Question

' n ' polarising sheets are arranged such that each makes an angle \(45^{\circ}\) with the proceeding sheet. An unpolarised light of intensity I is incident into this arrangement. The output intensity is I/ 64 . The value of \(n\) will be \(\left(\cos 45^{\circ}=1 / \sqrt{2}\right)\)

• A. 6
• B. 5
• C. 4
• D. 8

Hint:

The first sheet halves the intensity and every later sheet multiplies it by cos squared 45.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
Unpolarised light passing through the first sheet gives \(I_{1}=\dfrac{I}{2}\). Each next sheet applies Malus' law.

Step 2: Each further sheet
Each next sheet multiplies the intensity by \(\cos^{2}45^{\circ}=\dfrac{1}{2}\).

Step 3: Output intensity
After \(n\) sheets, \(I_{out}=\dfrac{I}{2}\left(\dfrac{1}{2}\right)^{n-1}=\dfrac{I}{2^{n}}\).

Step 4: Solve for n
\(\dfrac{I}{2^{n}}=\dfrac{I}{64}\Rightarrow 2^{n}=64\Rightarrow n=6\).

Final Answer:
There are 6 polarising sheets. \[ \boxed{\text{(A) }6} \]