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Find angle of polarizing sheet given original & transmitted intensity

  1. Oct 5, 2013 #1
    1. The problem statement, all variables and given/known data

    A beam of polarized light is sent through a system of two polarizing sheets. Relative to the polarization direction of that incident light, the polarizing directions of the sheets are at angles θ for the first sheet and 90° for the second sheet. If 0.10 of the incident intensity is transmitted by the two sheets, what is θ?

    2. Relevant equations

    I1 = Io /2

    I2 = I1 * cos^2(θ)

    3. The attempt at a solution

    The intensity of the beam after the first polarizing sheet is as follows:

    I1 = Io /2

    The intensity of the beam after the second polarizing sheet is as follows:

    I2 = I1 * cos^2(θ)

    Where θ is the angle difference between θ1 and θ2.

    The problem tells you that the transmitted beam is 0.10 of the original beam:

    I2 = 0.10 * Io

    Substituting with the equations above:

    I2 = 0.10 * Io = (Io/2) * cos^2(θ) and therefore:

    0.10 = cos^2(θ)/2

    Solving for θ:

    θ =arccos([itex]\sqrt{0.2}[/itex])= 63°

    Which means the angle of the first polarizing sheet is 63° out from 90° so the answer I get is:

    27° or 153°

    But this isn't correct according to the back of the book. Please help. Thanks in advance, MrMoose
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 5, 2013 #2

    haruspex

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    That would be true if the original beam were unpolarized, but it isn't.
     
  4. Oct 5, 2013 #3
    Thank you! I read that question over so many times and missed that part every time. I think I got it now:

    I1 = Io*cos^2(θ1)

    I2 = I1*cos^2(θ2)

    Where θ2 = 90 - θ1

    We also know that I2 = 0.10 * Io, therefore:

    I2 = 0.1 * Io = Io * cos^2(θ1) * cos^2(90-θ1)

    Io cancels out and you have:

    0.1 = cos^2(θ1) * cos^2(90-θ1) = (cos(θ1) * cos(90-θ1))^2

    Had to look up how to combine cosine functions for this next part:

    2*Sqrt(0.1) = cos(2*θ1- 90) + cos(90)

    Solving for θ1

    θ1 = (arccos(2*sqrt(0.1)) + 90) / 2

    θ1 = 70.38° which is one of the correct answers.

    My final question is this: The book also says that 20° is correct. How do they get this solution?
     
  5. Oct 6, 2013 #4

    haruspex

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    Whenever you execute an inverse trig function you need to be aware that there may be multiple valid answers. In this case, consider that cos x = cos -x.
     
  6. Oct 6, 2013 #5
    Great, Thanks Haruspex.

    So when I consider the following:

    θ1 = (arccos(-2*sqrt(0.1)) + 90) / 2

    I get θ1 = 109.62° , which is also 19.62° off the normal. That wraps it up!
     
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