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To minimize travel time between points 1 and 2 as light travels through two different media: say Air then Water.

Distance from ① to the interface: $$d_1 = \sqrt{a^2+x^2}$$ Distance from interface to point ②: $$d_2 = \sqrt{b^2 + (l-x)^2}$$

From basic kinematics, the time of travel is the distance divided by the speed ($t = d/v$), and we can say the speed of the light in each medium is $c/n$ where $c$ is the speed of light in vacuum and $n$ is the index of refraction in the particular medium. Thus the total time $t$ for the trip from ① to ② is: $$ t = \frac{d_1}{c/n_1} + \frac{d_2}{c/n_2} = \frac{n_1 \sqrt{a^2+x^2}}{c}+\frac{n_2 \sqrt{b^2 + (l-x)^2}}{c} $$

To find the least time, we need to see how $t$ changes with $x$, i.e. find $dt/dx$ and set it equal to 0: $$ \label{leasttime} \frac{dt}{dx} = \frac{n_1 x}{\sqrt{a^2+x^2} c} - \frac{n_2 (l-x)}{\sqrt{b^2 + (l-x)^2} c} = 0 $$

Noting that $\sin \left( \theta_1 \right) = \frac{x}{\sqrt{a^2+x^2}}$ and $\sin \left( \theta_2 \right) = \frac{l-x}{\sqrt{b^2+(l-x)^2}}$

Then, eq \eqref{leasttime} simplifies to: $$ n_1 \sin \theta_1 = n_2 \sin \theta_2 $$ which is Snell's Law.