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Question #43192

Difficulty: 8/10 | 🟢 Free

The Question

A monochromatic plane light wave of constant frequency f0f_0 is continuously incident from vacuum (z<0z < 0) onto a non-magnetic dielectric medium (z>0z > 0) at an angle of incidence θi=60∘\theta_i = 60^\circ. The boundary at z=0z=0 is stationary. The refractive index of the medium is spatially uniform everywhere in z>0z > 0, but varies slowly with time as n(t)=n0(1+αt)n(t) = n_0 (1 + \alpha t), where n0=3n_0 = \sqrt{3} and α>0\alpha > 0.

Consider a specific, short wave packet that crosses the boundary into the medium at t=0t = 0. At a later time t=t1t = t_1, the refractive index of the entire medium has become n(t1)=23n(t_1) = 2\sqrt{3}. Let f1f_1 be the instantaneous frequency of this specific wave packet at t=t1t = t_1, and θ1\theta_1 be the angle its wave vector makes with the z-axis at t=t1t = t_1. Assume the medium remains isotropic and dispersionless. Which of the following is correct?

Options

  • A.

    f1=f02,θ1=30∘f_1 = \frac{f_0}{2}, \theta_1 = 30^\circ

  • B.

    f1=f0,θ1=arcsin⁡(14)f_1 = f_0, \theta_1 = \arcsin\left(\frac{1}{4}\right)

  • C.

    f1=f02,θ1=arcsin⁡(14)f_1 = \frac{f_0}{2}, \theta_1 = \arcsin\left(\frac{1}{4}\right)

  • D.

    f1=f0,θ1=30∘f_1 = f_0, \theta_1 = 30^\circ

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Variable_Constraint

Develop pure symbolic logic skills by mastering algebraic constraints without relying on numerical data.

Concepts Behind This Question

Snell’s Law Vector Form|Optical Path Difference|Partial Differentiation

Why This Question Is Tricky

Common Mistakes

This question tests the fundamental symmetries underlying wave propagation.

Students typically memorize Snell's Law and the invariance of frequency across a boundary, assuming they hold universally.

However, these are consequences of spatial and temporal translation symmetries, respectively.

In a spatially uniform but time-varying medium, spatial momentum (the wave vector k⃗\vec{k}) is conserved, meaning the wave vector's direction (angle of refraction) remains locked to its initial value upon entry.

Conversely, the lack of time-translation symmetry means energy (frequency) is not conserved; the frequency scales inversely with the time-varying refractive index.

The trap is applying instantaneous Snell's law to a wave packet already inside the medium, incorrectly altering its trajectory and assuming frequency invariance.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterRay Optics and Optical Instruments - Class 12
ConceptSnell’s Law Vector Form
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree