light ray undergoing total internal reflection thrice inside a prism

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.

• What have you tried? Please show some effort, if for no other reason than we can determine you level of understanding of geometry and calculus. (I agree this is an interesting question, not unrelated to the behavior of light inside raindrops which leads to double rainbows). So,play around a bit with different triangles, i.e. collections of 3 vertex angles, along with different entrance angles, and see if you can find the limiting cases. (Hint: enter from the base of a very skinny isosceles) – Carl Witthoft Nov 2 '14 at 11:45
• What are the angles of this triangular prism? I mean, considering a transversal section through the prism, do you get a triangle with all the angles equal ? – Sofia Nov 7 '14 at 21:11

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).

• i get what you are saying but in your answer i think γ,β and α do not match with the diagram drawn..(i may be wrong) – ragvri Nov 22 '14 at 11:21

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

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

1. 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.