# Which is the right formula to evaluate the diameter of a divergent gaussian beam at a distance r?

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

2)$$d_f=d_0+2r\tan(\theta/2)$$

3)$$d_f=d_0+r\theta$$

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 $$z \gg z_0$$.

• 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! – muserock92 Jul 6 '19 at 11:02
• What does $T$ refer to exactly? – Puk Jul 6 '19 at 11:07
• 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! – muserock92 Jul 6 '19 at 11:11
• I'm confused, wouldn't that mean $T_{63} = 0.63$ and $T_{86} = 0.86$? – Puk Jul 6 '19 at 11:13
• 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. – muserock92 Jul 6 '19 at 11:32