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Question #43217

Difficulty: 8/10 | 🟢 Free

The Question

A transparent solid slab occupies the region 0≤x≤L0 \leq x \leq L and y≥0y \geq 0 in air (nair=1n_{\text{air}} = 1). The refractive index of the slab varies spatially and depends on the wavelength λ\lambda of light according to the relation: n(y,λ)=n02+k(λ)yn(y, \lambda) = \sqrt{n_0^2 + k(\lambda) y} where n0n_0 is a positive dimensionless constant and k(λ)k(\lambda) is a wavelength-dependent parameter with units of m−1\text{m}^{-1}. A narrow beam of white light is incident normally on the face x=0x=0 at the origin (0,0)(0,0), initially traveling along the positive xx-axis. The beam propagates through the slab and exits into the air from the face x=Lx=L. Assuming LL is sufficiently small such that the angles made by the ray with the xx-axis remain small (sin⁡θ≈tan⁡θ≈θ\sin \theta \approx \tan \theta \approx \theta), what is the angular dispersion Δθ=θV−θR\Delta \theta = \theta_V - \theta_R between the violet and red components of the light after it completely exits the slab? (Given: kVk_V and kRk_R are the values of k(λ)k(\lambda) for violet and red light, respectively).

Options

  • A.

    (kV−kR)L2n0\frac{(k_V - k_R) L}{2 n_0}

  • B.

    (kV−kR)L2n02\frac{(k_V - k_R) L}{2 n_0^2}

  • C.

    (kV−kR)Ln0\frac{(k_V - k_R) L}{n_0}

  • D.

    (kV−kR)Ln02\frac{(k_V - k_R) L}{n_0^2}

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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

Dispersion vs Deviation (703)|Refractive Index Relativity (693)|Calculus (Integration)

Why This Question Is Tricky

Common Mistakes

The core trap in this problem lies in neglecting the final refraction at the exit interface (x=Lx=L).

Students correctly apply the Eikonal equation or generalized Snell's Law to find the trajectory y(x)y(x) and calculate the internal slope dy/dxdy/dx at x=Lx=L.

However, they mistakenly assume this internal angle θin≈kL2n02\theta_{\text{in}} \approx \frac{k L}{2 n_0^2} is the final deviation angle.

They fail to recognize that the medium is bounded by air, and the ray must undergo refraction as it exits the gradient-index (GRIN) medium.

Applying Snell's Law at the exit boundary introduces a crucial factor of n(yL)≈n0n(y_L) \approx n_0, amplifying the exit angle to θout≈kL2n0\theta_{\text{out}} \approx \frac{k L}{2 n_0}.

Option B perfectly captures this missing refraction trap.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterRay Optics and Optical Instruments - Class 12
ConceptDispersion vs Deviation
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree