Q. The apparent wavelength of the light from a star moving away from the earth is 0.2% more than its actual wavelength. Then the velocity of the star is
A) $6 \times 10^{7}ms^{- 1}$
B) $6 \times 10^{6}ms^{- 1}$
C) $6 \times 10^{5}ms^{- 1}$
D) $6 \times 10^{4}ms^{- 1}$
Explanation
(c) $v = \frac{\mathrm{\Delta}\lambda}{\lambda} \times c = \frac{0.2}{100} \times 3 \times 10^{8}$
$= 6 \times 10^{5}\ ms^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.