JEE Advanced · 2016 · 22 May 2016 · Paper 1 · Geometrical Optics

Snell's Law at boundaries: At the top surface: $n_1\sin {\theta }_i=n(0)\sin {\theta }_0$

At the bottom surface: $n(d)\sin {\theta }_d=n_2\sin {\theta }_f$

Since the ray passes through the continuously varying medium, by the principle of continuity and Snell's law applied at each infinitesimal layer, we have:

$n_1\sin {\theta }_i=n(z)\sin \theta (z)=n_2\sin {\theta }_f$

Therefore, $n_1\sin {\theta }_i=n_2\sin {\theta }_f$. (A) is correct, (B) is incorrect.

The lateral displacement $l$ depends on the path the ray takes through the slab, which is determined by $n(z)$. However, once ${\theta }_i$ is fixed, ${\theta }_f$ is determined by the relation above, independent of the specific form of $n(z)$ or $n_2$ (as long as the ray emerges). The displacement $l$ depends on $n(z)$ but not on $n_2$. (C) and (D) are correct.