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If you split the circularly polarised light up into two perpendicular polarisations with a $\pi/2$ phase lag between them. You are free to choose which unit vectors to represent these perpendicular polarisations - so choose one in the plane of incidence and one parallel to it. Upon reflection at the Brewster angle the component polarised in the plane of ...


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One possibility is that your polarizer interacts with the other parts of your setup (for example, forms a resonant cavity with some other interfaces that enhances transmission). You can test this hypothesis by rotating your polarizer (is the intensity always brighter?). If you include a drawing of your setup, it would be easier to figure out the underlying ...


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The temperature changes it's optical activity. This depends on the wavelength of light used and the substance. From my own personal experience, a 633nm laser will result in little change in optical activity. A green (546.1nm) has a much more easily measurable change (if you are doing a lab). We have the equation $$[\alpha]=\frac{\alpha}{l c}$$ where ...


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In case your box was classical, it'd output either vertically polarized A and horizontally polarized B or vise-versa, a 45 degree polarizer placed in front of both would allow both to pass half of the time, and block them half of the time - but the pass\block statistics would not be correlated, while in the entangled case, if A passed, B will be blocked, and ...


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Electromagnetic field of light has two kind of angular momentum first spin angular momentum (SAM) and secondly orbital angular momentum (OAM). former one represent the dynamical rotation of electric (or magnetic) field of around propagation direction and indicate the polarization of beam. Later one represent the rotation of light around beam axes. The ...


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Basically, one needs to have a non-zero polarization when computing the Berry phase in traversing the Brillouin zone. It is difficult to say immediately if inversion-symmetry breaking is the only way to get a non-zero polarization, but at the present time this is the only way that we can conceive. That is not to say that in the future someone won't come up ...


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Actually you can argue that circular polarization is the "more natural" basis for a single photon. The photon carries one unit $\hbar$ of angular momentum, and circularly polarized light carries real angular momentum (an opportunity for me to mention one of my favorite experiments ever, using photon polarization to drive a pendulum). A single photon that's ...


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Many 3d glasses use circular polarization, where one lens uses left-hand polarization while the other uses right-hand polarization. This lets the viewer tilt their head a bit without losing the 3D effect, where linear polarization would let the image bleed through into the other eye.


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First of all, you need to understand from which root cause the spin appears in general. This root cause is a symmetry of the physical space-time. Particles with different spins (I mean, spin-0 particles, spin-½ particles, spin-1 particles, and so on) use different representations of a symmetry group to map geometry of the space-time to their quantum spin ...


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It seems that the definition of degree of polarization may be not so well-defined if $V=\frac{I_{max}-I_{min}}{I_{max}+I_{min}}$. For a elliptical polarized light, there is no natural polarized part, but still $V\neq 0$. This definition is used in measurements. Elliptical polarized light can be fully polarized, i.e. $V=1$ (theoretically). I really ...



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