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If you put unpolarized light in your Calcite crystal, then the beam will be divided into two beams, one for each polarization, with equal intensity. It you put polarized light (say, linearly), then the two beams will have an intensity that depends on the angle between the axis of polarization of the light and the crystal, following Malus' law. If the ...


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No, gravitational waves are not emitted isotropically. In the weak-field limit (i.e. far from the sources), the radiation emitted by a gravitational system is determined by the third time derivative of its quadrupole moment, which, being a tensor, needs to be projected along the line of sight to yield a (scalar) energy flux. This projection is what gives the ...


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Just as a supplement to ACuriousMind's answer, it is worth noting that buried in the bottom of their paper they actually show what the "spin 1/2" eigenstates are in terms of the regular basis: $|j=1/2\rangle=\frac{1}{\sqrt{2}}(|1, -1 \rangle + |0,1\rangle$) $|j=-1/2\rangle=\frac{1}{\sqrt{2}}(|-1, 1 \rangle + |0,-1\rangle$) where $|l, \sigma\rangle$ is the ...


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Nothing is happening. At least, nothing except that a new generalized quantity suggestively called "angular momentum" was defined and subsequently measured. But nothing we know about the usual angular momentum of photons is changed by this in any way. Standard total angular momentum is $J = L + S$, where $L$ is the orbital and $S$ the spin angular momentum. ...


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I have a site which also explains this phenomenon clearly here. In short here you would find that when an unpolarised light hit at the interface of the two medium the reflected light will be generated such that only that component of electric field is oscillating, which is perpendicular to the direction from point of incidence to the point of observation ...


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I'll try to build an answer based on the above comments. There are two things to consider : Temporal coherence, which has nothing to do with polarization. One simple way to modelize it is the following : imagine you have an incoherent pointlike source that emits EM radiation at some frequency $\omega$. This source can be a thermal one, or spectral for ...


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You are confused due to the vectorial form of the equation, so you should write it by components (I will use cartesian coordinates), and I will use $\partial_x = \dfrac{d}{dx}$ for comfort, that being said, your equation can be written as: $$ F=-\int{(dr)}{(\vec{\nabla} \cdot \vec{P}) \vec{E} }=-\int{(dr)}{(\partial_x P_x +\partial_y P_y + \partial_z P_z ) ...


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I am not able to understand your notation and even though your question is not clear enough I will try to answer it based on my understanding of what the question is. Question: Consider three linear polarizers $P_1, P_2 and P_3$ kept one after the other in front of a photon source (unpolarized). The axis of $P_1$ is vertical, that of $P_2$ is at at an ...


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Let the initial intensity of the light be $I_0$. After the first polarizer, the intensity will be $I_1=\frac12 I_0$. This is because the light was initially unpolarized. Now the light is horizontally polarized. To figure out the intensity through the second polarizer, remember that the intensity is proportional to the amplitude of the electromagnetic ...


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The key thing is that the surface have facets. That is, it has to have smooth flat parts that can reflect light like a mirror. If the surface is just amorphous then the scattering will tend to be too disorganized to see the polarization. I have seen polarized light coming off quite surprising surfaces. A manhole cover for example. It had been polished fairly ...


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In the classical theory of reflection (and refraction) of electromagnetic waves, there are equations which describe the reflection of light in two specific orientations. They are known as the Fresnel equations. However, the polarizations of light lie in a 2D vector space, so as long as you decompose any incoming wave of light into the two linearly ...


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Wether you get destructive interference or constructive will depend on the position. What you get is something in the image. Where the diagonal lines indicate maxima where constructive interference occurs while between them there is destructive interference.



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