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Intensity of Light by Polarizers

  1. Mar 1, 2007 #1
    1. The problem statement, all variables and given/known data

    Some biologically active molecules rotate the direction of polarization of linearly polarized light, either clockwise or anticlockwise. For example, 5 gm of l-leucine dissolved in 100 mL of water causes a rotation by 0.550o while the same concentration of d-glutamic acid causes a rotation of 0.620o. If these solutions are placed between crossed polarizers illuminated by a beam of unpolarized light of intensity I=13 W/m2, what is the intensity of light transmitted by:

    (a) the l-leucine solution? ___W/m^2
    (b) the d-glutamic acid solution? ___W/m^2

    2. Relevant equations


    3. The attempt at a solution

    To do this, for each solution, I plugged in the numbers in the equation:

    for l-leucine: I=(13 W/m^2)(cos^2(0.62))
    for d-glutamate: I=(13 W/m^2)(cos^2(0.55))

    That didn't work. So I then tried finding I by doing:
    I=(.5)(13)=6.75 then used this is the above equation and tried solving for Io then. That didn't work. A hint or suggestion would be greatly appreciated. Thank you!
  2. jcsd
  3. Mar 1, 2007 #2
    Hi deenuh,
    .55 degrees and .62 degrees are extremely small angles and the cosine of these angles squared will be much closer to one than the numbers which you are using.

    I hope that helps.
  4. Mar 2, 2007 #3
    Take a look at Malus Law again:
    Try to explain what I=Io*cos^2(theta) means.
    It helps if you first consider polarized light that is incident on a polarizer.

    What is theta?
    What is Io?
    Last edited: Mar 2, 2007
  5. Mar 2, 2007 #4
    Isn't Io equal to the beam of unpolarized light (in this case, 13 W/m^2)? Wouldn't theta be 45 degrees because the polarizers are crossed?
  6. Mar 2, 2007 #5
    Crossed filters are at a 90 degree angle. If the samples didnt rotate the polarisation of the light none would get through.

    You are going through two filters. First unpolarised light goes through a filter. Youve given the formula that tells us the intensity drop for this. Then this polarised light goes through a second filter which is 90 degrees to the first. Youve given the right formula for this as well you just need to think again about what is the right value for theta and Io for each sample. Theta is the angle between the plane of polarisation of the light and the plane of polarisation of the filter.
  7. Mar 2, 2007 #6
    Thanks for your suggestions, I was finally able to calculate the answer!
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