Question

In Young's double slit experiment, the fringe pattern is observed on a screen placed at a distance $D$. The slits are illuminated by light of wavelength $\lambda$. The distance from the central point where the intensity falls to half the maximum is

• A. $\frac{\lambda D}{4d}$
• B. $\frac{\lambda D}{d}$
• C. $\frac{\lambda D}{2d}$
• D. $\frac{\lambda D}{3d}$

Answer 1

The correct option is A $\frac{\lambda D}{4d}$

Resultant intensity is given by

$I=4I_0{\cos }^2\left(\frac{\varphi }{2}\right)$

$I_{max}=4I_0$

Here, $I=\frac{4I_0}{2}$

$\therefore \frac{4I_0}{2}=4I_0{\cos }^2\left(\frac{\varphi }{2}\right)$

$\Rightarrow \frac{1}{2}={\cos }^2\left(\frac{\varphi }{2}\right)$

$\Rightarrow \cos \left(\frac{\varphi }{2}\right)=\frac{1}{\sqrt{2}}$

$\Rightarrow \frac{\varphi }{2}=\frac{\pi }{4}$

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

Path difference, $\Delta x=\frac{\lambda }{2\pi }\varphi =\frac{\lambda }{2\pi }\frac{\pi }{2}$

$\Rightarrow \Delta x=\frac{\lambda }{4}$

Position of that point where intensity falls to half the maximum

$y=\frac{\Delta xD}{d}=\frac{\lambda D}{4d}$

Hence, option $\left(\text{A}\right)$ is correct.