Transcript INTERFERENCE

INTERFERENCE
DIVISION OF AMPLITUDE
Interference of waves occurs when waves overlap. There are
two ways to produce an interference pattern for light:
1. Division of amplitude
2. Division of wavefront
Both of these involve splitting the light from a single source into
two beams.
Division of amplitude
This involves splitting a single light beam into two beams, a
reflected beam and a transmitted beam, at a surface between
two media of different refractive index.
General Properties of Interference
Coherent Sources
Two coherent sources must have a constant phase difference.
Hence they will have the same frequency.
To produce an interference pattern for light waves the two, or
more, overlapping beams always come from the same single
source.
Light cannot be produced as a continuous wave but is made up
of photons which are continuously being emitted in bundles.
This is not the case for sound waves. We can have two separate
loudspeakers, connected to the same signal generator, emitting
the same frequency which will produce an interference pattern.
Path Difference
Sources S1 and S2 are two coherent sources in air.
Q
S1
S2
(S2Q - S1Q) = mλ
The path difference is
S2Q - S1Q. For constructive
interference to take place at
Q, the waves must be in
phase at Q. Hence the path
difference must be a whole
number of wavelengths.
where m = 0, 1, 2, 3, ... (an integer)
Similarly, for destructive interference to take place the waves
must be out of phase at Q by λ/2 (that is a ‘crest’ from S1 must
meet a ‘trough’ from S2).
(S2Q - S1Q) = (m +1/2 )λ
Optical Path difference
Consider two coherent beams S1 and S2 where S1P is in air and
S2P is in perspex of refractive index n = 1.5. Point P is in air.
The geometric path length (d) is the same for both sources.
So, the geometrical path difference
S1P - S2P = 0
But, the waves may not be in phase. This is because the speed
and wavelength of the light is reduced when it enters the
perspex.
n
air
 perspex
 perspex
air

1.5
If S1P is exactly 3 wavelengths then S2P would be 1.5 x 3 = 4.5
wavelengths
The optical path length must be considered not the geometrical
path length.
Optical path length = refractive index x geometrical path length
Optical path length = nd
The optical path difference can be calculated from:
Optical path difference = n1d – n2d
Conditions for maxima and minima are as follows:
(n1-n2)d = mλ
And
(constructive)
(n1-n2)d = (m+1/2)λ (destructive)
Example
Two beams of microwaves of wavelength 6.00x10-3m are emitted by a
source. They both travel 5cm to a detector, but one passes though air
while the other through quartz of refractive index 1.54. Do they cause
constructive or destructive interference?
Path difference = (nquartz - nair) x d
= (1.54 - 1) x 0.05
= 0.027m
Number of wavelengths = 0.027 /λ
= 0.027 / 6.00x10-3
= 4.5 wavelegnths
So destructive interference.
Phase difference
The phase difference is related to the optical path difference:
phase difference = 2π / λ x optical path difference
e.g. If the path difference was one wavelength then:
Phase difference = 2π / λ x λ
= 2π radians
When the optical path difference is a whole number of
wavelengths, the phase difference is a multiple of 2π, i.e. the
waves are in phase.