Reflected Light Interference
Consider a transparent film of thickness t and refractive index 
. A ray SA incident on the upper surface of the film is partly reflected along AT and partly reflected along AB. At B part of it is reflected along BC and finally emerges out along CQ. The difference in path between the two rays AT and CQ can be calculated. Draw CN normal to AT and AM normal to BC. The angle of incidence is i and the angle of refraction is r. Also produce CB to meet AE produced at P. Here ∠ APC = r.
The optical path difference,
In the Δ APM,
cos r = PM/AP
PM = AP.cos r = (AE + EP) cos r
= 2t cos r (∵ AE = EP = t)
This equation (i), in the case of reflected light does not represent the correct path difference but only the apparent. It has been established on the basis of electromagnetic theory that, when light is reflected from the surface of an optically denser medium (air-medium interface) a phase change π, equivalent to a path difference λ/2, occurs.
Therefore, the correct path difference in this case,
1. If the path difference x = nλ, where n = 0, 1, 2, 3, 4, …. etc. constructive interference takes place and the film appears bright.
2. If the path difference x = (2n + 1) λ/2 where n = 0, 1, 2, 3, 4, …. etc. destructive interference takes place and the film appears dark.
Here n is an integer only, therefore (n + 1) can also be taken as n.
Where, n = 0, 1, 2, 3, 4………etc.
It should be remembered that the interference pattern will not be perfect because the intensities of the rays AT and CQ will not be the same and their amplitudes are different. The amplitudes will depend on the amount of light reflected and transmitted through the films. It has been found that for normal incidence, about 4% of the incident light is reflected and 96% is transmitted. There is small difference in the amplitudes of the rays AT and CQ. Therefore, the intensity never vanishes completely and perfect dark fringes will not be observed for the rays AT and CQ alone. But in the case of multiple reflection, the intensity of the minima will be zero.
Consider reflected rays 1, 2, 3 etc. the amplitude of the incident ray is a. Let r be the reflection coefficient, t the transmission coefficient from rarer to denser medium and t’ the transmission coefficient from denser to rarer medium.
The amplitudes of the reflected rays are ar, atrt’, atr3’t’, ….etc. The ray 1 is reflected at the surface of a denser medium. It undergoes a phase change π. The rays 2, 3, 4 etc. are all in phase but out of phase with ray 1 by π.
The resultant amplitude of 2, 3, 4, etc. is given by
As r is less than 1, the terms inside the bracket form a geometric series.
According to the principle of reversibility.
tt’ = 1 – r2
Thus, the resultant amplitude of 2, 3, 4, ….etc. is equal in magnitude of the amplitude of ray 1 but out of phase with it. Therefore the minima of reflected system will be zero intensity.
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