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22

The problem with the suggestion of using polarization is that you now have the reflections off the polarizers to contend with. I think the short answer is "it depends on how 'black' you want it to be". "Truly black" = reflectance of 0. I am quite sure that is impossible - there will always be some probability of light scattering off a surface. All you can ...


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 ...


15

The video you show is of a liquid crystal display (LCD) monitor, also known as 'TFT'. The 'trick' depends on the specific design characteristics of these monitors (described below) and will not generally work with other types of displays such as LED and PLASMA displays. All liquid crystal displays (LCD) operate on the principle of being able to 'twist' ...


13

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 ...


12

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 ...


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

Magnet NNNNNNN SSSSSSS break into NNNN NNN SSSS SSS and is put together NNNNSSS SSSSNNN


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

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 ...


10

You might 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 ...


10

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 ...


9

With any simple harmonic oscillator there are two quantities we are interested in, the phase and the amplitude. Complex numbers are an easy way to represent both of these in a single value, especially as a complex number can be written in the form $Ae^{i\theta}$ where $A$ is the amplitude and $\theta$ is the phase. This doesn't mean light has some ...


9

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.


9

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 ...


9

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 ...


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 use of anything but properly designed sunglasses is very foolish and poses great risks to your long term sight, and maybe for reasons that many people do not wholly appreciate. First of all, let's write down what unpolarised light is. We choose two basis polarisation states: let's go with left and right polarised light in this case since you say that 3D ...


8

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 ...


7

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 ...


7

I generally agree with the above answer. But I can explain the same by simpler words IMHO. Linear polarization may be expressed as a superposition of two opposite circular polarizations. And the polarization direction depends on the phase difference of the two circular polarizations. In simple words: imagine a situation where you add two vectors that rotate ...


7

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 ...


7

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

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 ...


6

This link: http://alienryderflex.com/polarizer/ has an excellent explanation; much better than anything I could write here. Essentially, it says that this occurs because the 45 degree filter outputs a projection of the vertical rays at 45 degrees. This, in turn, has a horizontal component, which the final filter projects in its output.


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 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 ...



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