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I want to know what a standing wave of light would like and what properties it might have that are interesting.

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    $\begingroup$ en.wikipedia.org/wiki/Standing_wave#Visible_light $\endgroup$
    – hft
    Commented Dec 21, 2022 at 21:13
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    $\begingroup$ An 'intuitive' way is to think of something like the interference pattern except the distance between a peak and trough will be the wavelength of light, e.g., many hundreds of nm for visible light. So a standing wave of light will just be bands of light and dark, separated by it's wavelength. $\endgroup$
    – Kitchi
    Commented Dec 22, 2022 at 10:29

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The resonant cavity of a laser is a standing wave. It doesn't really look like anything in particular because the standing waves are not travelling to your eyes. However, you can let some of the wave escape the cavity to do all sorts of interesting things.

The most interesting property is the ability to control cats due to the tight collimation: cat and laser pointer from https://www.thepurringtonpost.com/laser-pointers-and-your-cat-the-good-bad-and-the-unattainable/

Other less interesting properties include that the light is coherent and monochromatic. It can also be very intense and easily focused. It can produce more heat at a target than in the cavity, thus allowing to thermally ablate materials.

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    $\begingroup$ Definitely. All that mumbo jumbo is useless fluff. Producing heat in a target, nah. Who cares that it is monochromatic? What even is coherency? Turning the real rulers of the planet into quivering masses of energy totally focused in that spot instead of on demanding belly rubs? Priceless $\endgroup$
    – CGCampbell
    Commented Dec 22, 2022 at 12:54
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    $\begingroup$ Cats react to non-coherent and polychromatic light as well. The important property in regard to cats is the beam colimation (resonant cavities are good at colimation, too). $\endgroup$
    – fraxinus
    Commented Dec 23, 2022 at 12:13
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    $\begingroup$ Why would standing waves not look like anything? Although they are not travelling, there is still an osscilating poynting vector that is non zero, so if I were to stick my head through one, I would still see something no? It's time average is obviously zero but that is very different than saying it is zero $\endgroup$
    – jensen paull
    Commented Dec 23, 2022 at 18:40
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    $\begingroup$ I am not going to encourage people to stick their head in a laser cavity nor write the answer to assume that they might stick their head in a laser cavity. At no point did I imply or state that it is because the Poynting vector is zero. $\endgroup$
    – Dale
    Commented Dec 23, 2022 at 20:30
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This answer is about how a technical diagram of such a wave looks like, since with the naked eye you won't be able to see a standing wave of light for the simple reason that you can only see the wave part travelling in your direction.

If the two wave components of the standing wave are of the same wavelength like the red and blue one, the standing wave looks like the green one:

standing wave 1

If we are moving with v=c/2 to the right relative to the setup so that the wavelengths get shifted and are no longer of the same frequency, the scene would be

standing wave 2

so the zero points of the standing green wave would have the lorentzcontracted distance to each other as they should and keep the correct position, but the rest of the wave beside the zeros would not look as uniform as if the observer was at rest relative to the two wave emitters.

Due to the relativity of simultaneity the hulls of the standing waveform (dashed) would be drawn in a wavier way like in the second image, where the shown setup (for example the inside of the laser pointer cavity from here) is moving to the left relative to the observer with c/2.

The colors are of course just for separation and do not represent the color of the lightbeams.

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    $\begingroup$ What are the two wave components? $\endgroup$
    – AnoE
    Commented Dec 22, 2022 at 12:10
  • $\begingroup$ @AnoE A standing wave requires two waves travelling in opposite directions. In the charts, the two waves here are shown in red (travelling to the right) and blue (travelling left). $\endgroup$
    – jjmontes
    Commented Dec 25, 2022 at 14:28
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A standing wave is formed when two (plane) waves of the same frequency travel in opposite directions. If you stand inside a plane wave of light, you see it as a "star": a point of light infinitely far away in the direction the wave is coming from. So if you filled a room up with an optical standing wave (not sure how you would do this... it'd be like building a room-sized lasing cavity), and then you stood in it, assuming the presence of your body doesn't ruin the wave, you would simply see a "star" in one direction and another "star" in exactly the opposite direction.

A laser is such a cavity where one wall is leaky, allowing one of the plane waves to exit as the laser beam. So it might be accurate to say a standing wave "looks like" two opposing laser beams.

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It arguably doesn't "look" like anything. The standing wave in an optical resonator (e.g., a laser cavity) is trapped in the cavity. It isn't entering your eye. If we're talking about a laser, then the part that escapes the cavity and ultimately enters your eye† arguably is no longer a standing wave.

A diagram that represents a standing wave of light would look the same as a diagram representing any other standing wave.

† Do not stare into beam with remaining eye.

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    $\begingroup$ Well, if you were an electron in that resonator though you'd get pretty excited! $\endgroup$
    – Peter - Reinstate Monica
    Commented Dec 22, 2022 at 12:14
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If you could stand at a node in a perfectly round monochromatic light two optic resonant cavity, you would see a small circle of light. If you could stand at the anti-node you would see a large circle of light.

No monochromatic light source is prefectly round so inside a real resonant cavity the light will have a central beam and what we call junk light around the central beam that will bounce around in the cavity by not produce a standing wave.

As the light leaves the resonant cavity it's no longer a standing wave and its properties are changed by the optics the light goes through or bounces off of, so what you see outside of the resonant cavity isn't what you would see inside the cavity.

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  • $\begingroup$ Aren't the nodes spaced apart by a wave length, i.e. something in the sub-micrometer range for visible light? How can you "stand at a node" or at an anti-node in this case? $\endgroup$
    – Paŭlo Ebermann
    Commented Dec 24, 2022 at 1:56
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The light in a laser cavity is standing wave. If we measure the intensity of light, we will get an image like this:

Chang H C, Kioseoglou G, Lee E H, et al. Lasing modes in equilateral-triangular laser cavities[J]. Physical Review A, 2000, 62(1): 013816.

This picture is from Chang H C, Kioseoglou G, Lee E H, et al. Lasing modes in equilateral-triangular laser cavities[J]. Physical Review A, 2000, 62(1): 013816.

But if we observe the cavity with eyes, in fact we just get the light escaping from the cavity, it looks like:

Guidry M A, Song Y, Lafargue C, et al. Three-dimensional micro-billiard lasers: the square pyramid[J]. EPL (Europhysics Letters), 2019, 126(6): 64004.

This picture is from Guidry M A, Song Y, Lafargue C, et al. Three-dimensional micro-billiard lasers: the square pyramid[J]. EPL (Europhysics Letters), 2019, 126(6): 64004.

Here we used a microscope camera to take the photo.

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