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 bri

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?

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Diễm Kiều 4 years 2021-08-07T19:27:06+00:00 1 Answers 7 views 0

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  1. minhkhue
    0
    2021-08-07T19:29:05+00:00
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    Answer:

    The thickness of the film is 4.32 μm.

    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 = \frac{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 = \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

    Therefore, the thickness of the film is 4.32 μm.

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