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Question #43601

Difficulty: 8/10 | 🟢 Free

The Question

A parallel plate capacitor of plate area $A$ is filled with a non-linear dielectric material. The electric displacement field is given by $\vec{D} = \epsilon_0 \vec{E} + \vec{P}$ . For this specific material, the macroscopic polarization is strongly non-linear and follows the relation $P = \alpha E^3$ , where $\alpha$ is a positive constant. Assume the linear term $\epsilon_0 E$ is negligible compared to $P$ . A time-varying displacement current $I_d(t) = 3 A \alpha \gamma^3 t^2$ is maintained across the capacitor for $t \ge 0$ (with $E=0$ at $t=0$ ), where $\gamma$ is a positive constant.

A linearly polarized laser beam of intensity $I_0$ passes through the dielectric parallel to the capacitor plates. Due to the electro-optic effect, its plane of polarization is rotated by an angle $\theta = \beta E$ (where $\beta$ is a positive constant) towards the transmission axis of an analyzer. The analyzer is placed after the capacitor, with its transmission axis initially at an angle of $45^\circ$ to the initial polarization direction of the laser beam. Which of the following expressions correctly represents the transmitted intensity $I(t)$ through the analyzer?

Options

A.
$\frac{I_0}{2} \left[ 1 + \sin\left( \frac{2\beta \alpha \gamma^3}{\epsilon_0} t^3 \right) \right]$
B.
$\frac{I_0}{2} \left[ 1 - \sin(2\beta \gamma t) \right]$
C.
$\frac{I_0}{2} \left[ 1 + \sin(2\beta \gamma t) \right]$
D.
$\frac{I_0}{2} \left[ 1 + \cos(2\beta \gamma t) \right]$

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Multi-Concept_Bridge

Master problems that require linking physics and math concepts to solve complex competitive questions.

Concepts Behind This Question

Malus's Law|Displacement Current Source|Dielectric Polarization Logic

Why This Question Is Tricky

Common Mistakes

The core trap in this problem is twofold.

First, students often mechanically apply $I_d = \epsilon_0 A \frac{\partial E}{\partial t}$ , ignoring that the fundamental definition of displacement current is based on the total displacement field $\vec{D}$ , not just the electric field $\vec{E}$ .

In a non-linear dielectric, $\vec{D}$ depends non-linearly on $\vec{E}$ , fundamentally altering the time evolution of the electric field (preventing an incorrect $E \propto t^3$ deduction).

Second, the problem tests the precise geometric application of Malus's Law.

Students frequently memorize $I = I_0 \cos^2\theta$ without defining $\theta$ as the relative angle between the current polarization state and the analyzer.

By rotating the polarization towards an analyzer initially at $45^\circ$ , the relative angle becomes $\pi/4 - \theta$ , which transforms the standard $\cos^2$ function into a $1 + \sin(2\theta)$ dependence.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Wave Optics - Class 12
Concept	Malus's Law
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free