Q. The wavelength of light observed on the earth from a moving star is found to decrease by 0.05%. the star is
A) Coming closer with a velocity of $1.5\ \times 10^{4}ms^{- 1}$
B) Moving away with a velocity of $1.5 \times 10^{4}ms^{- 1}$
C) Coming closer with a velocity of $1.5 \times 10^{5}ms^{- 1}$
D) Moving away with a velocity of $1.5 \times 10^{- 1}$
Explanation
(c) Using the relation for Doppler’s shift
$\mathrm{\Delta}\lambda = \frac{0.05}{100}\lambda\ \ \ \ \ (Given)$
Since,
$\mathrm{\Delta}\lambda = \frac{v}{c}\lambda$
$\therefore\frac{0.05}{100}\lambda = \frac{v}{c}\lambda\ or\ v = 5 \times 10^{- 4}c$
$\Longrightarrow v = 5 \times 10^{- 4} \times 3 \times 10^{8} = 1.5 \times 10^{5}ms^{- 1}$
Since, $\lambda$ decreases, the star is approaching the observer.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.