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It is given that index of refraction is proportional to function ${\left(\mathrm{y}-1\right)}^{-\frac{1}{2}}$.

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 $\mathrm{n}\left(\mathrm{x}\right)鈭�{\left(\mathrm{y}-1\right)}^{-\frac{1}{2}}$.

Use Fermat鈥檚 Principle,

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

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

Write the Euler鈥檚 equation and find it鈥檚 derivatives.

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

Obtain the first Euler integral.

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

Integrate the result in the above step.

$$\begin{gathered} \frac{\mathrm{dx}}{\mathrm{dy}}=\sqrt{\frac{\mathrm{C}\left(\mathrm{y}-1\right)}{1-\mathrm{C}\left(\mathrm{y}-1\right)}} \\ \mathrm{x}\left(\mathrm{y}\right)=鈭�\sqrt{\frac{\mathrm{C}\left(\mathrm{y}-1\right)}{1-\mathrm{C}\left(\mathrm{y}-1\right)}}\mathrm{dy} \\ \mathrm{x}\left(\mathrm{y}\right)=\frac{1}{\mathrm{C}}\left(\arcsin (\sqrt{\mathrm{C}\left(\mathrm{y}-1\right)}\right)-\sqrt{\mathrm{C}\left(\mathrm{y}-1\right)}\sqrt{1-\mathrm{C}\left(\mathrm{y}-1\right)})+{\mathrm{C}}_1 \end{gathered}$$

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