Optics Question 252
Question: A light source approaches the observer with velocity 0.8 c. The doppler shift for the light of wavelength $ 5500{AA} $ is
[MP PET 1996]
Options:
A) $ 4400{AA} $
B) $ 1833{AA} $
C) $ 3167{AA} $
D) $ 7333{AA} $
Show Answer
Answer:
Correct Answer: C
Solution:
According to Doppler’s principle $ \lambda ‘=\lambda \sqrt{\frac{1-v/c}{1+v/c}} $
for v = c $ \lambda ‘=5500\sqrt{\frac{(1-0.8)}{1+0.8}}=1833.3 $
$ \therefore $ Shift $ =55001833.3=3167\AA$
BETA
