Biprism, Thin Films & Other Devices question from JEE Main 2014 (12 Apr, Shift 1)

Answer: (c)

1. Direct path $SP = 2l$. Reflected ray meets R at the midpoint; each leg $= \frac{l}{\cos 30^\circ} = \frac{2l}{\sqrt{3}}$.
2. Reflected path $= \frac{4l}{\sqrt{3}}$, so $\Delta = \frac{4l}{\sqrt{3}} - 2l = \frac{2l(2-\sqrt{3})}{\sqrt{3}}$.
3. Reflection at R adds $\frac{\lambda}{2}$, so maxima need $\Delta = (2n-1)\frac{\lambda}{2}$.
4. $l = \frac{(2n-1)\lambda\sqrt{3}}{4(2-\sqrt{3})}$. So the answer is (c).