Interference fringe of light beams of different frequencies

[abcrelativity]

I have a question. If you have two light beams of different frequencies, would you observe interference fringes?

 

[cortiver]

This is an interesting question. It turns out that the answer is no. This is pretty straightforward to see mathematically, but for the benefit of intuition it's instructive to consider the case when the light beams have two frequencies f₁ and f₂ which are very close, but not quite equal. Then when adding these two light beams you have a beat frequency f_b = f₁ - f₂. If the two light beams are in phase at time t = 0 (so you have constructive interference), then at time t = pi/f_b the light beams will be 180 degrees out of phase (so you have destructive interference). However, unless f₁ and f₂ are really close (say within 10Hz of each other), then the beat frequency will be too fast the human eye to resolve. In that case the light intensity that we will see is the average of constructive interference and destructive interference, which amounts to no interference at all.

 

[technician]

With 2 different frequencies of sound interference shows itself as 'beats'
With radio waves of different frequencies beats are also produced. This is used in the superheterodyne principle of radio reception.
I am not aware of any similar effect with light.
Your question is a good question !

 

[abcrelativity]

Hi Cortiver, you seem to answer my question. This is for two light beams so frequency difference is really small (much less than 10 Hz). It is a difference of frequency of 1 part per 10'000 with respect to the average frequency of daylight. What I am not sure to understand is what would be the beat frequency? Would it increase when the difference in frequency is getting smaller or the reverse? Is there a way to quantify the beat frequency? Maybe if you have a reference of a book with this problem would be usefull..

 

[jtbell]

Hi Cortiver, you seem to answer my question. This is for two light beams so frequency difference is really small (much less than 10 Hz). It is a difference of frequency of 1 part per 10'000 with respect to the average frequency of daylight. What I am not sure to understand is what would be the beat frequency?

He told you in his post:

Then when adding these two light beams you have a beat frequency f_b = f₁ - f₂.

The frequency of visible light is around 10¹⁵ Hz. One part per 10000 of that would give you a beat frequency of 10¹¹ Hz which is much too rapid to see visually. I don't know whether electronic detection equipment is good enough to "see" it.

 

[cortiver]

Beat frequency is a pretty basic concept in physics so you should be able to find lots of good explanations on the web -- you could start with the Wikipedia article. The beat frequency is calculated as the difference in the frequencies of the two light beams, f_b = f₁ - f₂. So the smaller the difference between the frequencies of the two beams, the smaller the beat frequency (so the beats are slower and hence easier to detect). In the case that the frequencies are actually the same, f₁ = f₂, then the beat frequency is zero -- so that the interference fringes don't change with time at all.

I don't know how you're getting "1 part per 10,000". The frequency of visible light is measured in hundreds of terahertz, so a difference in 10Hz is more like one part in 10^14, which is pretty minuscule. That's why beats are much harder (impossible?) to observe with light waves than, say, sound waves.

 

[Cthugha]

Hi Cortiver, you seem to answer my question. This is for two light beams so frequency difference is really small (much less than 10 Hz). It is a difference of frequency of 1 part per 10'000 with respect to the average frequency of daylight.

Are you sure it is really 10 Hz? That is nothing in the optical domain. Even rather tiny polarization splittings of 10 microeV which are already hard to resolve in the spectral domain coreespond to a difference of 10^9 Hz. Also are we talking about two monochromatic beams or two spectrally broad beams with different center frequency? In the latter case beatings are washed out anyway.

Is there a way to quantify the beat frequency? Maybe if you have a reference of a book with this problem would be usefull..

You can see the beatings using a Michelson interferometer. If you record an interferogram of the combined beams you can directly see the beats in the signal. Depending on how good (and expensive) your translation stages are, you can resolve pretty fast beatings. If you use good piezos, temporal resolution in the upper attosecond range is possible.

 

[e.bar.goum]

This is an interesting question. It turns out that the answer is no. This is pretty straightforward to see mathematically, but for the benefit of intuition it's instructive to consider the case when the light beams have two frequencies f₁ and f₂ which are very close, but not quite equal. Then when adding these two light beams you have a beat frequency f_b = f₁ - f₂. If the two light beams are in phase at time t = 0 (so you have constructive interference), then at time t = pi/f_b the light beams will be 180 degrees out of phase (so you have destructive interference). However, unless f₁ and f₂ are really close (say within 10Hz of each other), then the beat frequency will be too fast the human eye to resolve. In that case the light intensity that we will see is the average of constructive interference and destructive interference, which amounts to no interference at all.

I think you're mixing up sound and light. The ear can resolve beat frequencies at 10Hz separation. Visible light is many orders of magnitude higher in frequency (THz). In addition, considering that a linewidth of 10 Hz is an extremely good laser, such a restriction would mean that in practice we would never see interference fringes. (And certainly not before a decade or so ago).

 

[cortiver]

I think you're mixing up sound and light. The ear can resolve beat frequencies at 10Hz separation. Visible light is many orders of magnitude higher in frequency (THz). In addition, considering that a linewidth of 10 Hz is an extremely good laser, such a restriction would mean that in practice we would never see interference fringes. (And certainly not before a decade or so ago).

There's no difference between sound and light, in principle. Yes, in practice it is much harder to observe beats with visible light due to the high frequencies, as I already noted.