Question

When a monochromatic light of wavelength 433 nm incident on a double slit of slit separation 6 µm, there are 5 interference fringes in its central maximum. How many interference fringes will be in the central maximum of a light of wavelength 632.9 nm for the same double slit?

Answer 1

Answer:

The number of interference fringes is  $n = 3$

Explanation:

From the question we are told that

     The wavelength is  $\lambda = 433 \ nm = 433 *10^{-9} \ m$

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

       The  order of maxima is m =  5

       

The  condition for constructive interference is

       $d sin \theta = n \lambda$

=>     $\theta = sin^{-1} [\frac{5 * 433 *10^{-9}}{ 6 *10^{-6}} ]$

=>    $\theta = 21.16^o$

So at  

      $\lambda_1 = 632.9 nm = 632.9*10^{-9} \ m$

   $6 * 10^{-6} * sin (21.16) = n * 632.9 *10^{-9}$

=>    $n = 3$