Question

As an admirer of Thomas Young, you perform a double-slit experiment in his honor. You set your slits 1.15 mm apart and position your screen 3.23 m from the slits. Although Young had to struggle to achieve a monochromatic light beam of sufficient intensity, you simply turn on a laser with a wavelength of 633 nm . How far on the screen are the first bright fringe and the second dark fringe from the central bright fringe?

Answer 1

Answer:

The distance of the first bright fringe is given as  $Y_C = 1.22 *10^{-3}m$

The distance of the second dark fringe from the central bright fringe is given as  $Y_D = 0.00192 \ m$

Explanation:

From the question we are told that

     The slit separation distance is  $d = 1.15 mm = \frac{1.15}{1000} =0.00115 m$

      The distance of the slit from the screen is  $D = 3.23 m$

        The wavelength is $\lambda = 633 nm$

For constructive interference to occur the distance between the two slit is mathematically represented as

            $Y_C =\frac{m \lambda D}{d}$

 Where m is the order of the fringe which has a value of 1 for first bright fringe

    Substituting  values

               $Y_C = \frac{1 * 633 *0^{-9} * 3.23}{0.00115}$

                $Y_C = 1.22 *10^{-3}m$

For destructive  interference to occur the distance between the two slit is mathematically represented as

            $Y_D = [n + \frac{1}{2} ] \frac{\lambda D}{d}$

      m = 2

so the formula to get the dark fringe is $n = \frac{1}{2} * 1$

                                                                 $n=1$

 Now substituting values

                 $Y_D = [ 1 + \frac{1}{2} ] * \frac{633 *10^{-9} * 3.23 }{0.00115}$

                   $Y_D =1.5 * \frac{633 *10^{-9} * 3.23 }{0.00115}$

                        $Y_D = 0.00192 \ m$