91Ó°ÊÓ

The velocity of the source,$v=0.9c$

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

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

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

The relationship between the observer and source frequency is given as follows.

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

Determine the wavelength equation for the observer.

${\mathrm{λ}}_{obs}=\frac{c}{f_{obs}}$

Substitute $f_{source}\sqrt{\frac{1−v/c}{1+v/c}}$for$f_{obs}$ into above equation.

$\begin{array}{l}{\mathrm{λ}}_{obs}=\frac{c}{f_{source}\sqrt{\frac{1−v/c}{1+v/c}}}\\ {\mathrm{λ}}_{obs}=\frac{c}{f_{source}}\sqrt{\frac{1+v/c}{1−v/c}}\end{array}$

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

$\begin{array}{c}{\mathrm{λ}}_{obs}={\mathrm{λ}}_{source}\sqrt{\frac{1+v/c}{1−v/c}}\\ \frac{{\mathrm{λ}}_{obs}}{{\mathrm{λ}}_{source}}=\sqrt{\frac{1+v/c}{1−v/c}}\end{array}$

Substitute $0.9c$for$v$ into above equation.

$\begin{array}{l}\frac{{\mathrm{λ}}_{obs}}{{\mathrm{λ}}_{source}}=\sqrt{\frac{1+0.9c/c}{1−0.9c/c}}\\ \frac{{\mathrm{λ}}_{obs}}{{\mathrm{λ}}_{source}}=\sqrt{\frac{1+0.9}{1−0.9}}\\ \frac{{\mathrm{λ}}_{obs}}{{\mathrm{λ}}_{source}}=\sqrt{\frac{1.9}{0.1}}\\ \frac{{\mathrm{λ}}_{obs}}{{\mathrm{λ}}_{source}}=4.358\end{array}$

Hence the factor by which wavelength would be change is$4.358$.