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Question #93384

Difficulty: 5/10 | 🟢 Free

The Question

A transparent solid sphere of radius RR is placed in a vacuum. The refractive index of the material of the sphere varies radially as n(r)=n0(Rr)kn(r) = n_0 \left(\frac{R}{r}\right)^k, where n0>1n_0 > 1 and 0<k<10 < k < 1 are constants, and rr is the distance from the center of the sphere. A light ray from the vacuum is incident on the surface of the sphere at an angle of incidence θ\theta. What is the total polar angle ΔΘ\Delta \Theta swept by the position vector of the light ray from its point of entry to its point of exit from the sphere?

Options

  • A.

    21−karccos⁡(sin⁡θn0)\frac{2}{1-k} \arccos\left(\frac{\sin\theta}{n_0}\right)

  • B.

    11−karccos⁡(sin⁡θn0)\frac{1}{1-k} \arccos\left(\frac{\sin\theta}{n_0}\right)

  • C.

    21+karccos⁡(sin⁡θn0)\frac{2}{1+k} \arccos\left(\frac{\sin\theta}{n_0}\right)

  • D.

    21−karcsin⁡(sin⁡θn0)\frac{2}{1-k} \arcsin\left(\frac{\sin\theta}{n_0}\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

Spherical Surface Refraction|Medium Dependency|Integration

Why This Question Is Tricky

Examiner Levers

Dimensional Analysis

Tests if the student verifies units/dimensions before selecting.

Difficulty Boost : +2

Common Mistakes

This question tests the application of Fermat's Principle in a spherically symmetric medium with a radial refractive index gradient.

The primary trap is the misapplication of the standard paraxial Snell's Law across a continuously varying medium.

Students must recognize that the generalized invariant is Bouguer's formula (rn(r)sin⁡ϕ=constantr n(r) \sin\phi = \text{constant}).

A secondary trap lies in the geometric setup of the differential equation for the ray trajectory; students often incorrectly define the path element or fail to account for the symmetry of the trajectory (missing the factor of 2 for the inbound and outbound legs).

Finally, recognizing the turning point boundary condition (sin⁡ϕ=1\sin\phi = 1) is required to correctly evaluate the integral.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterRay Optics and Optical Instruments - Class 12
ConceptSpherical Surface Refraction
Difficulty5
Cognitive Load6
Target Solve Time140 sec
AccessFree