The wavelength of light observed on the earth | vedclass

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Question

(B) According to the Doppler effect for light,the fractional change in wavelength is given by $\frac{\Delta \lambda}{\lambda} = \frac{v}{c}$,where $v$

Vedclass · Accepted Answer

Coming closer with a velocity of $1.5 	imes 10^5 \ m/s$

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(B) According to the Doppler effect for light,the fractional change in wavelength is given by $\frac{\Delta \lambda}{\lambda} = \frac{v}{c}$,where $v$ is the relative velocity of the source and $c$ is the speed of light $(3 	imes 10^8 \ m/s)$.Given that the wavelength decreases by $0.05\%$,we have $\frac{\Delta \lambda}{\lambda} = \frac{0.05}{100} = 5 	imes 10^{-4}$.Substituting the values: $\frac{v}{3 	imes 10^8} = 5 	imes 10^{-4}$.Solving for $v$: $v = (5 	imes 10^{-4}) 	imes (3 	imes 10^8) = 15 	imes 10^4 = 1.5 	imes 10^5 \ m/s$.Since the wavelength is decreasing (blue shift),the star must be moving towards the earth.Therefore,the star is coming closer with a velocity of $1.5 	imes 10^5 \ m/s$.

The wavelength of light observed on the earth,from a moving star is found to decrease by $0.05\%$. Relative to the earth,the star is:

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