A transparent slab of thickness d has a refractive index n(z) that increases with z. Here, z is the

Q: Why can we directly equate $n_1 \sin\theta_i$ and $n_2 \sin\theta_f$ without worrying about the slab?

Because the slab's interfaces are parallel! Snell's law applied successively across parallel layers ($n_1 \sin\theta_1 = n_2 \sin\theta_2 = n_3 \sin\theta_3 \dots$) proves that the product $n \sin\theta$ is conserved. The intermediate layers cancel out.

Q: Doesn't the bottom medium $n_2$ pull the ray and affect its path inside the slab?

No! Light doesn't "know" what medium is coming next. Its path at any depth $z$ is determined solely by the local refractive index $n(z)$ and the initial invariant $n_1 \sin\theta_i$.

Q: How do we actually calculate the exact value of the lateral displacement $l$?

You would need the exact mathematical function for $n(z)$. Once you have $n(z)$, you find $\sin\theta(z)$, convert it to $\tan\theta(z)$, and integrate $\int \tan\theta(z) \, dz$ from $0$ to $d$.

Q: What would happen if $n(z)$ decreased with depth instead of increasing?

The ray would bend away from the normal instead of towards it. If it bends enough to reach $\theta = 90^\circ$, it will suffer Total Internal Reflection (TIR) inside the slab and curve back upwards!

Q: Is the emergent ray always parallel to the incident ray?

Only if the top and bottom media are identical ($n_1 = n_2$). If $n_1 \neq n_2$, the emergent ray will be at a different angle, as dictated by $n_1 \sin\theta_i = n_2 \sin\theta_f$.

Question

A transparent slab of thickness $d$ has a refractive index $n(z)$ that increases with $z$. Here, $z$ is the vertical distance inside the slab, measured from the top. The slab is placed between two media with uniform refractive indices $n_1$ and $n_2(> n_1)$, as shown in the figure. A ray of light is incident with angle $	heta_i$ from medium 1 and emerges in medium 2 with refraction angle $	heta_f$ with a lateral displacement $l$.

Options:

(A) $l$ is dependent on $n(z)$

(B) $n_1 \sin 	heta_i = (n_2 - n_1) \sin 	heta_f$

(C) $n_1 \sin 	heta_i = n_2 \sin 	heta_f$

(D) $l$ is independent of $n_2$

Sigma Percentile · Accepted Answer

### The Magic of Continuously Varying Media

Imagine you are standing at the edge of a crystal-clear lake, looking down at a coin at the bottom. The light travels from the water to the air, bending at the interface. That is standard refraction. But what if the water wasn't uniform? What if it was perfectly pure at the top, but got progressively denser and saltier as you went deeper?

This is exactly the thrilling scenario we are exploring in this problem. We have a transparent slab where the refractive index $n(z)$ is not a single constant number. Instead, it increases continuously with the depth $z$.

This might look terrifying at first glance. How do we apply Snell's Law when there isn't just one interface, but an infinite number of them? Let's take a breath and break it down.

### The Master Equation: Generalized Snell's Law

When faced with a continuously varying medium, the best mental model is to slice the medium into an infinite number of incredibly thin, parallel layers. Each layer has a constant refractive index, but the index changes slightly from one layer to the next.

If we apply Snell's Law across the first boundary, we get:
$$n_1 \sin	heta_i = n_{slab, 1} \sin	heta_1$$

Across the second boundary:
$$n_{slab, 1} \sin	heta_1 = n_{slab, 2} \sin	heta_2$$

Do you see the beautiful pattern emerging? The product of the refractive index and the sine of the angle remains perfectly conserved throughout the entire journey!

$$n_1 \sin	heta_i = n(z) \sin	heta(z) = n_2 \sin	heta_f$$

This is the Generalized Snell's Law. It is a powerful invariant of the system. Because of this, we can completely ignore the messy, complicated path inside the slab if we only care about the final emergent angle. We can directly equate the initial state to the final state:

$$n_1 \sin	heta_i = n_2 \sin	heta_f$$

This immediately tells us that option (c) is absolutely correct, and option (b) is a trap!

### Decoding the Lateral Displacement

Now, let's tackle the lateral displacement $l$. This is the total horizontal distance the ray shifts while traveling through the slab.

To find this, we need to zoom in on a microscopic section of the ray's path. In a tiny vertical drop $dz$, the ray moves horizontally by a tiny amount $dx$. From simple trigonometry, we can see that:

$$dx = 	an	heta \, dz$$

To find the total lateral displacement $l$, we must integrate this horizontal shift over the entire thickness of the slab, from $z = 0$ to $z = d$:

$$l = \int_0^d 	an	heta \, dz$$

### The Final Verdict

Look closely at the integral for $l$. It depends on the angle $	heta$ at every depth $z$. And what determines $	heta$? Our master equation!

$$\sin	heta = \frac{n_1 \sin	heta_i}{n(z)}$$

This means $	heta$ is entirely dictated by the initial conditions ($n_1$ and $	heta_i$) and the refractive index profile of the slab $n(z)$.

Notice what is completely missing from this equation? The refractive index of the bottom medium, $n_2$! The ray's path inside the slab is already decided before it ever reaches the bottom interface. Therefore, the lateral displacement $l$ depends heavily on $n(z)$, but is completely independent of $n_2$.

This brilliant realization confirms that options (a) and (d) are also correct.

Physics is often about finding what changes and what stays the same. By identifying the conserved quantity in this varying medium, we unlocked the entire problem. Keep this invariant principle in your toolkit; it is a favorite concept for JEE Advanced!

Select Answer:

Visualized Solution

$Analyzing the Setup$

$Generalized Snell’s Law$

$Applying Snell’s Law$

$Lateral Displacement l$

$Dependency of l$

$Final Conclusion$

The Sigma Insight: Refraction and Total Internal Reflection

The Magic of Continuously Varying Media

The Master Equation

Decoding the Lateral Displacement

The Final Verdict

Similar Questions

A ray of light passes through four transparent media with refractive indices $μ_{1}$ , $μ_{2}$ , $μ_{3}$ and $μ_{4}$ as shown in the figure. The surfaces of all media are parallel. If the emergent ray $C D$ is parallel to the incident ray $A B$ , we must have

A ray of light $A O$ in vacuum is incident on a glass slab at angle $6 0^{\circ}$ and refracted at angle $3 0^{\circ}$ along $O B$ as shown in the figure. The optical path length of light ray from $A$ to $B$ is