Animated Solution for Physics - Optics: Monochromatic light is incident on a glass prism of angle $A$. If the refractive index of the material of the prism is $\mu$, a ray incident at an angle $\theta$, on the face $AB$ would get transmitted through the face $AC$ of the prism provided

Imagine a monochromatic light ray striking the first face, $AB$, of a glass prism. It refracts, bending towards the normal, and travels through the glass towards the second face, $AC$. Our ultimate goal is to ensure this ray successfully transmits through face $AC$ and emerges back into the air. This means it must absolutely avoid the trap of Total Internal Reflection (TIR) at the second boundary.

For the ray to emerge from face $AC$, the angle of incidence at this second face, which we will call $r_2$, must be strictly less than the critical angle of the glass-air interface, denoted as $i_C$. If it hits exactly at or above the critical angle, it gets trapped inside the prism!

Mathematically, we write this as:

From the fundamental geometry of a prism, we know a beautiful relation: the sum of the internal angles $r_1$ and $r_2$ is exactly equal to the prism angle $A$.

So, we can elegantly replace $r_2$ with $A-r_1$. Let's substitute this into our condition:

Now, let's take the sine on both sides. Since the sine function is strictly increasing for angles between $0^{\circ }$ and $90^{\circ }$, the inequality sign remains perfectly intact.

Recall that the sine of the critical angle, $\sin i_C$, is simply the ratio of the refractive indices, which in this case is $\frac{1}{\mu }$. So, our inequality transforms into:

Now, let's isolate $r_1$. First, we take the sine inverse on both sides:

Rearranging the terms to get $r_1$ on one side, we find that $r_1$ must be greater than a specific threshold:

Snell's Law

We are almost there! We need the condition on the initial incident angle $\theta$. Let's apply Snell's Law at the first face $AB$. The refractive index of air is $1$, so:

This gives us a direct expression for $\sin r_1$:

To make our substitution seamless, let's take the sine on both sides of our $r_1$ inequality:

Finally, replace $\sin r_1$ with $\frac{\sin \theta }{\mu }$:

Multiply by $\mu$ and take the sine inverse, and boom! We arrive at our final, elegant condition for $\theta$:

The incident angle $\theta$ must be greater than this nested expression to guarantee transmission.