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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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Just to add to the above answers, and since to did not limit your question to the visible range - if you define black as absence of light (photons emitted or reflected), then there is no such substance, because according to black body radiation model, everything with a temperature above absolute zero (which is essentially truly everything in the universe:) ...

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I assume you are talking about linear materials with dielectric constant greater than $1$. Say you have a free charge distribution $\rho$ in vacuum, which produces an E field. Now you introduce linear material. Then the free charges will be "weakened" because they will by partially screened by charges of the dipoles sticking on them. However, the opposite ...

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(For completely filled capacitors) Q = CV So, C = Q/V So, C is charge stored per unit Potential Difference applied. Now, V = Ed ,where d is distance between plates. $E = \dfrac{V}{d}$ Case 1) When you apply a constant V of 1V to capacitor E across capacitor is $\dfrac{1V}{d}$ which is constant independent of capacitance of capacitor or dielectric b/w ...

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Here's a simple explanation based on the dipolar nature of the medium: For most of the materials, we can assume that the source of the reflected and refracted waves are the induced tiny dipoles in the dielectric medium. In an isotropic medium, polarization vector is proportional to the (total) electric field vector with a constant (as opposed to a tensor ...

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The definition of s-polarised light is that the electric field is polarised so that it is perpendicular to the plane of incidence. Where there is a specular reflection, the plane of incidence contains the k-vector of the incoming wave and the reflected wave. Since the electric field of an EM wave must be perpendicular to the k-vector. This then leaves the ...

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There is a simple answer: Symmetry. Suppose the material is isotropic, and consider the initial condition of the p-polarized case, with a p-polarized light wave about to hit the surface. In this case, reflecting in the plane that contains the incident and scattered wave vectors leaves both the (vector) electric field and (pseudovector) magnetic field ...

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The observation you mention is a consequence of the fact that under the aforementioned conditions, the electric field of the s-polarisation (or, "TE") always remains parallel to the optical axis, no matter what the incidence angle is. If one substitutes the effective permittivity along the optical axis into the Snell law, it holds (for this polarisation ...

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LCD displays use three layers: a polarizer, which polarizes the light produced by the backlight system a liquid crystal, which has the ability to twist or not the direction of polarization of light, controlled by an electric field another polarizer, called analyzer, stops or not the light out of the liquid crystal, depending on it having rotated or not the ...

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Other answers paraphrase it well in technical terms. It might be easier to see if you remember that when two particles interact they must do so in a way so that the momentum, energy, spin, etc. are conserved. After the interaction the two particles still remain in a superposition state but if you measure one of them after an interaction you can find out ...

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Interference of electric field is basically addition of vectors. Two vectors sum up to a zero vector if, and only if, they are anti-collinear and they are of the same magnitude. The condition of (anti)collinearity is achieved only in special cases. One of these are linearly polarized waves of the same polarization. (They moreover have to propagate under a ...

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