Chapter 22, Problem 54P

To determine

Wavelength of reflected pulse measured by the first spaceship.

Answer to Problem 54P

Wavelength of reflected pulse will be $849.9\text{nm}$.

Explanation of Solution

Write the equation to find the frequency of pulse received by an observer in first spaceship.

    $f_{01}=f_{s1}\sqrt{\frac{1+v_{rel}/c}{1-v_{rel}/c}}$

Here, $f_{01}$ is the frequency of pulse received by an observer in first spaceship, $f_{s1}$ is the frequency of pulse produced in first spaceship, $v_{rel}$ is the speed of second spaceship relative to earth, and $c$ is the speed of light in vacuum.

Write the equation to find the frequency of pulse received by observer in second spaceship.

    $f_{02}=f_{s2}\sqrt{\frac{1+v_{rel}/c}{1-v_{rel}/c}}$

Here, $f_{02}$ is the frequency of pulse received by observer in second spaceship and $f_{s2}$ is the frequency of pulse produced in second spaceship.

Write the relation between $f_{s2}$ and $f_{01}$.

  $f_{s2}=f_{01}$

Rewrite the equation for $f_{02}$ by substituting the above relation.

  $f_{02}=f_{01}\sqrt{\frac{1+v_{rel}/c}{1-v_{rel}/c}}$

Rewrite the above equation by substituting the previous equation for $f_{01}$.

  $\begin{array}{c}{f}_{02}=\left(f_{s1}\sqrt{\frac{1+v_{rel}/c}{1-v_{rel}/c}}\right)\sqrt{\frac{1+v_{rel}/c}{1-v_{rel}/c}}\\ =f_{s1}\left(\frac{1+v_{rel}/c}{1-v_{rel}/c}\right)\end{array}$

Rewrite the above equation using binomial approximation.

  $\begin{array}{c}{f}_{02}\approx f_{s1}\left(1+\frac{v_{rel}}{c}\right)\left(1+\frac{v_{rel}}{c}\right)\\ \approx f_{s1}\left(1+{\left(\frac{v_{rel}}{c}\right)}^2\right)\\ \approx f_{s1}\left(1+\frac{2v_{rel}}{c}\right)\end{array}$                                                                           (I)

Write the relation between frequency and wavelength of pulse received by observer in second spaceship.

    $f_{02}=\frac{c}{{\lambda }_{02}}$                                                                                               (II)

Here, ${\lambda }_{02}$ is the wavelength of pulse received by observer in second spaceship.

Write the relation between frequency and wavelength of pulse produced in first spaceship.

    $f_{s1}=\frac{c}{{\lambda }_{s1}}$                                                                                              (III)

Here, ${\lambda }_{s1}$ is the wavelength of pulse produced in first spaceship.

Rewrite equation (I) by substituting equations (II) and (III).

  $\begin{array}{l}\frac{c}{{\lambda }_{02}}\approx \frac{c}{{\lambda }_{s1}}\left(1+\frac{2v_{rel}}{c}\right)\\ {\lambda }_{02}={\lambda }_{s1}{\left(1+\frac{2v_{rel}}{c}\right)}^{-1}\end{array}$

Conclusion:

Substitute $850.00\text{nm}$ for ${\lambda }_{s1}$, $24.6\text{km}$ for $v_{rel}$, and $3.0\times {10}^8\text{m}/\text{s}$ for $c$ in the above equation to find ${\lambda }_{02}$.

    $\begin{array}{c}{\lambda }_{02}=\left(850.00\text{nm}\left(\frac{{10}^{-9}\text{m}}{1\text{nm}}\right)\right){\left(1+\frac{2\left(24.6\text{km}\left(\frac{{10}^{-3}\text{m}}{1\text{km}}\right)\right)}{3.0\times {10}^8\text{m}/\text{s}}\right)}^{-1}\\ =\left(850.00\times {10}^{-9}\text{m}\right)\left(0.99\right)\\ =849.9\times {10}^{-9}\text{m}\left(\frac{1\text{nm}}{{10}^{-9}\text{m}}\right)\\ =849.9\text{nm}\end{array}$

Therefore, the wavelength of reflected pulse will be $849.9\text{nm}$.