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My question is pretty brief. When two sound waves of nearly same frequencies interfere, we get beats.

But, I have not observed something like that happening in the case of light. In fact, most of the light around us is a collection of continuous wavelength range which must be all nearly same frequencies.

Can we observe beats in light waves, as in case of sound waves?

If yes, how to observe them?

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Related meta post: meta.physics.stackexchange.com/q/6140/2451 –  Qmechanic Sep 21 at 12:52

4 Answers 4

up vote 14 down vote accepted

The math of beats is absolutely the same for light as it is for sound, but ...

  • Most light around us in incoherent, so does not form beats on macroscopic time or distance scales.

  • It is hard to select to sources that could conceivable give visible beats Look at it this way the frequency of green light is around $ c /(500\text{ nm}) \approx 6 \times 10^{15} \text{ Hz} $. All the colors have frequencies of the same order of magnitude, so it is very hard to the beat frequencies of two randomly selected colors tend to be around $10^{13}$--$10^{14} \text{ Hz}$: much faster than your eye can detect. To construct a pair of sources that you could see beat would require controlling their frequencies to about 1 part in $10^{14}$. That's not easy.

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+1. It might also be worth adding that the interference fringes observed in the double slit experiment are very closely related to the phenomenon of beats. –  Nathaniel Mar 7 '13 at 1:23
4  
+1 The key point is that most of the light around us is from incoherent sources. The easiest way to obtain beats is to start from a single source (a laser for instance), split it in two, shift the frequency of one the beams (using an acousto-optic modulator for instance) and have them interfere again. The signal generated by a photodiode will exhib a beat. –  Oli Mar 7 '13 at 3:36
    
@Oli Something similar was done by my Physics teacher. Those things looked like beats, but were not formed using the same procedure as for sound waves. That is why I preferred to take the forum's opinion. –  Cheeku Mar 7 '13 at 5:01
    
@Cheeku I'm not sure what you mean by "not the same procedure". It is the same physical effect, it is just that it is difficult (but not impossible) to make two independent sources interfere. –  Oli Mar 7 '13 at 6:26
    
Interference fringes, of course, arise from the same math in the spatial domain. Even there we use finely manufactured physically small features (or differences in feature length scales) to magnify their size to something we can see with the mark one eyeball. –  dmckee Mar 7 '13 at 17:27

It's a rather late answer, but the paper Visible optical beats at the hertz level has just appeared on the Arxiv and this describes exactly the phenomenon you ask about. This image from the paper shows the experimental setup:

Optical beating

The light frequency is modified using acousto-optic modulators (labelled AOM in the diagram), and to make it look pretty a lens is used to produce interference rings and a beam splitter is used to produce two images (labelled A and B) that are inverses of each other. That is, A is dark when B is bright and vice versa.

The paper reports that the brightness variation can be easily seen with the naked eye up to about a frequency difference of 20Hz, beyond which persistence of vision makes the beating disappear.

Later: a reviewer has pointed out this video on YouTube that shows a very similar experiment.

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I like these kind of setups, seems they give a lot of nice intro value (compared to the gear/setup effort involved) to engage students or researchers in the quantum fundamentals/optics areas. –  BjornW Sep 21 at 10:30

A laser gyroscope detects 'beats' of light.

In a laser gyro, laser light goes in two directions 'round a ring, clockwise and counterclockwise. If the gyro is rotating, one laser direction will be at a slightly different frequency than the other, and their interference will create a 'beat'.

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amazing !!!!!!!! –  Joe Blow Sep 24 at 16:26

The answer to your question is yes, we can observe beat notes between two different coherent sources of light. This fact underlies almost every precision laser experiment because it allows for lock-in detection. However, there is a subtle difference from audio beatnotes.

The difference is that with sound the oscillations are in the air pressure which is what you detect with your ear. The beatnote you get with two sound waves is in the volume of the field. Consider two sound waves with unit pressure combined at your ear, $$ P=\cos(\omega_1 t)+\cos(\omega_2 t)=\boxed{2\cos\left(\frac12(\omega_1-\omega_2)t\right) \cos\left(\frac12(\omega_1+\omega_2)t\right)}. $$ The first term is the beat note that you detect. It is a beat note in the volume of the strength of the signal. However, if you were to take the Fourier transform of this signal you would still find that there are only two frequencies: $\omega_1$ and $\omega_2$.

With light, the frequencies are so high, that all we can detect is the intensity of the field. This is an inherently non-linear process in which the beatnote actually becomes a component of the signal. Consider two optical fields of unit power (using complex notation this time), $$ E=e^{-i\omega_1 t}+e^{-i\omega_2t}. $$ When this field is incident upon a photodetector, the signal will be given by the time averaged intensity of the field $$ \langle E\rangle=E^*E=\boxed{2\left(1+\cos[(\omega_1-\omega_2)t]\right)}. $$ So, this signal actually has a component at the beatnote frequency. I.E. if you take the Fourier transform you will find a component at that frequency.

Why does the difference matter? With light you can combine two lasers whose frequencies are much too high to be detected directly and obtain a signal at the beatnote frequency. You can not, however, combine two sound fields with frequencies which are outside of the range of human hearing (above $\sim25\ \text{kHz}\ $) and be able to hear the beatnote between the two.

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