In Problems 5 to 7, use Fermat’s principle to find the path followed by a light ray if the index of refraction is proportional to the given function.

${\mathbf{x}}^{\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}}$

It is given that index of refraction is proportional to function ${\mathrm{x}}^{-\frac{1}{2}}$.

Calculus, sometimes known as infinitesimal calculus or calculus of infinitesimals, is the mathematical study of continuous change, similar to how geometry is the study of shape and algebra is the study of arithmetic operations generally.

Let index of refraction be $\mathrm{n}\left(\mathrm{x}\right)$. It is given $\mathrm{n}\left(\mathrm{x}\right)\propto {\mathrm{x}}^{-\frac{1}{2}}$.

Use Fermat’s Principle,

$$\begin{gathered} \mathrm{t}=\int \mathrm{dt} \\ \int \mathrm{dt}\propto \int \frac{\mathrm{n}}{\mathrm{c}}\mathrm{ds} \\ \int \frac{\mathrm{n}}{\mathrm{c}}\mathrm{ds}\propto {\int }_{{\mathrm{x}}_1}^{{\mathrm{x}}_2}\frac{1}{\sqrt{\mathrm{x}}}\sqrt{1+\mathrm{y}{\text{'}}^2}\mathrm{dx} \end{gathered}$$

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

Write the Euler’s equation and find it’s derivatives.

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

Obtain the first Euler integral.

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

Integrate the result in the above step.

$$\begin{gathered} \frac{dy}{dx}=\sqrt{\frac{Cx}{1-Cx}} \\ y\left(x\right)=\int \sqrt{\frac{Cx}{1-Cx}}dx \\ y\left(x\right)=\frac{1}{C}\left(arc\sin \left(\sqrt{Cx}\right)-\sqrt{Cx}\sqrt{1-Cx}\right)+C_1 \end{gathered}$$

Therefore, by Fermat’s principle the path followed by a light ray, if the index of refraction is proportional to the given function ${\mathrm{x}}^{-\frac{1}{2}}$, is $\mathrm{y}\left(\mathrm{x}\right)=\frac{1}{\mathrm{C}}\left(\arcsin \left(\sqrt{\mathrm{Cx}}\right)-\sqrt{\mathrm{Cx}}\sqrt{1-\mathrm{Cx}}\right)+{\mathrm{C}}_1$.