Question

You pass 633 nm laser light through a narrow single slit of width 0.24 mm and observe the diffraction pattern on a screen 6.0 m away. If the intensity at the central bright is Io , what is the intensity at a point on the screen 3.0 mm from the center of the pattern?

Answer 1

Answer:

The intensity is  $I  = 0.0175I_o$

Explanation:

From the question we are told that

   The wavelength is  $\lambda = 633 \ nm = 633*10^{-9} \ m$

   The width of the slit is $d = 0.24 \ mm = 0.00024 \ m$

   The distance from the screen is  $D = 6.0 \ m$

   The distance of the position considered from the center is $y = 3.0 \ mm = 0.003 \ m$

  Generally the intensity from at a point on the screen 3.0 mm from the center of the pattern is

       $I = I_o * \frac{sin^2 [\frac{\pi d * y}{ \lambda D } ]}{[\frac{\pi d * y }{\lambda D} ]^2}$

 Here  $I_o$ is the intensity of the central bright fringe

=>    $I  = I_o *  \frac{sin^2 [\frac{3.142  * 0.00024 * 0.003}{ 633*10^{-9} * 6 } ]}{[\frac{3.142*0.00024  * 0.003 }{633*10^{-9} * 6} ]^2}$

=>    $I  = 0.0175I_o$