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} $
BETA
