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 xx 10^(8)m cdot s^(-1))`

Options

A. `10.1` GHz
B. `12. 1 ` GHz
C. `17.3` GHz
D. `15.3` GHz

Answer

C

Explanation

Speed of the observer `=c/2`
`therefore ` Doppler effect in relevant with the theory of relativity as follows:
`v.=vsqrt((1+beta)/(1-beta))`
`beta = (" speed of observer " ) / (" speed of light " ) = 1/2`
`therefore v. = 10 sqrt((1+1/2)/(1-1/2))=10sqrt(3) = 10 xx 1.73`
`=17.3` GHz
The option (c) is correct.