INTERFERENCEINTERFERENCE

DIVISION OF AMPLITUDEDIVISION OF AMPLITUDE

Interference of waves occurs when waves overlap. Interference of waves occurs when waves overlap. There are There are

two ways to produce an interference pattern for light:two ways to produce an interference pattern for light:

1.1. Division of amplitudeDivision of amplitude2.2. Division of wavefrontDivision of wavefront

Both of these involve splitting the light from a single source intoBoth of these involve splitting the light from a single source intotwo beams. two beams.

Division of amplitudeDivision of amplitudeThis involves splitting a single light beam into two beams, aThis involves splitting a single light beam into two beams, areflected beam and a transmitted beam, at a surface betweenreflected beam and a transmitted beam, at a surface betweentwo media of different refractive index.two media of different refractive index.

General Properties of InterferenceGeneral Properties of Interference

Coherent SourcesCoherent SourcesTwo coherent sources must have a Two coherent sources must have a constant phase differenceconstant phase difference. . Hence they will have the Hence they will have the same frequencysame frequency..

To produce an interference pattern for light waves the two, or To produce an interference pattern for light waves the two, or more, overlapping beams always come from the more, overlapping beams always come from the same single same single sourcesource..

Light cannot be produced as a continuous wave but is made up Light cannot be produced as a continuous wave but is made up of photons which are continuously being emitted in bundles.of photons which are continuously being emitted in bundles.

This is not the case for sound waves. We can have two separate This is not the case for sound waves. We can have two separate loudspeakers, connected to the same signal generator, emitting loudspeakers, connected to the same signal generator, emitting the same frequency which will produce an interference pattern.the same frequency which will produce an interference pattern.

Path DifferencePath Difference

Sources SSources S11 and S and S22 are two coherent sources are two coherent sources in airin air..

S1

S2

Q The path difference is The path difference is SS22Q - SQ - S11Q. For Q. For constructive constructive

interferenceinterference to take place at to take place at Q, the waves must be in Q, the waves must be in phase at Q. Hence the path phase at Q. Hence the path difference must be a whole difference must be a whole number of wavelengths.number of wavelengths.

(S(S22Q - SQ - S11Q) = mλQ) = mλ where m = 0, 1, 2, 3, ... (an integer)where m = 0, 1, 2, 3, ... (an integer)

Similarly, for Similarly, for destructive interferencedestructive interference to take place the waves to take place the waves must be out of phase at Q by λ/2 (that is a ‘crest’ from Smust be out of phase at Q by λ/2 (that is a ‘crest’ from S11 must must

meet a ‘trough’ from Smeet a ‘trough’ from S22).).

(S2Q - S1Q) = (m +1/2 )λ(S2Q - S1Q) = (m +1/2 )λ

Optical Path differenceOptical Path differenceConsider two coherent beams SConsider two coherent beams S11 and S and S22 where S where S11P is in air and P is in air and

SS22P is in perspex of refractive index n = 1.5. Point P is in air.P is in perspex of refractive index n = 1.5. Point P is in air.

The The geometric path length (d)geometric path length (d) is the same for both sources. is the same for both sources.So, the So, the geometrical path difference geometrical path difference

SS11P - SP - S22P = 0P = 0

But, the waves may not be in phase. This is because the speed But, the waves may not be in phase. This is because the speed and wavelengthand wavelength of the light is reduced when it enters the of the light is reduced when it enters the perspex.perspex.

perspex

airn

5.1air

perspex

If S1P is exactly 3 wavelengths then S2P would be 1.5 x 3 = 4.5 wavelengths

The The optical path lengthoptical path length must be considered not the geometrical must be considered not the geometrical path length.path length.

Optical path length = refractive index x geometrical path lengthOptical path length = refractive index x geometrical path length

Optical path length = ndOptical path length = nd

The optical path difference can be calculated from:The optical path difference can be calculated from:

Optical path difference = nOptical path difference = n11d – nd – n22dd

Conditions for maxima and minima are as follows:Conditions for maxima and minima are as follows:

(n(n11-n-n22)d = mλ)d = mλ (constructive)(constructive)

AndAnd (n(n11-n-n22)d = (m+1/2)λ)d = (m+1/2)λ (destructive)(destructive)

ExampleExample

Two beams of microwaves of wavelength 6.00x10Two beams of microwaves of wavelength 6.00x10 -3-3m are emitted by a m are emitted by a source. They both travel 5cm to a detector, but one passes though air 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 while the other through quartz of refractive index 1.54. Do they cause constructive or destructive interference?constructive or destructive interference?

Path difference = (nPath difference = (nquartz quartz - n- nairair) x d) x d

= (1.54 - 1) x 0.05= (1.54 - 1) x 0.05

= 0.027m= 0.027m

Number of wavelengths = 0.027 /λNumber of wavelengths = 0.027 /λ

= 0.027 / 6.00x10= 0.027 / 6.00x10-3-3

= 4.5 wavelegnths = 4.5 wavelegnths

So So destructivedestructive interference. interference.

Phase differencePhase differenceThe phase difference is related to the optical path difference:The phase difference is related to the optical path difference:

phase difference = 2π / λ x optical path differencephase difference = 2π / λ x optical path difference

e.g. If the path difference was one wavelength then:e.g. If the path difference was one wavelength then:

Phase difference = 2Phase difference = 2π / λ x λπ / λ x λ

= 2π radians= 2π radians

When the optical path difference is a whole number of When the optical path difference is a whole number of wavelengths, the phase difference is a multiple of 2π, i.e. the wavelengths, the phase difference is a multiple of 2π, i.e. the waves are in phase.waves are in phase.