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Question #43236

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The Question

An optical medium is designed to model the combined scattering of a charged particle by a Coulombic field and an inverse-cube force field. The effective refractive index of the medium is given by $n(r) = \sqrt{1 + \frac{\alpha}{r} + \frac{\beta^2}{r^2}}$ , where $\alpha$ and $\beta$ are positive constants with units of length, and $r$ is the radial distance from the origin. A light ray enters this medium from vacuum ( $n=1$ ) at infinity with an impact parameter $b$ (where $b > \beta$ ). Which of the following expressions gives the total polar angle $\Phi$ swept by the ray's position vector from its entry at infinity to its exit at infinity?

Options

A.
$\frac{b}{\sqrt{b^2 - \beta^2}} \left[ \pi + 2 \arcsin\left( \frac{\alpha}{\sqrt{4(b^2 - \beta^2) + \alpha^2}} \right) \right]$
B.
$\frac{b}{\sqrt{b^2 + \beta^2}} \left[ \pi + 2 \arcsin\left( \frac{\alpha}{\sqrt{4(b^2 + \beta^2) + \alpha^2}} \right) \right]$
C.
$\frac{\beta}{\sqrt{b^2 - \alpha^2}} \left[ \pi + 2 \arccos\left( \frac{\beta}{\sqrt{4(b^2 - \alpha^2) + \beta^2}} \right) \right]$
D.
$\frac{\pi b}{\sqrt{b^2 - \beta^2}} + 2 \arcsin\left( \frac{\alpha}{\sqrt{4b^2 + \alpha^2}} \right)$

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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

Power Combination|Coulombic Central Force|Impact Parameter Geometry

Why This Question Is Tricky

Common Mistakes

This question tests the isomorphism between geometric optics in a graded-index medium and Hamiltonian dynamics in a central potential (Eikonal approximation).

The trap lies in the 'Power Combination' aspect: students often incorrectly assume that the total scattering angle (or optical power) of a combined potential is the linear sum of the scattering angles of the individual potentials (superposition), leading them to Option D.

However, in central force motion, the inverse-cube term modifies the effective centrifugal barrier, changing the angular frequency of the orbit, while the Coulombic term provides the radial turning point.

One must use the conservation of the optical invariant $L = n(r) r \sin \phi = b$ to set up the exact differential equation for the trajectory and integrate it simultaneously through the effective potential.

Quick Facts About This Question

Question Information

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