Q84 / 90 · JEE Main 2017 (02 Apr) · Physics

An observer is moving with half the speed of light towards a stationary microwave source emitting waves at frequency $10 GHz$ . What is the frequency of the microwave measured by the observer? (speed of light $= 3 \times 10^{8} m s^{- 1}$ )

A
$15.3 GHz$
B
$10.1 GHz$
C
$12.1 GHz$
Correct answer
D
$17.3 GHz$

Solution

\frac{c}{f^{'}} = \frac{c}{f} \sqrt{\frac{1 - β}{1 + β}}, [\begin{matrix} β = \frac{v}{c} \\ β = \frac{1}{2} \end{matrix}]

\Rightarrow \frac{1}{f^{'}} = \frac{1}{f} \sqrt{\frac{1 - \frac{1}{2}}{1 + \frac{1}{2}}} = \sqrt{\frac{1}{3}} \frac{1}{f}

\Rightarrow f^{'} = \sqrt{3} f

\Rightarrow f^{'} = 1.73 (10)

\Rightarrow f^{'} = 17.3 GHz

Hence, the correct answer is option: 'D'.

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