Doppler’s effect in light

Doppler’s effect in light

Doppler effect in light -1. Light waves also show the phenomenon of Doppler’s effect. If a light source is coming towards an observer then the apparent frequency of light is increased (or wavelength is decreased). Hence the spectral lines are shifted towards the violet end of spectrum. On the other hand, if the light source is moving away from the observer, then the spectral lines are shifted towards the red end of the spectrum.

Doppler effect in light
Doppler effect in light

2. Important difference between Doppler’s effect in sound and Doppler’s effect in light:
In sound Doppler’s effect not only depends upon the relative motion of the sound-source and the observer but also depends upon the manner of creation of their relative motion. For example, if the source is stationary and the observer is coming towards the source is coming with the same then the change in the frequency will be different from that when the observer is stationary and source is coming with the same velocity towards the observer. On the other hand, Doppler’s effect in the light depends only upon the relative motion of the light-source and the observer. It does not matter which one is moving. This can be understood by the principle of relativity.

3. Suppose the frequency of light emitted by a source is u. It can be shown on the basis of the theory of relativity that if either the source or the observer is moving with a velocity v such that the distance between them is decreasing, then the apparent frequency of the source is given by.

u’=u√ [{1+(v/C)}/{1-(v/C)}]
where C is the velocity of light. Above equation shows that apparent frequency of light will be increased.

4. If λ be the real wavelength of the light and λ’ be the apparent wavelength, then we can also determine the wavelength-shift.
u=C/λ and u’=C/λ’
Hence from equation (1),
C/ λ’=(C/ λ)

C/ λ’=(C/ λ)[{1+(v/C)}/{1-(v/C)}]
or
λ/λ’=√ [{1-(v/C)}/{1+(v/C)}] = [1-(v/C)]1/2 ×
[1+(v/C)]1/2
= [1-(v/2C+…)]×[1-(v/2C+…)]
Because, v<<C, hence heigher power terms can be neglected. Thus
λ/λ’=[1-(v/2C+…)]×[1-(v/2C+…)]=1-(v/C)
or 1-λ/λ’= v/C
or (λ- λ’)/λ= v/C
But
λ – λ’ = Δλ (wavelength-shift).

Δλ = (v/C)λ …………(2)

If, due to the motion of the light-source or the observer, the distance between them is increasing, then the apparent frequency of the source is given by :

u’=u√ [{1+(v/C)}/{1-(v/C)}]
i.e. in this case the apparent frequency of light will be decreased. . Again, in the form of wavelength shift, we have
Δλ = (v/C)λ
Important point:
It we move away from a band with the speed of sound, no sound is heard. Since we are moving away with the speed of sound, the sound wave emitted by the band are in a state of rest with respect to us. Hence, these waves do not impinge on our ears and are not able to create the sensation of hearing.
Alternativaly, n’=[(v-v0)]/v ×n=[(v-v)]/v ×n=0
A note of zero frequency would not produce any sound.