If two gamma laser beams with opposite phase and equal wavelengths interfere or if a gamma laser is divided into two beams in order to interfere with itself. Is it possible to obtain a totally destructive interference? Thanks.
No. Gamma photons exhibit the particle effect and collisions of gamma photons with atoms include Compton and photo electric effects even gamma photons at higher energies may disintegrate the nuclei. The pair-production of gamma photons above 1.02 MeV give out e_ -e+(electron -positron) particles. Therefore, I believe that photon interaction with matter strongly deals with the particle interaction. The wave effect such as interference usually needs proper conditions.
In fact, the interference requires temporal and spatial coherence to generate interference fringes on the screen. The gamma beam should be monochromatic i.e, the narrow energy width ( spectral width). However, it is doubtful to create coherent gamma photons using available optical components. Note that gamma is an ionizing particle and for such experiment , likely high vacuum is required to avoid the ionization with atmosphere!
No, it's impossible. It doesn't matter what wavelength (gamma, UV, visible, infrared, etc.), it is always impossible. High level: conservation of energy does not allow for energy to disappear. Lower level: imagine an interference setup (Mach Zehnder for example). Yes, you can totally cancel the output at one (!) output port. But not at the other output port(s), where the interference will be constructive. Some people might say: "I have only one output port". These people are incorrect.
What do you mean: "output port" ?
Yes energy conservation is not in contradiction of destructive interference. For instance, AR coating is based on the reduction of reflectance according to destructive interference at a certain wavelength at the expense of increase of transmittance. In the case of HR coating, transmission becomes negligible.
If you have a perfectly collimated laser beam with infinite coherence length (impossible), but for fun lets say we did, and we used a perfect beamsplitter to combine two of these beams (same magnitude) in a colinear fashion, then changed the phase of the first beam to be perfectly out of phase with the second, it would completely destructively interfere in the region where the beams are colinear after the beamsplitter, but much like the double slit experiment, the photons on the upstream side of the beam splitter know this and are reflected and transmitted in the other direction. the energy just gets redirected.
Note 1: no beam splitter can do this, as it requires a change in POL or wavelength to combine beams. Spatial beam multiplexing can never be perfectly colinear either.
Note 2: the E field energy does not get dumped into the B field. light is not an inductor.
Its always a good question, "where does the energy go in complete destructive interference?" the answer is "it is transmitted, reflected or absorbed"
Thanks to all, but I'm a little bit confused.
Can we say that it is possible in theory but not in practice? Or impossible even in theory?
Thanks again.
T+R+A=1
If A~0 , then T+R~1. When R-->0 then T-->1 and vice- versa to satisfy energy conrsevation law.
John L Hostetler and Martijn Heck are correct. It is a simple feature of waves, independent of wavelength and theoretically impossible (even if we had the perfect optics).
You have to start with two independent beams and somehow recombine them, otherwise the two beams would destructively interfere everywhere along their path, which is the same as not having any beams at all. You always have to ask yourself, where did the energy go? If you only consider the waves, you will find another wave somewhere (if not transmitted through a beam splitter, then reflected, sidebands or fringes if spatially multiplexed, where the fringes have a higher peak intensity but same average intensity as the sum of the two incident beams). One could consider more exotic interactions, such as photon-photon collisions, but this will produce other phtons/particles, and is not really destructive interference.
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