In the figure below

Solution:

For (A)

$\mathrm{x}_{\mathrm{P}}-\mathrm{x}_{\mathrm{Q}}=(\mathrm{d}+2.5)-(\mathrm{d}-2.5)$

$=5 \mathrm{~m}$

$\Delta \phi$ due to path difference $=\frac{2 \pi}{\lambda}(\Delta \mathrm{x})=\frac{2 \pi}{20}(5)$

$=\frac{\pi}{2}$

At $A, Q$ is ahead of $P$ by path, as wave emitted by $Q$ reaches before wave emitted by $P$. Total phase difference at $\mathrm{A}$

$=\frac{\pi}{2}-\frac{\pi}{2}$ (due to P being ahead of Q by $90^{\circ}$ )

$=0$

$\mathrm{I}_{\mathrm{A}}=\mathrm{I}_{1}+\mathrm{I}_{2}+2 \sqrt{\mathrm{I}_{1}} \sqrt{\mathrm{I}_{2}} \cos \Delta \phi$

$=\mathrm{I}+\mathrm{I}+2 \sqrt{\mathrm{I}} \sqrt{\mathrm{I}} \cos (0)$

$=4 \mathrm{I}$

For $\mathrm{C}$

$x_{Q}-x_{P}=5 m$

$\Delta \phi$ due to path difference $=\frac{2 \pi}{\lambda}(\Delta x)$

$=\frac{2 \pi}{20}(5)=\frac{\pi}{2}$

Total phase difference at $\mathrm{C}=\frac{\pi}{2}+\frac{\pi}{2}=\pi$

$\mathrm{I}_{\text {net }}=\mathrm{I}_{1}+\mathrm{I}_{2}+2 \sqrt{\mathrm{I}} \sqrt{\mathrm{I}_{2}} \cos (\Delta \phi)$

$=\mathrm{I}+\mathrm{I}+2 \sqrt{\mathrm{I}} \sqrt{\mathrm{I}} \cos (\pi)=0$

For B

$x_{P}-x_{Q}=0$

$\Delta \phi=\frac{\pi}{2}$ (Due to $\mathrm{P}$ being ahead of $\mathrm{Q}$ by $90^{\circ}$ )

$\mathrm{I}_{\mathrm{B}}=\mathrm{I}+\mathrm{I}+2 \sqrt{\mathrm{I}} \sqrt{\mathrm{I}} \cos \frac{\pi}{2}=2 \mathrm{I}$

$\mathrm{I}_{\mathrm{A}}: \mathrm{I}_{\mathrm{B}}: \mathrm{I}_{\mathrm{C}}=4 \mathrm{I}: 2 \mathrm{I}: 0$

$=2: 1: 0$

$\therefore$ correct option is (4)