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 $\frac{D\lambda }{3d}$

Resultant intensity
$\Rightarrow I_{net}=I_1+I_2+2\sqrt{I_1}\sqrt{I_2}\cos \varphi$
Given $I_1=I_2=I_0$
Let at a distance $y$, Resultant intensity $=I_0$
$\Rightarrow I_0=2I_0+2I_0\cos \varphi$
$\Rightarrow 1=2[1+\cos \varphi ]$
$\Rightarrow \frac{1}{2}=1+\cos \varphi$
$\Rightarrow -\frac{1}{2}=\cos \varphi$
$\Rightarrow \varphi =\frac{2\pi }{3}$
Relation between phase difference and path difference is
$\varphi =\frac{2\pi }{\lambda }\left(\Delta x\right)$
$\Rightarrow \frac{2\pi }{3}=\frac{2\pi }{\lambda }\left(\Delta x\right)$
$\Rightarrow \Delta x=\frac{\lambda }{3}$
$\Rightarrow \frac{dy}{D}=\frac{\lambda }{3}$
$\Rightarrow y=\frac{\lambda D}{3d}$