Gaussian Beam Radius at Distance z

This is the whole propagation of a Gaussian beam in one line, and it is a hyperbola. Near the waist $z \ll z_R$, the square term vanishes, $w \approx w_0$, and the beam looks collimated — doubling the distance barely widens it. Far out, $z \gg z_R$, the 1 becomes negligible, $w \to w_0 z/z_R = \lambda z/(\pi w_0)$, and the beam has straightened into a cone leaving the waist at the divergence half-angle. The Rayleigh range is exactly where the two behaviours cross over, and $w(z_R) = \sqrt2\,w_0$ is the corner of the hyperbola.

The practical uses run in both directions. Forwards, it tells you how wide the beam will be on the target, which is what you need for a safety calculation, a detector that must not be overfilled, or an aperture that must not clip. Backwards, it tells you where the waist actually is — and that is the harder and more useful question, because on a real bench the waist is very rarely where the drawing says it is.

Both inverse forms carry a genuine ambiguity, and it is not a defect in the algebra. The envelope is symmetric about the waist: $w(-z)$ and $w(z)$ are the same width. So a single measured radius names two planes, one on each side, and no arithmetic can tell you which one you are standing at. Two measurements can. Step the profiler along the axis and see whether the beam narrows or widens; if it narrows, the waist is still ahead of you.

The other thing this page cannot tell you from one reading is whether your "waist" is really a waist. A beam that is astigmatic has two waists at different planes, one for each axis, and a profiler reporting a single width is averaging two different beams. Astigmatism is ordinary in diode lasers, in beams that have passed through a tilted plate, and in anything with a cylindrical element in the path. If the numbers refuse to fit a single hyperbola, that is usually why.

A note on doing this properly. The two-point method above is a rough version of the real procedure. ISO 11146 asks for at least ten planes — five within one Rayleigh range of the waist and five beyond two Rayleigh ranges — precisely because two points can be fitted by a hyperbola that a third point would have ruled out. If the answer matters, take the ten.

And the usual caution: this is the perfect-Gaussian envelope. A real beam of quality $M^2$ follows the same shape everywhere but wider, with $z_R$ shortened by the same factor for a given waist. The beam-quality page is where that correction is made honest.