Question

A single slit of width 0.3 mm is illuminated by a mercury light of wavelength 254 nm. Find the intensity at an 11° angle to the axis in terms of the intensity of the central maximum.

Answer 1

Answer:

The  the intensity at an 11° angle to the axis in terms of the intensity of the central maximum is  

   $I_c = \frac{I}{I_o} =8.48 *10^{-8}$

Explanation:

From the question we are told that

   The  width of the slit is  $D = 0.3 \ mm = 0.3 *10^{-3} \ m$

    The  wavelength is  $\lambda = 254 \ nm = 254 *10^{-9} \ m$

     The angle is  $\theta = 11^o$

The intensity of at $11^o$ to the axis in terms of the intensity of the central maximum. is mathematically represented as

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

Where $\beta$ is mathematically represented as

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

substituting values

      $\beta = \frac{0.3 *10^{-3} sin (11 ) * 3.142}{254 *10^{-9} }$

     $\beta = 708.1 \ rad$

So

  $I_c = \frac{I}{I_o} = [ \frac{sin (708.1) }{(708.1)}] ^2$

   $I_c = \frac{I}{I_o} =8.48 *10^{-8}$