Analyzing Polarized Light Intensity Through Polarizers

This problem involves understanding how the intensity of light changes as it passes through a series of polarizers. We need to apply the principles of polarization and Malus's Law to find the final intensity relative to the initial intensity.

Understanding the Physics: Malus's Law

Key principles governing this problem are:

• When unpolarized light of intensity $I_{in}$ passes through a polarizer, the transmitted light is polarized, and its intensity is reduced by half: $I_{out} = \frac{1}{2} I_{in}$.
• When polarized light of intensity $I_{in}$ passes through a second polarizer, the transmitted intensity $I_{out}$ is given by Malus's Law: $I_{out} = I_{in} \cos^2 \theta$. Here, $\theta$ is the angle between the transmission axis of the first polarizer and the transmission axis of the second polarizer.

Step-by-Step Intensity Calculation

Let the initial intensity of the unpolarized light beam be $I_0$. We have three polarizers: $P_1$, $P_2$, and $P_3$. Let's track the intensity after each polarizer:

1. Intensity after Polarizer $P_1$

The unpolarized light with intensity $I_0$ is incident on the first polarizer, $P_1$. According to the first principle mentioned above, the intensity of the light transmitted by $P_1$ (let's call it $I_1$) is:

$I_1 = \frac{1}{2} I_0$

This transmitted light is now polarized along the pass axis of $P_1$.

2. Intensity after Polarizer $P_2$

The polarized light of intensity $I_1$ from $P_1$ is incident on the second polarizer, $P_2$. The pass axis of $P_2$ is inclined at an angle $\theta_{12} = 15^\circ$ to the pass axis of $P_1$. Applying Malus's Law, the intensity transmitted by $P_2$ (let's call it $I_2$) is:

$I_2 = I_1 \cos^2(\theta_{12})$

Substituting the expression for $I_1$ and the angle:

$I_2 = \left(\frac{1}{2} I_0\right) \cos^2(15^\circ)$

This light is now polarized along the pass axis of $P_2$.

3. Intensity after Polarizer $P_3$

The polarized light of intensity $I_2$ from $P_2$ is incident on the third polarizer, $P_3$. The problem states that the pass axis of $P_3$ is "crossed" with respect to $P_1$. This means the angle between the pass axis of $P_1$ and $P_3$ is $90^\circ$. Since the pass axis of $P_2$ is at $15^\circ$ relative to $P_1$, the angle between the pass axis of $P_2$ and $P_3$ is $\theta_{23} = 90^\circ - 15^\circ = 75^\circ$. Applying Malus's Law again, the final transmitted intensity (let's call it $I$) is:

$I = I_2 \cos^2(\theta_{23})$

Substituting the expression for $I_2$ and the angle $\theta_{23}$:

$I = \left[\left(\frac{1}{2} I_0\right) \cos^2(15^\circ)\right] \cos^2(75^\circ)$

$I = \frac{1}{2} I_0 \cos^2(15^\circ) \cos^2(75^\circ)$

Calculating the Ratio $(I_0 / I)$

We need to find the ratio $(I_0 / I)$. Rearranging the equation for $I$:

$\frac{I_0}{I} = \frac{I_0}{\frac{1}{2} I_0 \cos^2(15^\circ) \cos^2(75^\circ)}$

$\frac{I_0}{I} = \frac{2}{\cos^2(15^\circ) \cos^2(75^\circ)}$

Now, let's simplify the denominator using trigonometric identities:

We know that $\cos(75^\circ) = \cos(90^\circ - 15^\circ) = \sin(15^\circ)$.

So, the denominator becomes $\cos^2(15^\circ) \sin^2(15^\circ)$.

We can rewrite this using the identity $\sin(2\theta) = 2 \sin\theta \cos\theta$, which implies $\sin\theta \cos\theta = \frac{1}{2} \sin(2\theta)$.

Therefore, $\cos^2(15^\circ) \sin^2(15^\circ) = (\cos(15^\circ) \sin(15^\circ))^2 = \left(\frac{1}{2} \sin(2 \times 15^\circ)\right)^2$

= $\left(\frac{1}{2} \sin(30^\circ)\right)^2$

Since $\sin(30^\circ) = \frac{1}{2}$, we substitute this value:

= $\left(\frac{1}{2} \times \frac{1}{2}\right)^2 = \left(\frac{1}{4}\right)^2 = \frac{1}{16}$

Now, substitute this simplified value back into the ratio expression:

$\frac{I_0}{I} = \frac{2}{1/16} = 2 \times 16 = 32$

Final Result

The ratio $(I_0 / I)$ is exactly 32. This value matches one of the options provided.