JEE Main Solved Paper 2017 Question 27
Question: 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}} $ ) [JEE Main Solved Paper-2017]
Options:
A) 17.3 GHz
B) 15.3 GHz
C) 10.1 GHz
D) 12.1 GHz
Show Answer
Answer:
Correct Answer: A
Solution:
- Doppler effect in light (speed of observer is not very small compare to speed of light) $ f^{1}=\sqrt{\frac{1+V/C}{1-V/C}}f _{source}=\sqrt{\frac{1+1/2}{1-1/2}}(10,GHz) $ $ =17.3,GHz $
