Transcript Chapter 25: Interference and Diffraction

Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Chapter 25: Interference and
Diffraction
•Constructive and Destructive Interference
•The Michelson Interferometer
•Thin Films
•Young’s Double Slit Experiment
•Gratings
•Diffraction
•Resolution of Optical Instruments
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
§25.1 Constructive and Destructive
Interference
Two waves are coherent if they maintain a fixed phase
relationship (waves from the same source). Two waves
are incoherent otherwise (waves from different sources).
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Constructive interference occurs when two waves are in
phase. To be in phase, the points on the wave must have
=(2)m, where m is an integer.
When coherent waves are in phase, the resulting amplitude
is just the sum of the individual amplitudes. The energy
content of a wave depends on A2. Thus, IA2.
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The resulting amplitude and intensity are:
A  A1  A2
I  I1  I 2  2 I 1 I 2
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Destructive interference occurs when two waves are a half
cycle out of phase. To be out of phase the points on the
wave must have =(2)(m+½), where m is an integer.
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The resulting amplitude and intensity are:
A  A1  A2
I  I1  I 2  2 I1 I 2
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Coherent waves can become out of phase if they travel
different distances to the point of observation.
P
S1

d
S2
This represents the extra path length
(l) that the wave from S2 must
travel to reach point P.
l  d sin 
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When both waves travel in the same medium the
interference conditions are:
For constructive interference
l  m
where m = an integer.
1

For destructive interference l   m  
2

where m = an integer.
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Example (text problem 25.1): A 60.0 kHz transmitter sends
an EM wave to a receiver 21 km away. The signal also
travels to the receiver by another path where it reflects from
a helicopter. Assume that there is a 180 phase shift when
the wave is reflected.
(a) What is the
wavelength of
this EM wave?
c 3.0 105 km/sec
 
 5.0 km
3
f
60 10 Hz
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Example continued:
(b) Will this situation give constructive interference,
destructive inference, or something in between?
The path length difference is l = 10 km = 2, a whole
number of wavelengths. Since there is also a 180 phase
shift there will be destructive interference.
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§25.2 Michelson Interferometer
In the Michelson interferometer, a beam of coherent light is
incident on a beam splitter. Half of the light is transmitted
to mirror M1 and half is reflected to mirror M2.
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The beams of light are reflected by the mirrors, combined
together, and observed on the screen.
If the arms are of different lengths, a phase difference
between the beams can be introduced.
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Example (text problem 25.12): A Michelson interferometer is
adjusted so that a bright fringe appears on the screen. As
one of the mirrors is moved 25.8 m, 92 bright fringes are
counted on the screen. What is the wavelength of the light
used in the interferometer?
Moving the mirror a distance d introduces a path length
difference of 2d. The number of bright fringes (N)
corresponds to the number of wavelengths in the extra
path length.
N  2 d
2d

 0.561 m
N
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§25.3 Thin Films
When an incident light ray reflects from a boundary with a
higher index of refraction, the reflected wave is inverted (a
180° phase shift is introduced).
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A light ray can be reflected many times within a medium.
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Example (text problem 25.18): A thin film of oil (n=1.50) of
thickness 0.40 m is spread over a puddle of water (n=1.33).
For which wavelength in the visible spectrum do you expect
constructive interference for reflection at normal incidence?
Water
Oil
Air
Incident
wave
Consider the first two
reflected rays. r1 is
from the air-oil
boundary and r2 is from
the oil-water boundary.
r1 has a 180 phase shift (noil >nair),
but r2 does not (noil<nwater).
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Example continued:
To get constructive interference, the reflected waves must
be in phase. For this situation, this means that the wave
that travels in oil must travel an extra path equal to multiples
of half the wavelength of light in oil.
The extra path distance traveled is 2d, where d is the
thickness of the film. The condition for constructive
interference here is:
1
1  air 



