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11 September, 05:13

A thin layer of oil of refractive index 1.22 is spread on the surface of water (n = 1.33), If the thickness of the oil is 275 nm, then what is the wavelength of light in air that will be predomínantly reflected from the top surface of the oil?

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  1. 11 September, 07:09
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    6.71*10⁻⁷ m

    Explanation:

    Using thin film constructive interference formula as:

    2*n*t = m*λ

    Where,

    n is the refractive index of the refracted surface

    t is the thickness of the surface

    λ is the wavelength

    If m = 1

    Then,

    2*n*t = λ

    Given that refractive index pf the oil is 1.22

    Thickness of the oil = 275 nm

    Also, 1 nm = 10⁻⁹ m

    Thickness = 275*10⁻⁹ m

    So,

    Wavelength is:

    λ = 2*n*t = 2 * 1.22 * 275*10⁻⁹ m = 6.71*10⁻⁷ m
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