Why does EM destructive interference reflect on an resonnant cavity chamber but not on a piece of paper? Case 1) Light passing through a single or double slit with show an interference pattern on a screen with dark fringes. The observer eye wont receive light comming from the dark fringes (and therefore perceive them as dark) and therefore we can assume that no light is emitted or reflected from the dark fringes.
Case 2) Resonnant EM cavity, the EM potential at the wall of the cavity is 0 (similar to a dark fringe), but we know that light is reflected otherwise the cavity (laser or microwave oven) wouldnt resonate.
Why is it that, in case 1, a 0 EM potential (be it electric or magnetic) doesnt emit/reflect light but does so in case 2 ?
 A: I think we can understand this in the following simple terms:
Case 1 the double slit experiment:  The electric field at the wall $E_w$ is zero because of the contributions of two sources (the two slits) $S_1$, and $S_2$, $E_w = S_1 + S_2 = 0$. Note that here both $S_1$ and $S_2$ are propagating towards the wall, and thus an observer looking at the wall will have neither $S_1$ or $S_2$ propagating towards him. Thus he sees this portion of the wall as dark.
Case 2 EM cavity: Allow me to make the simplification to just a single source pointing at a metallic wall. It's the same problem really, we see a reflection (a shiny wall) even though $E_w = 0$. The reason for this is that $E_w$ here is the contribution of $S$ the source and $R$ the reflected wave. Thus we can write $E_w = S+R = 0$. But now $S$ and $R$ are propagating in opposite directions, in particular an observer looking at the wall has $R$ propagating towards him.
You might want to think of $R$ as being generated by the response of material to the sources. So in case 1 the wall has no response at all since the two sources cancel, (the wall feels no fields to responded to). However in case 2, the wall feels $S$, responds by $R$ (such that $E_w = 0$), and you observe this response.
A: Case(1) The interference pattern on the paper destroys the coherent phase information and scatters the light. If a polished mirror was used as the screen, various interference results would be observed as "speckle".
Case(2) The microwave oven is not usually resonant but the six walls are seen as polished mirrors (specular) to radio frequency waves.   There are two distinctly different types of  waves discussed in EM theory, standing waves and traveling waves. One form of standing wave results from the vector sum (interference) of a traveling wave reflecting off a mirror interfering with the traveling wave impinging on the mirror (speckle). Inside your microwave oven is lot of standing waves. the nature of the waves at the boundary of one of the walls is that the E field sums to zero at a conducting boundary but the B field sums constructively at the same boundary, the energy contained in the E field is converted to magnetic energy in the B field. this conversion continues back and forth along the standing wave, If you could observe these fields you would observe RF speckle.  Speckle is light and dark spots that move about as you the observer moves about in a sea of standing waves.  Observe supermarket scanners for speckle, if you move the speckle moves.  Fading of radio signals when driving slowly is another example. Standing waves are not dissipative but are reactive. but when you  put a chicken in the microwave. dielectric losses attenuate the waves changing the phase of the waves, the E and B come into phase indicating energy absorption is occurring (energy converted to power), the chicken tends towards a black body radiator at this point.  Somewhat remarkably,the peak of the approximate black body spectrum will be at a very much higher frequency than the initial microwave frequency, Many tens of thousands of GHz.
A: In the DSE from single photon experiments we know there are no photons in the dark bands, all the photons are in the bright bands.  An explanation for this is provided by considering the Feynman path integral theory, which basically states that a photon predetermines its path by considering all paths, the highest probability path that results is one that is the shortest as well as being an integer multiple of the light wavelength.  This mathematical theory produces the same result as the classical interference based theory, however it does not violate conservation of energy.
In a laser cavity we have the lasing medium which continually converts electrical energy to light. It requires that one of the mirrors be leaky otherwise there is no energy conversion in the medium. A way to think of this is that the leak allows photons to have acceptable paths where they can be absorbed by atoms/molecules.
We never know or can measure emitted light unless it is absorbed somewhere like in our detector.  Taking this further we can say light is never emitted until it has a known path to be absorbed.  Conceptually no field and zero field are different situations, we can have superimposed zero value EM fields in space but never observe them.  We only observe the full EM value when it is absorbed.
