A light source approaches the observer with velocity 0.5 cm. Doppler shift for light of wavelength 550 Å is
Answer & explanation
Correct answer: option 2
In case of light, $v'=\frac{c+v}{c}.v$ and $\frac{v'}{v}=\frac{c+v}{c}=\frac{λ}{λ'}$
or $λ'=\frac{λc}{c+v}=\frac{5500xc}{c+0.5c}=\frac{5500x2}{3}=3667Å$
$∴Δλ= (5500 −3667)Å =1833Å$