Re: Interference pattern - where does the energy g
With interference, you don't only have destructive interference, you have constructive interference as well. I assume you're talking about something like a double slit experiment, or something of that nature. Basically, in a very simplistic way, if in one place you have half the field, then in another place you have double the field. If there is less energy in some places, then there is more energy in other places. The interference "rearranges" how the energy is distributed, but if you add up all the photons hitting the wall in the light and dark areas of the interference pattern, it will add up to exactly the same number as if the photons from your light source hit the wall in such a way that there was no interference.
Second part, well, yes (as a thought experiment, kindof) and no (as in the real world view, it would be right near impossible). If you have 2 waves, exactly the same wavelength, going in exactly the same direction, but that are EXACTLY 1/2 of the wavelength phase shifted with respect to one another, then they will cancel out and you will get zero field at all points. the energy? it goes to the sources of the waves. Ok explanation:
http://www.physlink.com/Education/askExperts/ae155.cfm
What you may be thinking of it sending 2 waves directly at one another, head-on so to speak. In this case, you will get a standing wave as the result of the addition of 2 traveling waves, with constructive and destructive interference at different points. In this case, it's the same as the first question, the extra energy observed in constructive interference is the same as the energy "lost" in destructive interference.