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Intensity of natural light after polarization

  1. May 31, 2014 #1
    I read that if a beam of unpolarized light goes through a polarizer, the intensity of the polarized beam is equal to half the intensity of the original beam. Can someone explain me why? I thought the intensity would be the same.
     
  2. jcsd
  3. May 31, 2014 #2

    DrClaude

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    What do you think the polarizer does?
     
  4. May 31, 2014 #3

    jtbell

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    Unpolarized light is a mixture of light with all the different polarization angles. Consider what happens to the part that is polarized parallel to the polarizer, the part that is polarized perpendicularly, and the parts that are polarized at various in-between angles.
     
  5. May 31, 2014 #4
    A polarizer works by blocking light that has the "wrong" kind of polarization while letting the "right" kind through.
     
  6. May 31, 2014 #5
    Don't all nonparallel waves get blocked? Why would intensity get cut in half and not more? If polarisation is rather narrow as I suppose it usually is, should then there be much less light left after filtering nonparallel waves out?
     
  7. May 31, 2014 #6

    jtbell

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    Have you ever put two polarizers in front of an unpolarized light source and rotated one of them?

    When they are parallel (θ = 0), you get maximum intensity. When they are perpendicular (θ = 90°), you get zero intensity (rather, almost zero because polarizers are never perfect). When they have an angle θ between them, the intensity is proportional to cos2 θ.
     
  8. Jun 1, 2014 #7
    That makes more sense as I see proportionality, but I still don't see why a single polarizer would cut intensity in half. I'd expect intensity would cut in half if the polarizer allowed for wide polarization from 0 to 45 degrees for example. I witnessed such experiments some 20 years ago, never done it myself.
     
  9. Jun 1, 2014 #8

    Matterwave

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    The law of Malus states that the intensity of light, initially I, which has polarization at an angle ##\theta## with respect to a polarizing material will be ##I\cos^2(\theta)## after it passes through the polarizer.

    Unpolarized light has all different polarizations at all different angles in equal amounts. So, we integrate the above expression over the possible theta values and then divide by the total range of integration to get the average value:

    $$I'=\frac{I}{2\pi}\int_0^{2\pi}\cos^2(\theta)d\theta=\frac{I}{2}$$

    Really, the result is just that the average value of ##\cos^2(\theta)## over one full oscillation is 1/2.
     
  10. Jun 1, 2014 #9
    I wish that made sense in my head, I'll check the internet to see how this Malus fellow came up with that. Thanks.
     
  11. Jun 1, 2014 #10
    If you have difficulty understanding integrals, try calculating the intensity of a few angles and averaging them. Use for instance the angles 30°, 60°, 90°, all the way to 360°. Make sure the angles are evenly spaced.
     
  12. Jun 1, 2014 #11
    I'm not sure whether it is integral or Malus law itself I have trouble with. I watched a few YouTube videos and I kind of understand what they are saying, but it doesn't explain it in terms how I visualize it in my head.

    Say we have 36 photons, each polarized at 10 degrees more than previous one: 0, 10, 20, 30... 330, 340, and 350 degrees. I'm not sure whether 0 and 180, or 90 and 270 degree polarization is the same, but it doesn't really matter.

    Now we pass them all through vertical polarizer, and I conclude only two of those 36 photons should pass, 0 and 180 degree ones. What am I missing?
     
  13. Jun 1, 2014 #12
    That's the wrong conclusion. The 0° and 180° photons pass with 100% probability. The 90° and 270° pass with 0% probability - that is they don't pass. The ones with angle θ pass with |cosθ|2 probability. Welcome to quantum mechanics.
     
    Last edited: Jun 1, 2014
  14. Jun 1, 2014 #13
    Thanks, this made it clear.
     
  15. Jun 1, 2014 #14

    Nugatory

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    The problem is that you are thinking of photons as little teeny objects that travel through space and have a definite polarization. It's natural to think of them that way because we call them "particles", and the popular meaning of that word is basically "little teeny object that has some definite properties".... But that is not what the word "particle" means in physics, and it's not how photons behave.

    To understand polarized photons, you have to start with the classical picture of light as electromagnetic waves. If you grovel through all the math of a polarized wave impinging on a polarizer, you'll be able to derive Malus's Law. If you don't want to do that, you can get a decent intuitive understanding by imagining a wave oscillating in one plane (its polarization) as it passes through a narrow slit (the polarizer). If the waves are oscillating in the same plane as the slit, the wave will pass through unimpeded; if the slit is perpendicular none will make it through; and for in-between angles you'll get in-between transmission.

    OK, that's the behavior when we think of light as an electromagnetic wave. Get that clear in your mind before you move on to photons....

    Photons only come into the picture when the light interacts with matter. No matter how spread out the beam of light is when it reaches its target, it always delivers all of its energy in discrete lumps to single points on the target; the amount of energy in each lump is related to the frequency/wavelength of the light according to ##E=h\nu##; and the probability of a lump being delivered to any point is proportional to the intensity of the electromagnetic radiation at that point. When one of these lumps of energy appears at the target, we say that "a photon hit there". This is a quantum-mechanical phenomenon with no classical explanation or analogue.

    But what this means is that if (for example) 5% of the light energy, in the form of classical electromagnetic waves, makes it through the polarizer, then 5% of the energy of the light beam will be delivered to the target in the form of photons. (The other 95% of the energy will show up as photons hitting the polarizer).
     
    Last edited: Jun 1, 2014
  16. Jun 1, 2014 #15
    Got it. Thank you.
     
  17. Jun 1, 2014 #16

    Born2bwire

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    The strictly classical physics answer is that you can represent a polarized electromagnetic wave as the superposition of two waves that are polarized at orthogonal angles with a phase difference between the two components. If we choose the axis of the polarizer as one of these directions, then the component parallel to the polarizer passes through while the orthogonal component is completely blocked. Thus, for a wave of any polarization, the amplitude of the wave after passing through will be the projection of the original wave onto the axis of the polarizer ([itex]cos(\theta)[/itex]). Intensity is going to be the square of the amplitude and we get the [itex]cos^2(\theta)[/itex] factor. If we assume unpolarized light, then we simply take the average over all 360 degrees and we find that it should be 1/2. That is, as MatterWave explained already above, [tex]\frac{1}{2\pi} \int_0^{2\pi} d\theta \cos^2(\theta) = \frac{1}{2}[/tex]
     
    Last edited: Jun 1, 2014
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