Monochromatic light of wavelength 491 nm illuminates two parallel narrow slits 6.93 μm apart. Calculate the angular deviation of the third-o

Question

Monochromatic light of wavelength 491 nm illuminates two parallel narrow slits 6.93 μm apart. Calculate the angular deviation of the third-order (for m = 3) bright fringe (a) in radians and (b) in degrees.

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Hải Đăng 5 years 2021-08-01T04:30:16+00:00 1 Answers 17 views 0

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  1. Sigrid
    0
    2021-08-01T04:32:01+00:00
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    To solve this problem we will apply the concepts related to the double slit experiment. Here we test a relationship between the sine of the deviation angle and the distance between slit versus wavelength and the bright fringe order. Mathematically it can be described as,

    dsin\theta = m\lambda

    Here,

    d = Distance between slits

    m = Any integer which represent the order number or the number of repetition of the spectrum

    \lambda = Wavelength

    \theta = Angular deviation

    Replacing with our values we have,

    (6.93*10^{-6}) sin\theta = (3)(491*10^{-9})

    \theta = sin^{-1} (\frac{(3)(491*10^{-9}}{6.93*10^{-6}) })

    Part A)

    \theta = 0.2141rad

    PART B)

    \theta = 0.2141rad(\frac{360\°}{2\pi rad})

    \theta = 12.27\°

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