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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 ...


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 ...


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

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 ...


10

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


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

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

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 ...


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 ...


8

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

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 ...


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

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 ...


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

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 ...


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

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 ...


6

The confusion probably comes from the fact that E-fields and E-modes are entirely different (but are etymologically similar). A photon carries an intrinsic polarization. In classical E&M, we think of a single light wave as an oscillation of electric and magnetic fields. In vacuum, things are nice and simple, and the propagation direction, the electric ...


6

Many 3d glasses use circular polarization, where one lens uses left-hand polarization while the other uses right-hand polarization. This lets the viewer tilt their head a bit without losing the 3D effect, where linear polarization would let the image bleed through into the other eye.


5

There is no such mechanism. The probability for a photon to pass through a polarizer at an angle $\theta$ is $\cos^2(\theta)$, regardless of what has happened before, and regardless of how many photons "at once" try to pass through it. As Bell's theorem tells us, the quantum world is really random (or non-local).


5

The suggested experiment was performed by R. Beth already in 1936. In the experiment linearly polarized light was converted to circularly polarized light by means of a doubly refracting plate. The (macroscopic) reaction torque was measured and shown to conform with the angular momentum theory of the photon.



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