Animated Solution for Physics - Optics: 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 $n_1$ to $n_2$ over the height h. Which of the following statement(s) is (are) true about the light wave emerging out of the slab ?

Imagine you are standing on a plane wavefront of light, surfing your way through the vacuum of space. Suddenly, you hit a glass slab. But this isn't just any glass slab; it's a magical one where the optical density changes as you move from the bottom to the top. I know this might look terrifying at first glance, but let's take a breath and break it down. Notice how elegantly the physics unfolds when we track the journey of this wavefront.

We are given a glass slab of thickness $d$ and height $h$. The crucial piece of information is that the refractive index $n$ is not constant. It increases linearly from $n_1$ at the bottom to $n_2$ at the top, with $n_2>n_1$.

What does this mean physically? We know that the speed of light in a medium is inversely proportional to its refractive index, given by the relation $v=\frac{c}{n}$. Because the refractive index is higher at the top, light will travel slower there compared to the bottom. This differential speed is the engine that drives the entire phenomenon we are about to witness.

To understand what happens to the light wave, we need to invoke Huygens' Principle. Instead of tracking individual rays using a complex version of Snell's Law, we will track a single, continuous wavefront. A wavefront is simply a surface over which the light wave has the exact same phase.

Let's consider a perfectly vertical plane wavefront entering the left face of the slab at time $t=0$. Our goal is to find the shape and orientation of this wavefront the moment it completely emerges from the right face of the slab.

Let's focus on two extreme points on our wavefront: the very top point and the very bottom point.

The top point travels through a region with refractive index $n_2$. The time it takes to cross the thickness $d$ of the slab is:

Now, what about the bottom point? It travels through a region with a lower refractive index $n_1$, meaning it moves faster. It will cross the slab in a shorter time:

Because the bottom point crosses the slab faster, it emerges into the air while the top point is still struggling through the glass. By the time the top point finally emerges at $t=t_{top}$, the bottom point has been traveling freely in the air for an extra amount of time $\Delta t$:

In this extra time, how far does the bottom point travel in the air? Since it's in the air, it travels at the speed of light $c$. The extra distance $\ell$ it covers is:

Now, picture the emerging wavefront. The top point is just at the exit face of the slab, while the bottom point is a distance $\ell$ ahead of it. If we connect these two points, we get a straight line that is tilted forward at the bottom.

Let $\theta$ be the angle this tilted wavefront makes with the vertical. From simple right-triangle geometry, we can see that:

Substituting our expression for $\ell$, we get:

Therefore, the angle is $\theta ={\tan }^{-1}\left[\frac{(n_2-n_1)d}{h}\right]$.

Since light rays are always perpendicular to their wavefronts, the rays will deflect upwards by this exact same angle $\theta$. This perfectly matches option (B). Furthermore, looking at our final formula, we can clearly see that the deflection angle depends only on the difference $(n_2-n_1)$, and not on their individual absolute values. This confirms option (D) as well.