Question

Here, figure shows P and Q are two equally intense coherent sources emitting radiation of wavelength 20 m. The separation PQ is 5.0 m and phase of Q is ahead of that of P by ${90}^o$. A, B and C are three distant points of observation equidistant from the mid-point of PQ. The intensity of radiations at A, B and C in ratio form will be :

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

Answer 1

Answer: (C)

0 : 1 : 2

At A, PA-QA =PQ = 5m
$=\frac{20}{4}=\frac{\lambda }{4}=\Delta x$
$\Delta {\varphi }^{\prime }=\frac{2\pi }{\lambda }\Delta x=\frac{2\pi }{\lambda }\cdot \frac{\lambda }{4}=\frac{\pi }{2}$
$\because$ Phase of Q is greater by $\frac{\pi }{2}$ than P
$\therefore$ Net phase difference $\varphi =\frac{\pi }{2}-\frac{\pi }{2}=0$
$I_A=4I^{\prime }$
At $B,\varphi =\frac{\pi }{2},I_B=2I^{\prime }$
At $C\Delta x=QA-PA=\frac{\lambda }{4}$
${\varphi }^{\prime }=\frac{\pi }{2}$
Net phase difference $\varphi =\frac{\pi }{2}+\frac{\pi }{2}=\pi$
$I_c=0$
$I_A:I_B:I_C=4I^{\prime }:2I^{\prime }:0$
$=2:1:0$