Optics Question 147

Question: The time period of rotation of the sun is 25 days and its radius is $ 7\times 10^{8}m $ . The Doppler shift for the light of wavelength $ 6000{AA} $ emitted from the surface of the sun will be

[MP PMT 1994]

Options:

A) $ 0.04{AA} $

B) $ 0.40{AA} $

C) $ 4.00{AA} $

D) $ 40.0{AA} $

Show Answer

Answer:

Correct Answer: A

Solution:

$ \Delta \lambda =\lambda \frac{v}{c} $ and $ v=r\omega $

$ v=7\times 10^{8}\times \frac{2\pi }{25\times 24\times 3600},\ c=3\times 10^{8}m/s $

$ \therefore \Delta \lambda =0.04{AA} $



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