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 |

Interactive 16.3.1

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

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.