Animated Solution for Physics - Optics: A light beam is travelling from Region I to Region IV (Refer Figure). The refractive index in Regions I, II, III and IV are $n_0$, $\frac{n_0}{2}$, $\frac{n_0}{6}$ and $\frac{n_0}{8}$, respectively. The angle of incidence $\theta$ for which the beam just misses entering Region IV is

This problem is a beautiful demonstration of how light behaves when passing through multiple parallel transparent media. At first glance, calculating the angles of refraction at every single interface might seem like a tedious task. However, physics offers us an elegant shortcut.

Imagine a light beam traveling through a series of parallel regions, each with a progressively decreasing refractive index. We are given four regions: Region I, II, III, and IV, with refractive indices $n_0$, $\frac{n_0}{2}$, $\frac{n_0}{6}$, and $\frac{n_0}{8}$ respectively.

The question asks for the angle of incidence $\theta$ such that the beam "just misses" entering Region IV. Physically, this means the light ray must undergo Total Internal Reflection (TIR) exactly at the boundary between Region III and Region IV. When a ray "just misses" a medium, it grazes the interface, meaning the angle of refraction in that final medium becomes exactly $90^{\circ }$.

Generalized Snell's Law

Here is the powerful trick: for parallel interfaces, Snell's Law is conserved across all layers. You don't need to calculate the intermediate angles ${\theta }_2$ and ${\theta }_3$. The generalized form of Snell's Law states:

Because all these terms are equal, we can directly equate the term for the first region to the term for the last region:

Let's substitute the known values into our master equation. For Region I, the refractive index is $n_0$ and the angle of incidence is $\theta$. For Region IV, the refractive index is $\frac{n_0}{8}$ and the angle of refraction is $90^{\circ }$ (the grazing condition).

Since $\sin 90^{\circ }=1$, the equation simplifies beautifully. The $n_0$ terms cancel out on both sides:

Therefore, the required angle of incidence is:

Before we celebrate, there is a catch! We must ensure that Total Internal Reflection didn't accidentally happen prematurely in Region II or Region III. If it did, the ray would never even reach the final boundary.

We know the constant value for Snell's Law across all layers is $\frac{n_0}{8}$. Let's check the sine values for the intermediate regions:

For Region II:

For Region III:

Since both $\sin {\theta }_{II}$ and $\sin {\theta }_{III}$ are strictly less than $1$, TIR does not occur at the first or second interfaces. The ray safely reaches the boundary of Region IV, confirming our answer is perfectly correct.