Question

A thin film with an index of refraction of 1.50 is placed in one of the beams of a Michelson interferometer. If this causes a shift of 8 bright fringes in the pattern produced by light of wavelength 540 nm, what is the thickness of the film?

Answer 1

The thickness of the film is 4.32 μm.

Step-by-step explanation:

Given;

index of refraction of the thin film on one beam, n₂ = 1.5

number of bright fringes shift in the pattern produced by light, ΔN = 8

wavelength of the Michelson interferometer, λ = 540 nm

The thickness of the film will be calculated as;

`\delta N = (2L)/(\lambda) (n_2 -n_1)`

where;

n₁ and n₂ are the index of refraction on the beam

L is the thickness of the film

`\delta N = (2L)/(\lambda) (n_2 -n_1)\\\\L = (\lambda)/(2) ((N)/(n_2-n_1) )\\\\L = (540*10^(-9))/(2) ((8)/(1.5-1) )\\\\L = 4.32*10^(-6) \ m\\\\L = 4.32 \mu m`

Therefore, the thickness of the film is 4.32 μm.