Fermat’s principle of least time can be roughly stated as, “The path taken between two points by a ray of light is the path that can be traversed in the least time” (From Wikipedia). Fermat used this observation to derive the laws of reflection and refraction of light.

## Reflection of light

The law of reflection of light states that the angle at which a ray is incident on a reflective surface is equal to the angle at which it is reflected from the surface. In other words, in the following diagram (light travels from $A$ to $O$ to $B$), ${\theta}_i = {\theta}_r$.

Let’s try to derive this law from Fermat’s principle. The total distance that the ray of light has to travel from $A$ to $B$ with one reflection at $O$ is

$L = \overline{AO} + \overline{OB}$
$= \sqrt{a^2 + x^2} + \sqrt{b^2 + {\left( l - x \right)}^2}$

Now since the speed of light is constant, minimizing the time taken by light to take the path $AOB$ is equivalent to minimizing the path itself. To do this, we set $\frac{dL}{dx} = 0$.

$\frac{dL}{dx} = 0$
$\implies \frac{d}{dx} \left( \sqrt{a^2 + x^2} + \sqrt{b^2 + {\left( l - x \right)}^2} \right) = 0$
$\implies \frac{2x}{2\sqrt{a^2 + x^2}} + \frac{2 \left( l - x \right) \left( -1 \right)}{2\sqrt{b^2 + {\left( l - x \right)}^2}} = 0$
$\implies \frac{x}{\sqrt{a^2 + x^2}} = \frac{\left( l - x \right)}{\sqrt{b^2 + {\left( l - x \right)}^2}}$
$\implies \sin{ {\theta}_i } = \sin{ {\theta}_r }$
$\implies {\theta}_i = {\theta}_r \, \blacksquare$


## Refraction

The law of refraction of light, or Snell’s Law states that the ratio of the sines of the angles of incidence and refraction is equal to the inverse ratio of the indices of refraction of the two media.

With reference to the image above, Snell’s Law can be mathematically stated as,

$\frac{\sin {\theta}_i }{\sin {\theta}_r } = \frac{n_2}{n_1}$

Now to prove this using Fermat’s principle.

We know that,

$\text{index of refraction of a medium} = \frac{ \text{speed of light in vacuum} }{ \text{speed of light in the medium} }$

Let the speeds of light in our media be $v_1 = \frac{c}{n_1}$ and $v_2 = \frac{c}{n_2}$. Now the total time taken by light to travel from $A$ to $B$ with a refraction at $O$ is

$t = \frac{ \overline{AO} }{v_1} + \frac{ \overline{OB} }{v_2}$
$= \frac{ \overline{AO} \times n_1 }{c} + \frac{ \overline{OB} \times n_2 }{c}$
$= \frac{ n_1 \sqrt{ a^2 + x^2 } }{c} + \frac{ n_2 \sqrt{ b^2 + {\left( d - x \right) }^2 } }{c}$

To minimize $t$, we’ll set $\frac{dt}{dx} = 0$

$\frac{d}{dx} \left( \frac{ n_1 \sqrt{ a^2 + x^2 } }{c} + \frac{ n_2 \sqrt{ b^2 + {\left( d - x \right) }^2 } }{c} \right) = 0$
$\implies \frac{1}{c} \frac{d}{dx} \left( { n_1 \sqrt{ a^2 + x^2 } } + { n_2 \sqrt{ b^2 + {\left( d - x \right) }^2 } } \right) = 0$
$\implies \frac{2 x n_1}{ 2 \sqrt{ a^2 + x^2 } } + \frac{ 2 \left( d - x \right) n_2 \left( -1 \right) }{ 2 \sqrt{ b^2 + {\left( d - x \right) }^2 } } = 0$
$\implies n_1 \frac{x}{ \sqrt{ a^2 + x^2 } } = n_2 \frac{ \left( d - x \right) }{ \sqrt{ b^2 + {\left( d - x \right) }^2 } }$
$\implies n_1 \sin {\theta}_i = n_2 \sin {\theta}_r$
$\implies \frac{\sin {\theta}_i }{\sin {\theta}_r } = \frac{n_2}{n_1} \, \blacksquare$