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Question #43592

Difficulty: 8/10 | 🟢 Free

The Question

A circular aperture of a telescope is formed by a smart material membrane. Due to a heavily damped boundary condition, the effective diameter of the aperture decreases exponentially with time as D(t)=D0e−btD(t) = D_0 e^{-bt}, where D0D_0 is the initial diameter in meters (m\text{m}) at t=0t=0 and bb is a positive damping constant in inverse seconds (s−1\text{s}^{-1}). The telescope is used to resolve two distant point sources emitting light of wavelength λ\lambda (in m\text{m}). Assuming the standard Rayleigh criterion applies instantaneously, what is the time-averaged angular resolution limit ⟨θ⟩\langle \theta \rangle of the telescope over the time interval from t=0t=0 to t=1/bt=1/b?

Options

  • A.

    1.22λD0(e−1)\frac{1.22 \lambda}{D_0} (e - 1)

  • B.

    1.22λD0(ee−1)\frac{1.22 \lambda}{D_0} \left( \frac{e}{e - 1} \right)

  • C.

    1.22λD0(1−1e)\frac{1.22 \lambda}{D_0} \left( 1 - \frac{1}{e} \right)

  • D.

    1.22λD0e\frac{1.22 \lambda}{D_0} e

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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

Angular Resolution Limit|Boundary Condition Reflection|Damped Oscillation Equation

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to correctly apply time-averaging to a non-linear, time-dependent physical limit.

The primary trap lies in the mathematical order of operations regarding averages.

Students often incorrectly assume that the average of a function's reciprocal is the reciprocal of the function's average (i.e., ⟨1/D⟩=1/⟨D⟩\langle 1/D \rangle = 1/\langle D \rangle).

If a student calculates the time-averaged diameter ⟨D⟩\langle D \rangle first and then substitutes it into the Rayleigh criterion, they will arrive at the incorrect Option B.

The correct physical approach requires recognizing that the resolution limit itself is the quantity being averaged, necessitating the integration of ebte^{bt} rather than e−bte^{-bt}.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterWave Optics - Class 12
ConceptAngular Resolution Limit
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree