If I can find one, a photo demonstrating TIR.

When Snell's Law has no solution, what happens physically is that no light is transmitted into the new material. Instead, the light is *totally reflected* by the interface, in the usual manner.

By rearranging Snell's Law a little bit, we can see when this total reflection may occur. Because sines are always less than or equal to 1,
$$\sin\theta_t={n_i\over n_t}\sin\theta_i>1$$
during total reflection. What would make \({n_i\over n_t}\sin\theta_i\) larger than 1? Well, we know that \(\sin\theta_i\) must be less than 1 by itself, which means that the ratio \(n_i/n_t\) must be larger than 1. Total reflection only occurs when \(n_i>n_t\): that is, when light travels from a material of higher (slower) index to lower (faster) index. Total reflection only occurs when light tries to speed up, and fails. The most common example of this is when light leaves a material like glass or water and tries to enter air, and so this phenomenon is called total internal reflection: it only happens *inside* a material.

Total internal reflection is also more likely to occur when \(\theta_i\) is large, so that \(\sin\theta_i\) is already pretty close to 1. Rays of light which lie close to the normal have no trouble getting through. The dividing line occurs when \({n_i\over n_t\sin\theta_i}\) is exactly equal to 1: the incident angle that makes this so is called the critical angle \(\theta_c\). Rays with an incident angle less than the critical angle are transmitted; those greater than the critical angle are not. The critical angle is given by the formula

$${n_i\over n_t}\sin\theta_c=1$$
$$\implies \theta_c=\sin^{-1}{n_t\over n_i}$$

Find the critical angle when light tries to pass

(a) from water \((n=1.33)\) into air

The critical angle is the incident angle where
$$\begin{eqnarray}
{n_i\over n_t}\sin\theta_c&=&1\\
\implies \sin\theta_c&=&{n_t\over n_i}\\
\implies \theta_c&=&\sin^{-1}{n_t\over n_i}\\
&=&\sin^{-1}{1\over 1.33}=\boxed{48.7\degrees}\\
\end{eqnarray}
$$

(b) from water \((n=1.33)\) into glass \((n=1.5)\)

Using the formula derived in the previous section:
$$\theta_c=\sin^{-1}{n_t\over n_i}=\sin^{-1}{1.5\over 1.33}=\sin^{-1}1.13$$
But there *is* no inverse sine of 1.13, because sines are always less than or equal to 1. Thus there is no critical angle when light moves from water to glass, and no total internal reflection, because light moves more slowly in glass. Total internal reflection only occurs when light tries to speed up.

Total internal reflection is the reason diamonds are sparkly. Diamond possesses a high index of 2.4. Light can enter a diamond easily, but it can only re-enter the air if it hits the interface within \(\sin^{-1}\frc{2.4}=24.6\degrees\) of the normal. Assuming a random angle, the ray will exit the diamond less than one-third of the time. Most of the time, it will be reflected to hit another interior surface of the diamond, then another, and so forth, until it finally leaves. Gemcutters, aware of this property, cut the diamond into shapes which increase this effect. The end result: rays from an overhead lamp enter the diamond bounce around all over the place, and a small shift in the diamond's position can change the rays' exit points dramatically. The diamond emits light at all sorts of crazy angles, which we see as sparkles. Any transparent material will do this, including cut glass or even cut ice. But diamond's high index and correspondingly low critical angle heightens the effect.

But total internal reflection has uses more practical than jewelry. Suppose you have a very thin pipe of some transparent material, like glass. A ray of light enters into the pipe from one side, and travels in a straight line until it hits the side of the pipe. If the pipe is narrow enough and doesn't bend too much, the ray will hit the side of the pipe at a very large angle, and so is totally reflected back into the pipe: none of it escapes. The ray then travels until it hits another side of the pipe, bounces off of it, and so on until it reaches the end.

This is the principle behind fiber optics, and it is a way for light to travel long distances without being dissipated into the environment. If a doctor wants to inspect a patient's stomach lining, she can run a thin fiber of glass down the esophagus into the stomach, along with a small light source. Light reflected from the lining travels up through the fiber, allowing the doctor to see images from inside without invasive surgery or complicated electronics. Fiber optics are also used in communications, and many Internet and phone companies are switching over from electrical signals over copper wires, to pulses of laser light travelling through fiber.