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Polarization of light

  1. Jul 28, 2010 #1
    "Polarisation" is defined as the cofinement of the vibrations of the wave in only one plane and the removal of the vibrations in the other perpendicualar plane of the electromagnetic wave. But the e.m. wave is defined as a wave in which the electic vetors are restricted in one plane only where as the magnetic vector is vibrating in the other plane, i.e,. the em wave is itself is polasrised only as per above definition.
    In that case what does further polaristion of the wave mean? Removing the vibrations in the other perpendicular plane means removing the magnetic vector? Also if we assume that the electric field has components in all possible directions and out of these vibrations, those in one plane are removed and those in a plane perpendicular to it is retained, will it lead to the conclusion that the electric field vector has components parallel to the magnetic field vector also i.e.in the x-z plane ( assume the electric field is vibrating in the x-y plane and the magnetic field in the x-z plane, x-axis being the direction of propagation)?
    If that is so, will it not violate the very definition of the em wave which says that these are the waves in which electric filed, magnetic field and the direction of propagation all are mutually perpendicular to each other? Will any body kindly clarify these points?
     
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  3. Jul 28, 2010 #2

    sophiecentaur

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    You are describing an already polarised em wave - e.g. what gets radiated from a dipole. The polarisation is commonly referred to the plane of the E field.
    In the case of 'unpolarised' light' there is a mixture of light from many sources (e.g. the different parts of a hot lamp filament) and there will be a mixture of polarisations (i.e. E field directions).
    Passing the light through a polariser will select the E field components in a particular plane.
    Trying to explain this in terms of photons gets a bit tricky because, if you think of each one as having a particular polarisation, then its 'amplitude' can't be less than hf. So you have to think of the polariser as making each photon 'choose' which of two orthogonal polarisations it's going to be - half get through and half don't.
     
  4. Jul 28, 2010 #3
    True, but that means only that at one instant of time they are like that. there is nothing to stop the E and B fields (still mutually at right angles) rotating around the direction of propagation like the hands on a clock stuck at right angles.
     
  5. Jul 28, 2010 #4

    Andy Resnick

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    No, electromagnetic polarization is defined as the *instantaneous* direction of the electric field vector. Propagating radiation can be fully polarized, partially polarized, or unpolarized (I prefer the term 'randomly polarized' to unpolarized), and is related to the coherence properties of the radiation. Some common polarization states are linear, circular, and elliptical. Less common are radial and tangential.

    Non-propagating modes (e.g. the near field) still have an instantaneous polarization, but the situation gets complicated rather quickly. A simplified version is used to analyze FRET and dipole-dipole interactions.
     
  6. Jul 29, 2010 #5

    sophiecentaur

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    No problem with that. If you take an electric dipole (which produces 'linearly' polarised em waves) and rotate it or twist is (in a plane containing its axis) the E and H field planes of the received em wave will rotate accordingly, once the disturbance has arrived at the observer.

    You can produce Circularly Polarised waves with two dipoles at right angles, fed in quadrature (90o) and offset by a quarter of a wavelength. The sum of the two waves, on axis, is a wave with the E field plane rotating once every cycle of oscillation. Alternatively, you could rotate a single dipole at the transmitted frequency (not very practical, of course) and produce the same thing.

    A random array of transmitting dipoles, each at different angles and fed with signals of near but not the same frequency, switched on and off at random would produce a radio wave which would be much the same as the light that comes from an incoherent light source.

    There is, of course, just one value of E field at any point in space, comprising E fields of all the em waves passing through that point. Radio receivers, eyes and all other em detectors just select the particular E field variation (signal) of interest to them.
     
  7. Jul 30, 2010 #6
    Even though the E field of linearly polarized light is restricted on a plane, it can still be divided in two weaker perpendicular E fields. You can always polarize "further", but this will correspondingly reduce intensity.

    You don't have to think about the M field. Think in terms of the E field. The M field simply follows perpendicularly.
     
  8. Jul 31, 2010 #7

    sophiecentaur

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    "further polarisation" just involves selecting the component of an already linearly polarised wave which is in a different plane to its E field - which will be E Cos(Θ), where Θ is the angle between the original E field and the polariser. The E Sin(Θ) component will either be reflected or absorbed by the polariser, which reduces the intensity.
     
  9. Aug 6, 2010 #8
    Thank you for all the answers.

    Since in an unpolarized e.m. wave the E-vector will vibrate in all possible directions in its own plane, shall we assume that it is also possible for some of its vibrations could be present parallel to B-vector in the latter plane?

    Also. in the case of optical activity, it is defined as a phenomenon in which the plane of polarization is rotated through some angle by the substance placed in the path of the polarized light coming from a crystal. I don't understand why it is always defined in terms of the rotation of the plane of polarization? Is it technically wrong to say that "plane of vibration is rotated" ?
    Since the plane of vibration is always at right angles to the plane of polarization, is not true that the plane of vibration is also rotated through the same angle as that of the former?
     
  10. Aug 6, 2010 #9

    sophiecentaur

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    The E is always normal to the H plane. If one changes angle then so does the other.

    No one seriously uses the term "vibration", which is a mechanical phenomenon and it is confusing. You can rotate the polarisation plane of radio waves, too, very easily. In fact it's easier than to rotate light polarisation; just a bit of wire can do it!

    Quote "Since the plane of vibration is always at right angles to the plane of polarization, is not true that the plane of vibration is also rotated through the same angle as that of the former?"
    I can't think what you mean by this. You are using the term 'vibration' in an undefined way (what vibration?) - and my earlier comment, about using the term, applies. It's always a good idea to use the terms in as strict a way as possible if you want to understand and be understood with as little ambiguity as possible.
     
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