Show that for a source moving towards an observer equation (2-17) becomes$f_{obs}=f_{source}\sqrt{\frac{1+v/c}{1-v/c}}$

The observer frequency is $f_{obs}$.

The source frequency is$f_{source}$ .

The speed of the source is v.

The speed of the light is c .

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

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

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

The angle between the source and observer when the source moving towards the observe is equal to $180°$.

Calculate the observer frequency,

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

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

Solve further as,

$$\begin{gathered} f_{obs}=f_{source}\frac{\sqrt{\left(1-\frac{v}{c}\right)\left(1+\frac{v}{c}\right)}}{\sqrt{\left(1-\frac{v}{c}\right)\left(1-\frac{v}{c}\right)}} \\ {f}_{obs}=f_{source}\sqrt{\frac{\left(1+\frac{v}{c}\right)}{\left(1-\frac{v}{c}\right)}} \end{gathered}$$

Hence the required equation ${}_{f_{obs}=fsource\sqrt{\frac{\left({\displaystyle 1+\frac{v}{c}}\right)}{\left({\displaystyle 1-\frac{v}{c}}\right)}}}$is obtained.