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21 September, 20:53

Light of wavelength 633nm passes through a single slit of width 1.50*10^-5 m. At what angle does the second interference minimum (m=2) occur? °

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  1. 21 September, 22:21
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    Answer: The second interference is minimum at angle 5 degree approximately

    Explanation:

    Given that the

    Wavelength (λ) = 633nm

    Width d = 1.5 * 10^-5m

    n = 2

    The wavelength (λ) of the light passing through a single slit is related to the angle Ø by

    dsinØ = n (λ)

    Substitutes all the parameters into the above formula

    1.5*10^-5 * SinØ = 2 * 633 * 10^-9

    Make SinØ the subject of formula

    SinØ = 1.266*10^-6/1.5*10^-5

    SinØ = 0.0844

    Ø = Sin^-1 (0.0844)

    Ø = 4.84 degree.

    The second interference is minimum at angle 5 degree approximately
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