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Question #43124

Difficulty: 8/10 | 🟢 Free

The Question

A large tank contains a transparent liquid of refractive index $n = 2$ . A standing surface wave is established on the liquid, with its vertical displacement given by $z(x,t) = A \sin(kx) \cos(\omega t)$ . A narrow laser beam is incident from within the liquid onto the surface at the origin $(0,0)$ . The beam lies in the $xz$ -plane and makes a constant angle $\theta_0 = 60^\circ$ with the positive $z$ -axis. Due to the oscillation of the surface, the local normal vector tilts, causing the angle of incidence to vary. It is observed that Total Internal Reflection (TIR) of the laser beam is lost for exactly $\frac{1}{3}$ of the wave period during each cycle. A small cork of mass $m$ floats on the liquid surface at an antinode of the standing wave. Assuming deep water wave conditions ( $\omega^2 = gk$ ), what is the maximum kinetic energy of the cork? (Assume the cork moves only vertically and its dimensions are negligible. The unperturbed surface is $z=0$ .)

Options

A.
$\frac{2 m g}{3 k}$
B.
$\frac{\pi^2 m g}{18 k}$
C.
$\frac{4 m g}{3 k}$
D.
$\frac{2 m g}{9 k}$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Intuition_Breaker

Train your brain to resist "obvious" answers that are designed to deceive your intuition.

Concepts Behind This Question

Total Internal Reflection (TIR)|Vertical Oscillations|Time of Flight Logic

Why This Question Is Tricky

Common Mistakes

This problem tests the intersection of geometric optics, wave kinematics, and classical mechanics.

The core trap lies in recognizing that the angle of incidence is dynamically modulated by the slope of the wave, not just its height.

A common lethal mistake is applying the small-angle approximation ( $\theta_i \approx \theta_0 + Ak \cos(\omega t)$ ) for the normal's tilt, which leads to an incorrect algebraic condition ( $Ak \cos(\omega t) < -\pi/6$ ) because the critical angle difference ( $30^\circ$ ) is too large for $\tan \theta \approx \theta$ to hold.

This leads directly to the distractor $\frac{\pi^2 m g}{18 k}$ .

Furthermore, students often misinterpret the phase duration, failing to map '1/3 of the period' to the correct cosine threshold of $-1/2$ .

The problem rigorously demands exact trigonometric handling of the rotating normal vector before applying energy conservation.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Ray Optics and Optical Instruments - Class 12
Concept	Total Internal Reflection (TIR)
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free