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 $\frac{v'}{v}=\frac{c+v}{c}=\frac{\lambda}{\lambda'}$
or $\lambda'=\frac{\lambda c}{c+v}=\frac{5500 \times c}{c+0.5 c}=\frac{5500 \times 2}{3}$ = 3667 Å
∴ Δλ = (5500 - 3667) Å = 1833 Å