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Question #43189

Difficulty: 8/10 | 🟢 Free

The Question

A broad, uniform parallel beam of monochromatic light of intensity $I_0 \text{ W/m}^2$ is directed vertically upwards from deep within a liquid of refractive index $\mu = \frac{5}{3}$ . The liquid-air interface is disturbed by a standing wave, such that the surface profile at a given instant is $y(x) = \frac{3L}{2\pi} \sin\left(\frac{\pi x}{L}\right)$ , where $L = 2.0 \text{ m}$ is a constant length parameter. The light beam covers a large number of spatial periods along the $x$ -axis and extends uniformly in the $z$ -direction. Determine the fraction of the incident beam's cross-sectional area (in the $xz$ -plane) that undergoes Total Internal Reflection (TIR) at the liquid-air interface. Assume the refractive index of air is $\mu_{\text{air}} = 1$ .

Options

A.
$\frac{2}{3}$
B.
$\frac{1}{3}$
C.
$\frac{1}{2}$
D.
$\frac{2}{\pi} \arccos\left(\frac{2}{5}\right)$

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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

Refractive Index Relativity|Total Internal Reflection (TIR)|Spherical Surface Refraction

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to apply geometric optics to a dynamically curved surface using differential geometry.

The primary trap is treating the surface as flat, which would lead to an incidence angle of 0 degrees and 0% TIR.

A more lethal trap for advanced students is confusing the surface gradient $dy/dx$ with $\sin\theta_i$ instead of $\tan\theta_i$ .

Because the paraxial approximation ( $\sin\theta \approx \tan\theta$ ) is deeply ingrained, students often blindly equate the slope to the sine of the incidence angle, leading directly to the distractor option D.

Furthermore, the problem requires students to correctly identify that the 'cross-sectional area of the beam' corresponds to the projection on the x-axis ( $dx$ ), rather than mistakenly integrating the arc length ( $ds$ ) of the curved surface.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Ray Optics and Optical Instruments - Class 12
Concept	Refractive Index Relativity
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free