12
$\begingroup$

I am wondering if two or more specific (in terms of frequency, wavelength etc) light sources could exist and capable of transmitting visible or invisible 'light', and where the beams intersect each other, can have a visible (to the naked eye) indication of interference.

Obviously, this doesn't work with -for example- regular laser light. So I am looking if it is possible at all and if so, what kind of light is required.

Thank you.

*by mid-air I assume regular air as in our biosphere.

Very simply illustrated:interference?

$\endgroup$
  • 2
    $\begingroup$ "Visible interference" is a vague term. You'll have a superposition of the fields, but no interaction (discounting the photon-photon physics at super high energies). What exactly are you asking? $\endgroup$ – safesphere Jun 24 at 21:47
  • 1
    $\begingroup$ physics.stackexchange.com/questions/231050/… $\endgroup$ – BowlOfRed Jun 24 at 22:32
  • 2
    $\begingroup$ You might be interested in my question. In particular, one of the answers points out a very important problem, which is that even if you achieve the two beams to be in perfect synchronization/coherence and project them onto a common spot, you will have a problem with the fact that the fringes will be infinitesimally small, because the waves that are interfering are on the order of microns. Instead, you will need highly specialized equipment like an interferometer. $\endgroup$ – SpiralRain Jun 24 at 22:40
  • 1
    $\begingroup$ arxiv.org/pdf/1506.06668.pdf has some discussion of different methods (and demonstration of one method). $\endgroup$ – BowlOfRed Jun 24 at 22:44
  • 3
    $\begingroup$ Are your laser beams visible in the air, before they intersect? $\endgroup$ – nasu Jun 25 at 2:17
15
$\begingroup$

The word "interference" is often used for the patterns that result from waves adding up or cancelling each other when they are in or out of phase. Other answers to this question have mostly concentrated on something else: the possibility of one light wave changing the direction of another, which would only happen with very intense beams in air, and crazy-intense beams in vacuum. However, ordinary wave interference happens whenever two light waves overlap, so yes, this kind of interference does happen when two laser beams intersect. The only thing is, you won't detect it with human eyesight because the interference pattern is moving about too quickly, and the scale of the pattern is close to the wavelength of the light so our eyes cannot detect it without the aid of a microscope.

To get the interference pattern to stop moving about so quickly you need light sources with very well-defined frequency (or narrow bandwidth, as we say). The interference pattern moves through a wavelength in a time roughly given by the inverse of the frequency-spread of the light. For a typical laser the bandwidth is many megahertz so this time is sub-microsecond. In this case the interference is washed out on a timescale much faster than the human eye can respond, so it is not seen. In a high-precision optics and atomic physics lab you can get lasers with bandwidths of order hertz or less. Using these the interference pattern can be made sufficiently stable for the human eye to detect it with the aid of a microscope.

$\endgroup$
  • $\begingroup$ Interesting. Do you have a link to any video/photos of these types of experiments? Would love to see that for myself. $\endgroup$ – BarrettNashville Jun 25 at 13:43
  • $\begingroup$ @BarrettNashville sorry, no I don't, but I can tell you that the standard tool for measuring laser linewidth is to observe the "beat" pattern produced by this effect when two laser beams interfere on a photodiode with a fast response. Here one deliberately introduces a slight frequency offset between two laser beams, so the interference pattern moves in a regularly way, and the photodiode detects the peaks and troughs of the interference pattern. $\endgroup$ – Andrew Steane Jun 25 at 15:16
  • $\begingroup$ Thank you so much for the answer. This will help me to finetune my quest and be a bit more specific in a follow up question. $\endgroup$ – Hieronymus Jun 25 at 22:24
  • $\begingroup$ But even with the right equipment (lasers and detectors), you would need to put a screen or something where the lasers meet to detect this, right? In my reading of the question, it asks if you could detect anything from outside the path of either of the two laser beams just where they cross without interrupting them. Which in that case I don't think it's possible. $\endgroup$ – Andréas Sundström Jul 14 at 14:34
  • $\begingroup$ @AndréasSundström my reading of it is that you imagine a case where the beam paths are rendered visible by a low-density dust or other medium. In this case the interference pattern would be visible as wavelength-scale structure in the scattering. $\endgroup$ – Andrew Steane Jul 14 at 15:28
8
$\begingroup$

As I understand the question, you would like to know if interference fringes or other indications of interference can occur **in air* due to the intersection of two light beams. The answer is "yes".

