Question

In the figure below, $P$ and $Q$ are two equally intense coherent sources emitting radiation of wavelength $20\text{m}$. The separation between $P$ and $Q$ is $5\text{m}$ and the phase of $P$ is ahead of that of $Q$ by ${90}^{\circ }.A,B$ and $C$ are three distinct points of observation, each equidistant from the midpoint of $PQ.$ The intensities of radiation at $A,B,C$ will be in the proportion,

• A. $0:1:4$
• B. $2:1:0$
• C. $0:1:2$
• D. $4:1:0$

Answer 1

The correct option is B $2:1:0$

For $\left(A\right)$$x_p-x_Q=(d+2.5)-(d-2.5)=5\text{m}$

Phase difference $\Delta \varphi$ due to path difference
$=\frac{2\pi }{\lambda }\left(\Delta x\right)=\frac{2\pi }{20}\left(5\right)=\frac{\pi }{2}$
At $A,Q$ is a head of $P$ by path, as wave emitted by $Q$ reaches before wave emitted by $P$.
$\therefore$ Total phase difference at $A$
$=\frac{\pi }{2}-\frac{\pi }{2}=0$
(due to $P$ being ahead of $Q$ by ${90}^{\circ }$)
$I_A=I_1+I_2+2\sqrt{I_1}\sqrt{I_2}\cos \Delta \varphi$
$=I+I+2\sqrt{I}\sqrt{I}\cos \left(0\right)=4I$

For $\left(C\right)$,
Path difference, $x_Q-x_P=5\text{m}$
phase difference $\Delta \varphi$ due to path difference
$=\frac{2\pi }{\lambda }\left(\Delta x\right)=\frac{2\pi }{20}\left(5\right)=\frac{\pi }{2}$
Total phase difference at $C$
$=\frac{\pi }{2}+\frac{\pi }{2}=\pi$
$I_{\text{net}}=I_1+I_2+2\sqrt{I_1}\sqrt{I_2}\cos \left(\Delta \varphi \right)$
$=I+I+2\sqrt{I}\sqrt{I}\cos \left(\pi \right)=0$

For $\left(B\right)$,
path difference, $x_P-x_Q=0$
Phase difference, $\Delta \varphi =\frac{\pi }{2}$
Due to $P$ being ahead of $Q$ by ${90}^{\circ }$
$I_B=I+I+2\sqrt{I}\sqrt{I}\cos \frac{\pi }{2}=2I$

Therefore, intensities of radiation at $A,B$ and $C$ will be in the proportion $I_A:I_B:I_C=4I:2I:0=2:1:0$