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Question #43140

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The Question

A light source located at the origin $(0,0)$ inside a non-homogeneous transparent medium emits a ray of wavelength $\lambda$ . The medium occupies the region $y \ge 0$ . The ray is emitted such that it makes a fixed angle $\theta_0$ with the positive y-axis at $y=0$ . The refractive index of the medium varies with depth $y$ and wavelength $\lambda$ as $n(y, \lambda) = \sqrt{n_0^2(\lambda) + \alpha y}$ , where $\alpha$ is a positive constant with units of $\text{m}^{-1}$ , and $n_0(\lambda)$ is the refractive index at $y=0$ . The dispersive power of the medium at $y=0$ is characterized by $\frac{dn_0}{d\lambda} = -\beta$ (where $\beta > 0$ ). What is the rate of change of the lateral displacement $x$ with respect to wavelength $\lambda$ (i.e., $\frac{dx}{d\lambda}$ ) when the ray reaches a depth $y = H$ ?

Options

A.
$\frac{2\beta \sin\theta_0}{\alpha} \left( 2n_0 \cos\theta_0 - \frac{2n_0^2 \cos^2\theta_0 + \alpha H}{\sqrt{n_0^2 \cos^2\theta_0 + \alpha H}} \right)$
B.
$\frac{2\beta n_0 \sin\theta_0}{\alpha} \left( \frac{1}{\cos\theta_0} - \frac{n_0}{\sqrt{n_0^2 \cos^2\theta_0 + \alpha H}} \right)$
C.
$\frac{2\beta \cos\theta_0}{\alpha} \left( 1 - \frac{n_0 \cos\theta_0}{\sqrt{n_0^2 \cos^2\theta_0 + \alpha H}} \right)$
D.
$\frac{2\beta \sin\theta_0}{\alpha} \left( \frac{2n_0^2 \cos^2\theta_0 + \alpha H}{\sqrt{n_0^2 \cos^2\theta_0 + \alpha H}} + 2n_0 \cos\theta_0 \right)$

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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

Dispersion vs Deviation|Integration|Ray Optics

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to apply Fermat's principle and Snell's Law in a continuously varying medium, specifically focusing on the invariant $C = n(y)\sin\theta$ .

The primary trap lies in the differentiation step for dispersion.

Because the light source is located inside the medium and emits at a fixed angle $\theta_0$ , the invariant $C = n_0(\lambda)\sin\theta_0$ is explicitly a function of wavelength $\lambda$ .

A common misconception is treating the invariant $C$ as an absolute constant independent of $\lambda$ (which would only be true if the ray entered from a vacuum at a fixed external angle).

If a student incorrectly assumes $C$ is constant during differentiation, they will fail to apply the product rule correctly, leading directly to Option B.

Option C traps students who incorrectly set up the differential path length by dropping the $n_0 \sin\theta_0$ numerator when converting $dx/dy = \tan\theta$ into the integral.

Quick Facts About This Question

Question Information

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