INTERFERENCE DIVISION OF AMPLITUDE Interference of waves occurs

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 S 1 and S 2 are two coherent sources in air. Q The path difference is S 2 Q - S 1 Q. For constructive interference to take place at Q, the waves must be in S 1 phase at Q. Hence the path difference must be a whole S 2 number of wavelengths. (S 2 Q - S 1 Q) = mλ 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 S 1 must meet a ‘trough’ from S 2). (S 2 Q - S 1 Q) = (m +1/2 )λ

Optical Path difference Consider two coherent beams S 1 and S 2 where S 1 P is in air and S 2 P 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 S 1 P - S 2 P = 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. If S 1 P is exactly 3 wavelengths then S 2 P 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 = n 1 d – n 2 d Conditions for maxima and minima are as follows: (n 1 -n 2)d = mλ And (constructive) (n 1 -n 2)d = (m+1/2)λ (destructive)

Example Two beams of microwaves of wavelength 6. 00 x 10 -3 m are emitted by a source. They both travel 5 cm 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. 027 m Number of wavelengths = 0. 027 /λ = 0. 027 / 6. 00 x 10 -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.