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Question #43196

Difficulty: 8/10 | 🟢 Free

The Question

A light ray enters a non-homogeneous transparent medium at the origin (0,0)(0,0) of an xzxz-coordinate system. The zz-axis points vertically downwards into the medium, and the xx-axis lies along the surface boundary. The refractive index of the medium varies with depth zz as μ(z)=μ01−αz\mu(z) = \mu_0 \sqrt{1 - \alpha z} for z<1/αz < 1/\alpha, where μ0=1.5\mu_0 = 1.5 and α=0.1 m−1\alpha = 0.1 \text{ m}^{-1}. The ray enters the medium at an angle θ0=30∘\theta_0 = 30^\circ with respect to the positive zz-axis. The ray curves continuously, reaches a maximum depth, and eventually undergoes total internal reflection to return to the surface z=0z=0. What is the total horizontal distance xx traveled by the ray when it emerges back at the surface?

Options

  • A.

    103 m10\sqrt{3} \text{ m}

  • B.

    53 m5\sqrt{3} \text{ m}

  • C.

    203 m20\sqrt{3} \text{ m}

  • D.

    15 m15 \text{ m}

Check Your Answer

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

Total Internal Reflection (TIR)|Gravitational Field Variation|Integration

Why This Question Is Tricky

Common Mistakes

This question tests the application of Fermat's Principle and generalized Snell's Law in a medium with a continuous refractive index gradient.

A common trap is treating the gradient as discrete slabs and failing to construct the correct differential equation for the ray path (dx/dz=tan⁡θdx/dz = \tan\theta).

Students often incorrectly assume dx/dz=sin⁡θdx/dz = \sin\theta or fail to account for the symmetry of the trajectory, missing the factor of 2 for the return journey (leading to option B).

Advanced students can bypass the integration by recognizing the optomechanical analogy: the ray path is mathematically identical to the parabolic trajectory of a projectile in a uniform gravitational field, where the effective acceleration is derived from the gradient of μ2\mu^2.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterRay Optics and Optical Instruments - Class 12
ConceptTotal Internal Reflection (TIR)
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree