Chapter 35, Problem 84CP

(a)

To determine

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

Answer to Problem 84CP

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

Explanation of Solution

Write the expression to obtain the speed of light in medium $1$.

    $v_1=\frac{c}{n_1}$

Here, $v_1$ is the speed of light in medium $1$, $c$ is the speed of light in air and $n_1$ is the refractive index of the medium $1$.

Write the expression to obtain the speed of light in medium $2$.

    $v_2=\frac{c}{n_2}$

Here, $v_2$ is the speed of light in medium $2$, $c$ is the speed of light in air and $n_2$ is the refractive index of the medium $2$.

Write the expression to obtain the time when light arrive at point $Q$.

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

Here, $t$ is the time when light arrive at point $Q$, $r_1$ is the distance travelled by the light in medium $n_1$, $v_1$ is the velocity of light in medium $n_1$, $r_2$ is the distance travelled by the light in medium $n_2$ and $v_2$ is the velocity of light in medium $n_2$.

Substitute $\frac{c}{n_1}$ for $v_1$ and $\frac{c}{n_2}$ for $v_2$ in the above equation.

    $\begin{array}{c}t=\frac{r_1}{\left(\frac{c}{n_1}\right)}+\frac{r_2}{\left(\frac{c}{n_2}\right)}\\ =\frac{r_1{n}_1}{c}+\frac{r_2{n}_2}{c}\end{array}$                                                                                                     (I)

The light ray path in two mediums is as shown in the figure below.

Figure-(1)

Write the expression to obtain the value of $r_1$ based on Pythagoras theorem in triangle $PMN$.

    $r_1=\sqrt{a^2+x^2}$

Here, $r_1$ is the distance travelled by the light in medium $n_1$.

Write the expression to obtain the value of $r_2$ based on Pythagoras theorem in triangle $QNO$.

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

Here, $r_2$ is the distance travelled by the light in medium $n_2$.

Substitute $\sqrt{a^2+x^2}$ for $r_1$ and $\sqrt{{\left(d-x\right)}^2+b^2}$ for $r_2$ in equation (I).

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

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

(b)

To determine

The result is $\frac{n_{1}x}{\sqrt{a^2+x^2}}=\frac{n_2\left(d-x\right)}{\sqrt{b^2+{\left(d-x\right)}^2}}$ when the value of $t$ is differentiated with respect to $x$ and equate it to zero.

Answer to Problem 84CP

The result is $\frac{n_{1}x}{\sqrt{a^2+x^2}}=\frac{n_2\left(d-x\right)}{\sqrt{b^2+{\left(d-x\right)}^2}}$ when the value of $t$ is differentiated with respect to $x$ and equate it to zero.

Explanation of Solution

Consider the equation (II).

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

Differentiate the above equation with respect to $x$.

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

Substitute $0$ for $\frac{dt}{dx}$ in the above equation.

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

Therefore, the result is $\frac{n_{1}x}{\sqrt{a^2+x^2}}=\frac{n_2\left(d-x\right)}{\sqrt{b^2+{\left(d-x\right)}^2}}$ when the value of $t$ is differentiated with respect to $x$ and equate it to zero.

(c)

To determine

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

Answer to Problem 84CP

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

Explanation of Solution

Write the expression to obtain the value of $\sin {\theta }_1$ in triangle $PMN$.

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

Write the expression to obtain the value of $\sin {\theta }_2$ in triangle $QNO$.

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

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

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

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