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

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

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

${\mathit{f}}_{\mathbf{o}\mathbf{b}\mathbf{s}}\mathbf{=}{\mathit{f}}_{\mathbf{s}\mathbf{o}\mathbf{u}\mathbf{r}\mathbf{c}\mathbf{e}}\sqrt{\frac{\mathbf{1}\mathbf{+}{\displaystyle \frac{\mathbf{v}}{\mathbf{c}}}}{\mathbf{1}\mathbf{-}{\displaystyle \frac{\mathbf{v}}{\mathbf{c}}}}}$ 鈥︹�� (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.

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

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

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

Here, $\mathit{位}$is the wavelength,cis the speed of light and f is the frequency.

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

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

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

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

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

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

Solve further as,

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

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

Calculate the wavelength of the traffic signal.

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

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.