# Tag Info

## Hot answers tagged polarization

19

See the Wiki article on Polarized 3D glasses. Most likely, you have a pair of circularly polarized glasses. The mirror reverses the circular polarization. The article on Circular polarization does it better than I would be likely to achieve in less than an hour or two. Or Hyperphysics, or Google.

17

The key idea of how LCDs work is that a peculiar substance, a "liquid crystal", is placed between two linear polarizer filters. The filters are crossed, so normally a backlight behind the LCD isn't seen, appearing black. When voltage is applied to part of the device, such as one segment of a 7-segment digit, or one row and column defining one pixel in an ...

12

Cinema 3D glasses (at least those made by Read-D) are circular polarized. This has the advantage that the polarized light reflected from the screen doesn't depend on the angle between your eye and the screen and so you can move your head around while watching. But when you look in a mirror the rotation direction is reversed on reflection. The shutter ...

12

It's the same $\ell$ that indexes the spherical harmonics $Y_{\ell m}$ (or $Y_\ell^m$ if you prefer). We can decompose functions defined on the sphere (like anything defined on the sky) into a countably infinite sum of appropriately weighted spherical harmonics. $\ell$ counts the number of nodes, while different values of $m$, $0 \leq \lvert m \rvert \leq ... 11 It is possible to "block" light based on its polarization in a number of ways. In the situation you are describing, where the light hits a polarizing filter, it is simply absorbed by the filter. The filter does indeed heat up, and in fact if you put your hand near the screen you can usually feel that it is quite warm. It is also possible to have a polarized ... 11 Let's do some math in order not to be unsubstantiated. 1. Perpendicular polarizations. First wave$E_{1x} = E_0\,\cos\omega t$,second wave$E_{2y} = E_0\,\cos(\omega t+\Delta)$. Where$\Delta$-- is a phase difference between waves. Total field:$\vec{E} = E_0\left(\vec{i}\cos\omega t+\vec{j}\cos(\omega t+\Delta)\right)$. Intensity:$I_\perp\sim ...

9

Thank you so much for posting this. I had seen this a while ago and never got round to thinking it out in full, and now I have :). Those are not "cheap sunglasses": they are 3D glasses for use with 3D television screens and the like. This is a fantastic demonstration and I encourage anyone with a spare pair of these glasses at home to take them apart to play ...

8

The usual way linear polarisation is measured is by shining polarised light onto a polarising filter, rotating that filter and then using Malus' law to fit the data to a $I_0 cos^2(\theta_{beam} - \theta_{polariser})$ shape. By finding the angular position of the intensity peak we can infer the angle of polarisation of the incoming beam. Now, assume we ...

8

I can't improve on KDN's answer, but given Todd's comments this is an attempt to rephrase KDN's answer in layman's terms. A system is only in an eigenstate of spin around an axis if a rotation about the axis doesn't change the system. Take $z$ to be the direction of travel, then for a spin 1 system the $S_z$ = 0 state would be symmetric to a rotation about ...

7

You may start with understanding Rayleigh scattering, and then plane polarized light interacting with a simple anisotropic molecule before going onto chiral ones. A plane polarized light wave is propagating in the direction given by the right hand rule, so let's say it's electric ($E$) field is in the $\hat{i}$ direction, the magnetic ($B$) field in the ...

7

Yes. In fact, light will only interfere with light of the same polarization. If you take a Mach–Zehnder interferometer, for example, and put a polarization rotating optic (a waveplate) in one of the arms, the interference pattern will lose contrast. If the polarization is rotated 90 degrees, the pattern will vanish completely.

6

You don't need a vector field on the sphere - you just need vectors. Vectors don't have any intrinsic location, just a direction and a magnitude. The polarization of light is independent of the propagation direction of the light. Let's examine this with a simple experiment: Consider an ideal plane-wave laser beam, beam 1, propagating in the z-direction ...

6

The easy answer is to say that Brewster's law only applies to reflection from the interface with a transparent medium, and a mirror isn't transparent. Indeed for an ideal perfect mirror, all light of both polarizations is reflected perfectly, so there is nothing to say. For an actual real-world mirror, the metal mirror surface will have a finite skin ...

6

I'll try to hit the main points, but for details you need to read something longer than this post. I think I'd start with the various tutorials at Wayne Hu's web site. The Weiss report, which lays out the case for search for $B$ modes, might also be a good place to look. The general picture: (You may already know this, in which case skip ahead.) Both ...

