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. The smallest distance $x$ such that the $P$ is a maxima.

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

Answer 1

The correct option is B $\sqrt{8}D$



Here, $\Delta x=S_{1}P-S_{2}P$

$\Delta x=3\lambda \cos \theta$ and $\cos \theta =\frac{D}{\sqrt{x^2+D^2}}$

$\Rightarrow \Delta x=\frac{3\lambda D}{\sqrt{x^2+D^2}}$

For first maxima, $\Delta x=\lambda$

$\Rightarrow \lambda =\frac{3\lambda D}{\sqrt{x^2+D^2}}$

$\Rightarrow \sqrt{x^2+D^2}=3D$

$\Rightarrow x^2+D^2=9D^2$

$\therefore x=\sqrt{8}D$

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