In this section we will study how light rays get reflected off the surfaces of objects and what happens when light rays reach the boundary between two different materials.

The speed of light depends on the material where it travels:

• $v_x$: the speed of light in material $x$.
• $n_x$: the diffraction index of material $x$,

which tells you how much slower light is in that material.

  $v_x=c/n_x$.

When an incoming ray of light comes to the surface of a transparent object, part of it will be reflected and part of it will be transmitted. We measure all angles with respect to the normal, which is the direction perpendicular to the interface.

• $\theta_{i}$: The incoming or incidence angle.
• $\theta_{r}$: The reflection angle.
• $\theta_{t}$: The transmission angle:

the angle of the light that goes into the object.

Light that hits a reflective surface will bounce back exactly at the same angle as it came in on: $$\theta_{i} = \theta_{r}.$$

The transmission angle of light when it goes into a material with different refractive index can be calculated from Snell's law: $$n_i \sin\theta_{i} = n_t \sin \theta_{t}.$$

Light coming in from a medium with low refraction index into a medium with high refraction index gets refracted towards the normal. If the light travels in the opposite direction (from high $n$, to low $n$), then it will get deflected away from the normal. In the latter case, an interesting phenomenon called total internal refraction occurs, whereby light rays incident at sufficiently large angles with the normal get trapped inside the material. The angle at which this phenomenon starts to kick in is called the critical angle $\theta_{crit}$.

Consider a light ray inside a material of refractive indeed $n_x$ surrounded by a material with smaller refractive index $n_y$, $n_x > n_y$. To make this more concrete, think of a trans-continental underground optical cable made of glass $n_x=1.7$ surrounded by some plastic with $n_y=1.3$. All light at an angle greater than: $$\theta_{crit} = \sin^{-1}\left( \frac{n_y}{n_{x}} \underbrace{\sin(90^\circ)}_{=1} \right) = \sin^{-1}\!\left( \frac{n_y}{n_{x}} \right) = \sin^{-1}\!\left( \frac{1.3}{1.7} \right) = 49.88^\circ,$$ will get reflected every time it reaches the surface of the optical cable. Thus, if you shine a laser pointer into one end of such a fibre-optical cable in California, 100% of that laser light will come out in Japan. Most high-capacity communication links around the world are based around this amazing property of light. In other words: no total internal refraction means no internet.

Here is an illustration from one of René Descartes' books, which shows a man in funny pants with some sort of lantern which produces a light ray that goes into the water.

Q: What is wrong with the picture?

Hint: Recall that $n_{air}=1$ and $n_{water}=1.33$, so $n_i < n_t$.

Hint 2: What should happen to the angles of the light ray?

A: Suppose that the line $\overline{AB}$, is at $45^\circ$ angle, then after entering the water at $B$, the ray should be deflected towards the normal, i.e., it should pass somewhere between $G$ and $D$. If you wanted to be precise and calculate the transmission angle then we would use: $$n_i \sin\theta_{i} = n_t \sin \theta_{t},$$ filled in with the values for air and water $$1 \sin(45^\circ) = 1.33 \sin( \theta_{t} ),$$ and solved for $\theta_{t}$ (the refracted angle) we would get: $$\theta_{t} = \sin^{-1}\left( \frac{\sqrt{2}}{2\times1.33} \right) = 32.1^\circ.$$ The mistake apparently is due to Descartes' printer who got confused and measured angles with respect to the surface of the water. Don't make that mistake: remember to always measure angles with respect to the normal. The correct drawing should have the light ray going at an angle of $32.1^\circ$ with respect to the line $\overline{BG}$.

To understand refraction you need to imagine “wave fronts” perpendicular to the light rays. Because light comes in at an angle, one part of the wave front will be in material $n_i$ and the other will be in material $n_t$. Suppose $n_i < n_t$, then the part of the wavefront in the $n_t$ material will move slower so angles of the wavefronts will change. The precise relationship between the angles will depend on the refractive indices of the two materials:

$$n_i \sin\theta_{i} = n_t \sin \theta_{t}.$$

Whenever $n_i > n_t$, we reach a certain point where the formula: $$n_i \sin\theta_{i} = n_t \sin \theta_{t},$$ brakes down. If the transmitted angle $\theta_t$ becomes greater than $90^\circ$ (the critical transmission angle) it will not be transmitted at all. Instead, 100% of the light ray will get reflected back into the material.

To find the critical incident angle solve for $\theta_i$ in: $$n_i \sin\theta_{i} = n_t \sin 90^\circ,$$ $$\theta_{crit} = \sin^{-1}\left( \frac{n_t}{n_{i}} \right).$$

The summary of the “what happens when a light ray comes to a boundary”-story is as follows:

1. If $-\theta_{crit} < \theta_i < \theta_{crit}$, then some part of the light will be

transmitted at an angle $\theta_t$ and some part will be reflected at an angle $\theta_r=\theta_i$.

1. If $\theta_i \geq \theta_{crit}$, then all the light will get reflected at an angle $\theta_r=\theta_i$.

Note that when going from a low $n$ medium into a high $n$ medium, there is no critical angle – there will always be some part of the light that is transmitted.