Animated Solution for Physics - Optics: A wide slab consisting of two media of refractive indices $n_1$ and $n_2$ is placed in air as shown in the figure. A ray of light is incident from medium $n_1$ to $n_2$ at an angle $\theta$, where $\sin \theta$ is slightly larger than $1/n_1$. Take refractive index of air as 1. Which of the following statement(s) is(are) correct?

Imagine a ray of light navigating through a multi-layered optical slab. It starts in a medium with refractive index $n_1$, hits a second medium $n_2$, and eventually faces the ultimate boundary: air. The problem gives us one seemingly simple, yet profoundly powerful inequality: $\sin \theta >\frac{1}{n_1}$. This single mathematical statement is the master key that locks the light ray inside the slab forever. Let's unravel why.

Snell's Law Across Parallel Boundaries When dealing with multiple parallel optical interfaces, Snell's Law is our best friend. It tells us that the product of the refractive index and the sine of the angle of the ray with the normal is a conserved quantity across all parallel layers.

If we assume the ray manages to refract through medium $n_2$ and reaches the air interface (where $n_{\text{air}}=1$) at an angle ${\theta }_3$, we can write the continuous Snell's Law equation:

Notice something incredible here? The middle medium $n_2$ completely drops out of the relationship between the initial medium and the final medium! We can directly equate the first and last states:

We are given that $\sin \theta >\frac{1}{n_1}$. Let's substitute this inequality into our derived equation:

We arrive at the conclusion that $\sin {\theta }_3>1$. But wait, the sine of any real angle can never exceed 1! What does this mathematical impossibility mean physically? It means that our initial assumption—that the ray reaches and refracts into the air—is fundamentally flawed. The ray must undergo Total Internal Reflection (TIR) either before it reaches the air, or exactly at the air interface. It is physically trapped!

Armed with this absolute truth, let's evaluate the given options.

Option (A): What if $n_1=n_2$? If the two media have the same refractive index, the ray travels in a straight line until it hits the air interface. But as we just proved, it will face TIR at the air boundary because $\sin {\theta }_3>1$. The ray does not enter the air. Option A is incorrect.

Option (B): What if $n_2<n_1$? Here, the ray is traveling from a denser to a rarer medium at the first interface. It might suffer TIR right there if the angle is large enough. If it doesn't, it refracts into $n_2$ and travels to the air interface. But we already know it cannot escape into the air! It will suffer TIR at the top boundary, reflect back into $n_2$, and eventually refract back into $n_1$. In either case, it returns to $n_1$. Option B is correct.

Option (C): What if $n_2>n_1$? In this scenario, the ray travels from a rarer to a denser medium, so it bends towards the normal. TIR is impossible at the first interface. The ray safely reaches the air interface. However, the inescapable truth remains: $\sin {\theta }_3>1$. The ray undergoes TIR at the air boundary, reflects back into $n_2$, and then refracts back into $n_1$. Option C is correct.

Option (D): What if $n_2=1$? If $n_2$ is air, the very first interface is the $n_1$-air boundary. The critical angle for this interface is given by $\sin {\theta }_c=\frac{1}{n_1}$. Since we are explicitly given that $\sin \theta >\frac{1}{n_1}$, the incident angle is strictly greater than the critical angle. The ray undergoes TIR immediately and reflects back into $n_1$. Option D is correct.

By recognizing that Snell's Law is conserved across parallel layers, we bypassed complex intermediate calculations and went straight to the heart of the physics. The light ray, bound by the laws of optics, is forever destined to return to its origin.