Question

An observer is moving with half the speed of light towards a stationary microwave source emitting waves at frequency of $10GHz$. What is the frequency of the microwave measured by the observer? (Speed of light $=3\times {{10}^{8}}m/s$)
$A)\text{ }15.3GHz$
$B)\text{ }10.1GHz$
$C)\text{ }12.1GHz$
$D)\text{ }17.3GHz$

Answer 1

This problem can be solved by the application of the direct formula for the frequency measured by an observer due to his or the source’s motion in terms of the original frequency, the speed of the observer, that is, due to the relativistic Doppler effect.
Formula used:
$f={{f}_{0}}\sqrt{\dfrac{c+v}{c-v}}$

Complete answer:
We will solve this problem by using the relativistic Doppler effect formula.
The frequency $f$ of a wave as observed by an observer moving with a speed $v$ is given by
$f={{f}_{0}}\sqrt{\dfrac{c+v}{c-v}}$ --(1)
where $c=3\times {{10}^{8}}m/s$ is the speed of light in vacuum and ${{f}_{0}}$ is the original frequency of the wave.
Now, let us analyze the question.
Let the required frequency of the wave observed by the observer be $f$.
The original frequency of the wave is ${{f}_{0}}=10GHz$.
The speed of the observer is $v=\dfrac{c}{2}$
Where $c=3\times {{10}^{8}}m/s$ is the speed of light in vacuum.
Using (1), we get
$f=10 GHz\sqrt{\dfrac{c+\dfrac{c}{2}}{c-\dfrac{c}{2}}}=10\sqrt{\dfrac{\dfrac{3c}{2}}{\dfrac{c}{2}}}=10\sqrt{3}=17.3GHz$
Hence, the required frequency of the wave as measured by the observer is $17.3GHz$.

So, the correct answer is “Option D”.

Note:
Students often make the mistake that they do not realize that this is a question of relativistic Doppler effect and they use only the simpler Doppler effect formula that will not give the correct answer in this case. The student must realize that when the speed of the observer is comparable to the speed of light, the relativistic Doppler effect formula must be used.