# How “wide” is a beam of light? What is its half-diameter?

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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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 – Luboš Motl Jul 11 '13 at 5:54
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. – Luboš Motl Jul 11 '13 at 5:56