A star which is emitting radiation at a wavelength of 5000 Å is approaching the earth with a velocity of 1…
A star which is emitting radiation at a wavelength of 5000 Å is approaching the earth with a velocity of 1.50 × 10$^{6}$ ms^-1. Calculate the change in wavelength of the radiation as received on the earth.
Solution
Sol. The phenomenon of apparent change in frequency (or wavelength) of the light due to the relative motion between the source of light and the observer is called Doppler's effect in light.
So, $\Delta\lambda = \lambda \times \frac{v}{c}$
Given, wavelength, $\lambda = 5000\AA$
Velocity of source, $v = 1.5 \times 10^{6}$ ms$^{-1}$ and $c = 3 \times 10^{8}$ ms$^{-1}$
∴ $\Delta\lambda = 5000 \times \frac{1.5\times10^{6}}{3\times10^{8}} = 25\AA$
Answer: $25$ Å