Question

Figure here shows P and Q as two equally intense coherent sources emitting radiations of wavelength 20 m. The separation PQ is 5.0 m and phase of P is ahead of the phase of Q by ${90}^{\circ }$. A, B and C are three distant points of observation equidistant from the midpoint of PQ. The intensity of radiations at A, B, C will bear the ratio

[NSEP 1994]

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

Answer 1

The correct option is D

2 : 1 : 0

Since P is ahead of Q by ${90}^{\circ }$ and path difference between P and Q is $\lambda /4$​​​ (As $\lambda$ = 20m and PQ= 5m); therefore at A, phase difference induced due to path difference PQ is given by, $\Delta \varphi =\frac{2\pi }{\lambda }\Delta x$ = $\frac{2\pi }{\lambda }\left(\lambda /4\right)$ = $\pi /2$ = ${90}^{\circ }$ (it is negetive as wave from P will be behind wave from Q), hence the total phase difference is zero, so intensity is 4I. At C intensity is zero as total phase difference = ${180}^{\circ }$ and at B, the phase difference is ${90}^{\circ }$, so intensity is 4I${\cos }^2\left(\frac{{90}^{\circ }}{2}\right)$=2I. So, the ratio of intensities at A, B, C is 2:1:0.