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 $2\lambda$. The light is collected on a screen $S$ which is placed at a distance $D(>>\lambda )$ from the slit $S_1$ as shown in figure. Find the finite distance $x$ such that the intensity at $P$ is equal to the intensity at $O$.

• A. $x=\sqrt{2}D$
• B. $x=\sqrt{3}\lambda$
• C. $x=\sqrt{3}D$
• D. $x=\lambda$

Answer 1

The correct option is C $x=\sqrt{3}D$


Path difference $\Delta x$ at $O$,

$\Delta x=S_{1}O-S_{2}O=2\lambda \left(2^{nd}\text{order maximum}\right)$

Thus, maximum intensity is obtained at $O$.

Next maxima will be obtained at point $P$, when,

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

$\Rightarrow d\cos \theta =\lambda$

$\Rightarrow 2\lambda \cos \theta =\lambda$

$\Rightarrow \cos \theta =\frac{1}{2}$

$\Rightarrow \theta ={60}^0$

Now in $\Delta S_{1}PO$,

$\tan \theta =\frac{PO}{S_{1}O}=\frac{x}{D}$

$\Rightarrow \tan {60}^{\circ }=\frac{x}{D}$

$\therefore x=\sqrt{3}D$

Hence, option $\left(C\right)$ is correct.