# Tag Info

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If you setup a perfect cavity where no modes of light are possible, then the light will not be emitted in the first place (0 probability). You run into problems if you consider the emittor as a classical light-source and then combine it with a quantum-mechanical reasoning regarding interference in cases such as this.

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I know you asked for an analytical description, and ray diagrams, but seeing as this is homework related, I'd urge you to try the ray-tracing yourself and I think you'll find the answer without too much trouble. My suggestion would be to try and track how a couple significant rays move through your lens setup. For example, you might try one ray that passes ...

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It's a little unclear what you are asking, but sometimes one needs to ask an ill formed question of one's colleagues to help formulate a sound question. Plasmonic waveplates are different from waveplates relying on a homogeneous linear material's birefringence: the physics is different. The former's birefringence arises from engineered periodic nano-scale ...

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Great question -- it opens up a lot of physics. My favorite example is "why do some conducting metals look gold/copper/etc. rather than grey?" Turns out it's due to relativistic effects acting on the various electrons' orbitals. Then there's the blue jay, whose blue color is entirely interferometric rather than absorptive/reflective. and more :-)

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First, note that the way you have written the electrical wave-field isn't anything more than exactly a way to write a wave function in general. This is because the term $e^{i(kx-ωt)}$ can be written as: $$e^{i(kx-ωt)} = cos(kx-ωt) +i sin(kx-ωt)$$, and from here you can keep in general the real or the imaginary part as you wish. As for $κ$, the imaginary ...

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The point source keeps radiating light. Will the light undergo destructive interference completely? Point particle as a source of light is OK, but it would need to move with acceleration to produce EM radiation. Static source of light of zero size seems to be reduction ad absurdum, at least from the standpoint of common theory of light based on EM ...

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