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Question #43118

Difficulty: 8/10 | 🟢 Free

The Question

A point object is placed at the bottom of a transparent slab of thickness $t = 12 \text{ cm}$ . The refractive index of the slab is not uniform; it varies with the vertical distance $y$ from the bottom surface according to the relation $n(y) = \frac{n_0}{1 + \alpha y}$ , where $n_0 = 3$ and $\alpha = \frac{1}{12} \text{ cm}^{-1}$ . An observer looks at the object normally from above the slab (in air, $n_{\text{air}} = 1$ ). What is the apparent shift in the position of the object?

Options

A.
$6 \text{ cm}$
B.
$\frac{20}{3} \text{ cm}$
C.
$8 \text{ cm}$
D.
$4 \text{ cm}$

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

Apparent Depth/Shift|Dielectric Slab Displacement|Integration

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to transition from algebraic formulas to calculus-based optical path evaluation in non-homogeneous media.

The primary trap is calculating an average refractive index $n_{\text{avg}} = 2.25$ and applying the standard shift formula $\Delta t = t(1 - 1/n_{\text{avg}})$ , which incorrectly yields $20/3 \text{ cm}$ .

This assumes the optical path length scales linearly with the arithmetic mean of the refractive index.

By recognizing the analogy to a non-uniform capacitor (where equivalent air thickness $dy/n(y)$ acts like series capacitance $dy/\varepsilon(y)$ ), students must integrate the differential apparent depth to find the true shift.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Ray Optics and Optical Instruments - Class 12
Concept	Apparent Depth/Shift
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free