Question

In a young’s double slit experiment, $D$ equals the distance of screen and $d$ is the separation between the slit. The distance of the nearest point to the central maximum where the intensity is same as that due to a single slit, is equal to

• A. $\frac{D\lambda }{d}$
• B. $\frac{D\lambda }{2d}$
• C. $\frac{D\lambda }{3d}$
• D. $\frac{2D\lambda }{d}$

Answer 1

The correct option is C $\frac{D\lambda }{3d}$

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
$\Delta \phi =\frac{2\pi }{\lambda }\frac{dy}{D}\Rightarrow \cos \frac{\pi .dy}{\lambda D}=+\frac{1}{2}$
$\Rightarrow \frac{\pi .dy}{\lambda D}=\frac{\pi }{3}\Rightarrow y=\frac{\lambda D}{3d}$