You can do this, but most potentiometers can't handle the current. and typically have too high a resistance to be useful in serious lights. For the ROP, consider that you have a roughly 7.2V supply, and the ROP bulb is in the neighborhood of 1.5 ohms (yes, it's non-linear, but I'm looking for a quick approximation...), so when you're running half-voltage (= 1/4 power, more or less), your pot must also be 1.5 ohm, and drop 3.6V, so 2.4A * 3.6V = 8.6W. (Also, note that you have an efficiency of only 50% -- so if you're trying to stretch runtime, this might not be a good way...) So you'll need a pot going from 0 to no more than 10 ohms, and rated for 10A dissipation -- which exist, I'm sure, but are not common, and will probably be a little on the large side.
For the P7, the computations are a bit rougher, since all the (useful) current variation occurs in the top 25% or so of the voltage scale, but similar resistances should be useful. It will have less dissipation, though, which is good for efficiency.
In either case, the better solution is to use the potentiometer as a control signal for a variable-current (P7) or PWM (ROP or P7) driver. (Of course you could use variable-current drivers with incans, but there's less than no point.) Simplest way would be a serious MOSFET with the pot as a voltage divider driving the gate, but now the MOSFET has to dump the 10W instead, and you still lose efficiency.
Better yet is a circuit using the FET in switching mode (PWM), since the FET uses no power when fully off, and almost no power when fully on. Should be a pretty basic 555 project, if that route suits you.