Protip: Batteries are rated both by voltage and amperage. Link them in sequence (one negative to another positive) to add the volages together. Link them in parallel (a wire running across their positives and negatives) to add their amperages together.

Yeah I can't figure out where the positive ends go on this thing.

The negative clearly has that springy thingy, but the positive just touches the bottom of the case.

Yeah, you do realize that all you need to do is check the amperage of the batteries you are using (Galvanometer, or one of those fancy electric all in one doodads), then you can times them together to get the wattage, then you can find out how many batteries you need to get the wanted wattage.
v*a=w, w/v=a, w/a=v
(v*a)*x = w*x where x is number of batteries and v*a and w are for one battery.
or you can just put in the wattage you want and divide that by the wattage you have to find the amount of batteries w(wanted)/w(current) = number of times you need to multiply your batteries.

0,01 amps if you assume the wattage is 30 mW, proportionally lower if it's less than 30. Adding higher voltage would give more wattage (in your case, ~10 mW per volt if the amps don't change), buuuuuuuuut it might burn the components if they aren't built for higher voltage. In any case, you're probably going to shorten its lifespan by going above the intended values.

I asked my physics teacher about it (he's also a nuclear engineer, brilliant guy) and I found out I could add to it half of the input it already has.

So.
I took some tin foil and made a tube thingy around 3 AAA's.
Now it's brighter and the beam is more noticeable.

Definitely not, Roast Veg.
Amplitude is to do with wavelengths, Amps/Amperage is to do with electricity. Named after Mr Ampere.

Sources:
I'm English, and know basic physics.

damn it guys....
http://www.youtube.com/watch?v=-mHLvtGjum4

Alright, so I'm actually an electrical engineering and optical engineering double major (junior). Some circuit 101 stuff:

The power delivered to a load is the voltage times the current. If you know the resistance of the laser, then the power is also (V^2)/R or (A^2)*R, which is really just using Ohm's law.

Amplitude can be used to describe a lot of things. If you have an AC power source, then the amplitude is the maximum voltage of the AC signal, or half the total voltage swing. If you're talking about light, amplitude usually indicates the maximum of the electric field, or possibly the maximum intensity of the light if the light source varies in time. Basically, if it's a wave, it has an amplitude.

Power is measured in watts, or Joules/second. Current is measured in amperes, or electron flow per second. Electric potential is measured in Volts. They are referred to as power, current, and voltage, respectively, in my experience.

You cannot measure the power without knowing either the resistance of the laser or the current you're delivering to the laser. The resistance of the laser is a function of voltage to some extent, so the most reliable method will be to measure the current. Buy a cheap digital multimeter and hook it up in series with the batteries you're using and the laser and set it to measure current. Add more batteries until the voltage of the batteries (add them all together) times the current is more than 30mW, then go back one step.

The discussion is dead guys, he's already achieved what he wanted.

Ohms has nothing to do with it, ohms are the unit of resistance and you don't want to be adding resistance for no reason if you can just get the required voltage by adding more batteries, resistance is overall a pretty bad thing unless you want to reduce the current. Also if you watched that video and have read the replies to this thread you may be a little confused, we've all been writing current as A but the symbol for it is actually I and it's measured in Amps or Amperes (that used to get me quite a lot where I would write an A instead of I in equations).