Animated Solution for Physics - Optics: A rectangular glass slab $ABCD$ of refractive index $n_1$ is immersed in water of refractive index $n_2(n_1>n_2)$. A ray of light is incident at the surface $AB$ of the slab as shown. The maximum value of the angle of incidence ${\alpha }_{\text{max}}$, such that the ray comes out only from the other surface $CD$, is given by

A Ray's Journey

Imagine a ray of light entering a rectangular glass slab. It refracts at the first vertical face, travels through the dense glass, and strikes the top horizontal face. The problem presents a strict condition: we want the ray to emerge only from the opposite vertical face $CD$.

What does this mean physically? It implies that the ray must be completely trapped inside the slab when it hits the top face $AD$. It must not escape into the surrounding water. This phenomenon, where light is perfectly reflected back into the denser medium, is known as Total Internal Reflection (TIR).

Total Internal Reflection

For TIR to occur at the top face $AD$, the angle of incidence there—let's call it $r_2$—must be greater than or equal to the critical angle ${\theta }_C$ of the glass-water interface.

We are tasked with finding the maximum possible entry angle, ${\alpha }_{\text{max}}$. By Snell's Law, a larger entry angle $\alpha$ will result in a larger angle of refraction $r_1$. As we will see shortly, a larger $r_1$ forces $r_2$ to become smaller. Therefore, to push $\alpha$ to its absolute maximum limit, we must push $r_2$ to its absolute minimum limit, which is exactly the critical angle.

Now, let's look at the geometry of the refracted ray inside the slab. The normal at the first face $AB$ is horizontal, while the normal at the top face $AD$ is vertical. These two normals intersect at a perfect $90^{\circ }$ angle, forming a right-angled triangle with the ray's path.

Because the sum of angles in a triangle is $180^{\circ }$, the angle of refraction $r_1$ and the angle of incidence $r_2$ must be complementary. They add up to exactly $90^{\circ }$.

Since $r_1$ and $r_2$ are complementary, when $r_2$ is at its minimum, $r_1$ will be at its maximum. Let's substitute the critical angle for $r_2$ to find the maximum value of $r_1$.

Let's bring our focus back to the first face, $AB$. The ray is entering from water (refractive index $n_2$) into the glass (refractive index $n_1$). Applying Snell's Law at this interface gives us:

Now, we substitute our geometric expression for $(r_1{)}_{\text{max}}$ into Snell's Law.

From basic trigonometry, we know that $\sin (90^{\circ }-\theta )=\cos \theta$. This simplifies our equation beautifully.

We are almost at the finish line! We need to express $\cos {\theta }_C$ in terms of the given refractive indices. By definition, the sine of the critical angle is the ratio of the outer refractive index to the inner one.

This means the critical angle itself is:

Let's plug this definition of ${\theta }_C$ right into our cosine term.

Finally, to isolate ${\alpha }_{\text{max}}$, we take the sine-inverse of the entire expression on the right side.

And there we have it! This is the maximum angle of incidence that guarantees the ray will remain trapped inside the slab and only exit from the opposite face. This elegant interplay of Snell's law and geometry is the exact foundational principle behind optical fibers, which trap light inside them to transmit data over vast distances across the globe!