In the link you provided, two beams intersect to provide a high-intensity region where the light power density is high enough to ionize air. This is not interference in the sense meant by optical physicists. However, under certain conditions a very similar experiment will produce interference. Specifically:

  • the beams must be mutually coherent
  • the beams must be pulsed, in pulses that are brief enough that no significant changes occur due to heating of the air.
  • the ionization threshold must be rather sharp (a small change in power density must produce a large change in ionization)
  • the ionization lifetime - that is, the delay between absorption of light and emission of light - must be short compared to the distance air molecules (heated by the light beams) travel during the pulse duration.

If all those conditions are met, then the power of the two beams can be tuned so that ionization occurs where the beam amplitudes add (where the phases are the same), but not where the beam amplitudes subtract (where the phases are 180 degrees different). In that case, photoemission will occur only in the regions where ionization occurs, which will be in the planes of constructive interference.

Moreover, if the beams' powers are adjusted so that simply doubling the power of the incident beam cannot cause ionization and photoemission, but quadrupling the power will cause ionization and photoemission, then the presence of bright photoemission will be strong evidence that ionization is due to coherent interference of the two beams. That is because the power density at any point is proportional to the square of the summed amplitudes of the beams at that point.

Note that ionization is not necessary: any nonlinearly intensity-dependent photo-absorption process will produce "mid-air" interference fringes which can in turn produce visible effects.

$\endgroup$
8
$\begingroup$

Short answer: Probably not, with minor uncertainty emanating from the vagueness of your term "mid air".

Complete answer: I would make three points:

  1. In a vacuum, the answer would be "definitely no". If you solve Maaxwell's equations for the propagation of electromagnetic waves, the fields corresponding to each wave just superimpose linearly; there is no way for one wave to influence the path or intensity of the other.

  2. Same thing in a medium where the refractive index is independent of the intensity of the light passing through it. In practically all realistic circumstances, air is such a linear medium.

  3. But when light that passes through a medium gets extremely intense, the response to its electromagnetic field by the electrons and nuclei of the atoms of the medium can become nonlinear, resulting in a refractive index that depends on the intensity of the field. In this case, one light beam can change the refractive index "seen" by the other, resulting in a change of path. This phenomenon is called cross-phase modulation (XPM). I'm sorry I can't be more specific, but the gory details involved require a background in a field called nonlinear optics, which I'm unprepared to give here.

What would it take to accomplish this? As I said, extremely intense laser light, compressed into femtosecond pulses, might conceivably fit the bill in a sufficiently nonlinear medium. But air is almost certainly not that medium, so in practice, it's almost certainly not going to happen. Which brings us back to my short answer.

$\endgroup$
  • 1
    $\begingroup$ Nice answer. You might add, that many lasers do use nonlinear optics, either to double the output frequency (e.g. turn an infrared laser into a green beam), or to split a laser into two different wavelengths (e.g. convert a green laser into a red and a blue beam). These are just examples, there are more applications, one cooler than the other... $\endgroup$ – cmaster Jun 25 at 16:53
  • $\begingroup$ Thanks for your answer. Yes, I should add that the experiment would not take place in a vacuum. $\endgroup$ – Hieronymus Jun 25 at 22:27
  • $\begingroup$ This is a slightly obscure point, and another answer also mentioned this, but: you do in fact get some light-light scattering in a vacuum at sufficiently high energies. It doesn't happen classically, but quantum-ly it does, at very low rates, due to second-and-higher-order interactions / loop Feynman diagrams. $\endgroup$ – Glenn Willen Jun 26 at 1:17
  • $\begingroup$ Also at super high energies there are photon photon interactions at one loop order and above. Non linearity caused by vacuum polarisation of virtual electron positron pairs $\endgroup$ – lux Jun 26 at 3:52
4
$\begingroup$

For "ordinary" lasers (visible light, milliwatt power) they will pass through one another with no fringing or other interactions.

For powerful lasers that have wavelengths where the air is heated by their passage, then they will interact indirectly, by heating up the air along their paths and at their point of intersection, thereby changing its refractive index.

At super-high energy levels, there are ways for the photons to interact more directly, as pointed out by Safesphere above.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.