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The spin of a photon can be $+1$ or $-1$ but not zero. The spin is directly related to the circular polarisation of the light beam. I think spin $+1$ is clockwise polarisation and $-1$ is anticlockwise, though my success rate at remembering which way round it is rarely exceeds 50%. Linear polarisation is a sum of left and right circular polarisation i.e. it ...

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The polarization of photons is carried out by the same orientation of the magnetic and electric component of the EM field. In order to understand the spin, you have to bear in mind that the electric and the magnetic components are dipols and have a direction. Let one see in the direction of photon's movement on the two vectors representing the electric and ...

3

Wire grid polarisers allow radiation to pass that has it's electric field polarised perpendicularly to the direction of the wires. The explanation is that the component of the light polarised parallel to the wires sees the grid as if it were a solid conductor and therefore most of it is reflected and the rest absorbed in the first couple of skin depths. In ...

1

Light is the visible part of the electromagnetic radiation and consists of photons. Each photon has an oscillating electric and an oscillating magnetic field. In vacuum both fields are perpendicular to the direction of the photon's motion and perpendicular to each other too (see this sketch). There are mainly two used by people opportunities to increase the ...

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At any particular point in space, there is only one value for the electric field. Of course, if multiple electromagnetic fields are overlapping at that point, then their electric field components are added together to yield the total electric field (this is because electromagnetic fields combine with linearity). When multiple electromagnetic waves overlap ...

0

the general modern consensus is that sound does not have polarization possible due to lack of shear forces in air required for transverse wave propagation. however here is an old forgotten study that reports measuring a weak effect by a professor of physics at Ohio State university and published in Science magazine. wonder if there is any more modern ...

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As others have pointed out, the exact formula of the reflected light should be derived with Maxwell's Equation. If you are simply asking for the qualitative reason for the asymmetry of the polarization, it can be attributed to the asymmetry of the boundary geometry relative to the path of the incoming light. The size of the effect is determined by the ...

2

Firstly, the OP is forgetting that the classic microwave polariser experiment is done with EM radiation in a pure state, not a mixture. We simply have polarised light from, say, a Gunn diode and this pure quantum superposition is forced into a polarisation eigenstate by the polariser. So we begin with near to zero entropy light, absorb some of it (adding ...

4

As always the answer is a simple thing. You calculated the change in entropy using the definition of entropy $$\mathrm{d}S = \frac{\mathrm{d}Q_\mathrm{rev}}{T}$$ Note that this applies to heat transferred reversibly. More generally we must use Clausius theorem $$\mathrm{d}S \ge \frac{\mathrm{d}Q}{T}$$ ...

2

I think the factor you are ignoring is that the polariser will emit thermal radiation. If we continue with the ideal polariser, then it should only emit the polarisation which it absorbs (ideal components are weird). This means that there will still be a component of the absorbed polarisation in the beam after the polariser and so there will always be some ...

1

I believe the error is in assuming that the polarized beam is a pure state of zero entropy. If you characterize it in terms of polarization only, then the characterization is not complete. You need a complete set of commuting observables to charactherize a pure state. The macroscopic polarized beam is still compatible with many different quantum microstates ...

1

Perpendicular and parallel polarisation usually refer to linear polarisation of the E-field direction with respect to an interface between two media. Specifically, parallel refers to parallel to the plane of incidence - that plane containing a normal to the interface and the k-vector of the wave. Perpendicular polarisation means polarisation perpendicular ...

0

The electric field must be polarised so that it is perpendicular to the wave propagation. I think this is what option 3 is trying to say, though it is poorly and imprecisely phrased. This could be true whatever the polarisation state of the electric field if the plane referred to is perpendicular to the wave motion. But I think what is meant is a plane that ...

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What type of polarized light is he talking about. If it is linearly polarized then 3 is correct. If it is elliptically polarized light then 2 is correct. For elliptically polarized light if you are at one location then as time progress the electric vector would rotate thus its plane of vibration would keep on changing.

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If we try to polarize the same beam of light in two planes, or if we mix two planar polarized beams, the light will interfere. If the phases of two beams will be identical, then we get 45 degrees polarized light. If the phases of two beams will be different, then we will get so called circular polarized light In other words, any sort of polarized light ...

1

yes, you can. Actually an application is googles for 2d movies. You project on the screen two different images (that is why you see it blurred when watching without the googles), each one has a different polarization. Each plastic filter in the google is a polarizer, and each eye is tuned to a different polarization. So that each eye see only one of the two ...

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Yes, infrared light get pass through two polarized films (vertical & horizontal) even those films are 90 degree rotated.

3

Consider a one-particle state of a relativistic quantum field theory, and let this state be an eigenstate of the 4-momentum operator, $\hat P^\mu |p^\mu\rangle = p^\mu|p^\mu\rangle$. Other than the 4-momentum, what other quantum numbers can the state have, and how should they transform? That is, if there are any quantum numbers collectively labeled by $s$, ...

3

Indeed, the photon doesn't have spin, what it has is POLARIZATION. For a photon emitted from an atom the polarization is CIRCULAR. That means, if we look in a fixed plane perpendicular on its motion, we see that the electric vector does not have a fixed position, it rotates. If so, if the electric vector rotates counter-clockwise the circular polarization ...

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It's because there are alternating layers of + and - ions in the c-direction, yielding a dipole in the crystal. See Tasker -- http://www.surface.tulane.edu/teaching/classnotessurface/TaskerJPhysC79.pdf

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John Rennie's Answer, that complex numbers simplify calculations with sinusoidally varying quantities by letting you do linear operations with complex exponentials and then reverting back to sinusoids at the end of your calculation, is altogether correct and a summary of what is called the phasor method for dealing with any quantity that varies sinusoidally ...

0

I will give you my personal mental image of unpolarized light, maybe it will help. In a given point in space, the E field is a vector lying in the plane perpendicular to the propagation. In this plane, if you put the tail of the vector at the origin, then the tip of the vector is a point jerking in a random fashion around the origin. The important thing is ...

3

Adding two stokes vectors does not give you the stokes vector for the combination of the two beams. For example, adding a beam of horizontal and vertical polarization would make a beam of 45deg (linear) polarization. In order to add two beams you would have to come up with a Muller matrix $M_\vec{a}$ for adding $\vec{x}$ to $\vec{a}$. Unpolarized light has ...

1

Yes. A polarizing filter absorbs the wave polarized orthogonally to the polarization direction of the polarizer. Assume that the particles are photons in the polarization singlet, |Singlet> = (1/sqrt(2)) {|x>|x> + |y>|y>}. For simplicity let x be the direction of polarization transmitted by the filter for one of the particles. You see that in the ...

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