Question

The maximum intensity in Young's double-slit experiment is $I_0$. Distance between the slits is $d=5\lambda$, where $\lambda$ is the wavelength of monochromatic light used in the experiment. The intensity of light in front of one of the slits on a screen at a distance $D=10d$ is $\frac{I_o}{m}$.
Find $m$

Answer 1

Path difference between the waves from the two slits at point $P$ directly infront of one of the slits is given by :-
$\Delta x$ at $P=\frac{d}{D}x$
$\Rightarrow \Delta x=\frac{d}{D}\times \frac{d}{2}$
$\Rightarrow \Delta x=\frac{d^2}{2D}$
It is given in the question that :-
$d=5\lambda ,D=10d=10\times 5\lambda =50\lambda$
$\Rightarrow \Delta x=\frac{(5\lambda {)}^2}{2\times 50\lambda }$
$\Rightarrow \Delta x=\frac{25\lambda }{100\lambda }=\frac{\lambda }{4}$
Therefore, the phase difference between the waves at point $P$ is given by:-
$\Delta \varphi =\frac{2\pi }{\lambda }\Delta x$
$\Delta \varphi =\frac{2\pi }{\lambda }\times \frac{\lambda }{4}$
$\Rightarrow \Delta \varphi =\frac{\pi }{2}$
We know that the maximum intensity $\left(I_0\right)$ and intensity of source $\left(I\right)$ in YDSE are related as :-
$I_0=4I$
$\Rightarrow I=\frac{I_0}{4}$
Therefore, the net intensity of light infornt of one of the slits is given by :-
$I_{net}=I+I+2\sqrt{I}\sqrt{I}\cos \varphi$
$\Rightarrow I_{net}=2I+2I\cos \left(\frac{\pi }{2}\right)$
$\Rightarrow I_{net}=2I$
or, $\Rightarrow I_{net}=2\times \frac{I_0}{4}$
$\begin{array}{c}{I}_{net}=\frac{I_0}{2}\end{array}$
On comparing we get $m=2$