Question

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.

${\left(y-1\right)}^{\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}}$

Short answer

The path followed by a light ray by Fermat’s principle is $\mathrm{x}\left(\mathrm{y}\right)=\frac{1}{\mathrm{C}}\left(\arcsin \sqrt{\mathrm{C}(\mathrm{y}-1)}\right)-\sqrt{\mathrm{C}\left(\mathrm{y}-1\right)}\sqrt{1-\mathrm{C}(\mathrm{y}-1)})+{\mathrm{C}}_1$.

Step-by-step solution

Given Information

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

Definition of Calculus

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.

Fermat’s Principle

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

Use Fermat’s Principle,

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

Euler’s Equation

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

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

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

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 Euler’s result

Integrate the result in the above step.

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

Therefore, by Fermat’s 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 $\mathrm{x}\left(\mathrm{y}\right)=\frac{1}{\mathrm{C}}\left(\arcsin (\sqrt{\mathrm{C}\left(\mathrm{y}-1\right)}\right)-\sqrt{\mathrm{C}\left(\mathrm{y}-1\right)}\sqrt{1-\mathrm{C}\left(\mathrm{y}-1\right)})+{\mathrm{C}}_1$.