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Question #43232

Difficulty: 8/10 | 🟢 Free

The Question

An astronomical telescope consists of an objective lens and an eyepiece, both treated as thin lenses with vacuum focal lengths f0f_0 and fef_e respectively. The telescope is placed in a vacuum, but its cylindrical tube of length LL (the distance between the objective and eyepiece) is filled with a transparent gas of non-uniform density. The gas has a radially varying refractive index n(r)=n0(1−kr2)n(r) = n_0 (1 - k r^2), where rr is the radial distance from the optical axis, n0>1n_0 > 1, and k>0k > 0 is a small constant (kr2≪1k r^2 \ll 1). Assuming paraxial rays and that the ray trajectory r(x)r(x) inside the tube satisfies the differential equation ddx(ndrdx)=∂n∂r\frac{d}{dx} \left( n \frac{dr}{dx} \right) = \frac{\partial n}{\partial r}, determine the required tube length LL for the telescope to be in normal adjustment.

Options

  • A.

    L=12k[arctan⁡(n0f02k)+arctan⁡(n0fe2k)]L = \frac{1}{\sqrt{2k}} \left[ \arctan(n_0 f_0 \sqrt{2k}) + \arctan(n_0 f_e \sqrt{2k}) \right]

  • B.

    L=1k[arctan⁡(f0k)+arctan⁡(fek)]L = \frac{1}{\sqrt{k}} \left[ \arctan(f_0 \sqrt{k}) + \arctan(f_e \sqrt{k}) \right]

  • C.

    L=12k[arcsin⁡(n0f02k)+arcsin⁡(n0fe2k)]L = \frac{1}{\sqrt{2k}} \left[ \arcsin(n_0 f_0 \sqrt{2k}) + \arcsin(n_0 f_e \sqrt{2k}) \right]

  • D.

    L=n0(f0+fe)(1−k3(f02+fe2))L = n_0 (f_0 + f_e) \left( 1 - \frac{k}{3}(f_0^2 + f_e^2) \right)

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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

Telescope Normal Adjustment|Refractive Index Relativity|Orbital Velocity Energy Balance

Why This Question Is Tricky

Common Mistakes

This problem tests the student's ability to move beyond memorized geometric optics formulas (L=f0+feL = f_0 + f_e) and apply fundamental Hamiltonian optics principles.

The trap lies in treating the focal length as a constant property of the lenses.

In a medium with a gradient refractive index (non-uniform density), light rays follow curved trajectories analogous to a particle in a harmonic oscillator potential (conserving 'optical energy').

Students must integrate the paraxial ray equation to find the true path-dependent phase accumulation.

The normal adjustment condition requires the overall ray transfer matrix element to be zero, leading to an effective tube length governed by the addition of arctangent phase angles rather than simple linear distances.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterRay Optics and Optical Instruments - Class 12
ConceptTelescope Normal Adjustment
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree