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: Concept Identification:
This problem applies Malus's Law to determine light intensity through three polarizers (P1, P2, P3). P3 is oriented perpendicular to P1.

Step 2: Governing Equation:
Malus's Law: \( I = I_{inc} \cos^2\theta \), where \( \theta \) is the angle between transmission axes and \( I_{inc} \) is incident intensity.Let \( \theta_1, \theta_2, \theta_3 \) be the angles of P1, P2, and P3 respectively.

Step 3: Intensity Analysis:
Initial unpolarized light intensity: \( I_0 \).P1 transmits \( I_1 = I_0/2 \). Set \( \theta_1 = 0 \).P3 is crossed with P1, so \( \theta_3 = \pi/2 \).P2's axis is at angle \( \theta \) relative to P1, so \( \theta_2 = \theta \).Intensity progression:1. After P1: \( I_1 = I_0/2 \).2. After P2: Incident intensity \( I_1 \) at angle 0. Angle between P1 and P2 is \( \theta \). \( I_2 = I_1 \cos^2(\theta) = \left(\frac{I_0}{2}\right) \cos^2(\theta) \)3. After P3: Incident intensity \( I_2 \) at angle \( \theta \). P3 axis is at \( \pi/2 \). Angle between P2 and P3 is \( (\pi/2 - \theta) \). \( I_3 = I_2 \cos^2(\pi/2 - \theta) = I_2 \sin^2(\theta) \) Substituting \( I_2 \): \( I_3 = \left(\frac{I_0}{2}\right) \cos^2(\theta) \sin^2(\theta) \)To maximize \( I_3 \), use \( \sin(2\theta) = 2\sin\theta\cos\theta \):\( I_3 = \frac{I_0}{2} \left(\frac{\sin(2\theta)}{2}\right)^2 = \frac{I_0}{8} \sin^2(2\theta) \)\( I_3 \) is maximum when \( \sin^2(2\theta) = 1 \).This occurs when \( \sin(2\theta) = \pm 1 \), leading to \( 2\theta = \pi/2, 3\pi/2, \dots \), and thus \( \theta = \pi/4, 3\pi/4, \dots \).The relevant angle is \( \pi/4 \).

Step 4: Conclusion:
Maximum transmitted intensity occurs when the angle between P1 and P2 is \( \pi/4 \).