6

The absence of the $S_z=0$ spin projection is related to masslessness of the photon. Because the photon is massless, it propagates at the speed of light and has no rest-frame time evolution. This removes one of the allowed polarization states that would be present for massive bosons. Solving the eigenvalue problem for the spin operator S gives eigenvalues ...

6

The answers from KDN and John Rennie are right - I'll just try to illustrate how it works: The components of a massless spin 1 field satisfy $$\Box^2 A_{\mu}(x) = 0$$ Traditionally we perform the expansion in momentum variables $$A^{\mu}(x) = \int{\frac{1}{\sqrt{p^0}}A^{\mu}({\bf{p}})e^{-ip.x}}d^3{\bf{p}} + comp\ conj$$ If the particle is moving in the z ...

6

Here's my haltingly offered answer - I am not sure I am not overlooking a subtlety that you can see but I can't. I'm going to try to answer @dmckee 's take on your question: "How can we motivate the photon spin starting from the classical theory?" I relate to your worries that slipping between linear and circular polarized base states might just be ...

6

There is some evidence of polarization perception. Many people are able to perceive polarization of light. It may be seen as a yellowish horizontal bar or bow-tie shape (with "fuzzy" ends, hence the name "brush") visible in the center of the visual field against the blue sky viewed while facing away from the sun, or on any bright ...

5

Ernst Mach has once designed an experiment which nicely illustrates linear polarization using a glass cone. Polarized light falls on the cone from the top at Brewster angle. In case of unpolarized light the reflected light has symmetric distribution while with linearly polarized light two dark strips occur in the plane of polarization. It is demonstration ...

5

Yes, a photon in a polarized light is found in a pure state such as $|H\rangle$, $|V\rangle$, $|L\rangle$, $|R\rangle$, or any complex linear combination of them. A photon in (completely) unpolarized light is described by the density matrix  \rho = \frac{1}{2} \left( |L\rangle \langle L| + |R\rangle \langle R| \right) = \frac{1}{2} \left( |H\rangle ...

5

First, I just want to correct a minor misunderstanding. If anything, it would be the light with its electrical field oscillating in the plane parallel to the slit which would have more difficulty propagating, and even then, only in specific circumstances. The best way to explain this is to ignore, for a moment, the experiment you've described, and consider a ...

5

Summary: the energy change is negligible and if it is not, the energy difference comes from a frequency change of the photon. One must realize that unless the plate was already quickly rotating before the experiment, the energy stored in the rotation of the plate at the end is negligible relatively to the energy of the photon for the same reason why the ...

5

the original strategy of Feynman and wheeler was really about the desire to get rid of all self-interactions. In the modern language, it would eliminate most loop diagrams. In particular, consider an electron propagator, in the modern language. One may attach a photon propagator on it. That modifies the electron's self-energy, and this is the kind of a term ...

5

I agree with the answer given by sigoldberg1 -- it is most probably stress induced birefringence. Bragg scattering would not (at least to first order) change the polarisation of the incoming light beam, thus making the crossed polarisers pretty useless. Also, changes in colour due to Bragg scattering would be observed on reflection and would be angle ...

5

When light gets reflected from a dielectric surface (like the glass of your windshield), the two polarization components of the light don't get reflected by the same amount. The coefficients of reflections for both polarizations are called the Fresnel coefficients. More details here: http://en.wikipedia.org/wiki/Fresnel_coefficients According to these ...

5

Unpolarized light is not a superposition of polarized light (which would again be polarized) but a mixture of polarizations in all directions. It cannot be described by a single solution of the Maxwell equations. Instead one needs for proper modeling the statistical version of electrodynamics. See the book on quantum optics by Mandel and Wolf (its first ...

5

You are exactly correct in your first assertion: Each R stage in and RRR... sequence will diminish the light further. That's not because there is anything inherently different about circular basis states versus linear states, but because we cheat a bit in how we make circular polarizers, since they start with linear polarizers that are then followed by ...

5

Materials, and certainly materials transparent to light , have few magnetic properties. They are not composed out of atoms that have strong ferromagnetism. But all atoms have strong electric fields. This means that light, as it goes through a transparent medium has small probability to interact with its magnetic field component with the medium, which is ...

5

It sounds like your teacher's explanation might have been a little misleading. The reason sound can't be polarised is that it is a longitudinal wave, unlike light which is a transverse wave. (Those links have some animated diagrams that should help to make the difference clear.) "Transverse" means that if a beam of light is coming towards you, the ...

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