Question

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

Short answer

The required equation $f_{obs}=f_{source}\sqrt{\frac{1+v/c}{1-v/c}}$is obtained.

Step-by-step solution

Write the given data from the question.

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 .

Write the formulas to determine the equation which relates the frequency of observer and source.

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.

Determine the equation which relates the frequency of observer and source.

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.