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Question

(D) The Doppler effect for light states that the fractional change in wavelength is given by $\frac{\Delta \lambda}{\lambda} = \frac{v}{c}$,where $v$

Vedclass · Accepted Answer

$30$

Vedclass · Answer

(D) The Doppler effect for light states that the fractional change in wavelength is given by $\frac{\Delta \lambda}{\lambda} = \frac{v}{c}$,where $v$ is the velocity of the source and $c$ is the speed of light $(c \approx 3 	imes 10^8 \ m/s)$.Given that the apparent wavelength is $0.01\%$ more than the real wavelength,we have $\frac{\Delta \lambda}{\lambda} = 0.01\% = \frac{0.01}{100} = 10^{-4}$.Substituting this into the formula: $\frac{v}{c} = 10^{-4}$.Therefore,$v = c 	imes 10^{-4} = (3 	imes 10^8 \ m/s) 	imes 10^{-4} = 3 	imes 10^4 \ m/s$.Converting $m/s$ to $km/s$: $v = \frac{3 	imes 10^4}{10^3} \ km/s = 30 \ km/s$.

The apparent wavelength of the light from a star moving away from the Earth is $0.01\%$ more than its real wavelength. Then the velocity of the star is......$km/sec$.

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