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If you believe ray optics where a light ray is a straight line, a light beam is infinitely thin. If you think of fiber optics, you can guide a whole lot of photons down a fiber only one or a few wavelengths wide.

On the other hand, the Airy disk is typically many wavelengths wide and diminishes transversely as the inverse third power. It could in some sense be said to be infinitely broad, or to have infinitely wide tails.

A third concept is Feynman's idea that the photon goes from here to there by all possible paths, but destructive interference makes only the paths near the central axis count. If you do a naïve calculation, based on this idea, you end up with a rather fat ellipse of order the square root of the source to detector distance in wavelengths. This raises issues similar to those discussed in the recent popular tricky question about how we see the sun.

Let's make this question more precise: What is the half diameter of the light beam?

Assume we have an incoherent source of quasi-monochromatic light passing through a small circular hole in an opaque screen.

Similarly assume we have a small detector behind a second screen with a small circular hole only a few wavelengths in diameter. The source and the detector are far apart. Thousands or billions of wavelengths.

Make the diameter of both source and detector as small as possible, very close to just one wavelength. Now assume a third circular hole in an opaque screen midway between the source and the detector. However its diameter is larger and variable, or else we have a set of different sized third screens.

All three holes are coaxial and aligned perpendicular to the axis.

How big is the diameter of the third hole, if it passes fifty percent of the light that passes with the middle screen totally absent?

Going beyond this, can you tell me what is the formula for the amount of radiation received at the detector as a function of the diameter of the hole in the middle screen? Express the diameter in wavelengths. Express the amount of radiation as a percent of the radiation received if the middle screen is absent, or if the diameter of the middle hole is infinitely large.

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    $\begingroup$ At least at the beginning, you seem to mix several things that have nothing to do with each other. Light may stay in optical fibers because it gets reflected from the less dense material, the air. Otherwise light could surely not spontaneously stay in such a thin cylinder. The Airy disk appears otherwise. On the other hand, both the Airy disk and the fiber optics behaviors perfectly agree with the "Feynman path integral" calculations - which have been known for several centuries in this context as the Hyugens principle. In different situations, the "Feynman" calculation gives different results $\endgroup$
    – Luboš Motl
    Commented Jul 11, 2013 at 5:54
  • $\begingroup$ How thin a beam may get depends on the way how you prepare it, the source and the holes it gets through, and so on. If you make it too thin at one distance from the source, it will tend to spread more rapidly afterwords, and so on. I am afraid that if you want quantitative formulae, you should be more accurate in your description of "the" problem. $\endgroup$
    – Luboš Motl
    Commented Jul 11, 2013 at 5:56
  • $\begingroup$ With the conditions you set: pinholes approximately the size of the wavelength, the source will radiate practically isotropically. There will be no "beam" at all. The detector is behind a very small pinhole. What is it's purpose? If you ignore diffraction at the middle screen, a small coaxial detector will receive the same power regardless of the size of the central aperture, and regardless of the size of the source. $\endgroup$
    – garyp
    Commented Apr 6, 2018 at 17:44
  • $\begingroup$ Just a clarification: what you call as "Feynman's idea" or the Standard Model framework, a photon is an elementary particle, quanta of an electromagnetic field. The question of its substructure in these terms is hence baseless. $\endgroup$
    – topologically_astounded
    Commented Apr 6, 2018 at 20:03

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What is the half diameter of the light beam?

I might be mis-reading the question, because the experimental setup has be rather confused, so you may be asking something else...

However, there is a very well defined answer to your original question, quoted above. It's the half-power point. This is widely used, especially in antenna design.

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If you're using single mode fiber, then you will be launching the fundamental Gaussian mode. Then with an adjustable aperture and a power meter it is easy to adjust the aperture until you have the FWHM or whatever criterion you like.

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I'm a total amateur in this discussion, so forgive me asking dumb questions but sometimes dumb questions trun out to be not so dumb after all ;-). Several things don't seem to make sense to me. Firstly why are we fixated on wavelength instead of amplitude when we talk about wave width - wavelength is along the axis of the wave travel with no width component so surely irrelevant? Next, if light is an electromagentic wave and the electric field disspiates perpendicular to the direction of travel as the square of the distance, then surely there is a width of the wave that comprises say 99% of the electric field. AND for the magnetic field component where field strength diminishes in proportion to distance, surely there must be a height of the wave that contains 99% of the magnetic field? These seem to imply that zero width waves are a fallacy? If these perpendicular field strengths (electric and magnetic) do exist, then they seem to me to be the analogues to the water wave where gravity and flow against viscocity trying to level out the water causes the wave at the slit edge to spill out in all directions?

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The diameter of light mainly depends on the source of light, it's shape and structure. In the situation you have described above, if you use a stronger source of light (e.g. torch light) instead of the Sun, the diameter of third hole varies. The behavior of light can be explained only by measurement but not through physical interpretations (like the number of photons sent through a fiber) since it is not a physical thing.

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How “wide” is a beam of light?

It doesn't have a width in the usual sense. I know that sounds odd, in that you can see a beam of light in a dusty room, and if the gap in the blinds in 1" wide, the light beam is 1" wide. But a light beam is comprised of photons, and photons have an E=hf wave nature. You might think a wave in the ocean has a "width" of say 1m, because that's its height. But take a look at wind waves on Wikipedia. See the blue gif. The wave extends deep into the ocean. It has an amplitude of say 1m, but the displacement gets less and less as you get deeper and deeper. It's similar for a seismic wave. It might displace your fridge by 1m, but it isn't 1m wide.

enter image description here

GNUFDL image by Kraaiennest, see Wikepedia commons

What is its half-diameter?

That depends on the beam.

If you believe ray optics where a light ray is a straight line, a light beam is infinitely thin. If you think of fiber optics, you can guide a whole lot of photons down a fiber only one or a few wavelengths wide.

Yes, but don't forget the photon goes through both slits.

On the other hand, the Airy disk is typically many wavelengths wide and diminishes transversely as the inverse third power. It could in some sense be said to be infinitely broad, or to have infinitely wide tails.

Now you're talking.

A third concept is Feynman's idea that the photon goes from here to there by all possible paths

It's true. So does a seismic wave. If it propagates from A to B, it isn't only the houses sitting on top of the AB line that shake.

Let's make this question more precise: What is the half diameter of the light beam?

Sorry, it depends on the light beam. And uhnn, I've just noticed that this question dates from 2013!

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