Question

At rest, a light source emits $\mathbf{532}\mathit{n}\mathit{m}$light. (a) As it moves along the line connecting it and Earth. observers on Earth see$\mathbf{412}\mathit{n}\mathit{m}$ . What is the source's velocity (magnitude and direction)? (b) Were it to move in the opposite direction at the same speed. what wavelength would be seen? (c) Were it to circle Earth at the same speed. what wavelength would be seen?

Short answer

(a)The velocity of the source is $0.25c$and it is approaching towards the source.

(b)The wavelength of the light when the light is receding is$687\text{\hspace{0.17em}nm}$ .

(c)The wavelength of the light for circling the Earth$549\text{\hspace{0.17em}nm}$ .

Step-by-step solution

Write the given data from the question.

The wavelength of source,${\lambda }_{source}=532\text{\hspace{0.17em}nm}$

The wavelength of observer,${\lambda }_{obs}=412\text{\hspace{0.17em}nm}$

 Step 2: Determine the formulas to calculate the source velocity, wavelength in opposite direction and wavelength of light for circling Earth.

The expression to calculate the observer frequency in terms of source frequency, speed and speed of light is given as follows.

${\mathit{f}}_{\mathbf{o}\mathbf{b}\mathbf{s}}\mathbf{=}{\mathit{f}}_{\mathbf{s}\mathbf{o}\mathbf{u}\mathbf{r}\mathbf{c}\mathbf{e}}\frac{\sqrt{\mathbf{1}\mathbf{-}{\mathbf{\left(}\frac{\mathbf{v}}{\mathbf{c}}\mathbf{\right)}}^{\mathbf{2}}}}{\mathbf{1}\mathbf{+}\frac{\mathbf{v}}{\mathbf{c}}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{\theta }}$

Here,$\theta$is the angle between the source of light and observer.

The expression between the frequency, wavelength and speed of light is given as follows.

$\mathit{\lambda }\mathbf{=}\frac{\mathbf{c}}{\mathbf{f}}$

Here,$\mathit{\lambda }$is the wavelength, $\mathit{c}$is the speed of light and role="math" localid="1658810978725" $\mathit{f}$is the frequency.

(a) Calculate the source velocity.

Determine the wavelength equation for the observer.

${\lambda }_{obs}=\frac{c}{f_{obs}}$

Substitute$f_{source}\frac{\sqrt{1-{\left(\frac{v}{c}\right)}^2}}{1+\frac{v}{c}\cos \theta }$for$f_{obs}$into above equation.

$\begin{array}{l}{\lambda }_{obs}=\frac{c}{f_{source}\frac{\sqrt{1-{\left(\frac{v}{c}\right)}^2}}{1+\frac{v}{c}\cos \theta }}\\ {\lambda }_{obs}=\frac{c}{f_{source}}\frac{1+\frac{v}{c}\cos \theta }{\sqrt{1-{\left(\frac{v}{c}\right)}^2}}\end{array}$

Substitute${\lambda }_{source}$for $\frac{c}{f_{source}}$into above equation.

${\lambda }_{obs}={\lambda }_{source}\frac{1+\frac{v}{c}\cos \theta }{\sqrt{1-{\left(\frac{v}{c}\right)}^2}}$ …(i)

The observed wavelength is shorter than the wavelength in the rest frame. Therefore, light is approaching.The angle between the source and observer when the source moving towards the observe is equal to $180°$.

Substitute $180°$for$\theta$into equation (i).

$\begin{array}{c}{\lambda }_{obs}={\lambda }_{source}\frac{1+\frac{v}{c}\cos 180°}{\sqrt{1-{\left(\frac{v}{c}\right)}^2}}\\ {\lambda }_{obs}={\lambda }_{source}\frac{1-\frac{v}{c}}{\sqrt{1-{\left(\frac{v}{c}\right)}^2}}\\ {\lambda }_{obs}\sqrt{1-{\left(\frac{v}{c}\right)}^2}={\lambda }_{source}(1-\frac{v}{c})\end{array}$

Take a square of both the sides of the above equation.

