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

Answer 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

      $Z = 2 * 0.0212$

     $Z = 0.0424 \ m$