Question

A star moving away from earth with a speed of $0.02c$, while emitting light of frequency $6\times {10}^{14}\text{Hz}$. What frequency will be observed on earth?
(Take $c=3\times {10}^8\text{m/s}$)

• A. $0.12\times {10}^{14}\text{Hz}$
• B. $0.24\times {10}^{14}\text{Hz}$
• C. $5.88\times {10}^{14}\text{Hz}$
• D. $1.2\times {10}^{14}\text{Hz}$

Answer 1

The correct option is C $5.88\times {10}^{14}\text{Hz}$

The change in frequency is given by,
$\frac{\Delta f}{f}=\frac{v_r}{c}\Rightarrow \Delta f=f\times \frac{v_r}{c}$

$\therefore \Delta f=6\times {10}^{14}\times \frac{0.02c}{c}=0.12\times {10}^{14}\text{Hz}$

As the star is moving away from earth, the apparent frequency will be less than actual frequency, i.e. $f^{\prime }<f$

$\therefore f^{\prime }=f-\Delta f=6\times {10}^{14}-0.12\times {10}^{14}$
$=5.88\times {10}^{14}\text{Hz}$

Hence, $\left(B\right)$ is the correct answer.