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Question

Physics Question on Wave optics

A galaxy moves with respect to us so that sodium light of 589.0nm589.0 \,nm is observed at 589.6nm589.6 \,nm. The speed of the galaxy is

A

206kms1206\, km \,s^{-1}

B

306kms1306\, km \,s^{-1}

C

103kms1103\, km \,s^{-1}

D

51kms151\, km \,s^{-1}

Answer

306kms1306\, km \,s^{-1}

Explanation

Solution

The relation between υ,c\upsilon, c and λ\lambda are, υλ=c\upsilon\lambda = c
For small changes in υ\upsilon and λ\lambda,
Δυυ=Δλλ\frac{\Delta\upsilon}{\upsilon} = \frac{-\Delta\lambda}{\lambda}
As Δλ=589.6589.0=+0.6nm\Delta\lambda = 589.6 - 589.0 = + 0.6\, nm
Therefore, using doppler shift
Δυυ=Δλλ=vradialc\frac{\Delta\upsilon}{\upsilon} = \frac{-\Delta\lambda}{\lambda} = \frac{-v_{radial}}{c}
or vradial?+c(0.6589.0)v_{radial} ? +c\left(\frac{0.6}{589.0}\right)
=+3.06×105ms1= + 3.06 \times10^{5} m \,s^{-1}
=306kms1= 306 \,km\, s^{-1}