Polarisation question from MHT CET 2026 (13 Apr, Shift 2)

Answer: (b)

1. First sheet: $I/2$. Each of the next $(n-1)$ sheets multiplies by $\cos^2 45^\circ = \frac{1}{2}$.
2. $I_{out} = \frac{I}{2}\left(\frac{1}{2}\right)^{n-1} = \frac{I}{2^n}$
3. $\frac{I}{2^n} = \frac{I}{64} \Rightarrow n = 6$
4. So the answer is (b) $6$.