Auniform thin film of n=1.33 is on top of glass (n=1.5). When monochromatic light with tunable wavelength is directed from above, the reflected light is minimum when λ=505nm, and maximum when λ=650nm. What is the minimum thickness of the film?

There is reflection (two times) from upper and lower surface of the film. In both cases, reflection is from low to high density medium so there is change in phase of 180 twice.

So for constructive interference

2μd = n λ₁, d is thickness required, λ is wavelength n₁ is order of bright fringe

For destructive interference (minimum light)

2μd = (2n+1) λ₂/2

n λ₁ = (2n+1) λ₂/2

(2n+1) / 2n = λ₁ / λ₂

= 650 / 505

= 5 / 4 (approx)

2n = 4

n = 2

2μd = n λ₁

2 x 1.33 x d = 2 x 650 nm

d = 488.72 nm