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Physics Question on Wave optics

The wavelength of spectral line coming from a distant star shifts from 600 nm600\, nm to 600.1 nm600.1\, nm. The velocity of the star relative to earth is

A

50 kms−150\, km s^{-1}

B

100 kms−1100\, km s^{-1}

C

25 kms−125\, km s^{-1}

D

200 kms−1200\, km s^{-1}

Answer

50 kms−150\, km s^{-1}

Explanation

Solution

Here, λ=600 nm\lambda = 600\, nm
Δλ=600.1 nm−600 nm=0.1 nm\Delta\lambda = 600 .1\, nm - 600\, nm = 0.1 \,nm
As vsc=Δλλ\frac{v_{s}}{c} = \frac{\Delta\lambda}{ \lambda };
∴vs=Δλλc\therefore v_{s} = \frac{\Delta\lambda}{\lambda } c
=0.1 nm600 nm×3×108 ms−1=\frac{ 0.1 \,nm}{600 \,nm }\times 3\times 10^{8} \,ms^{-1}
∴vs=50×103ms−1\therefore v_{s} = 50 \times 10^{3} ms^{-1}
=50 kms−1= 50 \,km s^{-1}