Question

With reckless disregard for safety and the law, you set your high-performance rocket cycle on course to streak through an intersection at top speed . Approaching the intersection, you observe green (540 nm) light from the traffic signal. After passing through, you look back to observe red (650 nm) light. Actually, the traffic signal never changed color-it didn't have time! What is the top speed of your rocket cycle, and what was the color of the traffic signal (according to an appalled bystander)?

Short answer

The top speed of the rocket cycle is 0.092c and colour of traffic light us yellow.

Step-by-step solution

Write the given data from the question

The observer wavelength for green colour,${\lambda }_{obs}^{\text{'}}=540\mathrm{nm}$

The observer wavelength for red colour,${\lambda }_{obs}^{\text{'}}=640\mathrm{nm}$

Determination of the formulas to calculate the top speed of your rocket cycle, and the colour of the traffic signal.

The expression for the frequency when the rocket cycle is approaching towards the signal is given as follows.

${\mathit{f}}_{\mathbf{o}\mathbf{b}\mathbf{s}}\mathbf{=}{\mathit{f}}_{\mathbf{s}\mathbf{o}\mathbf{u}\mathbf{r}\mathbf{c}\mathbf{e}}\sqrt{\frac{\mathbf{1}\mathbf{+}{\displaystyle \frac{\mathbf{v}}{\mathbf{c}}}}{\mathbf{1}\mathbf{-}{\displaystyle \frac{\mathbf{v}}{\mathbf{c}}}}}$ …… (i)

Here,${\mathit{f}}_{\mathbf{o}\mathbf{b}\mathbf{s}}$ is the frequency of the observer, ${\mathit{f}}_{\mathbf{s}\mathbf{o}\mathbf{u}\mathbf{r}\mathbf{c}\mathbf{e}}$is the frequency of the source, c is the speed of light and v is the velocity of the source.

The expression for the frequency when the rocket cycle is receding from the signal is given as follows.

${\mathit{f}}_{\mathbf{obs}}\mathbf{=}{\mathit{f}}_{\mathbf{source}}\sqrt{\frac{\mathbf{1}\mathbf{-}\frac{\mathbf{v}}{\mathbf{c}}}{\mathbf{1}\mathbf{+}{\displaystyle \frac{\mathbf{v}}{\mathbf{c}}}}}$ …… (ii)

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,cis the speed of light and f is the frequency.

Calculate the top speed of your rocket cycle, and the colour of the traffic signal.

Determine the expression for the wavelength when rocket cycle is approaching towards the signal,

$$\begin{gathered} \mathrm{Substitute}\frac{c}{{\lambda }_{obs}}\mathrm{for}{f}_{obs}\mathrm{and}\frac{c}{{\lambda }_{source}}\mathrm{for}{f}_{source}\mathrm{into}\mathrm{equation}\left(\mathrm{i}\right). \\ \frac{c}{{\lambda }_{obs}}=\frac{\mathrm{c}}{{\mathrm{\lambda }}_{\mathrm{source}}}\sqrt{\frac{1+{\displaystyle \frac{v}{c}}}{1-\frac{v}{c}}} \\ \frac{1}{{\mathrm{\lambda }}_{\mathrm{obs}}}=\frac{1}{{\mathrm{\lambda }}_{\mathrm{source}}}\sqrt{\frac{1+{\displaystyle \frac{\mathrm{v}}{\mathrm{c}}}}{1-\frac{\mathrm{v}}{\mathrm{c}}}} \\ {\mathrm{\lambda }}_{\mathrm{source}}={\mathrm{\lambda }}_{\mathrm{obs}}\sqrt{\frac{1+{\displaystyle \frac{\mathrm{v}}{\mathrm{c}}}}{1-\frac{\mathrm{v}}{\mathrm{c}}}}\dots \dots .\left(\mathrm{iii}\right) \end{gathered}$$

