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Question #43125

Difficulty: 8/10 | 🟢 Free

The Question

A quantum particle of mass mm and total energy E=100 eVE = 100 \text{ eV} is projected from the origin (0,0)(0,0) in the x−yx-y plane. Its initial velocity vector makes an angle θ0\theta_0 with the positive yy-axis. The particle moves through a region with a conservative potential V(y)V(y) given by: V(y)=αyfor 0≤y≤dV(y) = \alpha y \quad \text{for } 0 \le y \le d V(y)=V1for y>dV(y) = V_1 \quad \text{for } y > d where α=10 eV/m\alpha = 10 \text{ eV/m}, d=6.4 md = 6.4 \text{ m}, and V1=36 eVV_1 = 36 \text{ eV}. Assuming the particle's trajectory can be described by the Eikonal approximation (where matter waves obey Fermat's principle), what is the maximum value of θ0\theta_0 such that the particle successfully enters the region y>dy > d?

Options

  • A.

    sin⁡−1(0.60)\sin^{-1}(0.60)

  • B.

    sin⁡−1(0.80)\sin^{-1}(0.80)

  • C.

    sin⁡−1(0.64)\sin^{-1}(0.64)

  • D.

    sin⁡−1(0.36)\sin^{-1}(0.36)

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Multi-Concept_Bridge

Master problems that require linking physics and math concepts to solve complex competitive questions.

Concepts Behind This Question

Total Internal Reflection (TIR)|Conservative Force Field|Dot Product Geometry

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to bridge geometric optics (Fermat's Principle/Snell's Law) with Hamiltonian mechanics in a non-homogeneous medium.

The primary trap exploits the assumption that the particle travels in a straight line to the interface.

Students commonly apply Snell's Law directly between the initial state and the final state (n0sin⁡θ0=nfsin⁡θfn_0 \sin\theta_0 = n_f \sin\theta_f) to find the critical angle for transmission, yielding sin⁡θ0=0.80\sin\theta_0 = 0.80.

However, they fail to account for the continuous gradient of the conservative potential field, which forces a curved trajectory.

If θ0>sin⁡−1(0.60)\theta_0 > \sin^{-1}(0.60), the particle's yy-velocity reaches zero, causing it to turn around (continuous total internal reflection) before it ever reaches the interface at y=dy = d.

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