Question

Two narrow slits spaced 100 microns apart are exposed to light of 600 nm. At what angle does the first minimum (dark space) occur in the interference pattern

Answer 1

Answer:

The angle is   $\theta = 0.1719^o$

Explanation:

From the question we are told that

   The  distance of separation is  $d = 100 * 10^{-6} \ m$

    The  wavelength of light is  $\lambda = 600 nm = 600 *10^{-9} \ m$

   

Generally the condition for destructive interference is mathematically represented as

         $dsin(\theta ) =[m + \frac{1}{2} ]\lambda$

Here  m is the order of maxima,  first minimum (dark space) m = 0

 So  

      $100 *10^{-6 } * sin(\theta ) =[0 + \frac{1}{2} ]600 *10^{-9}$

=>   $\theta = sin^{-1} [0.003]$

=>   $\theta = 0.1719^o$