# Can light be canceled by merging with an inverted wave?

Can light waves be canceled by merging them with their inverted waves? Seems like it would violate conservation of energy but waves are added together when they overlap, right? Where is the flaw in this logic? I'm thinking polarized laser light, added to its opposite, might becomes dark again. -- Appreciating your collective wisdom.

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Yes - light waves can destructively interfere. This is the principle behind interferometers. There is no violation of energy conservation because the energy of two waves doesn't add. The energy is proportional to the square of the amplitude, and the amplitudes add so $E \sim \left( A_1 + A_2 \right)^2 \sim A_1^2 + A_2^2 + 2 A_1 \cdot A_2$. The third term is an interference term between the two waves which can be negative and cancel the others. –  Michael Brown Jan 18 '13 at 2:21
@MichaelBrown that would make a pretty good answer. –  David Z Jan 18 '13 at 2:47
Thank you Michael. Wow, physics is fun. Thanks for the term interferometer. Thus, we can make a wave with no amplitude by having two waves merge - and then use it to detect minor shifts in gravity, rotation etc because of the resulting distortions to the overlapped waves. Coool. –  David Jan 18 '13 at 2:51
Possible duplicate: physics.stackexchange.com/questions/23930/… –  Steve B Jan 18 '13 at 3:19

Yes - light waves can destructively interfere. This is the principle behind interferometers. There is no violation of energy conservation because the energy of two waves doesn't add. The energy is proportional to the square of the amplitude, and the amplitudes add. So $E\sim\left(A_1+A_2\right)^2\sim A^2_1+A^2_2+2A_1\cdot A_2$. The third term is an interference term between the two waves which can be negative and cancel the other two terms.
There is a sense in which global cancellation is fine, but pointless. If you have one wave (take a scalar wave $\phi\left(x,y,x,t\right)$ instead of an electromagnetic wave for simplicity) $$\phi_1\left(x,y,x,t\right)$$ and another $$\phi_2\left(x,y,x,t\right)$$ which just happens to be $\phi_2 = -\phi_1$ everywhere at all times, then the sum of the two is clearly zero and you have global cancellation. But in a very trivial this is the same as there being no wave at all to start with. What you definitely cannot do is produce such a pair of waves from a source - a simple application of the wave equation in this instance shows that the source must vanish. So the whole idea of a global cancellation is trivial and pointless in practice. That is why I assumed you were referring to a local cancellation, but I should have been clear from the start.
Note that when $A_1$ and $A_2$ are in phase (both positive max or both negative max), you get extra energy. What ends up happening is that this doesn't cancel everywhere; it makes an interference pattern where it's higher energy some places but lower energy others. And if it interferes negatively when it strikes a surface, you get zero absorption on that surface - it all gets either reflected or it passes through. –  Will Cross Jan 18 '13 at 4:16