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Question #43733

Difficulty: 8/10 | 🟢 Free

The Question

A monochromatic light of wavelength λ=1.0×10−6 m\lambda = 1.0 \times 10^{-6} \text{ m} is incident normally on a single slit made in a large metallic plate. At time t=0t = 0, the slit width is a0=2.0×10−6 ma_0 = 2.0 \times 10^{-6} \text{ m} and the plate is at a uniform temperature T0T_0. The plate is then heated such that its temperature increases at a constant rate dTdt=10 K s−1\frac{dT}{dt} = 10 \text{ K s}^{-1}. The coefficient of linear expansion of the metal is α=2.0×10−5 K−1\alpha = 2.0 \times 10^{-5} \text{ K}^{-1}. Let θ\theta be the angular position of the first diffraction minimum on one side of the central maximum (where θ>0\theta > 0). Find the rate of change of the angular position, dθdt\frac{d\theta}{dt}, at t=0t = 0.

Options

  • A.

    −23×10−4 rad s−1-\frac{2}{\sqrt{3}} \times 10^{-4} \text{ rad s}^{-1}

  • B.

    +23×10−4 rad s−1+\frac{2}{\sqrt{3}} \times 10^{-4} \text{ rad s}^{-1}

  • C.

    −1.0×10−4 rad s−1-1.0 \times 10^{-4} \text{ rad s}^{-1}

  • D.

    +1.0×10−4 rad s−1+1.0 \times 10^{-4} \text{ rad s}^{-1}

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Correct Answer

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Explanation

How the Examiner Thinks

Multi-Concept_Bridge

Master problems that require linking physics and math concepts to solve complex competitive questions.

Concepts Behind This Question

Single Slit Diffraction Minima|Resolving Power Limits|Expansion of Cavities in Solids

Why This Question Is Tricky

Common Mistakes

This problem tests the intersection of wave optics and thermal physics, specifically targeting two common misconceptions.

First, students often incorrectly assume that a hole or slit in a solid shrinks upon heating due to the material expanding inward; however, the photographic enlargement principle dictates that the cavity expands proportionally (aa increases, leading to a negative rate of change for θ\theta).

Second, students habitually apply the small-angle approximation (sin⁡θ≈θ\sin \theta \approx \theta) for diffraction problems.

Here, the ratio λ/a0=0.5\lambda/a_0 = 0.5 yields a large angle (θ=30∘\theta = 30^\circ), making the approximation invalid and requiring exact implicit differentiation of the diffraction condition to avoid the trap answer of −1.0×10−4 rad s−1-1.0 \times 10^{-4} \text{ rad s}^{-1}.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterWave Optics - Class 12
ConceptSingle Slit Diffraction Minima
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree