Animated Solution for Physics - Optics: A parallel beam of light is incident from air at an angle $\alpha$ on the side $PQ$ of a right angled triangular prism of refractive index $n=\sqrt{2}$. Light undergoes total internal reflection in the prism at the face $PR$ when $\alpha$ has a minimum value of $45^{\circ }$. The angle $\theta$ of the prism is

The journey of a light ray through a prism is a beautiful interplay of Snell's Law and geometry. In this problem, we are tasked with finding the apex angle $\theta$ of a right-angled prism, given the limiting condition for Total Internal Reflection (TIR).

Imagine a parallel beam of light striking the vertical face $PQ$ of a right-angled prism. The ray enters from air into the denser glass medium (with refractive index $n=\sqrt{2}$) at an angle of incidence $\alpha$. It bends towards the normal, traveling through the prism until it hits the second face $PR$. Here, it undergoes Total Internal Reflection.

Let's focus on the first interface, $PQ$. Applying Snell's Law, we can relate the angle of incidence $\alpha$ to the angle of refraction $r_1$:

We are given that the minimum value of $\alpha$ for TIR to occur is $45^{\circ }$. Substituting $\alpha =45^{\circ }$ and $n=\sqrt{2}$:

Solving for $\sin r_1$: This tells us that the angle of refraction $r_1$ is exactly $30^{\circ }$.

Now, let's follow the ray to the second face, $PR$. For TIR to happen, the angle of incidence at this face, let's call it $r_2$, must be greater than or equal to the critical angle ${\theta }_c$.

The critical angle is determined by the refractive index:

Here is the crucial logical step: The problem states that $\alpha$ is at its minimum value. A smaller $\alpha$ results in a smaller $r_1$, which geometrically leads to a smaller $r_2$. Therefore, for the limiting case of TIR, $r_2$ must exactly equal the critical angle.

To connect $\theta$, $r_1$, and $r_2$, we look at the triangle $\Delta PNM$ formed by the apex $P$ and the two points $M$ and $N$ where the ray intersects the prism faces.

The sum of the interior angles of this triangle must be $180^{\circ }$. Let's determine these angles: 1. The top angle is simply the prism angle $\theta$. 2. At point $M$, the normal is horizontal, so the interior angle is $90^{\circ }+r_1$. 3. At point $N$, the normal is perpendicular to the face $PR$, making the interior angle $90^{\circ }-r_2$.

Summing them up:

Notice how the $90^{\circ }$ terms beautifully cancel out with the $180^{\circ }$, leaving us with a very elegant relation:

We have all the pieces of the puzzle! Substitute $r_2=45^{\circ }$ and $r_1=30^{\circ }$ into our geometric relation: