TITLE: Simple geometric proof for Snell's law of refraction
QUESTION [17 upvotes]: Snell's law of refraction can be derived from Fermat's principle
that light travels paths that minimize the time using simple
calculus. Since Snell's law only involves sines I wonder whether
this minimum problem has a simple geometric solution.

REPLY [22 votes]: Perhaps this will help, if you are looking at a non Calculus approach.
Consider two parallel rays $A$ and $B$ coming through the medium $1$ (say air) to the medium $2$ (say water). Upon arrival at the interface $\mathcal{L}$ between the two media (air and water), they continue their parallel course in the directions $U$ and $V$ respectively.
Let us assume that at time $t=0$, light ray $A$ arrives at the interface $\mathcal{L}$ at point $C$, while ray $B$ is still shy of the surface by a distance $PD$.  $B$ travels at the speed $v_{1}=\frac{c}{n_{1}}$ and arrives at $D$ in $t$ seconds. During this time interval, ray $A$ continues its journey through the medium $2$ at a speed $v_{2}=\frac{c}{n_{2}}$ and reaches the point $Q$. 
We can formulate the rest, geometrically (looking at the parallel lines) from the figure. Let $x$ denote the distance between $C$ and $D$.
\begin{eqnarray*}
x \sin\left(\theta_{i}\right) &=& PD \\
&=& v_{1} t \\
&=& \frac{c}{n_{1}} t  \\
x \sin\left(\theta_{r}\right) &=& CQ \\
&=& v_{2} t \\
&=& \frac{c}{n_{2}} t 
\end{eqnarray*}
Thus,
\begin{eqnarray*}
n_{1} \sin\left(\theta_{i}\right) &=& \frac{c}{x} t  \\
n_{2} \sin\left(\theta_{r}\right) &=& \frac{c}{x} t 
\end{eqnarray*}
Re arranging this will take us to the Snell's law as we know.
\begin{eqnarray*}
\frac{n_{2} }{n_{1}} &=& \frac{\sin\left(\theta_{i}\right) }{ \sin\left(\theta_{r}\right)}
\end{eqnarray*}