If I connected a 12V light-bulb to 24V, would the light be 4 times brighter

(assume i have a good light that didn't burn out) my idea behide this is to help me understand ohms law. i hypothesize that when voltage is doubled, then the current is doubled, thus, 4 times more power to the light, and the light is about 4 times brighter.

is this right? (i am also not taking to consideration that the light bulb filament stays the same resistance.)

(assuming an incandescent bulb, not an LED or CFL)

yes, the light consumes 4x the power with 2x the voltage. It's highly unlikely (in my view, impossible) to not burn out with 300% overload, though.

however, human vision is logarithmic. it may not look "4 times brighter"

Also, incandescent lights put out blackbody radiation. the colour will be shifted to blue or UV.

If you do this backwards, put a 24V bulb on 12V, it will consume 1/4 the power and glow dimly, putting out most energy in infra-red, and look much dimmer than 1/4 as bright.

Brightness of the light from the bulb depends on luminosity which decreases with distance away from the light source. In general, the more energy dissipated by a light source (your light bulb in this case), the more energy and thus 'more brightness' per unit area. That's luminosity (http://en.wikipedia.org/wiki/Luminosity), which describes how brightness is perceived albeit in terms of energy radiation.

Start with V=IR. Power dissipated by the resistor R is given by: P = IV; for a current I induced by the EMF source with the value V. V = IR or V/R = I so P = (I^2)R or P=V^2/R. Power dissipation is directly proportional to the square of the EMF source thus doubling the voltage would increase the power dissipated by the resistor by a factor of 4.

Basically, yes. The voltage doubles, which means the current doubles too. Thus, since power is voltage x current, the power goes up by 4 times. You can also use power = V²/R directly if you happen to know that one but it isn't really as intuitive.