Question

A rocket is moving away from the earth at a speed of $6\times {10^7}~\frac {m}{sec}$. The rocket has blue light in it. What will be the wavelength of light recorded by an observer on the earth (wavelength of blue light $ = 4600~\overset{O}{A}$)
A. $4460\mathop A\limits^0 $
B. $5520\mathop A\limits^0 $
C. \[3680\mathop A\limits^0 \]
D. $3920\mathop A\limits^0 $

Answer 1

Hint: Use the concept of Doppler shift for light to find the speed of the star. The formula of doppler shift is given by $\dfrac{{\Delta \lambda }}{\lambda } = \dfrac{v}{c}$, where c is the speed of light. The percent decrement in the wavelength is given in the question which can directly be used in the formula.

Formula used:
$\dfrac{{\Delta \lambda }}{\lambda } = \dfrac{v}{c}$

Complete answer:
The Doppler shift for light is given by the formula
$\dfrac{{\Delta \lambda }}{\lambda } = \dfrac{v}{c}$
Where,
\[\Delta \lambda\]= change in wavelength of light.
\[\lambda\]=actual wavelength of light.
v= velocity of the source.
c= velocity of light.

It is given in the question that,
$v =6\times {10^7}~\frac {m}{sec}$

From the formula we can say that,
$\dfrac{{\Delta \lambda }}{\lambda } = \dfrac{v}{c} = \dfrac{{6 \times {{10}^7}}}{{3 \times {{10}^8}}} = 0.2$

So change in wavelength is:
$ = {\lambda ^{'}} - \lambda = 0.2\lambda $
$ \Rightarrow {\lambda ^{'}} = 1.2\lambda $
$ \Rightarrow {\lambda ^{'}} = 1.2 \times 4600$

${\lambda ^{'}} = 5520\mathop A\limits^0 $. Hence option B is correct.

Additional Information:
When a light source moves toward or away from an observer, there are some certain shifts in the value of wavelength and frequency of light. There are different kind of shift like redshift and blueshift that is used to calculate the direction and speed of a distant star just by observing the light coming from it

Note: We know that, when the source moves toward the observer then the apparent wavelength is less than the actual wavelength which exactly came in the answer.
We cannot use the Doppler shift formula of sound waves for light waves. So, a general question arises here: If both sound and light travel as waves then why does the formula differ for them?
This is because the Doppler shift for sound waves is derived by taking a medium (in which sound is travelling) to be at rest and both source and observer moves relative to this medium. On the other hand, in the case of light, there is no such thing which is at universal rest. Everything is moving with respect to each other.