1. Intensity after the first polarizer

Unpolarized light of intensity $I$ passes through the first polarizing sheet.

For unpolarized light, the transmitted intensity becomes: $I_1 = \frac{I}{2}$

2. Effect of each subsequent polarizer

Each sheet makes an angle $45^\circ$ with the previous one.

By Malus' law, when polarized light passes through a polarizer at angle $\theta$, the transmitted intensity is: $I' = I \cos^2 \theta$

Here, for every successive sheet, $\theta = 45^\circ$ so $\cos^2 45^\circ = \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2}$

Thus, every polarizer after the first reduces the intensity by another factor of $\frac{1}{2}$.

3. Total intensity after $n$ sheets

• First sheet gives factor $\frac{1}{2}$
• Remaining $(n-1)$ sheets each give factor $\frac{1}{2}$

So, $I_{\text{out}} = I \left(\frac{1}{2}\right)^n$

4. Use the given output intensity

Given: $I_{\text{out}} = \frac{I}{64}$

So, $I\left(\frac{1}{2}\right)^n = \frac{I}{64}$

Cancelling $I$: $\left(\frac{1}{2}\right)^n = \frac{1}{64} = \left(\frac{1}{2}\right)^6$

Therefore, $n = 6$

5. Check options

• A: $4$ $\rightarrow \frac{I}{16}$
• B: $5$ $\rightarrow \frac{I}{32}$
• C: $3$ $\rightarrow \frac{I}{8}$
• D: $6$ $\rightarrow \frac{I}{64}$

Hence, the correct option is D.