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10 June, 15:30

Light of wavelength 630 nm is incident on a long, narrow slit. Determine the angular deflection of the first diffraction minimum if the slit width is (a) 0.020 mm, (b) 0.20 mm, and (c) 2.0 mm.

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  1. 10 June, 15:59
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    a) 1.8°

    b) 0.18°

    c) 0.018°

    Explanation:

    Wavelength (λ) = 630nm = 630 * 10^-9m

    The equation that describes the angular deflection of a dark band is

    Wsin (βm) = mλ

    w = width of the single slit

    λ = wavelength of the light

    βm = angular deflection of the mth dark band.

    a) In order to get the angular deflection of the first dark band for a slit with 0.02mm width, substitute w = 0.02mm = 0.02*10^-3, m = 1, λ = 630*10^-9

    0.02*10^-3 sin (β1) = 1 * 630*10^-9

    Sin (β1) = 630 * 10^-9 / 0.02*10^-3

    Sin (β1) = 0.0315

    β1 = Sin^-1 (0.0315)

    = 1.8°

    b) substitute w = 0.2mm = 0.2*10^-3, m = 1, λ = 630*10^-9

    0.2*10^-3 sin (β1) = 1 * 630*10^-9

    Sin (β1) = 630 * 10^-9 / 0.2*10^-3

    Sin (β1) = 0.00315

    β1 = Sin^-1 (0.00315)

    = 0.18°

    c) substitute w = 2mm = 2*10^-3, m = 1, λ = 630*10^-9

    2*10^-3 sin (β1) = 1 * 630*10^-9

    Sin (β1) = 630 * 10^-9 / 2*10^-3

    Sin (β1) = 3.15*10^-4

    β1 = Sin^-1 (3.15*10^-4)

    = 0.018°
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