A thin film with an index of refraction of 1.60 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 580 nm, what is the thickness of the film

3.867 μm

Explanation:

The index of refraction, μ = 1.6

Wavelength of the light, λ = 580 nm

N2 - N1 = (2L / λ) (n2 - n1), Making L subject of formula, we have

(N2 - N1) λ = 2L (n2 - n1)

L = [(N2 - N1) * λ] / 2(n2 - n1)

L = (8 * 580) / 2(1.6 - 1.0)

L = 4640 nm / 1.2

L = 3867 nm or 3.867 μm

Therefore we can come to the conclusion that the thickness of the film is 3.867 nm

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