Question

A rocket is receding away from earth with velocity $0.2\text{c}.$ The rocket emits signal of frequency $4\times {10}^7\text{Hz}.$ The apparent frequency of the signal produced by the rocket observed by the observer on earth will be

• A. $3\times {10}^6\text{Hz}$
• B. $4\times {10}^6\text{Hz}$
• C. $3.2\times {10}^7\text{Hz}$
• D. $5\times {10}^7\text{Hz}$

Answer 1

The correct option is C $3.2\times {10}^7\text{Hz}$

According to the Doppler effect in light,

$\frac{\delta \nu }{\nu }=\frac{v}{c}$
$\delta \nu =\frac{v}{c}\nu$
$\delta \nu =0.2\times 4.0\times {10}^7=0.8\times {10}^7{s}^{-1}$

this will be a red shift,

therefore ${\nu }^{\prime }=(4.0-0.8)\times {10}^7=3.2\times {10}^7$

Option "C" is correct.