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

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

Short answer

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

Step-by-step solution

Given Information.

The given function is$r^{-1}$.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

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

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.