How wide is the central diffraction peak on a screen 2.20 mm behind a 0.0328-mmmm-wide slit illuminated by 588-nmnm light?

Question

How wide is the central diffraction peak on a screen 2.20 mm behind a 0.0328-mmmm-wide slit illuminated by 588-nmnm light?

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Gerda 3 years 2021-08-09T17:58:40+00:00 1 Answers 63 views 0

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    2021-08-09T18:00:10+00:00

    Answer:

    y = 0.0394 \ m

    Explanation:

    From the question we are told that

            The  distance of the screen is  D  = 2.20 \ m

           The distance of separation of the slit is  d =  0.0328 \ mm  =  0.0328*10^{-3} \ m

            The  wavelength of light is  \lambda =  588 \ nm  =  588 *10^{-9} \ m

    Generally the condition for constructive interference is

                dsin\theta  =  n *  \lambda

    =>        \theta  =  sin^{-1} [ \frac{ n *  \lambda }{d } ]

    here n = 1 because we are considering the central diffraction peak

    =>        \theta  =  sin^{-1} [ \frac{ 1 *  588*10^{-9} }{0.0328*10^{-3} } ]

    =>       \theta  =  1.0274 ^o

    Generally the width of central diffraction peak on a screen is mathematically evaluated as

               y =  D tan (\theta )

    substituting values

            y = 2.20 *  tan (1.0274)

            y = 0.0394 \ m

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