Question

You illuminate a slit with a width of 77.7 μm with a light of wavelength 721 nm and observe the resulting diffraction pattern on a screen that is situated 2.83 m from the slit. What is the width, in centimeters, of the pattern’s central maximum

Answers

  1. Answer:

    The width is  Z = 0.0424 \ m

    Explanation:

    From the question we are told that

        The width of the slit is d = 77.7 \mu m = 77.7 *10^{-6} \ m

        The wavelength of the light is  \lambda = 721 \ nm

          The position of the screen is  D = 2.83 \ m

    Generally angle at which the first minimum  of the interference pattern the  light occurs  is mathematically  represented as

            \theta = sin ^{-1}[\frac{m \lambda}{d} ]

    Where m which is the order of the interference is 1

    substituting values

           \theta = sin ^{-1}[\frac{1 *721*10^{-9}}{ 77.7*10^{-6}} ]

          \theta = 0.5317 ^o

     Now the width of first minimum  of the interference pattern is mathematically evaluated as

           Y = D sin \theta

    substituting values

           Y = 2.283 * sin (0.5317)

           Y = 0.02 12 \ m

     Now the width of  the  pattern’s central maximum is mathematically evaluated as

            Z = 2 * Y

    substituting values

          Z = 2 * 0.0212

         Z = 0.0424 \ m

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