Practicing Success

Target Exam

CUET

Subject

Physics

Chapter

Wave Optics

Question:

A light source approaches the observer with velocity 0.5 cm. Doppler shift for light of wavelength 5500 Å is 

Options:

616 Å 

1833 Å

5500 Å 

6160 Å

Correct Answer:

1833 Å

Explanation:

$ \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$