Start Learning Free

Before You Begin

Question #43240

Difficulty: 8/10 | 🟢 Free

The Question

In a modified Young's Double Slit Experiment, the space between the slit plane (at x=0x = 0) and the screen (at x=Dx = D) is completely filled with a non-uniform transparent medium. The refractive index of this medium varies spatially as n(y)=n0+kyn(y) = n_0 + k y, where yy is the transverse coordinate parallel to the slit plane, n0n_0 is a dimensionless constant, and kk is a positive constant in m−1\text{m}^{-1}. The two slits S1S_1 and S2S_2 are located at y=d/2y = d/2 and y=−d/2y = -d/2 respectively. Assuming d≪Dd \ll D and considering only paraxial rays (y≪Dy \ll D) such that the geometric paths can be approximated as straight lines, find the yy-coordinate of the central bright fringe on the screen.

Options

  • A.

    kD22n0\frac{k D^2}{2 n_0}

  • B.

    kD2n0\frac{k D^2}{n_0}

  • C.

    0

  • D.

    kD24n0\frac{k D^2}{4 n_0}

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

Phase and Path Difference|Optical Path Difference|Calculus/Integration

Why This Question Is Tricky

Common Mistakes

This question tests the transition from algebraic optics to wave optics in a non-homogeneous medium.

Students often fall into the trap of assuming the path difference is generated solely by the local refractive index at the slits, effectively treating the medium as two separate uniform blocks of n1=n0+kd/2n_1 = n_0 + kd/2 and n2=n0−kd/2n_2 = n_0 - kd/2.

This incorrect algebraic approach yields Δx=−kdD+n0yd/D=0\Delta x = -kdD + n_0 y d / D = 0, leading to an erroneous shift of kD2/n0k D^2 / n_0 (Option B).

The correct physical execution requires integrating the spatially varying phase accumulation ∫n(r)dl\int n(r) dl along the wavefront's path, revealing that the average refractive index along each specific ray dictates the true optical path length.

Quick Facts About This Question

Question Information

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