# Calculating measured intensity of a Gaussian beam

Consider a light source which emits a Gaussian beam with total power $$P_0$$, initial width $$d$$, and divergence half-angle $$\theta$$ (in radians). The beam radius, measured from the waist of the beam, can be given as $$\widetilde{w}(z) = w_0 \sqrt{1+\left(\frac{z}{z_R}\right)^2},$$ where $$z$$ is the distance measured from the waist, $$w_0$$ is the waist radius $$w_0 = \frac{ \lambda}{\pi n \theta},$$ and $$z_R$$ is the Rayleigh range $$z_R = \frac{\pi w_0^2 n}{\lambda}.$$ Let $$z_0$$ be the distance of the origin of the beam measured from the waist, i.e., $$\widetilde{w}(z_0) = \frac{d}{2}.$$ Then the radius of the beam can be given as $$w(z) = \widetilde{w}(z+z_0),$$ where $$z$$ is the distance from the origin, not from the waist.

The intensity profile of the beam can be given as $$\mathcal{I}(r, z) = \frac{2P_0}{\pi w_0^2} \left( \frac{w_0}{w(z)} \right)^2 \mathrm{exp}\left(-\frac{2r^2}{w(z)^2}\right),$$ where $$r$$ is the radial distance from the center axis of the beam, and $$z$$ is the distance from the origin of the beam.

At a distance of $$z = R$$ from the origin, there is a disc-shaped detector with diameter $$D$$. I am trying to calculate the measured intensity at the detector using the Gaussian beam model.

The power through a surface $$S$$ can be calculated as the surface integral $$\int_{S} \mathcal{I} dS,$$ so the power at the detector can be calculated as an integral in polar coordinates, $$P:=\int_{r \le D/2} \mathcal{I}(r, R) dA = \int_{r = 0}^{D/2} \int_{\theta = 0}^{2\pi} \mathcal{I}(r, R)r d\theta dr.$$ This evaluates to $$P = P_0 \left(1- \mathrm{exp} \left( -\frac{D^2}{2 w(R)^2}\right)\right).$$

The measured intensity is the power through the detector divided by the area of the detector, i.e., $$\mathcal{I}_{measured} = \frac{P}{\pi (D/2)^2} = \frac{4P_0 \left(1- \mathrm{exp} \left( -\frac{D^2}{2 w(R)^2}\right)\right)}{ \pi D^2}.$$

However, I do not see how this is in line with the inverse square law, i.e., the measured intensity should be proportional to $$R^{-2}$$. With this model, I get results which are in completely different orders of magnitude. Alternatively, one would calculate the measured intensity simply as $$\frac{P_0}{2\pi (1-\cos{\theta}) R^2} \propto R^{-2}.$$

• You get the inverse square law for spherical waves. In leading order, you have exactly that (because for $z\gg z_R$, the beam radius grows linearly): $1-\exp(-1/R^2) \approx 1/R^2$. Commented Jul 14, 2023 at 9:29
• @Samuel But shouldn't the order of magnitude be similar? Commented Jul 14, 2023 at 9:31
• I edited my comment. Commented Jul 14, 2023 at 9:32
• If your disk always integrates all the beam's intensity, why would it measure a different value for different distances? Your disk integrates the intensity, ie your disk absorbs all the energy and shows always the same value. If you have an intensity discrimination detector, eg a camera, then you can look at the actual intensity profile Commented Jul 14, 2023 at 11:30

It doesn't exactly follow an inverse square law, though that is an excellent approximation in the far field.

First, there are a couple better pictures of a Gaussian beam from the RP Photonics Encyclopedia article on Gaussian Beams below.

Rays are used in the approximation to light propagation that ignores diffraction. Rays are lines perpendicular to the wave fronts. If you have a point source of light, wave fronts are expanding spheres, and rays are lines outward from the source. In that case, you get inverse square law intensity.

One approximation to a laser beam is that it is perfectly collimated. In that case, intensity is constant as a function of distance from the laser.

For a Gaussian beam, "rays" follow hyperbolic paths. Far from the laser, they are very close to straight lines in an expanding cone. Typically the divergence angle is a few milliradians. The wave fronts are very close to sections of a sphere, though the intensity drops off as you get farther from the beam axis.

At the beam waist, the perfectly collimated approximation is better.

In the near field, neither one is correct.