Question

An unpolarized light of certain intensity passes through a combination of two polarizers whose transmission axes are at \(30^\circ\) and \(90^\circ\), respectively, with respect to the horizontal axis. A third polarizer with its transmission axis at \(60^\circ\) with the horizontal axis is placed between the two existing polarizers. The ratio of the output intensities with and without the third polarizer is:

• A. 3/4
• B. 4/3
• C. 9/4
• D. 4/9

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
When unpolarized light passes through the first polarizer, its intensity is halved, and it becomes polarized parallel to the polarizer's axis. As it passes through subsequent polarizers, the transmitted intensity is governed by Malus's Law, which depends on the angle between the transmission axes of adjacent polarizers.
Step 2: Key Formula or Approach:
Unpolarized light through first polarizer: $I_1 = \frac{I_0}{2}$.
Malus's Law for subsequent polarizers: $I_{out} = I_{in} \cos^2(\theta)$, where $\theta$ is the relative angle between the two consecutive transmission axes.
Step 3: Detailed Explanation:
Let the initial intensity of the unpolarized light be $I_0$.
Case 1: Without the third polarizer (Only $P_1$ and $P_2$)
Light passes through $P_1$ (at $30^\circ$):
$I_1 = \frac{I_0}{2}$.
Light then passes through $P_2$ (at $90^\circ$). The relative angle is $\theta_1 = 90^\circ - 30^\circ = 60^\circ$.
$I_{out1} = I_1 \cos^2(60^\circ) = \frac{I_0}{2} \left( \frac{1}{2} \right)^2 = \frac{I_0}{2} \times \frac{1}{4} = \frac{I_0}{8}$.
Case 2: With the third polarizer $P_3$ (at $60^\circ$) inserted between $P_1$ and $P_2$
Light passes through $P_1$ (at $30^\circ$):
$I_1 = \frac{I_0}{2}$.
Light passes through $P_3$ (at $60^\circ$). The relative angle is $\theta_2 = 60^\circ - 30^\circ = 30^\circ$.
$I_2 = I_1 \cos^2(30^\circ) = \frac{I_0}{2} \left( \frac{\sqrt{3}}{2} \right)^2 = \frac{I_0}{2} \times \frac{3}{4} = \frac{3I_0}{8}$.
Light passes through $P_2$ (at $90^\circ$). The relative angle is $\theta_3 = 90^\circ - 60^\circ = 30^\circ$.
$I_{out2} = I_2 \cos^2(30^\circ) = \frac{3I_0}{8} \left( \frac{\sqrt{3}}{2} \right)^2 = \frac{3I_0}{8} \times \frac{3}{4} = \frac{9I_0}{32}$.
Ratio:
We need the ratio of intensity with the third polarizer to that without it:
$\text{Ratio} = \frac{I_{out2}}{I_{out1}} = \frac{\frac{9I_0}{32}}{\frac{I_0}{8}} = \frac{9}{32} \times \frac{8}{1} = \frac{9}{4}$.
Step 4: Final Answer:
The ratio is 9/4.