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Question:
As we know light are electromagnetic wave, and has a wave properties. I learned in high school that when two wave cross each other, they can either superimpose (increase in amplitude) or cancel each other out (amplitude decrease). Suppose you have two laser of exactly same amplitude and wavelength. Can you position it in such a way that the two wave cancle each other out completely ?. But then what would you observe ? since there won't be any wave (zero amplitude)



Replies:
You would observe just what you said -- nothing. You would have canceled out all the energy there.

It is a bit hard to swallow.

Maybe its easier to picture in your head if you go back to sound waves, not light waves, tho the ideas are the same. If you put a speaker behind you that plays some music, but put another speaker somewhere else that plays the same sounds, but in a way that the sound waves go one way while the other waves go the other way, then they would cancel out and you would hear nothing. One speaker would be trying to compress the air, to make that part of a sound wave (sound waves are compressed then less then compressed then less air). The other speaker would be trying to make the air less dense. Both speakers at the same time, and they cancel each other out, and the air "feels" nothing, you hear nothing.

Where it gets even more hard to picture is that in these cases, there is no power coming out of the speakers in total. (ideal world, where things exactly and truely cancel -- hard to do in practice). But one speaker is trying to put out some sound power, the other is putting out the opposite. No net sound comes out, so no power is coming out on a steady basis, tho some may have been put out at the very beginning, when the waves were first put out.

Complete cancellation like this is a bit confusing, and also hard to do. But you can partially cancel out the sound or light, in a small area, like where your waves cross and destructively interfere. You can see this every time you see ripples in a pond that cross each other.

S. Ross



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