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Yes, you have the right idea. For example, in a linear medium (with electric susceptibility $\chi$ and background polarization $\mathbf{P_0}$), we can write $$\mathbf{D} \equiv \varepsilon_0 \mathbf{E} + \mathbf{P} = \varepsilon_0(1+\chi) \mathbf{E} + \mathbf{P}_0,$$ we can rearrange the macroscopic Maxwell's equations to say: $$\nabla\cdot\mathbf{E} =\frac{\...
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I think that for this case it is worthwhile to go back to Maxwell. Even before Maxwell had formulated the set of equations that today we know as 'Maxwell's equations' he had already worked out far reaching ramifications.
Wikisource has a transcript of the 1861 paper On physical lines of force
As we know, Maxwell worked with the supposition of a mediator of ...
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The angle of refraction is not 90${}^\circ$. The angle between the refracted ray and what would be the reflected ray is 90${}^\circ$. The reason is that the source of reflected light is the polarization of the medium. The direction of the reflected ray is fixed, determined by the direction of the incident ray. Brewster' angle is that incident angle that ...
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This appears to be a materials issue rather than a pure physics issue, so I will be speculating here.
There are two ways to make a more powerful lens: grind it to a greater curvature or use a material with a higher index of refraction.
If you use a greater curvature, then the thick part of the lens must be thicker. The lens will have more weight, and the ...
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Sure, light could have a different speed in the same direction; that’s what happens at normal incidence (where there’s no refraction). The thing that’s actually conserved is the in-plane component of momentum, $k_x$. At normal incidence, $k_x=0$ regardless of the refractive index, so there’s no need to refract. But at oblique incidence, without refraction in ...
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At Brewster's angle there is still reflection of TE polarised light. Also, by Snell's law the angle of reflection equals the angle of incidence (+$\pi$).
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It was a software bug that purely by accident gave the correct result for the soap plot ($n$ was $\frac{1}{n}$). It seemed strange at first that there was interference with layers of $90 nm$ for blue light waves of $460nm$. But considering that path length is at least twice the thickness depending on angle and optically much longer when $n=2.6$, so at least $...
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