The apparent wavelength of light from a star | vedclass

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(B) The Doppler effect for light is given by the formula $\frac{\Delta \lambda}{\lambda} = \frac{v}{c}$,where $\Delta \lambda$ is the change in wavele

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$60$

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(B) The Doppler effect for light is given by the formula $\frac{\Delta \lambda}{\lambda} = \frac{v}{c}$,where $\Delta \lambda$ is the change in wavelength,$\lambda$ is the actual wavelength,$v$ is the velocity of the star,and $c$ is the speed of light.Given that the apparent wavelength is $0.02 \%$ more than the actual wavelength,we have $\frac{\Delta \lambda}{\lambda} = 0.02 \% = \frac{0.02}{100} = 2 	imes 10^{-4}$.Substituting the values into the formula: $2 	imes 10^{-4} = \frac{v}{3 	imes 10^8 \ m/s}$.Solving for $v$: $v = (2 	imes 10^{-4}) 	imes (3 	imes 10^8 \ m/s) = 6 	imes 10^4 \ m/s$.Converting to $km/s$: $v = \frac{6 	imes 10^4}{10^3} \ km/s = 60 \ km/s$.

The apparent wavelength of light from a star moving away from the earth is $0.02 \%$ more than the actual wavelength. The velocity of the star is $[c = 3 \times 10^8 \ m/s]$. (in $km/s$)

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