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Question #43242

Difficulty: 8/10 | 🟢 Free

The Question

Two identical coherent sources $S_1$ and $S_2$ are placed at the ends of a narrow 1D waveguide of length $d = 10\text{ m}$ . They emit waves in phase with the same initial amplitude. The waveguide is filled with a dissipative medium where the amplitude of a wave decays exponentially with distance, halving every $5\text{ m}$ . A detector moves along the waveguide to observe the interference pattern. Assuming the wavelength $\lambda$ is very small compared to the attenuation length, what is the fringe visibility $V = \frac{I_{\text{max}} - I_{\text{min}}}{I_{\text{max}} + I_{\text{min}}}$ of the local interference pattern at a distance of $x = 2.5\text{ m}$ from $S_1$ ?

Options

A.
$0.80$
B.
$0.60$
C.
$\frac{2\sqrt{2}}{3}$
D.
$1.00$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Intuition_Breaker

Train your brain to resist "obvious" answers that are designed to deceive your intuition.

Concepts Behind This Question

Phase and Path Difference|Damped Oscillation Equation|Approximations

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to adapt the standard interference model to a dissipative medium.

The primary trap lies in confusing the attenuation of amplitude with the attenuation of intensity, or incorrectly applying the visibility formula directly to the individual intensities ( $V = \frac{I_1 - I_2}{I_1 + I_2}$ ) rather than using the superposition of amplitudes ( $I_{\text{max/min}} = (\sqrt{I_1} \pm \sqrt{I_2})^2$ ).

Students who mistakenly assume intensity halves every $5\text{ m}$ will arrive at $\frac{2\sqrt{2}}{3}$ , while those who forget the cross-term in interference will get $0.60$ .

This forces a rigorous application of wave superposition where spatial amplitude decay dictates the local contrast of the interference fringes.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Wave Optics - Class 12
Concept	Phase and Path Difference
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free