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Question #95174

Difficulty: 5/10 | 🟢 Free

The Question

A plane electromagnetic wave is normally incident from vacuum ( $z < 0$ ) onto a non-magnetic dielectric medium ( $z > 0$ ) of refractive index $n$ . The incident electric field is given by $\vec{E}_i = E_0 \cos(kz - \omega t) \hat{i}$ . The dielectric medium is forced to oscillate transversely with a non-uniform acceleration, such that its instantaneous velocity is $\vec{v}(t) = v_0 \sin(\Omega t) \hat{i}$ , where $v_0 \ll c$ and $\Omega \ll \omega$ . Due to the motion of the medium, the transmitted electromagnetic wave develops a time-dependent longitudinal electric field component $E_{tz} \hat{k}$ to satisfy Gauss's law. Assuming the boundary remains at $z=0$ , find the maximum amplitude of this longitudinal component, $E_{tz, \text{max}}$ , to first order in $v_0/c$ .

Options

A.
$\frac{2(n-1)}{n} \frac{v_0}{c} E_0$
B.
$\frac{n^2-1}{n^2} \frac{v_0}{c} E_0$
C.
$\frac{2(n^2-1)}{n+1} \frac{v_0}{c} E_0$
D.
$\frac{n-1}{n+1} \frac{v_0}{c} E_0$

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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

Transverse Nature of EM Waves|Non-Uniform Acceleration Constraint|Vector Decomposition Ambiguity

Why This Question Is Tricky

Examiner Levers

Dimensional Analysis

Tests if the student verifies units/dimensions before selecting.

Difficulty Boost : +2

Common Mistakes

Students typically assume that an electromagnetic wave normally incident on a dielectric interface remains strictly transverse ( $E_z = 0$ ), applying standard static Fresnel equations.

The trap lies in ignoring the non-inertial, dynamic nature of the medium.

When the dielectric oscillates transversely, Minkowski's electrodynamic constitutive relations ( $\vec{D} = \epsilon \vec{E} + \frac{n^2-1}{c^2} \vec{v} \times \vec{H}$ ) couple the transverse magnetic field to the longitudinal displacement field.

To satisfy Gauss's Law ( $\nabla \cdot \vec{D} = 0 \implies D_z = 0$ ), the electric field must develop a compensating longitudinal component $E_z$ .

Failing to account for the relativistic $\vec{v} \times \vec{H}$ term leads to the incorrect assumption that the wave remains purely transverse.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Electromagnetic Waves - Class 12
Concept	Transverse Nature of EM Waves
Difficulty	5
Cognitive Load	6
Target Solve Time	140 sec
Access	Free