CUET Preparation Today
A light source approaches the observer with velocity 0.5 cm. Doppler shift for light of wavelength 5500 Å is |
616 Å 1833 Å 5500 Å 6160 Å |
1833 Å |
$ \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$
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