Q. The intensity at the maximum in a Young’s double slit experiment is $I_0$. Distance between two slits is $d = 5\lambda $, where $\lambda$ is the wavelength of light used in the experiment. What will be the intensity in front of one of the slits on the screen placed at a distance D = 10 d ?
Solution:
Path difference
$= S_2 P - S_1P$
$ = \sqrt{D^2 + d^2 } - D $
$ = D \left( 1+ \frac{1}{2} \frac{d^2}{D^2} \right) - D$
$ = D \left[ 1 + \frac{d^2}{2D^2} - 1 \right] = \frac{d^2}{2D}$
$\Delta x = \frac{d^2}{2 \times 10d} = \frac{d}{20} = \frac{5 \lambda}{20} = \frac{\lambda}{4}$
$ \Delta \phi = \frac{2\pi}{\lambda}. \frac{\lambda}{4} = \frac{\pi}{2}$
So, intensity at the desired point is
$I = I_0 \, cos^2 \frac{\phi}{2} = I_0 cos^2 \frac{\pi}{4} = \frac{I_0}{2}$
