Question

'n' polarizing sheets are arranged such that each makes an angle \(45^{\circ}\) with the preceding sheet. An unpolarized light of intensity \(I\) is incident into this arrangement. The output intensity is found to be \(I/64\). The value of \(n\) will be

• A. \(3\)
• B. \(6\)
• C. \(5\)
• D. \(4\)

Hint:

The first sheet halves the intensity; each next sheet multiplies by \(\cos^2 45^{\circ}=\frac12\).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Unpolarised light passing through the first sheet loses half its intensity: \(I_1 = \frac I2\). For every later sheet, Malus's law gives \(I' = I\cos^2\theta\).

Step 2: Each later sheet:
With \(\theta = 45^{\circ}\), \(\cos^2\theta = \frac12\). So each of the other \((n-1)\) sheets halves the intensity.

Step 3: Solve:
\[ I_{\text{out}} = \frac I2\left(\frac12\right)^{n-1} = \frac{I}{2^n} \]
\(\frac{I}{2^n} = \frac{I}{64}\), so \(2^n = 64\) and \(n = 6\).

Final Answer:
The number of sheets is \(6\), option (B). \[ \boxed{6} \]