Question

the wavelength of light​coming from a distant star is found to be 0.001% more than thelight comingfrom the earth. Calculate the velocity of the star.

Answer 1

$$\begin{gathered} \mathrm{Wavelength}\mathrm{of}\mathrm{light}\mathrm{coming}\mathrm{from}\mathrm{the}\mathrm{earth}=\mathrm{\lambda } \\ \mathrm{Wavelength}\mathrm{of}\mathrm{light}\mathrm{coming}\mathrm{from}\mathrm{the}\mathrm{distant}\mathrm{star},\mathrm{\lambda }\text{'}=\mathrm{\lambda }+\frac{0.001}{100}\mathrm{\lambda } \\ \mathrm{So},∆\mathrm{\lambda }=\mathrm{\lambda }\text{'}-\mathrm{\lambda } \\ =\frac{0.001}{100}\mathrm{\lambda } \\ \mathrm{Now}\mathrm{using}, \\ \frac{∆\mathrm{\lambda }}{\mathrm{\lambda }}=\frac{\mathrm{v}}{\mathrm{c}} \\ \mathrm{or}\frac{\frac{0.001}{100}\mathrm{\lambda }}{\mathrm{\lambda }}=\frac{\mathrm{v}}{\mathrm{c}} \\ \mathrm{or}\frac{0.001}{100}=\frac{\mathrm{v}}{3\times {10}^8} \\ \mathrm{or}\mathrm{v}=\frac{3\times {10}^8\times 0.001}{100}=\frac{3\times {10}^8\times {10}^{-3}}{{10}^2}=3\times {10}^3\mathrm{m}/\mathrm{s} \\ \mathrm{This}\mathrm{is}\mathrm{the}\mathrm{velocity}\mathrm{of}\mathrm{the}\mathrm{star}. \end{gathered}$$