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

12. ${\mathit{y}}^{\mathbf{-}\mathbf{1}}$

The given function is$y^{‐1}$.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

$$\begin{gathered} \int nds=\int n\sqrt{d{x}^2+d{y}^2} \\ =\int n\sqrt{1+y{\text{'}}^2}dx \end{gathered}$$

Here $n=y^{‐1}$

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

Since $F=y^{‐1}\sqrt{1+y{\text{'}}^2}$includes both $y$and $y\text{'}$. Variables are required to be changed.

Let

$$\begin{gathered} dx=x\text{'}dy \\ y\text{'}=\frac{1}{x\text{'}} \end{gathered}$$

Thus from above two equations

$$\begin{gathered} \int \frac{\sqrt{1+y{\text{'}}^2}}{y}dx=\int \frac{\sqrt{1+y{\text{'}}^2}}{y}x\text{'}dy \\ =\int \frac{\sqrt{1+x{\text{'}}^2}}{y}x\text{'}dy \\ =\int \frac{\sqrt{x{\text{'}}^2+1}}{y}dy \end{gathered}$$

Now, let $F=\frac{\sqrt{x{\text{'}}^2+1}}{y}$

Euler equation for coordinates$\left(y,x\right)$is$\frac{d}{dy}\left(\frac{\partial F}{\partial x\text{'}}\right)-\frac{\partial F}{\partial x}=0$

Calculate the required derivatives

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

Therefore,

$$\begin{gathered} \frac{d}{dy}\left[\frac{x\text{'}}{y\sqrt{1+x{\text{'}}^2}}\right]=0 \\ \frac{x\text{'}}{y\sqrt{1+x{\text{'}}^2}}=C \\ x{\text{'}}^2=\frac{C^2{y}^2}{1-C^2{y}^2} \\ x\text{'}=\frac{Cy}{\sqrt{1-C^2{y}^2}} \end{gathered}$$

Where $C$is constant.

Integrate $x=\frac{Cy}{\sqrt{1-C^2{y}^2}}$to get the desired result

$$\begin{gathered} x=\int \frac{Cy}{\sqrt{1-C^2{y}^2}}dy \\ =-\frac{\sqrt{1-C^2{y}^2}}{C}+B \end{gathered}$$

Move $B$to the left side and square both sides

$y^2+{\left(x-B\right)}^2=\frac{1}{C^2}$

It corresponds to circle with radius $\frac{1}{C}$and $B$offset on $x‐$axis

Therefore, $y^2+{\left(x-B\right)}^2=\frac{1}{C^2}$, where$C$ is a constant and$B$ is the integration constant.