Let us consider a situation where a swimmer is drowning at one end of the river. You are on the bank of the other side of the river. You try to save the swimmer and for that you have to traverse both bank and river in order to reach the drowning swimmer.

Which one would you choose – the shortest path or the fastest route?

Obviously, the fastest route!!

Humans have been found to follow this “Least Time” rule when forced to cross between two surfaces. When scientific experiments were conducted on the little fire ant – *Wasmannia auropunctata, *it was observed that the ants resembled humans in the same way – the same “Least Time” rule.

The ants were forced to walk on different surfaces, to reach their food. The speed of ants was different on different surfaces. Given the average speed of ants, and the path followed by these ants, the researchers found out that the path followed was not of shortest distance but of the fastest route.

The ants accounted for the different walking speeds on the different surfaces and traveled a longer distance on the surface where they could walk faster.Let us try to find the shortest time taken by ants to reach food $Q'$ from their nest $Q$. The ants have to traverse through two surfaces– Surface 1 and Surface 2. The distance to be travelled on Surface 1 is $d$ and the distance to be travelled on Surface 2 is $d'$.

Let the speed of ants in the surface 1 be $\frac{1}{n}$ and the speed of ants on surface 2 be $\frac{1}{n'}$. Then the distance travelled by ants from their nest to the food is $\Delta$. $$\Delta = \frac{d}{\frac{1}{n}} + \frac{d'}{\frac{1}{n'}}$$ $$\Delta = nd + n' d'$$

Let $h$ and $h'$ represent perpendicular distances to the surface and $p$ the total length intercepted by these perpendiculars, then we can write $$d^2 = h^2 +(p-x)^2 $$ $$d' ^2 = h' ^2 + x^2$$

When $d$ and $d'$ are substituted in $\Delta$, we obtain $$\Delta = n \sqrt{[h^2 + (p-x)^2]} + n' \sqrt{[h'^2 + x^2]}$$

Since we considered that the ants travel through the fastest route, $$\frac{d \Delta}{dx} = 0$$ $$ \frac{d \Delta}{dx} = \frac{n}{2 \sqrt{h^2 + (p-x)^2}}(-2p +2x) + \frac{n'}{2\sqrt{h'^2 + x^2}}2x = 0 $$

On simplyfying, we get: $$ n \frac{p-x}{x} = n' \frac{x}{d'}$$

With reference to the figure, we can say $$ n \sin{\phi} = n' \sin{\phi'} $$

To our surprise, Light seems to follow the same “Least Time” rule. In optics this “Least Time” rule is called the “Fermat’s Principle of Least time”. Fermat himself stated that the time required for the light to traverse between two points is a minimum. His justification for this principle is that “Nature is Economical”. But he was unaware of circumstances where exact opposite is true.

To be more general * “The Path taken by a light ray going from one point to other through any set of media is such that optical path is either a minimum, a maximum, or a stationary point at the position of the true ray.” *The modern day definition involves Calculus of Variation and is beyond the scope of this article.

The optical path $\Delta = nd = ct$ represents the distance light travels in vacuum in the same time it travels a distance d in the medium. If you noticed, The equation $ n \sin{\phi} = n' \sin{\phi'} $ is analogous to Snell's law or Descartes law in optics. $$ \mu \sin{i} = \mu' \sin{r} $$

If we compare both of the equations, we can see that $n$ (the inverse of the ant's velocity), and $\mu$ are analogous, as are $\phi , i$ and $\phi', r$. $\Delta$ is analogous to the optical path.

Using Fermat’s Principle of Least time, we can derive the Snell’s law or Descartes law. The derivation is similar to the derivation discussed above. It is left as an exercise for the reader to try.