A) stationary with respect to the earth
B) approaching the earth with velocity of light
C) receding from the earth with velocity of Light
D) receding from the earth with a velocity equal to \[1.5\times {{10}^{6}}m/s\]
Correct Answer: D
Solution :
\[\frac{\Delta \lambda }{\lambda }=\frac{v}{c}\] Now, \[\Delta \lambda =\frac{0.5}{100}\] \[\Rightarrow \] \[\frac{\Delta \lambda }{\lambda }=\frac{0.5}{100}\] \[\therefore \] \[v=\frac{0.5}{100}\times c=\frac{0.5}{100}\times 2\times {{10}^{8}}\] \[=1.5\times {{10}^{6}}m/s\] Increase in\[\lambda \]indicates that the star is receding.
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