Question

' $n$ ' polarizing sheets are arranged such that each makes an angle $45^{\circ}$ with the preceeding 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. 6
• B. 5
• C. 3
• D. 4

Answer 1

The Correct Option is A

After passing through the first sheet
\(I_1=\frac{I}{2}\)
After passing through the second sheet
\(I_2 = I_1\text{ }cos^{2}(45\degree)= \frac{I}{4}\)
After passing through n^th sheet
\(I_n=\frac{I}{2^n}=\frac{I}{64}\)
\(n=6\)

Answer 2

After passing through the first sheet: \[ I_1 = \frac{I}{2} \] After passing through the second sheet: \[ I_2 = I_1 \cos^2(45^\circ) = \frac{I}{4} \] After passing through the \( n \)th sheet: \[ I_n = \frac{I}{2^n} = \frac{I}{64} \] Solving for \( n \), \[ n = 6 \]