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

let at distance y this happens$.$

$\Rightarrow I_0=I_0+I_0+2\sqrt{I_0}\sqrt{I_0}\cos \varphi$

$\Rightarrow I_0=2I_0+2I_0\cos \varphi$

$\Rightarrow 1=2\left[1+\cos \varphi \right]$

$\Rightarrow \frac{1}{2}=1+\cos \varphi$

$\Rightarrow \frac{-1}{2}=\cos \varphi$

$\Rightarrow \varphi =\frac{2\pi }{3}$

$\Rightarrow \frac{2\pi }{3}=\frac{2\pi }{\lambda }\left(\Delta x\right)$

$\Rightarrow \Delta x=\frac{\lambda }{3}$

$\Rightarrow \frac{dy}{D}=\frac{\lambda }{3}$

$\Rightarrow y=\frac{\lambda D}{3d}$

Hence,

option $\left(C\right)$ is correct answer.