Question

Two coherent point sources $S_1$ and $S_2$ vibrating in phase emit light of wavelength $\lambda$. The separation between the sources is $2\lambda$. Consider a line passing through $S_2$ and perpendicular to the line $S_1{S}_2$. Find the position of farthest minima.

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

Answer 1

The correct option is C $D=\frac{15\lambda }{4}$



The maximum and minimum path difference between the waves are, $2\lambda$ (when point $P$ is very close to $S_2$) and zero (when point $P$ is at infinity), respectively.
For any minima, the path difference is equal to odd multiple of $\frac{\lambda }{2}$.

$\therefore$ Only two minima obtained from $S_2$ till infinity, along the line $S_{2}P$.

Close to $S_2$, the path difference is $2\lambda$. As we move away from $S_2$, along the line $S_{2}P$, the path difference starts to decrease.

$\therefore$ The farthest minima will have a path difference $\lambda /2$ and the nearest minima will have a path difference of $\frac{3\lambda }{2}$.

Now, for the farthest minima,
$S_{1}P-S_{2}P=\frac{\lambda }{2}$

$\Rightarrow \sqrt{4{\lambda }^2+D^2}-D=\frac{\lambda }{2}$

$4{\lambda }^2+D^2={(D+\frac{\lambda }{2})}^2$

$\Rightarrow 4{\lambda }^2+D^2=\frac{{\lambda }^2}{4}+D^2+D\lambda$
$\Rightarrow D=4\lambda -\lambda /4=\frac{15\lambda }{4}$

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