Start Learning Free

Before You Begin

Question #43560

Difficulty: 8/10 | 🟢 Free

The Question

A transparent medium occupies the region y≥0y \ge 0. The refractive index of the medium varies with temperature TT as n(T)=T0Tn(T) = \sqrt{\frac{T_0}{T}}, where T0T_0 is a constant. The temperature profile in the medium is given by T(y)=T01+kyT(y) = \frac{T_0}{1 + ky}, where k=13 m−1k = \frac{1}{\sqrt{3}} \text{ m}^{-1}. A narrow light ray traveling in the xyxy-plane from vacuum (y<0y < 0, n=1n=1) enters the medium at the origin (0,0)(0,0). The incident ray makes an angle θ0=60∘\theta_0 = 60^\circ with the positive yy-axis and has a positive xx-component of velocity. An observer is located in the medium at a height H=23 mH = 2\sqrt{3} \text{ m}. When the observer looks back along the tangent to the ray at y=Hy=H, the ray appears to originate from a virtual point source on the xx-axis at x=Xappx = X_{\text{app}}. Find the value of XappX_{\text{app}}.

Options

  • A.

    1 m1 \text{ m}

  • B.

    2 m2 \text{ m}

  • C.

    3 m3 \text{ m}

  • D.

    1.5 m1.5 \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

Virtual Object Trap|Variation of Density with Temperature|Integration

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to transition from discrete geometric optics to continuous gradient-index (GRIN) optics.

The primary trap lies in treating the medium as a single block and calculating the apparent position using only the final emergent angle θH\theta_H.

A student falling for this misconception will calculate θH=30∘\theta_H = 30^\circ and incorrectly assume the apparent origin is simply Xapp=Htan⁡θH=2 mX_{\text{app}} = H \tan\theta_H = 2 \text{ m}.

This fundamentally ignores the lateral shift x(H)x(H) accumulated during the curved trajectory.

The correct approach requires integrating the path equation to find the exact coordinates of the observer and then extrapolating the local tangent backward.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterRay Optics and Optical Instruments - Class 12
ConceptVirtual Object Trap
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree