Question

How wide is the central diffraction peak on a screen 2.30 m behind a 0.0368-mm-wide slit illuminated by 558-nm light

Answer 1

Answer:

The  value $y = 0.0349 \ m$

Explanation:

From the question we are told that

   The  distance of the screen is  $D = 2.30 \ m$

   The  width of the slit is  $d = 0.0368 \ nm = 0.0368 *10^{-3} \ m$

   The  wavelength is  $\lambda = 558 \ nm = 558 *10^{-9} \ m$

The  width of the central diffraction peak is  mathematically represented as

        $k = 2 * y$

Where  y is the distance from the center to the high peak which  is mathematically represented as

       $y = \frac{\lambda * D }{d }$

substituting values

      $y = \frac{ 558 *10^{-8} * 2.30 }{0.0368 *10^{-3} }$

      $y = 0.0349 \ m$