2d   m  oil   m  
2
2  noil 


2dnoil
 air 
Only the wavelengths that
1

m  
satisfy this condition will have
2


constructive interference.
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Example continued:
Make a table:
m air(m)
0
1
2
3
4
2.40
0.80
0.48
0.34
0.27
All of these wavelengths will show
constructive interference, but it is
only this one that is in the visible
portion of the spectrum.
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§25.4 Young’s Double-Slit
Experiment
Place a source of coherent light
behind a mask that has two vertical
slits cut into it. The slits are L tall,
their centers are separated by d,
and their widths are a.
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The slits become sources of
waves that, as they travel
outward, can interfere with
each other.
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The pattern seen on the screen
There are
alternating
bright/dark
spots.
An
intensity
trace
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The bright spots occur where there is constructive
interference:
l  d sin   m
where m is an integer
and is called the “order”.
The dark spots occur where there is destructive interference:
1

l  d sin    m  
2

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Example (text problem 25.28): Show that the interference
fringes in a double-slit experiment are equally spaced on a
distant screen near the center of the interference pattern.
The condition for constructive
interference isL
l  d sin   m
m
 sin  
d
From the geometry of the
problem,
h
tan  
D
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Example continued:
The screen is far away compared to the distance between
the slits (D>>d) so tan  sin  . Here,
m
h
sin    
and tan    
d
D
m h


d
D
mD
h
d
The distance between two adjacent minima is:
h2  h1 
D
d
m2  m1  
D
d
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§25.5 Gratings
A grating has a large
number of evenly spaced,
parallel slits cut into it.
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Example (text problem 25.38): Red light with =650 nm can
be seen in three orders in a particular grating. About how
many rulings per cm does this grating have?
For each of the maxima
d sin   m
d sin  0  0
d sin 1  1
d sin  2  2
d sin  3  3
d sin  4  4
Third order is observed.
This order is not
observed.
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Example continued:
Since the m = 4 case is not observed, it must be that
sin4>1. We can then assume that 390°. This gives
d  3  1.95 10 6 m
and
1
N   510,000 lines/m  5100 lines/cm.
d
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§25.6-7 Diffraction
Using Huygens’s principle: every
point on a wave front is a source
of wavelets; light will spread out
when it passes through a narrow
slit.
Diffraction is appreciable only when
the slit width is nearly the same size
or smaller than the wavelength.
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The
intensity
pattern on
the screen.
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The minima occur when:
a sin   m
where m = 1, 2,…
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Example (text problem 25.49): Light from a red laser passes
through a single slit to form a diffraction pattern on a distant
screen. If the width of the slit is increased by a factor of two,
what happens to the width of the central maximum on the
screen?
The central maximum occurs between =0 and  as
determined by the location of the 1st minimum in the
diffraction pattern:
a sin   m


Let m =+1 and assume
that  is small.
a
From the previous picture,  only determines the half-width
of the maximum. If a is doubled, the width of the
maximum is halved.
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
§26.8 Resolution of Optical
Instruments
The effect of diffraction is to spread light out. When viewing
two distant objects, it is possible that their light is spread out
to where the images of each object overlap. The objects
become indistinguishable.
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For a circular aperture, the
Rayleigh criterion is:

a sin   1.22
where a is the aperture size of your instrument,  is
the wavelength of light used to make the observation,
and  is the angular separation between the two
observed bodies.
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a sin   1.22
To resolve a pair of objects, the
angular separation between
them must be greater than the
value of .
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Example (text problem 25.56): The radio telescope at
Arecibo, Puerto Rico, has a reflecting spherical bowl of 305 m
diameter. Radio signals can be received and emitted at
various frequencies at the focal point of the reflecting bowl.
At a frequency of 300 MHz, what is the angle between two
stars that can barely be resolved?
a sin   1.22

8
3
.
0

10
m/s
1.22
1.22
sin  

a
sin   4.110 3
  0.23 degrees

300 10 Hz
6
300 m
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Summary
•Conditions for Constructive/Cestructive Interference
•Thin Films
•Young’s Experiment
•Gratings
•Diffraction
•Rayleigh Criterion
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