JEE Main · 2020 · 06 Sep 2020 · Shift 1 · Wave Optics

The two sources are coherent with $P$ leading $Q$ by a fixed phase of $\frac{\pi }{2}$, and the extra phase from a path difference $\Delta$ is $\frac{2\pi }{\lambda }\Delta$ with $\lambda =20\text{m}$. At the point where $Q$ is nearer by $\Delta =\frac{5}{2}\text{m}$, the path contributes $\frac{2\pi }{20}\cdot \frac{5}{2}=\frac{\pi }{2}$ that cancels the source lead, giving zero net phase and maximum intensity $4I_0$. At the point equidistant from both sources only the $\frac{\pi }{2}$ source lead survives, so the intensity is $2I_0\left(1+\cos \frac{\pi }{2}\right)=2I_0$. At the point where $P$ is nearer, the path phase adds to the source lead to make $\pi$, giving complete cancellation and zero intensity. The three intensities are in the ratio $4I_0:2I_0:0=2:1:0$.