Question

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

Short answer

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

Step-by-step solution

Given Information.

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

Definition of Euler equation

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.

Use Euler equation

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.