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

13. $\sqrt{\mathbf{y}}$

The givenfunctionis$\sqrt{y}$.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 $n=\sqrt{y}$

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

Since $F=\sqrt{y}\sqrt{1+y{\text{'}}^2}$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

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

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

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

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

Therefore,

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

Where $C$is constant.

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

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

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

${\left(x-B\right)}^2=4C^2\left(y-C^2\right)$

It corresponds to parabola

Therefore, ${\left(x-B\right)}^2=4C^2\left(y-C^2\right)$, where $C$is a constant and

$B$is the integration constant.