Coherent light with wavelength 601 nm passes through two very narrow slits, and the interference pattern is observed on a screen a distance

Question

Coherent light with wavelength 601 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 a distance of 4.84 mm from the center of the central bright fringe. For what wavelength of light will thefirst-order dark fringe be observed at this same point on the screen?

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Ladonna 1 year 2021-08-09T00:07:57+00:00 1 Answers 122 views 0

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    2021-08-09T00:09:04+00:00

    Answer:

    The wavelength is  [tex]\lambda = 1805 nm[/tex]

    Explanation:

    From the question we are told that

        The wavelength of the light is  [tex]\lambda = 601 \ nm = 601 *10^{-9} \ m[/tex]

         The  distance of the screen is  D  =  3.0  m

         The  fringe width is  [tex]y = 4.84 \ mm = 4.84 *10^{-3} \ m[/tex]

         

    Generally the fringe width for a bright fringe  is mathematically represented as

              [tex]y = \frac{ \lambda * D }{d }[/tex]  

    =>     [tex]d = \frac{ \lambda * D }{ y }[/tex]

    =>     [tex]d = \frac{ 601 *10^{-9} * 3}{ 4.84 *10^{-3 }}[/tex]

    =>     [tex]d = 0.000373 \ m[/tex]

    Generally the fringe width for a dark fringe  is mathematically represented as

          [tex]y_d = [m + \frac{1}{2} ] * \frac{\lambda D }{d }[/tex]

    Here  m = 0  for  first order dark fringe

       So  

             [tex]y_d = [0 + \frac{1}{2} ] * \frac{\lambda D }{d }[/tex]

    looking at which we see that   [tex]y_d = y[/tex]

             [tex]4.84 *10^{-3} = [0 + \frac{1}{2} ] * \frac{\lambda * 3 }{ 0.000373 }[/tex]

    =>    [tex]\lambda = 1805 *10^{-9} \ m[/tex]

    =>    [tex]\lambda = 1805 nm[/tex]

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