diffraction

[Deejaybee]

you are talking about the free-electron soup that is most metals, then you are partially correct; this is the basis of reflection from metals (at visible wavelengths) and, as an interesting aside, also the reason why polarisers do not remove the reflection from metallic surfaces in the same way that they do from dielectrics.

however the effect then is one of reflection.

how does your treatment acocunt for dielectric materials? Then there are no free electrons to interact; excited states do exist, as pointed out by one commenter earlier, but these (except in rather special cases) result in different frequencies being generated by either up-conversion, or partial conversion to phonons followed by re-radiation.

None of these effect describe diffraction.

regards

--
DeeJayBee

deejaybee.smugmug.com

 

[Deejaybee]

just to add Joe,

interference is effectively the result of mixing the electrtic field of the incoming light- hence othogonal polarisation cannot interfere as the E vectors are...orthogonal...

To the previous poster, at the risk of appearing to be patronising (which is not intended), think of the X and Y axis on a graph as being two states of polarisation which happen to be orthogonal. Then, trying to add +1 to -1 on the X axis of a graph results in zero (destructive interference). But add +1 on the X axis to -1 on the Y axis, net result is not zero (interference of othogonal components has not occured)

Hence if interference is occuring, it is "automatically" selecting aligned states of polarisation; an additional polariser will therefore have no additional effect.

Regards,

--
DeeJayBee

deejaybee.smugmug.com