits.
*Q37.5
Answer (c). Underwater, the wavelength of the light decreases according to
water
air
water
=
n
. Since
the angles between positions of light and dark bands are proportional to
,
the underwater fringe
separations decrease.
Q37.6
Every color produces its own pattern, with a spacing between the maxima that is characteristic
of the wavelength.
With white light, the central maximum is white.
The F
rst side maximum is
a full spectrum with violet on the inner edge and red on the outer edge on each side.
Each side
maximum farther out is in principle a full spectrum, but they overlap one another and are hard to
distinguish.
Using monochromatic light can eliminate this problem.
*Q37.7
With two F
ne slits separated by a distance
d
slightly less than
, the equation
d
sin
θ
=
0 has
the usual solution
=
0.
But
d
sin
=
1
has no solution. There is no F
rst side maximum.
d
sin
=
(1/2)
has a solution. Minima ﬂ
ank the central maximum on each side.
Answer
(b).
353
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document354
Chapter 37
Q37.8
As water evaporates from the “soap” bubble, the thickness of the bubble wall approaches
zero. Since light reﬂ
ecting from the front of the water surface is phaseshifted 180
°
and light
reﬂ
ecting from the back of the soap F
lm is phaseshifted 0
°
, the reﬂ
ected light meets the
conditions for a minimum. Thus the soap F
lm appears black, as in the textbook illustration
accompanying this question.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '11
 williams,frank
 Light, Wavelength, Sin, Light waves, NM

Click to edit the document details