Determine the expression for the wavelength when rocket cycle is receding from the signal,

$$\begin{gathered} \mathrm{Substitute}\frac{c}{{\lambda }_{obs}^{\text{'}}}\mathrm{for}{f}_{{}_{obs}}^{\text{'}}\mathrm{and}\frac{c}{{\lambda }_{source}^{\text{'}}}\mathrm{for}{f}_{source}^{\text{'}}\mathrm{into}\mathrm{equation}\left(\mathrm{ii}\right). \\ \frac{c}{{\lambda }_{obs}^{\text{'}}}=\frac{\mathrm{c}}{{\lambda }_{source}^{\text{'}}}\sqrt{\frac{1-{\displaystyle \frac{v}{c}}}{1+\frac{v}{c}}} \\ \frac{1}{{\lambda }_{obs}^{\text{'}}}=\frac{1}{{\lambda }_{source}^{\text{'}}}\sqrt{\frac{1-\frac{\mathrm{v}}{\mathrm{c}}}{1+\frac{\mathrm{v}}{\mathrm{c}}}} \\ {\lambda }_{source}^{\text{'}}={\lambda }_{obs}^{\text{'}}\sqrt{\frac{1-{\displaystyle \frac{\mathrm{v}}{\mathrm{c}}}}{1+\frac{\mathrm{v}}{\mathrm{c}}}}\dots \dots .\left(\mathrm{iv}\right) \end{gathered}$$

Since the wavelength of the two sources are equal. Therefore, equate the equation (iii) and (iv).

$$\begin{gathered} {\lambda }_{obs}^{}\sqrt{\frac{1+{\displaystyle \frac{v}{c}}}{1-\frac{v}{c}}}={\lambda }_{obs}^{\text{'}}\sqrt{\frac{1-{\displaystyle \frac{v}{c}}}{1+\frac{v}{c}}} \\ {\lambda }_{obs}^{}\left(1+\frac{v}{c}\right)={\lambda }_{obs}^{\text{'}}\left(1-\frac{v}{c}\right) \\ {\lambda }_{obs}^{}+{\lambda }_{obs}^{}\frac{v}{c}={\lambda }_{obs}^{\text{'}}-{\lambda }_{obs}^{\text{'}}\frac{v}{c} \\ \frac{v}{c}({\lambda }_{obs}^{}+{\lambda }_{obs}^{\text{'}})={\lambda }_{obs}^{}-{\lambda }_{obs}^{\text{'}} \end{gathered}$$

Solve further as,

$$\begin{gathered} \frac{v}{c}=\frac{{\lambda }_{obs}-{\lambda }_{obs}^{\text{'}}}{{\lambda }_{obs}+{\lambda }_{obs}^{\text{'}}} \\ \mathrm{Substitute}650nm\mathrm{for}{\lambda }_{obs}^{\text{'}}540nm\mathrm{for}{\lambda }_{obs}\mathrm{and}3\times {10}^{8}m/s\mathrm{for}c\mathrm{in}\mathrm{the}\mathrm{equation} \\ \frac{v}{c}=\frac{650-540}{650+540} \\ \frac{v}{c}=\frac{110}{1190}c \\ v=0.092c \end{gathered}$$

Hence the top speed of the rocket cycle is 0.092c.

Calculate the wavelength of the traffic signal.

$$\begin{gathered} \mathrm{Substitute}540\mathrm{nm}\mathrm{for}{\mathrm{\lambda }}_{\mathrm{obs}}\mathrm{and}0.092c\mathrm{for}v\mathrm{into}\mathrm{equation}\left(\mathrm{iii}\right). \\ {\lambda }_{source}=540\sqrt{\frac{1+{\displaystyle \frac{0.092c}{c}}}{1-\frac{0.092c}{c}}} \\ {\lambda }_{source}=540\sqrt{\frac{1+0.092}{1-0.092}} \\ {\lambda }_{source}=540\times 1.096 \\ {\lambda }_{source}=592\mathrm{nm} \end{gathered}$$

By the bystander sees a wavelength of 592 nm which could make it yellow light.

Hence the top speed of the rocket cycle is 0.092c and colour of traffic light us yellow.