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I know how photons are diffracted round a corner in quantum mechanics. But is there also an explanation in classical electromagnetic theory (i.e. by Maxwell's laws?)

Imagine there was an EM wave traveling purely normal to a slab of material which is highly attenuating, but which has a small hole. (I've drawn a picture, so you can see what I mean.)

Maxwell's equations in vacuum are:

[tex] \nabla \cdot \vec{E} = 0 [/tex]

[tex] \nabla \cdot \vec{B} = 0 [/tex]

[tex] \nabla \wedge \vec{E} = - \frac{d \vec{B} }{dt} [/tex]

[tex] \nabla \wedge \vec{B} = \mu_{0} \epsilon_{0} \frac{d \vec{E} }{dt} [/tex]

And inside the slab, E and B would equal zero, since it is highly attenuating.

So then, am I right in thinking Maxwell's equations predict the wave coming out of the hole to be traveling normal to the slab (i.e. not diffracted)?

But obviously, light is diffracted in this situation. Is there some kind of edge effect that makes the wave change direction? Or are Maxwell's equations simply unable to predict this phenomenon?

Imagine there was an EM wave traveling purely normal to a slab of material which is highly attenuating, but which has a small hole. (I've drawn a picture, so you can see what I mean.)

Maxwell's equations in vacuum are:

[tex] \nabla \cdot \vec{E} = 0 [/tex]

[tex] \nabla \cdot \vec{B} = 0 [/tex]

[tex] \nabla \wedge \vec{E} = - \frac{d \vec{B} }{dt} [/tex]

[tex] \nabla \wedge \vec{B} = \mu_{0} \epsilon_{0} \frac{d \vec{E} }{dt} [/tex]

And inside the slab, E and B would equal zero, since it is highly attenuating.

So then, am I right in thinking Maxwell's equations predict the wave coming out of the hole to be traveling normal to the slab (i.e. not diffracted)?

But obviously, light is diffracted in this situation. Is there some kind of edge effect that makes the wave change direction? Or are Maxwell's equations simply unable to predict this phenomenon?