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What could cause a polarizer to not block 100% light?

  1. Jul 19, 2013 #1
    i have a cheap polarizing shade i bought from walmart. i aligned my polarizing shades with the polarized light coming from my laptop, however not all the light was blocked. just to be extra sure that it wasn't my lcd that was being weird, i rotated a laser beam through my polarizing shades, and still some light was able to get through. the amount of light reduced but not all of it was blocked. i don't understand. the polarizer did indeed decrease the amount of light going through it when aligned in the same direction as the E-field oscillation, however not all of the light was blocked. how can light not be blocked in this case? i held my polarizers up against the visible spectrum displayed on my lcd screen and it appeared to block the blue/violet end the least and the red/orange end the most. this implies that the space between each conducting material is not that small. However, my laser light was red and so this does not really explain why my laser light was able to get through.
     
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  3. Jul 19, 2013 #2

    Andy Resnick

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    The light emitted by your screen is not 100% polarized, and the extinction ratio of your polarizer is low.
     
  4. Jul 19, 2013 #3
    why wouldn't it be 100% polarized? in order for light to make it out of the screen it has to be aligned with the second polarizer.
     
  5. Jul 20, 2013 #4

    Baluncore

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    Linear Polarisation or circular polarisation ?
    Shades will be linear.
    Is the screen circular ?
     
  6. Jul 20, 2013 #5

    Claude Bile

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    It's common for polarizers to not block light with 100% efficiency.

    Claude.
     
  7. Jul 21, 2013 #6
    if light indeed IS a transverse wave of an electric field, and when the null point of a wave's E-field whose direction of oscillation is parallel to the conductive bars, hits the conductive bars, does it pass through?

    is this why it is not 100% efficient?
     
  8. Jul 22, 2013 #7

    sophiecentaur

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    If there is any cross-polar component in the output of your polariser then the analyser will let it through and you will never get a zero. How many things have you ever bought that were 'perfect'?
     
  9. Jul 22, 2013 #8
    what do you mean by a cross-polar compoenent in the output of a polariser? are you referring to the direction perpendicular to the polarization axis? and what analyzer are we talking about?
     
  10. Jul 23, 2013 #9

    sophiecentaur

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    I am talking Practical systems. When you either generate or 'select' a particular plane of polarisation (with Light or Radio Waves), there will be components of the Electric field that are not exactly in that plane. (It is always worth while doing a Google search when you come across an unfamiliar term; in this case it will produce many useful hits.)

    Light is produced by a huge number of atoms, emitting their photons independently. The resulting light will be unpolarised and consist of a finite range of frequencies. In a laser, the range of frequencies will still be finite (not zero). Pass this through a Polariser and you will select a range of angles (again, finite and not zero).

    Again, if you had looked in Google, you would have seen that the term 'Analyser' is commonly used to describe the second polariser in the system. (Used to 'analyse' the light). Any analyser will also have a finite range of selected polarisation angles. "Cross-polar" refers to the component of the nominally polarised wave that will be passed by the analyser when its orientated at right angles to the nominalplane of polarisation.

    Polarisation is a subject that is full of misapprehensions and fuzzy reasoning. In many ways, it is best approached from the direction of Radio Antennae and waves because the actual sources are easier to control. You can physically tilt a transmit and receive dipole and make a polariser with a screen of parallel wires. You can also treat the signals as coherent and from a single source (rather than a load of different atoms). Once you have sorted this out, you can more safely move over into optical polarisation and avoid a lot of common mistakes. There are some really glaring faults in the way the light polarisation is described in books because the vector nature of the fields is often side-tracked.
     
  11. Jul 26, 2013 #10
  12. Jul 26, 2013 #11

    sophiecentaur

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    In principle, EM of all frequencies is produced in the same way. The main difference is the charge systems which change energy levels to cause photon emission. RF photon energy is much lower and RF sources can normally be looked on as coherent, making it easier to get to grips with. It's a good idea, imo, not to dismiss RF in your learning journey.
    Which polarised did you want an explanation for? An RF dipole is a good start.
     
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