Question

A polaroid sheet is rotated between two crossed polarizers. The intensity of transmitted light would be maximum, when the angle between the axes of the first polarizer and the polaroid sheet is

• A. \(\pi\)/2
• B. \(\pi\)/4
• C. \(\pi\)
• D. \(\pi\)/3

Hint:

For a system of three polarizers where the first and last are crossed, the maximum transmission always occurs when the middle polarizer is placed at 45° (\(\pi/4\) radians) to both of them. This is a standard result worth remembering.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
This problem involves Malus's Law, which describes the intensity of light transmitted through a series of polarizers. We have three polarizers: a first polarizer (P1), a middle polaroid sheet (P2), and a third polarizer (P3) which is crossed with the first.

Step 2: Key Formula or Approach:
Malus's Law states that if the angle between the transmission axes of two polarizers is \(\theta\), the intensity of the transmitted light \(I\) is related to the incident polarized light intensity \(I_{inc}\) by: \[ I = I_{inc} \cos^2\theta \] Let's denote the angles of the transmission axes of P1, P2, and P3 with respect to some reference (e.g., the vertical) as \(\theta_1, \theta_2, \theta_3\).

Step 3: Detailed Explanation:
Let the initial unpolarized light have intensity \(I_0\). The first polarizer (P1) polarizes the light, and its transmitted intensity is \(I_1 = I_0/2\). Let's set the axis of P1 to be at \(\theta_1 = 0\). The third polarizer (P3) is "crossed" with the first one. This means their axes are perpendicular. So, \(\theta_3 = \pi/2\) (or 90°). The middle polaroid sheet (P2) is rotated. Let its axis be at an angle \(\theta\) with respect to the first polarizer's axis. So, \(\theta_2 = \theta\). Now, let's trace the intensity: 1. Intensity after P1: \(I_1 = I_0/2\). 2. Intensity after P2: The light incident on P2 has intensity \(I_1\) and is polarized at angle 0. The angle between P1 and P2 is \(\theta\). Using Malus's Law: \[ I_2 = I_1 \cos^2(\theta) = \left(\frac{I_0}{2}\right) \cos^2(\theta) \] 3. Intensity after P3: The light incident on P3 has intensity \(I_2\) and is polarized at angle \(\theta\). The axis of P3 is at \(\pi/2\). The angle between P2 and P3 is \((\pi/2 - \theta)\). Using Malus's Law again: \[ I_3 = I_2 \cos^2(\pi/2 - \theta) = I_2 \sin^2(\theta) \] Substituting the expression for \(I_2\): \[ I_3 = \left(\frac{I_0}{2}\right) \cos^2(\theta) \sin^2(\theta) \] We want to find the angle \(\theta\) that maximizes this final intensity \(I_3\). We can use the trigonometric identity \(\sin(2\theta) = 2\sin\theta\cos\theta\). \[ I_3 = \frac{I_0}{2} (\sin\theta\cos\theta)^2 = \frac{I_0}{2} \left(\frac{\sin(2\theta)}{2}\right)^2 = \frac{I_0}{8} \sin^2(2\theta) \] The intensity \(I_3\) will be maximum when \(\sin^2(2\theta)\) is maximum. The maximum value of \(\sin^2(2\theta)\) is 1. This occurs when: \[ \sin(2\theta) = \pm 1 \] \[ 2\theta = \frac{\pi}{2}, \frac{3\pi}{2}, \dots \] \[ \theta = \frac{\pi}{4}, \frac{3\pi}{4}, \dots \] The angle given in the options is \(\pi/4\).

Step 4: Final Answer:
The transmitted intensity is maximum when the angle between the first polarizer and the middle polaroid is \(\pi/4\).