# Gaussian laser spot intensity calculation

Here's a problem in a laser spectroscopy class I have been trying to figure out for quite some time:

A typical dielectric mirror has a damage threshold of $I_\textrm{threshold} = 5\times 10^8~{\rm W/cm^2}$ for 20 nanosecond pulses. What is the smallest beam diameter ($2\omega_0$) that can be used with a 1Hz beam of 3W? Assume a gaussian temporal and spatial intensity distribution, and that the FWHM is 20ns.

First, some relevant equations.

Gaussian temporal intensity distribution: $$I(t) = I_\textrm{peak}e^{-4\ln2 \frac{t^2}{\tau^2}}$$

Gaussian spatial intensity distribution: $$I(r) = \frac{2P}{\pi \omega^2}e^{-\frac{2r^2}{\omega^2}}$$ with $$\omega (z) = \omega_0 \sqrt{1+\left( \frac{\lambda (z-z_0)}{\pi \omega_0^2} \right)^2}$$

The way I thought of going about this problem, was to calculate the total intensity (energy) over the spatial distribution:

$$E_\textrm{spatial} = \int_{0}^{\infty} I(r)~\mathrm dr$$

And then calculate the total energy over the duration of the pulse

$$E_\textrm{pulse} = \int_{-\infty}^{\infty} I(t)~\mathrm dt$$

And use these to calculate the total spatial energy during a pulse for a circular area $A_\textrm{circle} = \pi r^2$

$$\frac{I_\textrm{total}}{\pi r^2} = I_\textrm{threshold}$$

which when solved for the double of the diameter gives

$$2d = \sqrt{\frac{8 I_\textrm{total}}{\pi I_\textrm{threshold}}}$$

However, I don't know what the $\omega$ is (should I also integrate of $\omega\,?$), and I am quite sure that my way of doing this is correct. The question asks specifically for $2\omega_0$, not some $d$, so I feel I should use the definition of $\omega$ for something. I would appreciate some guiding on how to solve the problem.

Intensity is energy per area per time. This means that if we integrate over the intensity profile of one pulse over time and beam profile we will get the pulse energy of that pulse. As formula \begin{align}E_p &= \int ~\mathrm d^2r \int ~\mathrm dt \;I(r,t) \\ &= \int ~\mathrm d^2r \int~\mathrm dt \;I_\text{peak} \; \exp\left({-\frac{-4\text{ln2}t^2}{\tau^2}}\right)\exp\left({-\frac{-2r^2}{w_0^2}}\right)\\ &= 2\pi \int ~\mathrm d r \int~\mathrm dt \;I_\text{peak} \; r\exp\left({-\frac{-4\text{ln2}t^2}{\tau^2}}\right)\exp\left({-\frac{-2r^2}{w_0^2}}\right) \end{align} The pulse energy $E_p$ is know to be 3 J from your problem. You only have to solve the integral and then solve for $I_\text{peak}$. Afterwards just compare $I_\text{peak}$ as a function of $w_0$ to $I_\text{threshold}$ and you are done. I don't really understand what you have done because you don't define $I_\text{total}$ but it looks wrong because the units don't add up (at least if $I_\text{total}$ is also an intensity).
EDIT: Full solution: When you solve the integral you find $$E_p = \frac{\pi^{3/2} \tau\; w_0^2 }{4\sqrt{\ln2}}I_\text{peak}.$$ I just pluged the integrals into wolfram alpha. Now we find for the peak intensity $$I_\text{peak} = \frac{ 4\sqrt{\ln2}}{\pi^{3/2} \tau\; w_0^2 }E_p$$ and we set $I_\text{peak} = I_\text{threshold}$ to find the smallest $w_0$ $$\frac{ 4\sqrt{\ln2}}{\pi^{3/2} \tau\; w_0^2 }E_p = I_\text{threshold}\\ 2w_0 =2\sqrt{ \frac{ 4\sqrt{\ln(2)} E_p}{\pi^{3/2} \tau I_\text{threshold}}}$$
• Why do you use $\text{d}^2r$ for integration of the spatial part? – Yoda Oct 12 '16 at 8:27
• Because it is a two dimensional integration over a plane. For a simple gaussian you can of course use $\int \rm{d}^2r \rightarrow 2\pi \int r\rm{d}r$ – Jannick Oct 12 '16 at 8:50
• Oh, okay. Also, did you forget the factor $\frac{2P}{\pi \omega^2}$? – Yoda Oct 12 '16 at 9:38