A source produces a light beam of intensity $I_0$ polarized along the $x$-direction. The beam is sent along the $z$-direction. It enters a polaroid $P_1$ with its polaroid axis aligned along the $y$-direction so that no light exits the polaroid. When another polaroid $P_2$ is placed in between the source and $P_1$, the intensity measured after $P_1$ is $3I_0/16$. Which among the following is a possible value of $\theta$, the angle of the polaroid axis measured from the $x$-axis?
Show Hint
The output intensity formula for three polaroids where the outer two are crossed is always $I = \frac{I_0}{4} \sin^2(2\theta)$.
Setting this equal to $\frac{3I_0}{16}$ gives $\sin(2\theta) = \frac{\sqrt{3}}{2}$ immediately.
Remembering this general formula saves time in multiple-choice exams.
Step 1: Understanding the Question:
This question requires calculating the orientation angle of an intermediate polaroid inserted between two crossed polaroids to achieve a given final light intensity. Step 2: Key Formulas and Approach:
1. Malus's Law:
\[ I = I_{\text{in}} \cos^2 \phi \]
where $\phi$ is the angle between the polarization of the incident light and the transmission axis of the polaroid.
2. Trigonometric identity:
\[ \sin(2\theta) = 2 \sin \theta \cos \theta \] Step 3: Detailed Explanation:
The incident light has intensity $I_0$ and is polarized along the $x$-axis (angle $0^\circ$).
Polaroid $P_2$ is placed at an angle $\theta$ relative to the $x$-axis. By Malus's Law, the intensity of light transmitted through $P_2$ is:
\[ I_2 = I_0 \cos^2 \theta \]
The light emerging from $P_2$ is now linearly polarized along the axis of $P_2$ (angle $\theta$).
Polaroid $P_1$ is oriented along the $y$-axis (angle $90^\circ$). The angle between the polarization of light leaving $P_2$ and the axis of $P_1$ is:
\[ \phi = 90^\circ - \theta \]
By Malus's Law, the intensity of light transmitted through $P_1$ is:
\[ I_1 = I_2 \cos^2(90^\circ - \theta) = I_0 \cos^2\theta \sin^2\theta \]
Using the identity $\cos\theta \sin\theta = \frac{1}{2} \sin(2\theta)$:
\[ I_1 = I_0 \left( \frac{\sin(2\theta)}{2} \right)^2 = I_0 \frac{\sin^2(2\theta)}{4} \]
We are given that the final intensity is $I_1 = \frac{3I_0}{16}$:
\[ I_0 \frac{\sin^2(2\theta)}{4} = \frac{3I_0}{16} \]
\[ \sin^2(2\theta) = \frac{3}{4} \implies \sin(2\theta) = \frac{\sqrt{3}}{2} \quad (\text{for } 0^\circ<\theta<90^\circ) \]
