Question

An interference pattern is produced by light with a wavelength 550 nm from a distant source incident on two identical parallel slits separated by a distance (between centers) of 0.500 mm .
a. If the slits are very narrow, what would be the angular position of the second- order, two-slit interference maxima?
b. Let the slits have a width 0.300 mm. In terms of the intensity lo at the center of the central maximum, what is the intensity at the angular position in part "a"?

Answer 1

Answer:

a

 $\theta = 0.0022 rad$

b

 $I = 0.000304 I_o$

Explanation:

From the question we are told that  

   The  wavelength of the light is $\lambda = 550 \ nm = 550 *10^{-9} \ m$

    The  distance of the slit separation is  $d = 0.500 \ mm = 5.0 *10^{-4} \ m$

 

Generally the condition for two slit interference  is  

     $dsin \theta = m \lambda$

Where m is the order which is given from the question as  m = 2

=>    $\theta = sin ^{-1} [\frac{m \lambda}{d} ]$

 substituting values  

      $\theta = 0.0022 rad$

Now on the second question  

   The distance of separation of the slit is  

       $d = 0.300 \ mm = 3.0 *10^{-4} \ m$

The  intensity at the  the angular position in part "a" is mathematically evaluated as

      $I = I_o [\frac{sin \beta}{\beta} ]^2$

Where  $\beta$ is mathematically evaluated as

       $\beta = \frac{\pi * d * sin(\theta )}{\lambda }$

  substituting values

     $\beta = \frac{3.142 * 3*10^{-4} * sin(0.0022 )}{550 *10^{-9} }$

    $\beta = 0.06581$

So the intensity is  

    $I = I_o [\frac{sin (0.06581)}{0.06581} ]^2$

   $I = 0.000304 I_o$