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Intensity of light vs amplitude

  1. Jun 24, 2010 #1
    Hi all,

    It is common knowledge that unpolarized light, when passing through an ideal polarizer, suffers a drop of half its original intensity.

    But since intensity is proportional to square of the amplitude, can we then infer to say that the new amplitude of light through the polarizer is now 1/[tex]\sqrt{}2[/tex] of original amplitude?

    Thanks in advance. Physicsforums rock!
     
  2. jcsd
  3. Jun 24, 2010 #2
    No,you can't.
    The amplitude is the same,provided that the polarizer does not absorb.
     
  4. Jun 25, 2010 #3
    Hi netheril,

    I don't quite understnad your reply. If amplitude does not change, how come intensity is reduced by half after passing through the polarizer? (My understanding is that the ratio of 2 intensities - i.e. before or after passing through polarizer intensities - is equal to square of the corresponding 2 amplitudes.)

    Fellow forummers, pls advise.
     
  5. Jun 25, 2010 #4
    The amplitude does not change as light passes through a polarizer. The intensity reduces because not all polarizations survive.
     
  6. Jun 25, 2010 #5

    sophiecentaur

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    The OP is totally correct.
    The component of the E field in the polarised direction is the same and the component at right angles is zero. Result, overall, is half power, when you add up all the random contributions to the incident light.
    Post 4 is not really correct in using the term "survive" it's a matter of selection of vector components then integrating the result.
     
  7. Jun 25, 2010 #6
    In fact,my previous answer is not so correct.

    There are two ways to treat unpolarized light:

    One is to regard it as a combination of two incoherent light whose polarization directions are orthogonal with same amplitude A.Because they are not coherent,the total intensity is the sum of the intensity of the two components,that is, 2A2
    Having passed through a polarizer,the light orthogonal to the direction of the polarized disappears.Thus the amplitude is the same while the intensity is reduced to A2,half the original.

    But what happens in reality is that the unpolarized light consists of light of different direction in a whole circle.They are not coherent either.Having passed through a polarizer,the light vector will turn into its projection over the polarizer direction.So the amplitude of different components are no longer the same--They are now distributed over (0,A).Integrating their intensity and you will get half the original.
     
  8. Jun 26, 2010 #7
    Thank you all for the wonderful replies. Very much appreciated.

    So can i summarrize my own understanding as shown (pls correct me if there are any mistakes):

    1. Upolarized light passing through a single polarizer. Amplitude the same. But intensity is halved.

    2. Polarized light passing through a polarizer (tilted at an angle "theta"). Amplitude will be Acos(theta). But intensity will be Icos(theta) squared. [This is also known as Malus law]

    Am i right?
     
  9. Jun 30, 2010 #8
    Hi forummers,

    Can kindly please enlighten me on the post before this?

    Thanks!

    ( A high school physics teacher here) - Someone told me this is a good way to solicit response from the forum :tongue2:
     
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