Question

A system of three polarizers $P_1, P_2, P_3$ is set up such that the pass axis of $P_2$ is inclined at $30^\circ$ to the pass axis of $P_1$ and the pass axis of $P_3$ is inclined at $60^\circ$ to pass axis of $P_1$. When a beam of unpolarized light of intensity $I_0$ is incident on $P_1$, the intensity of light transmitted by the three polarizers is $I$. The ratio $(I_0/I)$ equals:

• A. $\frac{32}{9}$
• B. $\frac{8}{3}$
• C. $\frac{32}{3}$
• D. $\frac{32}{1}$

Answer 1

Answer: (A)

$\frac{32}{9}$

Answer 2

1. When unpolarized light of intensity $I_0$ passes through the first polarizer ($P_1$), the transmitted intensity is reduced by half and becomes plane-polarized. So, the intensity after $P_1$ is $I_1 = \frac{I_0}{2}$.

2. The light transmitted by $P_1$ is incident on $P_2$. The pass axis of $P_2$ is inclined at $30^\circ$ to the pass axis of $P_1$. According to Malus' Law ($I_{out} = I_{in} \cos^2\theta$), the intensity transmitted by $P_2$ is $I_2 = I_1 \cos^2(30^\circ)$. $I_2 = \frac{I_0}{2} \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{I_0}{2} \times \frac{3}{4} = \frac{3I_0}{8}$. This light is now polarized along the pass axis of $P_2$.

3. The light transmitted by $P_2$ is incident on $P_3$. The pass axis of $P_3$ is inclined at $60^\circ$ to the pass axis of $P_1$. The pass axis of $P_2$ is at $30^\circ$ to $P_1$. Therefore, the angle between the pass axis of $P_2$ (which is the polarization direction of light incident on $P_3$) and the pass axis of $P_3$ is $\theta = |60^\circ - 30^\circ| = 30^\circ$.

4. According to Malus' Law, the intensity transmitted by $P_3$ is $I = I_2 \cos^2(30^\circ)$. $I = \frac{3I_0}{8} \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3I_0}{8} \times \frac{3}{4} = \frac{9I_0}{32}$.

5. The ratio $\frac{I_0}{I}$ is $\frac{I_0}{9I_0/32} = \frac{32}{9}$.