Question

A source of light approaching the Earth emits a light of wavelength \(600\, {nm}\). If this light is observed at a wavelength of \(599\, {nm}\) by an observer on Earth, then the speed of the source of light in km/s is: \[ {(speed of light in vacuum } c = 3 \times 10^8 \, {m/s)} \]

• A. 250
• B. 300
• C. 400
• D. 500

Hint:

For small relative velocities, use the approximation \( \frac{\Delta \lambda}{\lambda} = \frac{v}{c} \) to estimate Doppler shifts in wavelength.

Answer 1

The Correct Option is D

Since the source is approaching, we use the relativistic Doppler effect formula for light: \[ \frac{\Delta \lambda}{\lambda} = \frac{v}{c} \] Where: \[ \Delta \lambda = \lambda_{{source}} - \lambda_{{observed}} = 600 - 599 = 1\, {nm} \] \[ \lambda = 600\, {nm}, c = 3 \times 10^8\, {m/s} \] Now, substitute into the formula: \[ \frac{1}{600} = \frac{v}{3 \times 10^8} \Rightarrow v = \frac{1}{600} \times 3 \times 10^8 = 5 \times 10^5\, {m/s} \] Convert to km/s: \[ v = 500\, {km/s} \]