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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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§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, IA2.
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The resulting amplitude and intensity are:
2121
21
2 IIIII
AAA
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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:
2121
21
2 IIIII
AAA
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Coherent waves can become out of phase if they travel different distances to the point of observation.
This represents the extra path length (l) that the wave from S2 must travel to reach point P.
sindl
P
d
S2
S1
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For constructive interference ml
where m = an integer.
For destructive interference
2
1ml
where m = an integer.
When both waves travel in the same medium the interference conditions are:
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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?
km 0.5Hz 1060
km/sec 100.33
5
f
c
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(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.
Example continued:
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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.
m 561.02
2
N
d
dN
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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?
AirWater Oil
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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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:
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2
2
1
2
12
oilair
oil
airoil
m
dn
nmmd
Only the wavelengths that satisfy this condition will have constructive interference.
Example continued:
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Make a table:
m air(m)
0 2.40
1 0.80
2 0.48
3 0.34
4 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.
Example continued:
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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:
mdl sin
where m is an integer and is called the “order”.
The dark spots occur where there is destructive interference:
2
1sin mdl
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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
d
m
mdl
sin
sin
From the geometry of the problem,
D
htan
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The screen is far away compared to the distance between the slits (D>>d) so tan sin . Here,
d
Dmh
D
h
d
mD
h
d
m
tan and sin
The distance between two adjacent minima is:
d
Dmm
d
Dhh
1212
Example continued:
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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
4sin
3sin
2sin
1sin
0sin
4
3
2
1
0
d
d
d
d
d
This order is not observed.
Third order is observed.
md sin
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Since the m = 4 case is not observed, it must be that sin4>1. We can then assume that 390°. This gives
m 1095.13 6 d
and lines/cm. 5100lines/m 000,5101
d
N
Example continued:
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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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ma sinThe minima occur when:
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?
a
ma
sin
The central maximum occurs between =0 and as determined by the location of the 1st minimum in the diffraction pattern:
Let m =+1 and assume that is small.
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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§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:
22.1sin a
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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22.1sin aTo 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?
degrees 23.0
101.4sin
m 300Hz 10300
m/s 100.322.122.1sin
22.1sin
3
6
8
a
a
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Summary
•Conditions for Constructive/Cestructive Interference
•Thin Films
•Young’s Experiment
•Gratings
•Diffraction
•Rayleigh Criterion
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