Obtain the equaitons for constructive and destructive interference for transmitted and reflected waves in thin films.
Interference in thin films:

Explanation

(i) Consider a thin film of transparent material of refractive index `mu` (not to confuse with order of fringe n) and thickness d. A parallel beam of light is incident on the film at an angle i. (ii) The wave is divided into two parts at the dark fringes upper surface, one is reflected and the other is refracted. (iii) The refracted part, which enters into the film, again gets divided at the lower surface into two parts, one is transmitted out of the film and the other is reflected back in to the film.
For transmitted light: (i) The light transmitted may interfere to produce a resultant intensity. Consider the path difference between the two light waves transmitted from B and D. (ii) The two waves moved together and remained in phase up to B where splitting occurred. The extra path travelled by the wave transmitted from D is the path inside Interference fringe pattern the film, BC + CD. (iii) If we approximate the incidence to be nearly normal (i = 0), then the points B and D are very close to each other. (iv) The extra distance travelled by the wave is approximately twice thickness of the film, BC + CD = 2d. As this extra path is traversed in a medium of refractive index `mu` , the optical path difference is, `delta = 2 mu d ` (v) The condition for constructive interference in transmitted ray is,
` 2mu d = n lamda`
(vi). Similarly, the condition for destructive interference in transmitted ray is,
` 2 mu = (2n - 1) lamda/2`
For reflected light: ` 2mu d + lamda/2 = n lamda " or " 2 mu d = (2n - 1) lamda/2`
(vi) The additional path difference `lamda/2` is due to the phase change of it in rarer to denser reflection taking place at A. (vii) The condition for destructive interference for reflected ray is,
` 2 mu d + lamda/2 = (2n + 1) lamda/2 " or " 2 mu d = n lamda` .