estimationhomework-and-exerciseslaserthermodynamics
How strong would a laser beam have to be (and what frequency would be optimal) for it to be able to make a hole in snow?
Here I am looking for a laser beam that could make the hole after less than 2 seconds of hitting the snow. This would be outside, in open air, where the ambient temperature is about 14 degrees Fahrenheit (about -10 Celsius).
For the purposes of this question, the thermal conductivity of the snow would be (from some example numbers obtained from here), 0.12 W/mK.
There is probably a lot more to this if you consider drafts and air turbulence, but lets keep it simple. Since the warmest place is in the building I would think opening the windows would only help with the cooling. Either it would let hot air escape, or let cold air come in. Sounds like a win-win.
Now if you want to consider humidity that a whole nother ball game.
First, some questions about your setup:
Second, here's an experiment you can try to see if the beam divergence actually is a problem. Repeat your spot size measurements at various distance, but instead of melting a brick directly, put your focusing lens in front of the beam. Then, move the brick back and forth until you find the distance from the lens with the smallest melted spot. Measure the size of this spot and the distance from the lens versus the distance between the laser and the lens. If the variation in spot size and focus distance is too large, read on for a possible solution.
If you want to transport a beam a long distance, the trick is to start with a larger beam diameter. Figure 5.5 in the PDF linked to in @akhmeteli's comment to his answer shows that large beams don't diverge as fast as small beams. The equation for the diameter of a beam as a function of distance from its narrowest point is $$\omega(z) = \omega_0 \sqrt{1+\left(\frac{z\lambda}{\pi \omega_0^2 n}\right)^2}$$ where $z$ is the distance from the narrowest spot (which should be the laser aperture), $\omega_0$ is the diameter of the beam at its narrowest, $n$ is the index of refraction of the material the laser travels through (which is 1 for air), and $\lambda$ is the wavelength of the laser (10.6 microns for a CO$_2$ laser). Far from the laser, the waist is approximately $$\omega(z) \approx \frac{z\lambda}{\pi \omega_0 n}.$$ So, if you double the diameter of the beam ($\omega_0$), your halve the divergence angle (the asymptotic cone in your plot).
The setup I envision is shown in the diagram below:
I chose to use off-axis paraboloid mirrors instead of lenses since they have a higher damage threshold than lenses (at least the ones from ThorLabs do). But, you can also substitute lenses and flat mirrors to achieve the same effect. In any case, use two mirrors or two lenses in a confocal arrangement (distance between elements equals the sum of their focal lengths) to substantially increase the diameter of the beam and collimate it so that you can transport it any distance without worrying about divergence. Then, use a third mirror or lens mounted on your motion system to direct and focus the beam on the cutting surface.
Some advantages to this setup:
Best Answer
Lets start with some assumptions. You probably have the beam from a laser pointer in mind so the hole size you want to burn is about $4\,\text{mm}$ in diameter. Lets assume that it's roughly $-5^\circ\text{C}$ ($23^\circ\text{F}$) outside. One final assumption, and this one is just a guesstimate, lets assume that due to thermal conductivity you need to supply a factor of 10 more energy than is actually required to melt the ice over the course of 1 second.
A sphere of ice with a $2\,\text{mm}$ radius contains $3.4\,\text{g}$ of frozen water assuming it is densely packed snow with $20\%$ the density of liquid water. It takes $35\,\text{J}$ of energy to raise the temperature to $0^\circ\text{C}$ and $1120\,\text{J}$ to convert it to liquid water. Assuming our factor of 10 more power applied over 1 second for thermal conductivity and this becomes $11.5\,\text{kW}$ of laser power. As a comparison, typical laser pointers output $1\,\text{mW}$ so you would need a laser with $10,000,000$ times more power. Lasers that powerful certainly exist, but you're not going to find one at Office Depot.
The problem with that calculation is the large beam size. If you are willing to accept a much smaller hole, then the amount of power is greatly reduced. For a hole of $0.2\,\text{mm}$ diameter you could get away with a laser with only $28\,\text{W}$ of power.
These calculations are pretty rough since the effect of thermal conductivity wasn't treated carefully.