Biprism, Thin Films & Other Devices question from MHT CET 2025 (20 Apr, Shift 2)

Answer: (b)

1. Ray AB (reflected at A): $I_1 = 0.25I = \frac{16I}{64}$.
2. Ray A'B': transmitted $0.75I$, reflected at C $0.25\times0.75I$, transmitted at A': $I_2 = 0.75\times0.25\times0.75I = \frac{9I}{64}$.
3. $\frac{I_{max}}{I_{min}} = \left(\frac{\sqrt{16}+\sqrt{9}}{\sqrt{16}-\sqrt{9}}\right)^2 = \left(\frac{7}{1}\right)^2 = 49$.
4. So the answer is (b) $49 : 1$.