Question 263: Waves Optics

Question

The apparent wavelength of light from a star moving away from the earth is $0.02 \%$ more than the actual wavelength. The velocity of star is $\left[\mathrm{c}=ight.$ velocity of light $\left.=3 	imes 10^8 \mathrm{~m} / \mathrm{s}ight]$

Accepted Answer

When a star moves away from the Earth, the observed wavelength $(\lambda')$ becomes longer than the actual wavelength $(\lambda)$. This phenomenon is known as the
redshift
due to the
Doppler effect
.
According to the Doppler effect for light (for velocities much less than $c$):
$$ \frac{\Delta \lambda}{\lambda} = \frac{v}{c} $$
where
$\Delta \lambda = \lambda' - \lambda$ is the
change in wavelength
,
$v$ is the
velocity of the star away from the Earth
,
$c$ is the
speed of light
.
Given:
The apparent wavelength is $0.02\%$ more than the actual wavelength.
So,
$$ \frac{\Delta \lambda}{\lambda} = 0.02\% = \frac{0.02}{100} = 0.0002 $$
Substitute in the Doppler formula:
$$ 0.0002 = \frac{v}{3 	imes 10^8} $$
Solving for $v$:
$$ v = 0.0002 	imes 3 	imes 10^8 $$
$$ v = 6 	imes 10^4~\mathrm{m/s} $$
$$ v = 60~\mathrm{km/s} $$
Correct option:
Option B
$60~\mathrm{km}/\mathrm{s}$

Answer

$30 \mathrm{~km} / \mathrm{s}$

Answer

$60 \mathrm{~km} / \mathrm{s}$

Answer

$90 \mathrm{~km} / \mathrm{s}$

Answer

$120 \mathrm{~km} / \mathrm{s}$

Waves Optics

The apparent wavelength of light from a star moving away from the earth is 0.02 \% more than the actual wavelength. The velocity of star is \left[\mathrm{c}=\right. velocity of light \left.=3 \times 10^8 \mathrm{~m} / \mathrm{s}\right]

Choose the correct answer: