If energy can not be created or destroyed then what happens to the energy of 2 photons when they destructively intefere with each other?

It pops up somewhere else.

The bright bits between interference fringes are twice as bright as you would expect if the interference didn't happen.

If energy can not be created or destroyed then what happens to the energy of 2 photons when they destructively interfere with each other?

The interference picture is not destructive in all the space. In some regions the fields cancels each other, in other they add. The energy of EMF is an integral over the whole space from the energy density (E² + B²). It remains constant despite the bright and dark region may travel in space.

Edited September 7, 201015 yr by Bob_for_short

My original question made an assumption which may have been wrong so to clarify here are two statements.

 

A) A photon can only interferre constructively and destructively with itself.

B) Two photons can interferre constructively and destructively with each other.

 

Which one is correct or are they both correct?

 

Statement 'A' makes sense in Young's slits experiment and I understand the conservation of energy in this case. However the peeks on the track of a CD are a height that is 1/4 the wavelength of the light used to read it therefore there is a phase difference of 1/2 a wavelength between peeks and troughs which cancels out the light at this point. So in this case what happens to the cancled light? Is it just bad wording and the light is only cancled in the direction it would otherwise have been read and is actually defracted in other directions?

 

And is statement 'B' how lasers work or is constructive interferrence not the same as coherent light?