Question

The heavenly body is receding from earth, such that the fractional change in $\lambda $ is $1$ , then its velocity is:
(A) $c$
(B) $\dfrac{{3c}}{5}$
(C) $\dfrac{c}{5}$
(D) $\dfrac{{2c}}{5}$

Answer 1

Hint: In order to solve this question the knowledge of Doppler Effect for Light is needed. Here it is given that the heavenly body is receding from the earth and due to this movement the wavelength is changing for the observer. Use the relation of frequency, velocity and wavelength in order to get the answer.

Complete step by step solution:
When a star or a planet or any heavenly body starts receding away from the earth.
If $f$ is the frequency of the vibration.
Then, $f = \dfrac{c}{\lambda }$
If $v$ is the velocity of body moving away
${\lambda ^{'}}$ is the apparent wavelength to an observer on the earth
${\lambda ^{'}} = \dfrac{{(c + v)}}{f}$
Here, $c$ and $v$ are in opposite to each other.
Putting $f = \dfrac{c}{\lambda }$ we get,
 ${\lambda ^{'}} = \dfrac{{(c + v)}}{c}\lambda $
This can be simplified as,
${\lambda ^{'}} = \left( {1 + \dfrac{v}{c}} \right)\lambda $
On simplifying it further we get,
$\dfrac{{{\lambda ^{'}} - \lambda }}{\lambda } = \dfrac{v}{c}$
Fractional change in wavelength is given in the question that is $1$
Putting it in the above equation we have,
$\dfrac{v}{c} = 1$
Here it is clear from the above equation that $v = c$
So the velocity is $c$ .

So, option $A$ is the correct answer.

Note: In relativistic Doppler Effect is the change in frequency as well as in the wavelength of light that is caused by the relative motion of the source of light and the observer. This phenomenon can also be described using the theory of relativity. It should be kept in mind that the relativistic Doppler Effect is different from the non-relativistic Doppler effect as the equations here include the time dilation effect of the special theory of relativity given by Einstein and do not involve the medium of propagation as a reference point.