Question

The angle between the axes of a polariser and an analyser is \( 45^\circ \). If the intensity of the unpolarized light incident on the polariser is \( I \), then the intensity of the light emerged from the analyser is:

• A. \( 2I \)
• B. \( \frac{I}{2} \)
• C. \( I \)
• D. \( \frac{I}{4} \)

Hint:

Malus’s Law states that the transmitted intensity through a polariser-analyser setup is given by \( I = I_0 \cos^2 \theta \), where \( \theta \) is the angle between the two axes.

Answer 1

The Correct Option is D

Step 1: Malus's Law and Polarization
When unpolarized light of intensity \( I_0 \) is incident on a polarizer, the transmitted intensity after the first polarizer is given by: \[ I' = \frac{I_0}{2} \] Step 2: Intensity after the Analyser
According to Malus's Law, the intensity of light transmitted through an analyser is: \[ I = I' \cos^2 \theta \] where \( \theta \) is the angle between the transmission axis of the analyser and the polarizer.
Step 3: Substituting Values
Given that \( \theta = 45^\circ \), we substitute: \[ I = \frac{I_0}{2} \cos^2 45^\circ \] Since \( \cos 45^\circ = \frac{1}{\sqrt{2}} \), we get: \[ I = \frac{I_0}{2} \times \left( \frac{1}{\sqrt{2}} \right)^2 \] \[ I = \frac{I_0}{2} \times \frac{1}{2} \] \[ I = \frac{I_0}{4} \] Step 4: Conclusion
Thus, the intensity of the light emerging from the analyser is \( \frac{I}{4} \).