Good morning, I have to evaluate the diameter of a gaussian beam at a distance $r$ which has a divergence $\theta$ and a starting diameter $d_0$. I found these three ways:

1) $ d_f=d_0+r\tan(\theta) $



I was wondering which is the right one between these three formulas and if they are different approximations of the same formula. Many thanks.


I think this can be seen as a matter of how you define the divergence angle $\theta$. Different authors seem to define it slightly differently, however in practice this is rarely an issue since $\theta$ is typically small enough that $2\tan(\theta/2) \approx \tan(\theta) \approx \theta$.

The Gaussian beam as a solution to the paraxial Helmholtz equation has the intensity $$I(\rho, z) = I_0 \left[\frac{W_0}{W(z)}\right]^2 \exp \left[-\frac{2\rho^2}{W^2(z)}\right],$$ $$W(z) = W_0 \sqrt{1 + (z/z_0)^2},$$ $$W_0 = \sqrt{\frac{\lambda z_0}{\pi}},$$ where $\rho$ is the distance from the beam axis, $z$ is the position along the beam axis, $W_0$ is called the waist radius and $z_0$ the Rayleigh range (from Saleh & Teich). The fraction of the power of the beam contained by a circle of radius $\rho$ about the beam that is a distance $z$ away from the beam waist is $$1 - \exp \left[-\frac{2\rho^2}{W^2(z)}\right].$$ If you define the beam diameter (as is usually done) as the diameter containing $1 - e^{-2}$ of the power, then the beam diameter is $$D(z) = 2W(z) = 2W_0 \sqrt{1 + (z/z_0)^2} = D_0 \sqrt{1 + (z/z_0)^2}$$ with $D_0 = 2W_0$. Far away from the beam waist ($z \gg z_0$), we have $$D(z) \approx D_0 \frac{z}{z_0}.$$ This diameter is defined by a cone along the beam axis with the vertex at the center of the beam. The half-angle of this cone is $$ \theta_0 = \tan^{-1}\left(\frac{W_0}{z_0}\right) $$

Now, if we define the divergence to be the full vertex angle of this cone, i.e. $\theta = 2\theta_0$, $$ D(z) = \sqrt{D_0^2 + \left(2\frac{W_0}{z_0}z\right)^2} = \sqrt{D_0^2 + \left(2z\tan\theta_0\right)^2} = \sqrt{D_0^2 + \left[2z\tan(\theta/2)\right]^2} $$

If we define the divergence as $\theta = \tan^{-1}(D_0/z_0)$ (although I don't see an obvious reason for doing so), $$ D(z) = \sqrt{D_0^2 + \left(\frac{D_0}{z_0}z\right)^2} = \sqrt{D_0^2 + \left(z\tan\theta\right)^2} $$

If we define the divergence as $\theta = D_0/z_0$ so that $D(z) \approx \theta z$ for $z\gg z_0$ (this is twice the definition in Saleh & Teich), $$ D(z) = \sqrt{D_0^2 + \left(\frac{D_0}{z_0}z\right)^2} = \sqrt{D_0^2 + \left(\theta z\right)^2}. $$

I believe the formulas you gave are approximations for these expressions that are valid for $z \approx 0$ and $z \gg z_0$.

  • $\begingroup$ Many thanks fot this high quality answer! Now it's all clear to me. It seems you're very familiar with gaussian beams so I have another little question: I have this formula for evaluate power of a gaussian beam trasmitted through an aperture having diameter $d$ if I have the waist diamater containing 63% of the power: $T_{63}=1-e^{-(\frac{d}{w})^2}$. If I have the diameter 86% one the formula became $T_{86}=1-e^{-(\frac{d}{2w})^2}$. Do you know the origin of this difference by a square root of 2? Many thanks again! $\endgroup$ – muserock92 Jul 6 '19 at 11:02
  • $\begingroup$ What does $T$ refer to exactly? $\endgroup$ – Puk Jul 6 '19 at 11:07
  • $\begingroup$ I know the $d_{63}$ or $d_{86}$ diameter which are the value of the diameter of the beam where I can found $63/%$ or $86/%$ of the power respectively. I need to know how much power pass from an aperture, so $T$ is intended as transmitted power i guess. Many thanks! $\endgroup$ – muserock92 Jul 6 '19 at 11:11
  • $\begingroup$ I'm confused, wouldn't that mean $T_{63} = 0.63$ and $T_{86} = 0.86$? $\endgroup$ – Puk Jul 6 '19 at 11:13
  • $\begingroup$ Sorry for this, I try to explain me better. I know that the diameter value of my gaussian beam where I have the $86$ % of the power is e.g. $6 mm$. I have to compute the power which pass through an aperture e.g. of $7 mm$. I should go for the $T_{86}$ formula. I was wondering why I have to use two different formulas if I have the $d_{86}$ or $d_{63}$ and how do you think about this. Yes, $T_{63}=0.63$ of the power. Many thanks. $\endgroup$ – muserock92 Jul 6 '19 at 11:32

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