Question

At two points $P$ and $Q$ on a screen in Young's double slit experiment, waves from slits $S_1$ and $S_2$ have a path difference of $0$ and $\frac{\lambda }{4}$ respectively. The ratio of intensities at $P$ and $Q$ will be:

• A. $2:1$
• B. $\sqrt{2}:1$
• C. $4:1$
• D. $3:2$

Answer 1

The correct option is A $2:1$

Given, path difference at point $P$,$\Delta x_1=0$
$\Rightarrow \Delta {\varphi }_1=0^{\circ }$
$\therefore I_1=I_0+I_0+2I_0\cos 0^{\circ }=4I_0$
Similarly, path difference at point $Q$,
$\Delta x_2=\frac{\lambda }{4}$
$\Rightarrow \Delta {\varphi }_2=\frac{2\pi }{\lambda }.\frac{\lambda }{4}=\left(\frac{\pi }{2}\right)$
$\therefore I_2=I_0+I_0+2I_0\cos \frac{\pi }{2}=2I_0$
$\therefore \frac{I_1}{I_2}=\frac{4I_0}{2I_0}=\frac{2}{1}$

Hence, $\left(A\right)$ is the correct answer.