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Question #76742

Difficulty: 8/10 | 🟢 Free

The Question

In a Young's Double Slit Experiment (YDSE), the slit separation is d=1.5 mmd = 1.5 \text{ mm} and the screen distance is D=2.0 mD = 2.0 \text{ m}. A monochromatic light of wavelength λ=6000 A˚\lambda = 6000 \text{ \AA} and intensity I0=10 W/m2I_0 = 10 \text{ W/m}^2 is used. A transparent thin film of thickness t=15.0 \mumt = 15.0 \text{ \mu m} and initial refractive index μ=1.50\mu = 1.50 is placed in front of one of the slits. The position of the central maximum on the screen is measured using a traveling microscope with a least count of LC=0.02 mmLC = 0.02 \text{ mm}. If the refractive index of the film is slowly varied while keeping its thickness strictly constant, what is the minimum change in the refractive index (Δμ\Delta \mu) that can be confidently detected as a shift in the central maximum by this microscope?

Options

  • A.

    1.0×10−31.0 \times 10^{-3}

  • B.

    4.0×10−24.0 \times 10^{-2}

  • C.

    1.331.33

  • D.

    3.0×10−53.0 \times 10^{-5}

Check Your Answer

Correct Answer

Step-by-Step Solution

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

Least Count and Resolution Limits|Central Fringe Shift

Why This Question Is Tricky

Examiner Levers

Extraneous Data

Provides unnecessary variables (Red Herrings) to test conceptual focus.

Difficulty Boost : +3

Dimensional Analysis

Tests if the student verifies units/dimensions before selecting.

Difficulty Boost : +2

Common Mistakes

This question tests the student's ability to bridge deterministic wave optics with experimental error analysis.

A common trap is mechanically associating 'minimum detectable shift' with the wavelength λ\lambda or the fringe width β\beta, ignoring the macroscopic measuring instrument's limitations.

Another lethal trap is confusing the optical path difference change (Δx=tΔμ\Delta x = t \Delta \mu) with the physical geometric shift on the screen (Δy\Delta y), leading students to equate the least count directly to the path difference (Option C).

The correct logic requires realizing that physical observability is bounded by the measuring device's least count, and one must use the geometric magnification factor (D/dD/d) to translate optical path changes into measurable screen distances.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterUnits and Measurements - Class 11 Part 1
ConceptLeast Count and Resolution Limits
Difficulty8
Cognitive Load7
Target Solve Time170 sec
AccessFree