Biprism, Thin Films & Other Devices question from MHT CET 2024 (4 May, Shift 2)

Answer: (a)

1. $I_1 = \frac{I_0}{4}$, $I_2 = 0.5 \times \frac{3I_0}{4} = \frac{3I_0}{8}$.
2. Amplitudes $\propto \sqrt{I}$: $\frac{1}{2}$ and $\sqrt{\frac{3}{8}}$.
3. $\frac{I_{max}}{I_{min}} = \left[\frac{\frac{1}{2} + \sqrt{\frac{3}{8}}}{\frac{1}{2} - \sqrt{\frac{3}{8}}}\right]^2$. So the answer is (a).