Transmission

Index of Refraction

When light travels in a vacuum, it travels at a speed of $$c=3\ten8\u{m/s}$$ Light travels slower in other media, however, sometimes significantly slower, due to the interaction between the light and the atoms in the material. We write light's speed in a material as $$v={c\over n}$$ where \(n\) is the index of refraction of the material. For example, in a material with an index of \(n=2\), light travels at half the speed it does in vacuum. The index of refraction is always greater than 1 (or else light would travel faster than c), and is equal to 1 in vacuum. Here are some sample values:
Material n
Vacuum 1 \(v=c\) in vacuum
Air 1.0003 We'll just call this 1.
Water 1.33 \(\approx {4\over 3}\hbox{, so }v=\frac{3}{4}c\)
Glass 1.5 Approximately; index varies for different types
Diamond 2.4 Diamond has a rather high index
Remember: the higher the index of refraction is, the slower light moves through that material. (It may help to visualize the index as some sort of "friction" or "resistance".)

Refraction

When a ray of light passes from one material to another, so that it slows down or speeds up, it will usually change direction. This is called refraction, and we'll explain the reason for this later. When light slows down, it refracts towards the normal of the surface; when it speeds up, it refracts away from the normal. (Because "when life speeds up, it becomes less normal.") The larger the discrepancy between the indices and the larger the angle the ray makes with the normal, the more it refracts.
Interactive 16.3.1

Snell's Law

The precise direction that the refracted ray takes can be calculated using Snell's Law: $$n_i\sin\theta_i=n_t\sin\theta_t$$
Light, travelling through air, hits a pane of glass (n=1.5) 20° above the surface of the glass. What direction does the ray go once it is in the glass?
We can solve Snell's Law to find the transmission angle, that the ray of light makes inside the glass. $$n_i\sin\theta_i=n_t\sin\theta_t \implies \theta_t=\sin^{-1}\left({n_i\over n_t}\sin\theta_i\right)$$ We're given that \(n_t=1.5\), and we know that the index of refraction in air is \(n_i=1\) (more or less). The incident angle \(\theta_i\) must be measured from the normal, not the interface, and so \(\theta_i=\)90°-20°=70°. Thus $$\theta_t=\sin^{-1}\left(1\over 1.5\sin 70°\right)=\boxed{38.8°}$$ Notice that the ray of light slows down when it enters the glass, and so the transmitted angle 38.8° is smaller than the incident angle 70°:

Interactive 16.3.2
Let's do one more Snell's Law calculation, which will lead us into the next section.
A flashlight is under water (n=1.33) and shines up into the air. What angle does the transmitted ray make with the normal, if the incident angle is
(a) \(\theta_i=\)45°
$$ \begin{eqnarray} 1.3\sin45\degrees&=&1\sin\theta_t\\ \implies \sin\theta_t&=&1.33\sin45\degrees=0.940\\ \implies \theta_t&=&\sin^{-1}0.940\\ \implies &=&\boxed{70.1\degrees}\\ \end{eqnarray} $$
(b) \(\theta_i=\)50°
$$ \begin{eqnarray} 1.3\sin50\degrees&=&1\sin\theta_t\\ \implies \sin\theta_t&=&1.3\sin50\degrees=1.02\\ \end{eqnarray} $$ If you tried to solve this on your calculator and you got an "Error", that's right! We're looking for an angle \(\theta_t\) whose sine is 1.02, but sine has a maximum value of 1! So what does the ray actually do? We'll talk about that in the next section.

Optional: Dispersion