Refractive index of a thin soap film of a uniform thickness is 1.34. Find the smallest thickness of the film that gives in interference maximum in the reflected light when light of wavelength `5360 Å` fall at normal incidence.

Explanation

Condition for maximum interference in the reflected light, in case of thin film interference, can be expressed as
`2 mu t cos r = (2 n - 1) lamda//2, n = 1, 2,…`
[for plane parallel films]
where `mu` is the refractive index of film relative ot the surrounding:
`t` is the thickness of film , and
r is the angle of refraction.
For normal incidence, `r = 0`
`:. 2mu t = (2n - 1) lambda//2, n = 1, 2, 3 ,...`
For minimum thickness, `n = 1`
`:. 2 mu t_(min) = lambda//2` or `t_(min) = (lambda)/(4 mu)`
Given: `lambda = 5360 Å` and `mu = 1.34`
`:. t_(min) = (5360)/(4 xx 1.34)` or `t_(min) = 1000`.