Optics Question 150
Question: The periodic time of rotation of a certain star is 22 days and its radius is 7 ’ 108 metres. If the wavelength of light emitted by its surface be $ 4320{AA} $ , the Doppler shift will be (1 day = 86400 sec)
[MP PET 2001]
Options:
A) $ 0.033{AA} $
B) $ 0.33{AA} $
C) $ 3.3{AA} $
D) $ 33{AA} $
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Answer:
Correct Answer: A
Solution:
$ \Delta \lambda =\lambda .\frac{v}{c} $ where $ v=r\omega =r\times ( \frac{2\pi }{T} ) $
$ \therefore \Delta \lambda =\frac{4320\times 7\times 10^{8}\times 2\times 3.14}{3\times 10^{8}\times 22\times 86400}=0.033{AA} $
BETA
