Light of wavelength 614 nm is incident perpendicularly on a soap film (n = 1.33) suspended in air. What are the (a) least and (b) second lea

Question

Light of wavelength 614 nm is incident perpendicularly on a soap film (n = 1.33) suspended in air. What are the (a) least and (b) second least thicknesses of the film for which the reflections from the film undergo fully constructive interference?

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Thu Hương 2 months 2021-08-01T07:19:56+00:00 1 Answers 3 views 0

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    2021-08-01T07:21:34+00:00

    Answer:

    The least and second least thicknesses of the film are 0.115 um and 0.346 um respectively.

    Explanation:

    Optical path length ===> 2n * t = (m + 0.5) * λ

    λ = 614 nm , n = 1.33

    Substitute in the parameters in the equation.

    ∴ 2(1.33) * t = (m + 0.5) * 614

      2.66 * t = 614m + 307

      t = (614m + 307) / 2.66 ………….(1)

    (a) for m = 0

        t = (614m + 307) / 2.66

        t = (614(0) + 307) / 2.66

        t = 307 / 2.66

        t = 115 nm == 0.115 um

    (b) for m = 1

        t = (614(1) + 307) / 2.66

        t = (614 + 307) / 2.66

        t = 921 / 2.66

        t = 346.24 nm = 0.346 um

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