A light source approaches the observer with velocity 0.5 cm. Doppler shift for light of wavelength 5500 Å is
Answer & explanation
Correct answer: option 2
$ \frac{\nu'}{\nu} = \frac{c+v}{c} = \frac{\lambda'}{\lambda}$
$\lambda'= \frac{\lambda c}{c+v}= \frac{5500\times c}{c+0.5c} = \frac{11000 c}{3c} = 3667A^o$
$ \Delta \lambda = \lambda - \lambda'= 1883A^o$