JEE Challenger
More from Ray Optics and Optical Instruments
PhyMedAdv 2016 • Paper 1

Refraction and Lateral Displacement in Variable Index Slab

A transparent slab with thickness dd possesses a refractive index n(z)n(z) which increases along zz, where zz represents the vertical depth within the slab measured from its top surface. The slab is situated between two media of constant refractive indices n1n_1 and n2(>n1)n_2 (> n_1), as depicted in the figure. A light ray enters from medium 1 at an angle of incidence θi\theta_i and exits into medium 2 at a refraction angle θf\theta_f, undergoing a lateral shift ll.

Which of the following statement(s) is(are) true?

Question Diagram 1

Options

A

n1sin⁡θi=n2sin⁡θfn_1 \sin \theta_i = n_2 \sin \theta_f

Correct
B

n1sin⁡θi=(n2−n1)sin⁡θfn_1 \sin \theta_i = (n_2 - n_1) \sin \theta_f

C

ll is independent of n2n_2

Correct
D

ll is dependent on n(z)n(z)

Correct

Topics & Concepts

Ray Optics and Optical InstrumentsReflection and Refraction

Step-by-Step Solution

To determine the correct statements, we analyze the refraction of light through a continuously varying medium stratified in parallel layers along the zz-axis.

1. Application of Snell's Law

For a medium whose refractive index n(z)n(z) varies only along the depth zz (parallel planar layers), Snell's law holds across all infinitesimally thin parallel layers: n(z)sin⁡θ(z)=constantn(z) \sin \theta(z) = \text{constant}

At the top boundary, as the light enters from medium 1 into the slab at depth z=0z = 0: n1sin⁡θi=n(0)sin⁡θ(0)n_1 \sin \theta_i = n(0) \sin \theta(0)

At any arbitrary depth zz within the slab: n(z)sin⁡θ(z)=n1sin⁡θin(z) \sin \theta(z) = n_1 \sin \theta_i

At the bottom interface (z=dz = d), as the light exits the slab into medium 2: n(d−)sin⁡θ(d−)=n2sin⁡θfn(d^-) \sin \theta(d^-) = n_2 \sin \theta_f

By equating the conserved quantity throughout the path: n1sin⁡θi=n2sin⁡θfn_1 \sin \theta_i = n_2 \sin \theta_f

Thus, Statement (A) is TRUE and Statement (B) is FALSE.

2. Derivation of Lateral Shift ll

The lateral shift ll is the total horizontal distance traversed by the ray as it travels through the slab from depth z=0z = 0 to z=dz = d.

For an infinitesimal vertical displacement dzdz, the corresponding horizontal displacement dxdx is given by: dx=tan⁡θ(z) dzdx = \tan \theta(z) \, dz

From Snell's law, we have: sin⁡θ(z)=n1sin⁡θin(z)\sin \theta(z) = \frac{n_1 \sin \theta_i}{n(z)}

Using the trigonometric identity tan⁡θ=sin⁡θ1−sin⁡2θ\tan \theta = \frac{\sin \theta}{\sqrt{1 - \sin^2 \theta}}: tan⁡θ(z)=n1sin⁡θin(z)1−(n1sin⁡θin(z))2=n1sin⁡θi[n(z)]2−n12sin⁡2θi\tan \theta(z) = \frac{\frac{n_1 \sin \theta_i}{n(z)}}{\sqrt{1 - \left(\frac{n_1 \sin \theta_i}{n(z)}\right)^2}} = \frac{n_1 \sin \theta_i}{\sqrt{[n(z)]^2 - n_1^2 \sin^2 \theta_i}}

Integrating from z=0z = 0 to z=dz = d, the total lateral shift ll is: l=∫0ddx=∫0dn1sin⁡θi[n(z)]2−n12sin⁡2θi dzl = \int_0^d dx = \int_0^d \frac{n_1 \sin \theta_i}{\sqrt{[n(z)]^2 - n_1^2 \sin^2 \theta_i}} \, dz

From this expression:

  • The lateral shift ll depends directly on the function n(z)n(z) within the slab. Thus, Statement (D) is TRUE.
  • The lateral shift ll depends only on n1n_1, θi\theta_i, dd, and n(z)n(z), and is completely independent of the refractive index n2n_2 of the medium below the slab. Thus, Statement (C) is TRUE.

Conclusion

The correct options are A, C, and D.

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