In the figure below, $P$ and $Q$ are two equally intense coherent sources emitting radiation of wavelength $20$ m. The separation between $P$ and $Q$ is $5$ 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 ratio:
Answer: (B) $2:1:0$
A path difference $\Delta x$ gives a phase difference $\dfrac{2\pi}{\lambda}\Delta x$; for $\Delta x = 5$ m, $\lambda = 20$ m this is $\dfrac\pi2$. With $I = 4I_0\cos^2(\phi/2)$:
**At A** (on the side of Q): the wave from P travels $5$ m more, lagging by $\pi/2$, which cancels its $\pi/2$ lead. $\phi = 0$, $I_A = 4I_0$.
**At B** (on the perpendicular bisector): no path difference, $\phi = \pi/2$, $I_B = 2I_0$.
**At C** (on the side of P): the wave from Q travels $5$ m more, lagging a further $\pi/2$. $\phi = \pi$, $I_C = 0$.
$$I_A : I_B : I_C = 4 : 2 : 0 = 2 : 1 : 0$$
Solution by Sreeraj P, M.Sc Physics