Question

Two coherent point sources $S_{1}$ and $S_{2}$ vibrating in phase emit light of wavelength $\lambda$. The separation between the sources is $2 \lambda$. The smallest distance from $S_{2}$ on a line passing through $S_{2}$ and perpendicular to $S_{1} S_{2}$, where minimum of intensity occurs, is

• A. $\frac{7 \lambda}{12}$
• B. $\frac{15 \lambda}{4}$
• C. $\frac{\lambda}{2}$
• D. $\frac{3 \lambda}{2}$

Answer 1

Answer: (A)

Given, separation between sources $S_{1}$ and $S_{2}=2 \lambda$. For minimum intensity at $P$, destructive interference must take place at $P$.

So, $S_{1} P-S_{2} P=\Delta x$ (path difference)
$=(2 n+1) \frac{\lambda}{2}$ (for destructive interference)
For minimum distance,
$S_{1} P-S_{2} P=\frac{3 \lambda}{2} \neq \frac{\lambda}{2}$ ...(i)
$\Rightarrow \sqrt{x^{2}+(2 \lambda)^{2}}-x=\frac{3 \lambda}{2}$
$\Rightarrow x^{2}+(2 \lambda)^{2}=\left[x+\left(\frac{3 \lambda}{2}\right)\right]^{2}$
$\Rightarrow x^{2}+4 \lambda^{2}=x^{2}+\frac{9 \lambda^{2}}{4}+2 \cdot x \cdot \frac{3 \lambda}{2}$
$\Rightarrow x \cdot 3 \lambda=4 \lambda^{2}-\frac{9 \lambda^{2}}{4}=\frac{7 \lambda^{2}}{4}$
$\Rightarrow x=\frac{7 \lambda}{12}$
If we proceed with Eq. (i) taking $S_{1} P-S_{2} P=\frac{\lambda}{2}, x=\frac{15 \lambda}{4}$ which is more than $\frac{7 \lambda}{12}$.