Chapter 35, Problem 35.84CP

Pierre de Fermat (1601–1665) showed that whenever light travels from one point to another, its actual path is the path that requires the smallest time interval. This statement is known as Fermat’s principle. The simplest example is for light propagating in a homogeneous medium. It moves in a straight line because a straight line is the shortest distance between two points. Derive Snell’s law of refraction from Fermat’s principle. Proceed as follows. In Figure P34.54, a light ray travels from point P in medium 1 to point Q in medium 2. The two points are, respectively, at perpendicular distances a and b from the interface. The displacement from P to Q has the component d parallel to the interface, and we let x represent the coordinate of the point where the ray enters the second medium. Let t = 0 be the instant the light starts from P. (a) Show that the time at which the light arrives at Q is

$t=\frac{r_1}{v_1}+\frac{r_2}{v_2}=\frac{n_1\sqrt{a^2+x^2}}{c}+\frac{n_2\sqrt{b^2+{(d-x)}^2}}{c}$

(b) To obtain the value of x for which t has its minimum value, differentiate t with respect to x and set the derivative equal to zero. Show that the result implies

$\frac{n_{1}x}{\sqrt{a^2+x^2}}=\frac{n_2(d-x)}{\sqrt{b^2+{(d-x)}^2}}$

(c) Show that this expression in turn gives Snell’s law.

$n_1\sin {\theta }_1=n_2\sin {\theta }_2$

Figure P34.54 Problems 54 and 55.

To determine

To show: The time at which the light arrives at Q is $t=\frac{r_1}{v_1}+\frac{r_2}{v_2}=\frac{n_1\sqrt{a^2+x^2}}{c}+\frac{n_2\sqrt{b^2+{\left(d-x\right)}^2}}{c}$.

Answer to Problem 35.84CP

The time at which the light arrives at Q is $t=\frac{r_1}{v_1}+\frac{r_2}{v_2}=\frac{n_1\sqrt{a^2+x^2}}{c}+\frac{n_2\sqrt{b^2+{\left(d-x\right)}^2}}{c}$. It is valid.

Explanation of Solution

The given figure is shown below.

Figure (1)

In right angle triangle PAB,

$r_1=\sqrt{a^2+x^2}$

In right angle triangle BCQ,

$r_2=\sqrt{b^2+{\left(d-x\right)}^2}$

The time taken by light to travel from point P to Q is,

$t=\frac{r_1}{v_1}+\frac{r_2}{v_2}$ (1)

Here,

$v_1$ is the velocity of light to travel from point P to B.

$v_2$ is the velocity of light of travel from point B to Q.

The expression for refractive index is,

$n=\frac{c}{v}$

Substitute $n_1$ for $n$ and $v_1$ for $v$ in above equation

$\begin{array}{c}{n}_1=\frac{c}{v_1}\\ v_1=\frac{c}{n_1}\end{array}$

Substitute $n_2$ for $n$ and $v_2$ for $v$ in above equation

$\begin{array}{l}{n}_2=\frac{c}{v_2}\\ v_2=\frac{c}{n_2}\end{array}$

Substitute $\frac{c}{n_1}$ for $v_1$, $\frac{c}{n_2}$ for $v_2$, $\sqrt{a^2+x^2}$ for $r_1$ and $\sqrt{b^2+{\left(d-x\right)}^2}$ for $r_2$ in equation (1),

$t=\frac{\sqrt{a^2+x^2}}{\frac{c}{n_1}}+\frac{\sqrt{b^2+{\left(d-x\right)}^2}}{\frac{c}{n_2}}$

$t=\frac{n_1\sqrt{a^2+x^2}}{c}+\frac{n_2\sqrt{b^2+{\left(d-x\right)}^2}}{c}$ (2)

Conclusion:

Therefore, the time at which the light arrives at Q is $t=\frac{r_1}{v_1}+\frac{r_2}{v_2}=\frac{n_1\sqrt{a^2+x^2}}{c}+\frac{n_2\sqrt{b^2+{\left(d-x\right)}^2}}{c}$. It is valid

To determine

To show: The result $\frac{n_{1}x}{\sqrt{a^2+x^2}}=\frac{n_2\left(d-x\right)}{\sqrt{b^2+{\left(d-x\right)}^2}}$ is implied to obtain the value of $x$ for which $t$ has its minimum value.

Answer to Problem 35.84CP

Yes, the result $\frac{n_{1}x}{\sqrt{a^2+x^2}}=\frac{n_2\left(d-x\right)}{\sqrt{b^2+{\left(d-x\right)}^2}}$ is implied to obtain the value of $x$ for which $t$ has its minimum value.

Explanation of Solution

Apply the condition of maxima and minima,

$\frac{dt}{dx}=0$

Substitute $\frac{n_1\sqrt{a^2+x^2}}{c}+\frac{n_2\sqrt{b^2+{\left(d-x\right)}^2}}{c}$ for $t$ in above equation,

$\begin{array}{l}\frac{d}{dx}\left(\frac{n_1\sqrt{a^2+x^2}}{c}+\frac{n_2\sqrt{b^2+{\left(d-x\right)}^2}}{c}\right)=0\\ \frac{n_{1}x}{c\sqrt{a^2+x^2}}-\frac{n_2\left(d-x\right)}{c\sqrt{b^2+{\left(d-x\right)}^2}}=0\\ \frac{n_{1}x}{c\sqrt{a^2+x^2}}=\frac{n_2\left(d-x\right)}{c\sqrt{b^2+{\left(d-x\right)}^2}}\end{array}$

$\frac{n_{1}x}{\sqrt{a^2+x^2}}=\frac{n_2\left(d-x\right)}{\sqrt{b^2+{\left(d-x\right)}^2}}$ (2)

Conclusion:

Therefore, the result $\frac{n_{1}x}{\sqrt{a^2+x^2}}=\frac{n_2\left(d-x\right)}{\sqrt{b^2+{\left(d-x\right)}^2}}$ is implied to obtain the value of $x$ for which $t$ has its minimum value.

To determine

To show: The expression of Snell’s law, $n_1\sin {\theta }_1=n_2\sin {\theta }_2$.

Answer to Problem 35.84CP

The expression of Snell’s law is $n_1\sin {\theta }_1=n_2\sin {\theta }_2$.

Explanation of Solution

In right angle triangle PAB,

$\sin {\theta }_1=\frac{x}{\sqrt{a^2+x^2}}$

Similarly, in right angle triangle BCQ,

$\sin {\theta }_2=\frac{\left(d-x\right)}{\sqrt{b^2+{\left(d-x\right)}^2}}$

Substitute, $\sin {\theta }_1$ for $\frac{x}{\sqrt{a^2+x^2}}$ and $\sin {\theta }_2$ for $\frac{\left(d-x\right)}{\sqrt{b^2+{\left(d-x\right)}^2}}$ in equation (2),

$n_1\sin {\theta }_1=n_2\sin {\theta }_2$

Conclusion:

Therefore, the expression of Snell’s law is $n_1\sin {\theta }_1=n_2\sin {\theta }_2$.