Question

A star is receding from the earth with a velocity of ${10^5}m/s$. If the wavelength of its spectral line is $5700\,A^\circ $, then Doppler shift will be:
A) $200\,A^\circ $
B) $1.9\,A^\circ $
C) $20\,A^\circ $
D) $0.2\,A^\circ $

Answer 1

Hint: Here, we will use the formula of Doppler’s shift. Just keep in mind that the Doppler shift also as the Doppler Effect is the change in the wavelength or the frequency of sound waves when the source is moving relative to the observer.

Formula used:
$\dfrac{{\Delta \lambda }}{\lambda } = \dfrac{v}{c}$
Here, $\Delta \lambda $ is the wavelength shift, $\Delta f$ is the frequency shift, $v$ is the speed of the object, and $c$ is the speed of light.

Complete step by step solution:
As we know, Doppler’s effect in light is given by
$\dfrac{{\Delta \lambda }}{\lambda } = \dfrac{v}{c}$
Here, $\Delta \lambda $ is the wavelength shift, $\Delta f$ is the frequency shift, $v$ is the speed of the object, and $c$ is the speed of light.
Here, the above equation can be written as
$\Delta \lambda = \dfrac{{\lambda v}}{c}$
Now, putting the values, we get
$\Delta \lambda = \dfrac{{5700 \times {{10}^5}}}{{3 \times {{10}^8}}}$
$ \Rightarrow \lambda = 1.9A^\circ $

Therefore, the Doppler shift will be $1.9A^\circ $. Hence, option (B) is the correct option.

Additional Information:
Now, we will derive the formula of the Doppler shift.
For this, let
$c = \dfrac{{{\lambda _S}}}{T}$
Here, $c$ is the wave velocity, ${\lambda _S}$ is the wavelength of the source, and $T$ is the time taken by the wave.
Now, the above equation can be written as
$T = \dfrac{{{\lambda _S}}}{c}$
Now, we know that the distance between the source and the observer is given by
$d = {v_S}T$
Here, ${v_S}$ is the velocity of the source when it is moving towards the observer and $d$ is the distance covered by the source.
Now, putting the value of $T$ in the above equation, we get
$d = \dfrac{{{v_S}{\lambda _S}}}{c}$
Now, the observed wavelength is given by
${\lambda _0} = {\lambda _S} - d$
Now, putting the value of $d$ in the above equation, we get
${\lambda _0} = {\lambda _S} - \dfrac{{{v_S}{\lambda _S}}}{c}$
Now, taking ${\lambda _S}$ common in the above equation
${\lambda _0} = {\lambda _S}\left( {1 - \dfrac{v}{c}} \right)$
$ \Rightarrow \,{\lambda _0} = {\lambda _S}\left( {\dfrac{{c - v}}{c}} \right)$
Now, the Doppler shift is given by
$\Delta \lambda = {\lambda _S} - {\lambda _0}$
Now, putting the value of ${\lambda _0}$ , we have
$\Delta \lambda = {\lambda _S} - {\lambda _S}\left( {\dfrac{{c - v}}{c}} \right)$
Taking ${\lambda _S}$ common from the above equation, we have
$ \Rightarrow \,\Delta \lambda = {\lambda _S}\left( {1 - \dfrac{{c - v}}{c}} \right)$
On further solving, we have
$ \Rightarrow \,\Delta \lambda = {\lambda _S}\left( {\dfrac{{c - c + v}}{c}} \right)$
$\Delta \lambda = \dfrac{{{\lambda _S}v}}{c}$
Which is the expression of the Doppler shift when the source is moving and the observer is stationary.

Note: For the sound waves, the equation for the Doppler shift differs depending on whether it is the source, the observer, or the air, which is moving. Also, we know that the light did not require any medium. Hence, the Doppler shift for the light having no medium depends only on the relative speed of the observer and the source.