Q. 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
A) Moving away with a velocity of $1.5 \times 10^{5}m/s$
B) Coming closer with a velocity of $1.5 \times 10^{5}m/s$
C) Moving away with a velocity of $1.5 \times 10^{4}m/s$
D) Coming closer with a velocity of $1.5 \times 10^{4}m/s$
Explanation
(b) $\frac{\mathrm{\Delta}\lambda}{\lambda} = \frac{v}{c} \Rightarrow \frac{0.05}{100} = \frac{v}{3 \times 10^{8}} \Rightarrow v = 1.5 \times 10^{5}m/s$
(Since wavelength is decreasing, so star is coming closer)
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.