Q 12-08-224JEE MainJEE Main 2017 (2 Apr)Medium
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 $= 3\times10^8\ \text{m s}^{-1}$)
Answer: (D) $17.3$ GHz
At $v = c/2$ the relativistic Doppler formula for light must be used (source and observer approaching, $\beta = 0.5$):
$$f' = f\sqrt{\frac{1+\beta}{1-\beta}} = 10\sqrt{\frac{1.5}{0.5}} = 10\sqrt3 \approx 17.3\ \text{GHz}$$
Solution by Sreeraj P, M.Sc Physics