light ray undergoing total internal reflection thrice inside a prism

Question

Can it ever be possible that a light ray going inside a triangular prism of refractive index n having refracting angle A undergo total internal reflection thrice before emerging out?

Note:I know it "may" be possible if the medium surrounding the prism from the three sides is different but I am asking for the case when the medium is same on all the three sides.

Answer 1

The answer is YES. See diagram:

The key here is that you can write down the expressions for the angles $a_2, a_3, a_4$ in terms of the angles of the prism $\alpha, \beta, \gamma$ and the first angle $a_1$. Now $\alpha + \beta + \gamma = 180$ and you know you must have total internal reflection with the first three angles, but not with $a_4$.

$$a_2 = \gamma - a_1\\ a_3 = \beta - a_2\\ a_4 = \alpha - a_3$$

So expressing $a_4$ in terms of $a_1$ we get

$$\begin{align}\\ a_4 &= \alpha - (beta - (\gamma - a_1))\\ &= \alpha - \beta + \gamma - a_1\\ &= \alpha + \beta + \gamma - (a_1 + 2\beta)\\ &= 180 - (a_1 + 2\beta)\end{align}$$

Playing around a bit with these equations, I find that the optimal solution is obtained with

$$\alpha = 36\\ \beta = 72\\ \gamma = 72\\ a_1 = 36$$

In this case, you have $a_2 = 36, a_3 = 36, a_4 = 0$. In other words, as long as you can get total internal reflection with an angle of incidence of 36 degrees, you can do it - and as a bonus, you go "straight in" and come "straight out" - see below (approximately accurate):

This requires a refractive index just greater than 1.7 ($1/\sin(36)$ - definitely possible.

If you are willing to have the exit direction ($a_4$) be something other than zero, then you can improve on the above solution (make it possible with lower refractive index).

Answer 2

This is not a complete answer, just a possible way to handle the problem.

I assume that the transversal section of your prism is a triangle with all the angles equal to 60⁰. Let's denote by A, B, C, the three corners of this triangle. Let's denote by $P_1$ the point of incidence on the prism, and then by $P_2$, $P_3$, $P_4$ the points where the ray falls on the internal surfaces of the prism. Let $θ_i$ be the incidence angle of the ray on your prism.

1. Then $θ_1$ is the refraction angle, given by

   $$\sin(θ_1) = \sin(θ_i)/n$$

2. Now, look at the triangle $P_1 A P_2$. Its angles are equal to: $90⁰-\arcsin(\sin(θ_i)/n)$, $60⁰$, and $30⁰+\arcsin(\sin(θ_i)/n)$. Therefore, on the second internal surface of the prism your ray falls under an incident angle

   $$θ_2 = 90⁰-30⁰-\arcsin(\sin(θ_i)/n) = 60⁰-\arcsin(\sin(θ_i)/n)$$

3. You have to check if this angle satisfies the requirement of total reflection, which is $\sin(θ_2) \ge 1/n$. To help you a bit, according to simple trigonometric rules

$$\sin(60⁰-\arcsin(\sin(θ_i)/n)) = \sin(60⁰)\cos(\arcsin(\sin(θ_i)/n))) - \cos(60⁰)\sin(\arcsin(\sin(θ_i)/n))) \\ =(\sqrt{3}/2)\cos(\arccos(90⁰-\sin(θ_i)/n)) - (1/2)\sin(θ_i)/n \\ ={(\sqrt{3})[(π/4)-\sin(θ_i)] - \sin(θ_i)}/(2n) \\ = {(π/8)x\sqrt{3} - [(1+\sqrt{3})/2]\sin(θ_i)}/n $$

4. If the last line is greater or equal to $1/n$, then repeat the steps 2 and 3, until you find your answer. But, it seems to me that the answer is negative. Even if you take $θ_i = 0$, you get that the last line is smaller than $1/n$.

So, maybe you try with a triangle which doesn't have all the angles equal, but I don't believe that it will do better.