Question

A beam of unpolarized light of intensity \(I_0\) falls on a system of four identical linear polarizers placed in a line as shown in the figure. The transmission axes of any two successive polarizers make an angle of \(30^\circ\) with each other. If the transmitted light has intensity \(I\), the ratio \(\dfrac{I}{I_0}\) is: 

• A. \(\dfrac{81}{256}\)
• B. \(\dfrac{9}{16}\)
• C. \(\dfrac{27}{64}\)
• D. \(\dfrac{27}{128}\)

Hint:

Each polarizer reduces intensity by \(\cos^2 \theta\) relative to the previous one. Start with \(\frac{I_0}{2}\) for unpolarized light.

Answer 1

The Correct Option is D

Step 1: Initial intensity after first polarizer.
Unpolarized light passing through the first polarizer transmits half the intensity: \[ I_1 = \frac{I_0}{2} \]

Step 2: Use Malus's law for each subsequent polarizer.
For each polarizer making \(30^\circ\) angle with the previous one: \[ I_2 = I_1 \cos^2 30^\circ = \frac{I_0}{2} \times \frac{3}{4} = \frac{3I_0}{8} \] \[ I_3 = I_2 \cos^2 30^\circ = \frac{3I_0}{8} \times \frac{3}{4} = \frac{9I_0}{32} \] \[ I_4 = I_3 \cos^2 30^\circ = \frac{9I_0}{32} \times \frac{3}{4} = \frac{27I_0}{128} \]

Step 3: Conclusion.
\[ \frac{I}{I_0} = \frac{27}{128} \]