Question

Unpolarized light of intensity \( I_0 \) passes through a polarizer \( P_1 \). The light coming out of the polarizer falls on a quarter-wave plate with its optical axis at 45° with respect to the polarization axis of \( P_1 \) and then passes through another polarizer \( P_2 \) with its polarization axis perpendicular to that of \( P_1 \). The intensity of the light coming out of \( P_2 \) is \( I \). The ratio \( \frac{I_0}{I} \) is:

Hint:

When light passes through a polarizer, the intensity is halved. A quarter-wave plate changes the polarization but doesn't affect intensity. The intensity passing through the second polarizer is determined by the angle between the polarization axis and the axis of the polarizer.

Answer 1

Step 1: Analyze the light passing through polarizers.
When unpolarized light passes through a polarizer, the intensity is reduced by half: \[ I_1 = \frac{I_0}{2} \] Next, the light passes through a quarter-wave plate with its optical axis at 45° to the polarization axis of \( P_1 \). A quarter-wave plate changes the polarization of the light without affecting the intensity. After passing through the quarter-wave plate, the light remains polarized but is now at a 45° angle to its original polarization direction.
Step 2: Analyze the second polarizer.
The second polarizer \( P_2 \) has its axis perpendicular to the initial polarization direction of the light. As the light enters the second polarizer, the intensity is reduced by a factor of \( \cos^2(45^\circ) \). Therefore, the intensity after passing through \( P_2 \) is: \[ I = \frac{I_0}{2} \cos^2(45^\circ) = \frac{I_0}{2} \times \frac{1}{2} = \frac{I_0}{4} \] Step 3: Conclusion.
Thus, the ratio \( \frac{I_0}{I} \) is: \[ \frac{I_0}{I} = 4 \]