Coherent light with wavelength of 580 nm passes through two very narrow slits, and the interference pattern is observed on a screen a distan

Question

Coherent light with wavelength of 580 nm passes through two very narrow slits, and the interference pattern is observed on a screen a distance of 3.00 m from the slits. The first-order bright fringe is at 4.81 mm from the center of the central bright fringe. For what wavelength of light will the first-order dark fringe be observed at this same point on the screen?

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Farah 4 years 2021-09-01T01:08:38+00:00 1 Answers 4 views 0

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  1. nguyetanh
    0
    2021-09-01T01:10:22+00:00
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    Answer:

    the required wavelength is 1.15815 μm

    Explanation:

    Given the data in the question;

    The position of bright fringes y_m on screen in double slit experiment is expressed as follows;

    y_m = mλD / d

    solving for d, we substitute 1 for m

    y₁ = (1)λD / d

    d = λD / y₁

    given that λ = 580 nm = 5.8 × 10⁻⁷ m,  D = 3.00 m and y₁= y₀ = 4.81 mm = 0.00481 m

    so we substitute

    d = λD / y₁

    d = ( 5.8 × 10⁻⁷ m × 3.00 m ) / 0.00481 m

    d = 0.00000174 m² / 0.00481 m

    d = 3.6117 × 10⁻⁴ m

    Now, position of dark fringe  y_m on screen in double slit experiment is expressed as;

    y_m = ( m + 1/2 )λD / d

    we substitute 0 for m

    y₀ = ( 0 + 1/2 )λD / d

    y₀ = λD / 2d

    2y₀d = λD

    λ =  2y₀d  / D

    we substitute

    λ =  ( 2(0.00481 m) ( 3.6117 × 10⁻⁴ m) )  / 3.0 m

    λ = 1.15815 × 10⁻⁶ m

    λ = 1.15815 μm

    Therefore, the required wavelength is 1.15815 μm

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