When a monochromatic light of wavelength 433 nm incident on a double slit of slit separation 6 µm, there are 5 interference fringes in its c

Question

When a monochromatic light of wavelength 433 nm incident on a double slit of slit separation 6 µm, there are 5 interference fringes in its central maximum. How many interference fringes will be in the central maximum of a light of wavelength 632.9 nm for the same double slit?

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Nick 5 years 2021-08-10T07:44:41+00:00 1 Answers 26 views 0

Answers ( )

  1. thuthuy
    0
    2021-08-10T07:46:18+00:00
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    Answer:

    The number of interference fringes is  n = 3

    Explanation:

    From the question we are told that

         The wavelength is  \lambda = 433 \ nm = 433 *10^{-9} \ m

          The distance of separation is  d = 6 \mu m = 6 *10^{-6} \ m

           The  order of maxima is m =  5

           

    The  condition for constructive interference is

           d sin \theta = n \lambda

    =>     \theta = sin^{-1} [\frac{5 * 433 *10^{-9}}{ 6 *10^{-6}} ]

    =>    \theta = 21.16^o

    So at  

          \lambda_1 = 632.9 nm = 632.9*10^{-9} \ m

       6 * 10^{-6} * sin (21.16) = n * 632.9 *10^{-9}

    =>    n = 3

       

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