I was screwing around earlier with my mini mag LED (rebel LED). Anyway I had this really old set of panasonic alkalines that had basically no juice in them. Not paying attention I accidently put the last cell in backwards. Turned the light on, nothing. Removing the cells I realized my mistake. I put the batteries back in correctly, it works, although extremely dim because the cells are so weak. I then put new duracells back in, everything is fine, no issues. I don't understand, this light isn't supposed to have reverse polarity yet everything is normal, nothing harmed. I've read that if you mix up the cells you will fry the LED. So why didn't that happen? Is it ok because the cells were so drained and dead? Or does one have to put BOTH cells in backwards to do any damage? What gives?
What you describe wouldn't hurt the light even if it has no reverse polarity protection.
It's because you only put one battery in backwards.
I don't know how many batteries your light takes (you didn't say) but let's just say it takes 2 batteries. Let's say each battery has 1 volt in it right now. Essentially a backwards battery = negative voltage (because it wants to make current flow in the opposite direction to what we consider the correct direction). If you put both batteries in the right way, then the voltage being supplied to the circuit is 1 volt + 1 volt = 2 volts.
If you put one battery in the right way, and one backwards, now you have 1 volt - 1 volt = 0 volts. So you see there is no negative voltage, so it doesn't matter if the light has reverse polarity protection. This would have no effect on any flashlight regardless of the circuitry.
If you had put both in backwards you would be supplying -1 + -1 volts = -2 volts. Now you are supplying negative (reverse) voltage (with reference to the circuit's correct polarity). This might damage the light depending on the circuitry.
Another example of what wouldn't damage the light is if the light takes 3 batteries and you put 2 in the right way and 1 in backwards. 1 + 1 - 1 volts = 2 volts. The light would be dim but work, and no reverse voltage would be applied to the circuit, no danger of hurting the light.