Question

Two coherent narrow slits emitting light of wavelength $\lambda$ in the same phase are placed parallel to each other at a small separation of $3\lambda$. The light is collected on a screen S which is placed at a distance $D(>>\lambda )$ from the slits. What is the smallest distance $x$ such that the intensity at $P$ is equal to the intensity at $O$?

• A. $\sqrt{3}D$
• B. $\sqrt{8}D$
• C. $\sqrt{5}D$
• D. $\sqrt{5}\frac{D}{2}$

Answer 1

The correct option is D $\sqrt{5}\frac{D}{2}$

The path difference between the two waves is given by
$\Delta x=2\lambda =S_{1}P-S_{2}P$
$⟹2\lambda =S_1{S}_{2}cos\theta$
Given $S_1{S}_2=3\lambda$ and $cos\theta =\frac{D}{\sqrt{x^2+D^2}}$
Hence $3\lambda \frac{D}{\sqrt{x^2+D^2}}=2\lambda$
$⟹9D^2=4(x^2+D^2)$
$⟹x=\frac{\sqrt{5}D}{2}$