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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?

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onyebuchinnaji

Answer:

The thickness of the film is 4.32 ÎŒm.

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;

[tex]\delta N = \frac{2L}{\lambda} (n_2 -n_1)[/tex]

where;

n₁ and n₂ are the index of refraction on the beam

L is the thickness of the film

[tex]\delta N = \frac{2L}{\lambda} (n_2 -n_1)\\\\L = \frac{\lambda}{2} (\frac{N}{n_2-n_1} )\\\\L = \frac{540*10^{-9}}{2} (\frac{8}{1.5-1} )\\\\L = 4.32*10^{-6} \ m\\\\L = 4.32 \mu m[/tex]

Therefore, the thickness of the film is 4.32 ÎŒm.

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