Question

A beam of light traveling horizontally consists of an unpolarized component with intensity \( I_0 \) and a polarized component with intensity \( I_p \). The plane of polarization is oriented at an angle \( \theta \) with respect to the vertical. The figure shows the total intensity \( I_{\text{total}} \) after the light passes through a polarizer as a function of the angle \( \alpha \), that the axis of the polarizer makes with respect to the vertical. Identify the correct statement(s). 

• A. \( \theta = 125^\circ \)
• B. \( I_p = 5 \, \text{W/m}^2 \)
• C. \( I_0 = 17.5 \, \text{W/m}^2 \)
• D. \( I_0 = 10 \, \text{W/m}^2; \, I_p = 20 \, \text{W/m}^2 \)

Hint:

In mixed light (unpolarized + polarized), half the unpolarized intensity always transmits, while the polarized part varies with \( \cos^2(\alpha - \theta) \).

Answer 1

The Correct Option is D

Step 1: Expression for transmitted intensity. 
The total transmitted intensity through a polarizer for a beam containing both unpolarized and polarized light is given by: \[ I_{\text{total}} = \frac{I_0}{2} + I_p \cos^2(\alpha - \theta). \] Step 2: Identify maximum and minimum intensities from the graph. 
From the graph: \[ I_{\text{max}} = 25 \, \text{W/m}^2, \quad I_{\text{min}} = 5 \, \text{W/m}^2. \] Step 3: Use intensity relations. 
At maximum: \( I_{\text{max}} = \frac{I_0}{2} + I_p \) At minimum: \( I_{\text{min}} = \frac{I_0}{2} \) Subtracting, \[ I_{\text{max}} - I_{\text{min}} = I_p \Rightarrow 25 - 5 = 20 \Rightarrow I_p = 20 \, \text{W/m}^2. \] Substitute into minimum intensity equation: \[ 5 = \frac{I_0}{2} \Rightarrow I_0 = 10 \, \text{W/m}^2. \] Step 4: Determine angle \( \theta \). 
The maxima occur when \( \alpha - \theta = 0 \), i.e., when polarizer’s axis aligns with the plane of polarization. From the graph, the peak occurs near \( \alpha = 35^\circ \) and trough near \( 125^\circ \), implying \( \theta \approx 35^\circ \). 
Step 5: Final Answer. 
Thus, \( I_0 = 10 \, \text{W/m}^2 \) and \( I_p = 20 \, \text{W/m}^2 \).