Question

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?

Answer 1

Answer:

Explanation:

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