$\begin{array}{c}{\lambda }_{0bs}^2[1-{\left(\frac{v}{c}\right)}^2]={\lambda }_{source}^2{(1-\frac{v}{c})}^2\\ {\lambda }_{0bs}^2\left[(1-\frac{v}{c})(1+\frac{v}{c})\right]={\lambda }_{source}^2{(1-\frac{v}{c})}^2\\ {\lambda }_{0bs}^2(1+\frac{v}{c})={\lambda }_{source}^2(1-\frac{v}{c})\\ {\lambda }_{0bs}^2+\frac{v}{c}{\lambda }_{0bs}^2={\lambda }_{source}^2-{\lambda }_{source}^2\frac{v}{c}\end{array}$

Solve further as,

$\begin{array}{c}\frac{v}{c}{\lambda }_{0bs}^2+{\lambda }_{source}^2\frac{v}{c}={\lambda }_{source}^2-{\lambda }_{0bs}^2\\ \frac{v}{c}({\lambda }_{0bs}^2+{\lambda }_{source}^2)={\lambda }_{source}^2-{\lambda }_{0bs}^2\\ \frac{v}{c}=\frac{{\lambda }_{source}^2-{\lambda }_{0bs}^2}{{\lambda }_{0bs}^2+{\lambda }_{source}^2}\end{array}$

Substitute$532\text{\hspace{0.17em}nm}$ for ${\lambda }_{source}$and$412\text{\hspace{0.17em}nm}$ for${\lambda }_{obs}$ into above equation.

$\begin{array}{c}\frac{v}{c}=\frac{{532}^2-{412}^2}{{532}^2+{412}^2}\\ \frac{v}{c}=\frac{113280}{452768}\\ \frac{v}{c}=0.25\\ v=0.25c\end{array}$

Hence the velocity of the source is $0.25c$and it is approaching towards the source.

(b) Determine the wavelength of the light for the reverse direction.

The speed of the light would be the same in the opposite direction but the angle between the observer and source is$0°$.

Calculate the wavelength of the light,

Substitute $532\text{\hspace{0.17em}nm}$for${\lambda }_{source}$, $0°$for $\theta$and$0.25c$for $v$into equation (i).

$\begin{array}{l}{\lambda }_{source}=532\times {10}^{-9}\frac{1+\frac{0.25c}{c}\cos 0}{\sqrt{1-{\left(\frac{0.25c}{c}\right)}^2}}\\ {\lambda }_{source}=532\times {10}^{-9}\frac{1+0.25}{\sqrt{1-0.0625}}\\ {\lambda }_{source}=532\times {10}^{-9}\times 1.29\\ {\lambda }_{source}=687\text{\hspace{0.17em}nm}\end{array}$

Hence the wavelength of the light when the light is receding is$687\text{\hspace{0.17em}nm}$ .

(c) Determine the wavelength of the light for It is circling the Earth.

The speed of the light would be the same but for circling the Earth the angle between the observer and source is $90°$.

Calculate the wavelength of the light,

Substitute $532\text{\hspace{0.17em}nm}$for${\lambda }_{source}$ ,$90°$for$\theta$ and$0.25c$ for$v$ into equation (i).

$\begin{array}{l}{\lambda }_{source}=532\times {10}^{-9}\frac{1+\frac{0.25c}{c}\cos 90°}{\sqrt{1-{\left(\frac{0.25c}{c}\right)}^2}}\\ {\lambda }_{source}=532\times {10}^{-9}\frac{1}{\sqrt{1-0.0625}}\\ {\lambda }_{source}=532\times {10}^{-9}\times 1.032\\ {\lambda }_{source}=549\text{\hspace{0.17em}nm}\end{array}$

Hence the wavelength of the light for circling the Earth $549\text{\hspace{0.17em}nm}$.