Question

A sodium lamp in space was emitting waves of wavelength \[{\rm{2880}}\mathop {\rm{A}}\limits^{\rm{0}} \]. When observed from a planet, its wavelength was recorded at \[{\rm{2886}}\mathop {\rm{A}}\limits^{\rm{0}} \]. Find the speed of the planet.
A. \[4.25 \times {10^5}m/s\]
B. \[6.25 \times {10^5}m/s\]
C. \[2.75 \times {10^5}m/s\]
D. \[3.75 \times {10^5}m/s\]

Answer 1

Hint:In order to proceed into the problem, it is important to know the definition of doppler effect of light. Doppler effect of light is described as the apparent change in the frequency of the light which is observed by the observer due to relative motion between the source of light and the observer. Now let us solve the problem step by step considering the definition.

Formula Used:
According to the doppler effect of we have a formula given by,
\[\dfrac{{\Delta \lambda }}{\lambda } = \dfrac{{{V_{rel}}}}{c}\]
Where,
\[\Delta \lambda \] is the change in wavelength.
\[\lambda \] is the wavelength of sodium lamps in space.
\[{V_{rel}}\] is the relative velocity i.e., with respect to the space observer on the Earth.
\[c\] is speed of light and the value is \[{\rm{3 \times 1}}{{\rm{0}}^{\rm{8}}}{\rm{m/s}}\].

Complete step by step solution:
\[\dfrac{{\Delta \lambda }}{\lambda } = \dfrac{{{V_{rel}}}}{c}\]……. (1)
We know that \[\Delta \lambda = \lambda ' - \lambda \]
Here \[\lambda '\]is the wavelength which is observed by the observer from the planet.
By data it is given that \[{\rm{\lambda ' = 2886}}\mathop {\rm{A}}\limits^{\rm{0}} \] and \[{\rm{\lambda = 2880}}\mathop {\rm{A}}\limits^{\rm{0}} \]

Now substituting the value of \[\Delta \lambda \]in equation (1) we get,
\[\dfrac{{\lambda ' - \lambda }}{\lambda } = \dfrac{{{V_{rel}}}}{c}\]
\[ \Rightarrow \dfrac{{2886 - 2880}}{{2880}} = \dfrac{{{V_{rel}}}}{{3 \times {{10}^8}}}\]
Since the observer is on the earth, rearranging the equation for \[{V_{rel}}\]
\[ \Rightarrow {V_{rel}} = \dfrac{{6 \times 3 \times {{10}^8}}}{{2880}}\]
\[ \Rightarrow {V_{rel}} = 0.00625 \times {10^8}\]
\[\therefore {V_{rel}} = 6.25 \times {10^5}m/s\]
Therefore, the speed of the planet is \[6.25 \times {10^5}m/s\].

Hence, Option B is the correct answer.

Note:The doppler effect can also be applied to light waves. When a light coming from a moving source experiences the Doppler effect, depending on the direction of motion, there can be a red or blue shift in the frequency of the light.