Animated Solution for Physics - Optics: AB and CD are two slabs. The medium between the slabs has refractive index 2. Find the minimum angle of incidence at Q, so that the ray is totally reflected by both the slabs.

The phenomenon of Total Internal Reflection (TIR) is one of the most fascinating concepts in optics. It’s the reason diamonds sparkle and optical fibers can transmit data across oceans. In this problem, we are challenged to find the minimum angle of incidence that guarantees a light ray is trapped between two parallel interfaces. Let's dive into the physics!

Imagine you are a light ray travelling through a medium with a refractive index of $2$. Above you is a slab with a refractive index of $\sqrt{2}$, and below you is another slab with a refractive index of $\sqrt{3}$. You strike the top interface at an angle $i$.

Our goal is to ensure that you don't escape into the top slab, nor do you escape into the bottom slab after bouncing off the top. You must be totally internally reflected at both boundaries.

For Total Internal Reflection to occur, the light must be travelling from a denser medium to a rarer medium. Furthermore, the angle of incidence must be greater than or equal to the critical angle ${\theta }_C$ for that specific pair of media.

The critical angle is determined by Snell's Law and is given by the elegant relation:

If our angle of incidence $i$ is greater than ${\theta }_C$, the ray is perfectly reflected back into the denser medium.

Let's first evaluate the top interface, where the ray strikes point $Q$. The ray is trying to cross from the middle medium (${\mu }_2=2$) into the top medium (${\mu }_1=\sqrt{2}$).

Applying our master equation, we find the critical angle ${\theta }_{C_1}$ for this interface:

Taking the inverse sine, we get:

This means that to prevent the ray from escaping through the top slab, the angle of incidence $i$ must be at least $45^{\circ }$.

Assuming the ray is successfully reflected at $Q$, it travels downwards and strikes the bottom interface at point $P$. Because the two interfaces are perfectly parallel, the geometry of alternate interior angles dictates that the angle of incidence at $P$ is exactly the same as the angle $i$ at $Q$.

Now, the ray is trying to cross from the middle medium (${\mu }_2=2$) into the bottom medium (${\mu }_3=\sqrt{3}$). Let's find the critical angle ${\theta }_{C_2}$ for this second boundary:

Taking the inverse sine, we find:

To prevent the ray from escaping through the bottom slab, the angle of incidence $i$ must be at least $60^{\circ }$.

We now have two strict conditions that must be satisfied simultaneously for the ray to be trapped:

1. $i\ge 45^{\circ }$ (to reflect off the top slab) 2. $i\ge 60^{\circ }$ (to reflect off the bottom slab)

To satisfy both conditions, $i$ must be greater than or equal to the larger of the two critical angles. Therefore, the minimum angle of incidence that guarantees Total Internal Reflection at both interfaces is:

And there we have it! By simply evaluating the critical angle at each boundary and finding the most restrictive condition, we've successfully trapped the light ray.