A monochromatic light wave is incident normally on a glass slab of thickness d, as shown in the figure. The refractive index of the slab increases linearly from n₁ to n₂ over the height h. Which of the following statements is/are true about the light wave emerging out of the slab?

Consider the 6 x 6 square in the figure. Let A1, A2, ..., A49 be the points of intersections (dots in the picture) in some order. We say that A and A are friends if they are adjacent along a row or along a column. Assume that each point A has an equal chance of being chosen.

Correct Options:

1. It will deflect up by an angle
${tan}^{-1}\left[\frac{(n_2-n_1)d}{h}\right]$

2. The deflection angle depends only on n₂ − n₁ and not on the individual values of n₁ and n₂.

Step-by-step Explanation:

1. Understanding the Physical System:
As shown in the figure:

A plane, monochromatic light wave with vertical wavefronts is incident normally from the left onto a glass slab of thickness d and height h. The refractive index of the slab increases linearly from n₁ at the bottom to n₂ at the top (where n₂ > n₁).

2. Optical Path Difference Across the Height:
The optical path traveled by light through the slab at any height y is given by:
$\mathrm{\Delta }x\left(y\right)=n\left(y\right)·d$
At the bottom boundary (y = 0), the refractive index is n₁, so the optical path length is:
$\mathrm{\Delta }{x}_1=n_{1}d$
At the top boundary (y = h), the refractive index is n₂, so the optical path length is:
$\mathrm{\Delta }{x}_2=n_{2}d$

3. Phase Difference and Tilt of the Wavefront:
Since n₂ > n₁, the light wave at the top travels through a higher optical path and therefore suffers a greater effective retardation/delay compared to the light ray at the bottom.
The total optical path difference between the top ray and the bottom ray upon emerging from the slab is:
$\mathrm{\Delta }x=\mathrm{\Delta }{x}_2-\mathrm{\Delta }{x}_1=(n_2-n_1)d$

4. Calculating the Angle of Deflection (θ):
Because the wavefront tilts by a spatial path displacement of Δx over a total transverse height h, the emerging wavefront is tilted at an angle θ relative to the vertical.
Since rays propagate perpendicular to the wavefronts, the direction of propagation tilts upward by the exact same angle θ:
$tan\theta =\frac{\mathrm{\Delta }x}{h}=\frac{(n_2-n_1)d}{h}$
Taking the inverse tangent on both sides gives:
$\theta ={tan}^{-1}\left[\frac{(n_2-n_1)d}{h}\right]$

5. Analyzing Dependence:
From the formula derived above, the deflection angle θ depends directly on the difference (n₂ − n₁), thickness d, and height h. It does not depend independently on the individual baseline refractive index values of n₁ or n₂.

Thus, both correct statements are:
• It will deflect up by an angle ${tan}^{-1}\left[\frac{(n_2-n_1)d}{h}\right]$.
• The deflection angle depends only on n₂ − n₁ and not on the individual values of n₁ and n₂.