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Question #43243

Difficulty: 8/10 | 🟢 Free

The Question

A plane monochromatic wave of wavelength λ\lambda is incident normally on a rectangular slit of width aa. A transparent glass plate of uniform thickness tt is placed immediately behind the slit, covering it completely. The refractive index of the glass varies with temperature TT as μ(T)=μ0+γT\mu(T) = \mu_0 + \gamma T, where μ0\mu_0 and γ\gamma are positive constants. A steady temperature gradient is established along the width of the slit such that the temperature varies linearly from T0T_0 at the lower edge to T0+ΔTT_0 + \Delta T at the upper edge. Assuming thermal expansion of the glass and the slit is negligible, what is the minimum positive temperature difference ΔT\Delta T required so that the intensity of light at the geometric center of the diffraction pattern (i.e., at an angle θ=0∘\theta = 0^\circ) becomes zero?

Options

  • A.

    λγt\frac{\lambda}{\gamma t}

  • B.

    λ2γt\frac{\lambda}{2 \gamma t}

  • C.

    2λγt\frac{2 \lambda}{\gamma t}

  • D.

    λγa\frac{\lambda}{\gamma a}

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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|Diffraction vs Interference|Angular Resolution Limit

Why This Question Is Tricky

Common Mistakes

This question tests the understanding of Fraunhofer diffraction and the formation of minima in a single slit.

A common and lethal misconception is applying the destructive interference condition for two discrete point sources (where a phase difference of π\pi or path difference of λ/2\lambda/2 causes cancellation) to a continuous aperture.

For a continuous single slit, the first minimum occurs when the phase difference between the extreme edges is 2π2\pi, allowing the slit to be conceptually divided into two halves that perfectly cancel each other.

Students who incorrectly assume a π\pi phase difference is sufficient for the entire slit will fall into the trap and choose the option corresponding to half the required temperature difference.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterWave Optics - Class 12
ConceptPhase and Path Difference
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree