Use Fermat’s principle to find the path followed by a light ray if the index of refraction is proportional to the given function

14.${\mathit{r}}^{\mathbf{-}\mathbf{1}}$

The given function is$r^{-1}$.Path followed by light is to be found out using Euler equations.

The Euler equations are a set of second-order ordinary differential equations that are stationary points of the given action functional in the calculus of variations and classical mechanics.

To find the path traversed by light in a given medium, the path taken by the light is to be minimized (time wise). Velocity of light is scaled by a factor $n^{‐1}$in a refractive medium, then the time required to travel from point A to point B is

$$\begin{gathered} t={\int }_A^{B}dt \\ ={\int }_A^{B}vds \\ =c^{‐1}{\int }_A^{B}nds \end{gathered}$$

Therefore, following integral needs to be minimized

$$\begin{gathered} \int nds=\int n\sqrt{d{r}^2+r^2{\theta }^2} \\ =\int n\sqrt{1+r^2\theta {\text{'}}^2}dr \end{gathered}$$

Here $n=r^{‐1}$

Therefore $F=r^{‐1}\sqrt{1+r^2\theta {\text{'}}^2}$is to be minimized

Euler equation for coordinates $\left(r,\theta \right)$is $\frac{d}{dr}\left(\frac{\partial F}{\partial \theta \text{'}}\right)-\frac{\partial F}{\partial \theta }=0$

Calculate the required derivatives

$$\begin{gathered} \frac{\partial F}{\partial \theta \text{'}}=\frac{r^2\theta \text{'}}{r\sqrt{1+r^2\theta {\text{'}}^2}} \\ \frac{\partial F}{\partial \theta }=0 \end{gathered}$$

Therefore,

$$\begin{gathered} \frac{d}{dr}\left[\frac{r^2\theta \text{'}}{r\sqrt{1+r^2\theta {\text{'}}^2}}\right]=0 \\ \frac{r^2\theta \text{'}}{r\sqrt{1+r^2\theta {\text{'}}^2}}=C \\ \theta {\text{'}}^2=\frac{C^2}{r^2\left(1-C^2\right)} \\ \theta \text{'}=\frac{C}{r\sqrt{1-C^2}} \end{gathered}$$

Where $C$is constant.

Integrate $\theta \text{'}=\frac{C}{r\sqrt{1-C^2}}$to get the desired result

$$\begin{gathered} \theta =\int \frac{C}{r\sqrt{1-C^2}}dr \\ =\frac{C}{\sqrt{1-C^2}}\int \frac{dr}{r} \\ =\frac{C}{\sqrt{1-C^2}}\mathrm{In}r+B \end{gathered}$$

Move $B$to the left side, multiply by constants to isolate $r$and take the exponential of the whole expression to get

$e^{\frac{C}{\sqrt{1-C^2}}\left[\theta -B\right]}=r$

Take$a=e^{\frac{BC}{\sqrt{1-C^2}}}$and $b=\frac{C}{\sqrt{1-C^2}}$to get

$r=a{e}^{b\theta }$

It corresponds to two-dimensional spiral

Therefore, $r=a{e}^{b\theta }$, where $a=e^{\frac{BC}{\sqrt{1-C^2}}}$and $b=\frac{C}{\sqrt{1-C^2}}$, where $C$ is a constant and $B$is the integration constant.