Answer: (2)

Solution

For (A):

$x_P - x_Q = (d + 2.5) - (d - 2.5) = 5$ m

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

At A, Q is ahead of P by path, as wave emitted by Q reaches before wave emitted by P.

$\therefore$ Total phase difference at A: $\frac{\pi}{2} - \frac{\pi}{2} = 0$

$I_A = I_1 + I_2 + 2\sqrt{I_1}\sqrt{I_2}\cos\Delta\phi$

$= I + I + 2\sqrt{I}\sqrt{I}\cos(0) = 4I$

For C: Path difference, $x_Q - x_P = 5$ m

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

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

$I_{\text{net}} = I_1 + I_2 + 2\sqrt{I_1}\sqrt{I_2}\cos(\Delta\phi)$

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

For B: Path difference, $x_P - x_Q = 0$

Phase difference, $\Delta\phi = \frac{\pi}{2}$ (due to P being ahead of Q by 90$°$)

$I_B = I + I + 2\sqrt{I}\sqrt{I}\cos\frac{\pi}{2} = 2I$

Therefore intensities of radiation at A, B and C will be in the ratio $4I : 2I : 0 = 2 : 1 : 0$