Chapter 38, Problem 98PQ

A Fermat’s principle of least time for refraction. A ray of light traveling in a medium with speed v₁ leaves point A and strikes the boundary between the incident and transmitted media a horizontal distance x from point A as shown in Figure P38.98. The refracted ray travels with speed v₂ in the second medium, eventually reaching point B. The horizontal distance between points A and B is L. a. Calculate the time t required for the light to travel from A to B in terms of the parameters labeled in the figure. b. Now take the derivative of t with respect to x. What is the condition for which the ray of light will take the shortest time to travel from A to B?

Figure P38.98

To determine

The time required for the light ray to travel from the points A to B.

Answer to Problem 98PQ

The time taken for the light to travel from point $A$ to $B$ is $t=\frac{\sqrt{x^2+y_1^2}}{v_1}+\frac{\sqrt{y_2^2+{\left(L-x\right)}^2}}{v_2}$.

Explanation of Solution

Write the expression for the time taken for the light ray in the second medium.

    $t_2=\frac{d_2}{v_2}$

Here, $v_2$ is the speed of the light in the second medium, $d_2$ is the distance travelled by the light in the second medium, $t_2$ is the time taken for the light ray in the second medium.

Write the expression for distance travelled by the light in the second medium.

    $d_2=\sqrt{y_2^2+{\left(L-x\right)}^2}$

Here, $y_2$ is the vertical distance of the light in the second medium, $L$ is the distance between the point $A$ to the point $B$, and $x$ is the horizontal distance from point $A$ to the transmitted media as shown in the Figure-(1).

Write the expression for the time taken for the light ray in the first medium.

    $t_1=\frac{d_1}{v_1}$

Here, $v_1$ is the speed of the light in the first medium, $d_1$ is the distance travelled by the light in the first medium, $t_1$ is the time taken for the light ray in the first medium.

Write the expression for distance travelled by the light in the first medium.

    $d_1=\sqrt{x^2+y_1^2}$

Here, $y_1$ is the vertical distance of the light in the first medium, and $x$ is the horizontal distance from point $A$ to the transmitted media as shown in the Figure-(1).

Write the equation for total time taken for the light is.

    $t=t_1+t_2$                           (I)

Conclusion:

Substitute $\frac{d_1}{v_1}$ for $t_1$ and $t_2=\frac{d_2}{v_2}$ for $t_2$ in equation (I).

    $t=\frac{d_1}{v_1}+\frac{d_2}{v_2}$                            (II)

Substitute $\sqrt{x^2+y_1^2}$ for $d_1$ and $\sqrt{y_2^2+{\left(L-x\right)}^2}$ for $d_2$ in the above equation.

    $t=\frac{\sqrt{x^2+y_1^2}}{v_1}+\frac{\sqrt{y_2^2+{\left(L-x\right)}^2}}{v_2}$                   (III)

Therefore, the time taken for the light to travel from point $A$ to $B$ is     $t=\frac{\sqrt{x^2+y_1^2}}{v_1}+\frac{\sqrt{y_2^2+{\left(L-x\right)}^2}}{v_2}$

To determine

The required condition to travel for getting the shortest time to reach the point $A$ to point $B$.

Answer to Problem 98PQ

The required condition for getting the minimum time to reach the point $A$ to point $B$ is.    $\frac{v_2}{v_1}=\left(\frac{L-x}{x}\right)\left(\frac{\sqrt{\left(x^2+y_1^2\right)}}{\sqrt{\left(y_2^2+{\left(L-x\right)}^2\right)}}\right)$.

Explanation of Solution

Take the derivative of $t$ with respect to $x$ in the equation (III).

    $\begin{array}{c}t=\frac{\sqrt{x^2+y_1^2}}{v_1}+\frac{\sqrt{y_2^2+{\left(L-x\right)}^2}}{v_2}\\ \frac{dt}{dx}=\frac{d}{dx}\left(\frac{\sqrt{x^2+y_1^2}}{v_1}+\frac{\sqrt{y_2^2+{\left(L-x\right)}^2}}{v_2}\right)\end{array}$

From the formula $x^n=n{x}^{n-1}$ simplify the above equation.

    $\begin{array}{c}\frac{dt}{dx}=\frac{1}{v_1}\left(\frac{1}{2}{\left(x^2+y_1^2\right)}^{\frac{-1}{2}}\left(2x\right)\right)+\frac{1}{v_2}\left(\frac{1}{2}{\left(y_2^2+{\left(L-x\right)}^2\right)}^{\frac{-1}{2}}2\left(L-x\right)\left(-1\right)\right)\\ =\frac{1}{v_1}\left(\frac{1}{2}\left(\sqrt{\left(x^2+y_1^2\right)}\right)\left(2x\right)\right)+\frac{1}{v_2}\left(\frac{1}{2}\left(\sqrt{\left(y_2^2+{\left(L-x\right)}^2\right)}\right)2\left(L-x\right)\left(-1\right)\right)\\ =\frac{x}{v_1\sqrt{\left(x^2+y_1^2\right)}}-\frac{L-x}{v_2\sqrt{\left(y_2^2+{\left(L-x\right)}^2\right)}}\end{array}$            (IV)

The light ray in which its shortest time taken is.

    $\frac{dt}{dx}=0$

Conclusion:

Substitute $0$ for $\frac{dt}{dx}$ in equation (IV).

    $\begin{array}{c}\frac{dt}{dx}=\frac{x}{v_1\sqrt{\left(x^2+y_1^2\right)}}-\frac{L-x}{v_2\sqrt{\left(y_2^2+{\left(L-x\right)}^2\right)}}\\ 0=\frac{x}{v_1\sqrt{\left(x^2+y_1^2\right)}}-\frac{L-x}{v_2\sqrt{\left(y_2^2+{\left(L-x\right)}^2\right)}}\\ \frac{L-x}{v_2\sqrt{\left(y_2^2+{\left(L-x\right)}^2\right)}}=\frac{x}{v_1\sqrt{\left(x^2+y_1^2\right)}}\\ \frac{L-x}{x}=\frac{v_2\sqrt{\left(y_2^2+{\left(L-x\right)}^2\right)}}{v_1\sqrt{\left(x^2+y_1^2\right)}}\end{array}$

Simplify the above equation.

    $\begin{array}{c}\frac{L-x}{x}=\frac{v_2\sqrt{\left(y_2^2+{\left(L-x\right)}^2\right)}}{v_1\sqrt{\left(x^2+y_1^2\right)}}\\ \frac{L-x}{x}=\left(\frac{v_2}{v_1}\right)\left(\frac{\sqrt{\left(y_2^2+{\left(L-x\right)}^2\right)}}{\sqrt{\left(x^2+y_1^2\right)}}\right)\\ \frac{v_2}{v_1}=\left(\frac{L-x}{x}\right)\left(\frac{\sqrt{\left(x^2+y_1^2\right)}}{\sqrt{\left(y_2^2+{\left(L-x\right)}^2\right)}}\right)\end{array}$

Therefore, the required condition for getting the minimum time to reach the point $A$ to point $B$ is. $\frac{v_2}{v_1}=\left(\frac{L-x}{x}\right)\left(\frac{\sqrt{\left(x^2+y_1^2\right)}}{\sqrt{\left(y_2^2+{\left(L-x\right)}^2\right)}}\right)$.