Question

In a young’s double slit experiment, $D$ equals the distance of screen from the plane of slits and $d$ is the separation between the slits. The distance of the nearest point from the central maximum on the screen, where the intensity is same as that due to a single slit, is equal to
Given that $\frac{d}{D}={10}^{-4},\lambda =6000\dot{A}$

Answer 1

Lets consider intensity if light through a slit is $I,$
In YDSE resultant intensity when phase difference is $\Delta \phi$ is given by
$I^{\prime }=4I{\cos }^2\frac{\Delta \phi }{2}$
Given that $I^{\prime }=I$
$\Rightarrow {\cos }^2\frac{\Delta \phi }{2}=\frac{1}{4}\Rightarrow \cos \frac{\Delta \phi }{2}=\pm \frac{1}{2}$
We know that
$\Delta \phi =\frac{2\pi }{\lambda }\Delta x$ and $\Delta x=\frac{dy}{D}$
$\Delta \phi =\frac{2\pi }{\lambda }\frac{dy}{D}\Rightarrow \cos \frac{\pi .dy}{\lambda D}=+\frac{1}{2}$
$\Rightarrow \frac{\pi .dy}{\lambda D}=\frac{\pi }{3}\Rightarrow y=\frac{\lambda D}{3d}$
$y=\frac{6000\times {10}^{-10}}{3\times {10}^{-4}}=2\text{mm}$