Question

When illuminated with monochromatic light, a double slit produces a pattern that is a combination of single-slit diffraction and double-slit interference. This can be easily seen if the separation between the slits and the size of slits are related by simple fractions. Find the ratio of the width of the slits to the separation between them, if the first minimum of the single slit pattern falls on the fifth maximum of the double slit pattern.

Answer 1

Answer:

The ratio is  $k:d = 1 : 5$

Explanation:

From the question we are told that

   The first minimum of the single slit pattern falls on the fifth maximum of the double slit pattern.

Generally the condition for constructive interference for as single slit is  

     $ksin(\theta) = n\lambda$

Here  k is the width of the slit  and n is the order of the fringe and for single slit n =  1 (cause we are considering the first maxima)

Generally the condition for constructive interference for as double slit is    

        $dsin\theta = m\lambda$

Here  d is the separation between the  slit  and m is the order of the fringe and for double slit  m  =  5  (cause we are considering the first maxima)

=>     $dsin\theta = 5\lambda$

So

       $\frac{ksin(\theta)}{dsin(\theta)} = \frac{\lambda}{5\lambda}$

=>    $\frac{k }{d} = \frac{1}{5}$

So  

      $k:d = 1 : 5$