Chapter 25, Problem 41P

(a)

To determine

The time taken by the light to reach the earth surface from the atmosphere at the distance of $100\text{km}$.

Answer to Problem 41P

The time taken by the light to travel $100\text{km}$ is $334\mu \text{s}$.

Explanation of Solution

Given info: The distance $h$ between the atmosphere and the surface of earth is $100\text{km}$, the index of refraction where the light enters the atmosphere is 1 and refractive index varies linearly with distance.

The value of refractive index at the Earth’s surface is $1.000293$.

The expression for the refractive index at any distance $x$ from where the light enter is,

$\begin{array}{c}n\left(x\right)=1+\left(\frac{1.000293-1}{h}\right)x\\ =1+\frac{0.00293}{h}x\end{array}$

Formula to find the time required to transverse the atmosphere is,

$\Delta t={\displaystyle \underset{0}{\overset{h}{\int }}\frac{dx}{v}}$ (1)

Here,

$v$ is the speed of light.

$\Delta t$ is the time taken by light to cover distance $dx$.

The expression for index of any medium is,

$n(x)=\frac{c}{v}$

Here,

$c$ is the speed of light in vacuum.

$v$ is the speed of light in the medium.

Rearrange the above equation for $v$.

$v=\frac{c}{n(x)}$ (2)

Substitute $\frac{c}{n\left(x\right)}$ for $v$ from equation (2) in equation (1).

$\Delta t={\displaystyle \underset{0}{\overset{h}{\int }}\frac{n\left(x\right)}{c}dx}$

Substitute $1+\frac{0.00293}{h}x$ for $n\left(x\right)$ in above equation.

$\begin{array}{c}\Delta t={\displaystyle \underset{0}{\overset{h}{\int }}\frac{\left(1+\frac{0.00293}{h}x\right)}{c}dx}\\ =\frac{1}{c}{\displaystyle \underset{0}{\overset{h}{\int }}\left[1+\left(\frac{0.00293}{h}\right)x\right]dx}\\ =\frac{1}{c}{\left[x+\frac{0.00293}{h}\cdot \frac{x^2}{2}\right]}_0^h\\ =\frac{1}{c}\left[h+\frac{0.00293}{h}\cdot \frac{h^2}{2}\right]\end{array}$

Further simplify the above equation.

$\begin{array}{c}\Delta t=\frac{1}{c}\left[h+\frac{0.00293}{h}\cdot \frac{h^2}{2}\right]\\ =\frac{h}{c}+\frac{0.00293h}{2}\\ =\frac{h}{c}\left[\frac{2.00293}{2}\right]\end{array}$

Substitute $100\text{km}$ for $h$ and $3\times {10}^8\text{m}/\text{s}$ for $c$ in above equation.

$\begin{array}{c}\Delta t=\frac{100\text{km}\times \frac{{\text{10}}^3\text{m}}{1\text{km}}}{3\times {10}^8\text{m}/\text{s}}\cdot 1.001465\\ =\frac{100146.5\text{m}}{3\times {10}^8\text{m}/\text{s}}\\ =3.3382\times {10}^{-4}\text{s}\left(\frac{1\text{μs}}{{10}^{-6}\text{s}}\right)\\ \approx 334\text{μs}\end{array}$

Thus, the time taken by the light to travel $100\text{km}$ is $3.3382\times {10}^{-4}\text{s}$.

Conclusion:

Therefore, the time taken by the light to travel $100\text{km}$ is approximately $334\mu \text{s}$.

(b)

To determine

The percentage increase in time when the light travels in the absence of Earth’s atmosphere.

Answer to Problem 41P

The percentage increase in the time when atmosphere is absent is $0.015\%$.

Explanation of Solution

Given info: The distance $h$ between the atmosphere and the surface of earth is $100\text{km}$, the index of refraction where the light enters the atmosphere is 1.

Formula to calculate the time required when light travels in absence of atmosphere is,

$t=\frac{h}{c}$ (3)

Here,

$h$ is distance between the atmosphere and the surface of Earth.

$c$ is the speed of light in the vacuum.

$t$ is time taken.

Substitute $3\times {10}^8\text{m}/\text{s}$ for $c$ and $100\text{km}$ for $h$ in equation (3).

$\begin{array}{c}t=\frac{100\text{km}\times \frac{{\text{10}}^3\text{m}}{1\text{km}}}{3\times {10}^8\text{m}/\text{s}}\\ =\frac{{10}^5\text{m}}{\begin{array}{l}3\times {10}^8\text{m}/\text{s}\\ =3.33\times {10}^{-4}\text{s}\end{array}}\end{array}$

The time taken by the light to travel $100\text{km}$ in vacuum is $3.33\times {10}^{-4}\text{s}$.

Write the expression for percentage increase.

$\begin{array}{c}\%\text{increase}=\frac{3.3382\times {10}^{-4}\text{s}-3.33\times {10}^{-4}\text{s}}{3.3382\times {10}^{-4}\text{s}}\\ =0.015\end{array}$

Thus, the percentage increase in the time when atmosphere is absent is $0.015\%$.

Conclusion:

Therefore, the percentage increase in the time when atmosphere is absent is $0.015\%$.