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)
$=(2n+1)\frac{\lambda }{2}$ (for destructive interference)
For minimum distance,
$S_{1}P-S_{2}P=\frac{3\lambda }{2}\ne \frac{\lambda }{2}$ ...(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 $\frac{7\lambda }{12}$.