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Question #91072

Difficulty: 5/10 | 🟢 Free

The Question

In a modified Young's Double Slit Experiment, two coherent point sources S1S_1 and S2S_2 emit light of wavelength λ\lambda. The amplitude of light from S1S_1 decays exponentially with time as A1(t)=A0e−γtA_1(t) = A_0 e^{-\gamma t} (where γ>0\gamma > 0), while the amplitude of light from S2S_2 remains constant at A2(t)=A0A_2(t) = A_0. At t=0t=0, both sources have the same intensity I0I_0. A photographic plate placed on the observation screen is exposed to the interference pattern from t=0t=0 to t=ln⁡2γt = \frac{\ln 2}{\gamma}. The visibility of the fringes on the developed photographic plate is defined as V=Emax−EminEmax+EminV = \frac{E_{max} - E_{min}}{E_{max} + E_{min}}, where EE is the total time-integrated light energy per unit area. Which of the following gives the correct value of VV?

Options

  • A.

    83+8ln⁡2\frac{8}{3 + 8\ln 2}

  • B.

    8−421+2ln⁡2\frac{8 - 4\sqrt{2}}{1 + 2\ln 2}

  • C.

    45\frac{4}{5}

  • D.

    43+8ln⁡2\frac{4}{3 + 8\ln 2}

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

Intensity Distribution in YDSE|Damped Oscillation Equation

Why This Question Is Tricky

Examiner Levers

Dimensional Analysis

Tests if the student verifies units/dimensions before selecting.

Difficulty Boost : +2

Common Mistakes

This question tests the student's ability to handle non-stationary interference patterns by integrating the time-varying intensity over the exposure duration.

A primary trap is the misapplication of the intensity-amplitude relation; students often incorrectly assume intensity decays at the same rate as amplitude (I∝e−γtI \propto e^{-\gamma t} instead of e−2γte^{-2\gamma t}), leading to Option B.

Another common mistake is calculating the instantaneous visibility at the end of the exposure time (t=Tt=T) instead of using the time-integrated energy fluence, which leads to Option C.

The problem enforces the physical reality that photographic detectors record the time-integral of the Poynting vector, not a steady-state snapshot.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterWave Optics - Class 12
ConceptIntensity Distribution in YDSE
Difficulty5
Cognitive Load6
Target Solve Time140 sec
AccessFree