Apparent wavelength of light coming from a st | vedclass

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(B) The apparent wavelength $\lambda'$ for a source moving away from the observer is given by the Doppler effect formula: $\lambda' = \lambda \sqrt{\f

Vedclass · Accepted Answer

$60$

Vedclass · Answer

(B) The apparent wavelength $\lambda'$ for a source moving away from the observer is given by the Doppler effect formula: $\lambda' = \lambda \sqrt{\frac{c+v}{c-v}}$.Given that $\lambda' = \lambda + 0.02\% 	ext{ of } \lambda = \lambda(1 + 0.0002) = 1.0002\lambda$.Thus,$\sqrt{\frac{c+v}{c-v}} = 1.0002$.Squaring both sides,$\frac{c+v}{c-v} = (1.0002)^2 \approx 1.0004$.Using the approximation for $v \ll c$,$\frac{\Delta \lambda}{\lambda} \approx \frac{v}{c}$.Here,$\frac{\Delta \lambda}{\lambda} = \frac{0.02}{100} = 0.0002$.Therefore,$v = 0.0002 	imes c$.Given $c = 3 	imes 10^5 \, km/s$,we have $v = 0.0002 	imes 3 	imes 10^5 = 60 \, km/s$.

Apparent wavelength of light coming from a star moving away from Earth is $0.02 \%$ more than its actual wavelength. The velocity of the star is ........ $km/s$.

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