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

11.$\mathit{x}\mathbf{+}\mathbf{1}$

The givenfunctionis$x+1$.Path followed by light is to be found out using Euler equations.

In the calculus of variations andclassical mechanics, the Euler equations is a system of second-orderordinary differential equations whose solutions arestationary points of the givenaction functional.

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 $n=x+1$

Therefore$F=\left(x+1\right)\sqrt{1+y{\text{'}}^2}$is to be minimized

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

Calculate the required derivatives

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

Therefore,

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

Where$C$is constant.

Square both sides of the equation to get$y\text{'}$

$$\begin{gathered} {\left(x+1\right)}^{2}y{\text{'}}^2=C^2\left(1+y{\text{'}}^2\right) \\ \Rightarrow \left({\left(x+1\right)}^2-C^2\right)y{\text{'}}^2=C^2 \\ \Rightarrow y\text{'}=\frac{C}{\sqrt{{\left(x+1\right)}^2-C^2}} \end{gathered}$$

Integrate$y\text{'}=\frac{C}{\sqrt{{\left(x+1\right)}^2-C^2}}$to get$y$

$$\begin{gathered} y=\int \frac{C}{\sqrt{{\left(x+1\right)}^2-C^2}}dx \\ =Cln\left(\sqrt{{\left(x+1\right)}^2-C^2}+\left(x+1\right)\right)+B \end{gathered}$$

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