Question

Light of wavelength 476.1 nm falls on two slits spaced 0.29 mm apart. What is the required distance from the slits to the screen if the spacing between the first and second dark fringes is to be 4.2 mm?

Answer 1

Answer:

The distance is $D = 2.6 \ m$

Explanation:

From the question we are told that

    The wavelength of the light is  $\lambda = 476.1 \ nm = 476.1 *10^{-9} \ m$

      The  distance between the slit is  $d = 0.29 \ mm = 0.29 *10^{-3} \ m$

       The  between the first and second dark fringes is  $y = 4.2 \ mm = 4.2 *10^{-3} \ m$

Generally  fringe width is mathematically represented as

       $y = \frac{\lambda * D }{d}$

Where D is the distance of the slit to the screen

   Hence

        $D = \frac{y * d}{\lambda }$

substituting values

       $D = \frac{ 4.2 *10^{-3} * 0.29 *10^{-3}}{ 476.1 *10^{-9} }$

        $D = 2.6 \ m$