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Homework Help: Linear Polarizers and Light Intensity

  1. Dec 6, 2013 #1
    1. The problem statement, all variables and given/known data

    Initially unpolarized light is shining on a series of six polarizers in a row. Each polarizer is rotated by the same angle, θ, relative to the previous polarizer. Choose a fraction (anything less than 1) as your final intensity.

    What is the angle, θ, such that the final intensity(Sout) equals the initial intensity(Sin) that you chose? Give your answer in degrees.

    f = chosen fraction = 1/6

    2. Relevant equations

    Sout = Sin*cos^2θ

    3. The attempt at a solution

    I think that the intensity after each polarizer can be found using the equation Sout = Sin*f. I think I might be more confused about the wording of the problem than anything, but wanted to post on here to see if anyone had any insight for me. Thank you!
  2. jcsd
  3. Dec 6, 2013 #2
    No the intensity after each polarizer can be found using the equation you stated in the "relevant equations" section. The intensity after all 6 polarizers is Sout = Sin*f . Can you make any progress?
  4. Dec 6, 2013 #3
    So do I need to use the equation I posted at all? Or is Sout=Sin*f enough to figure out the answer?
  5. Dec 6, 2013 #4
    You need both equations. Sout=Sin*f tells you how the intensity changes after all 6 polarisers, it's easier if you write it as S6=S0*f where S6 is the intensity after passing through 6 polarisers and S0 is the initial intensity (0 polarisers). The other equation can be written as S(n+1) = S(n)*cos^2θ i.e. the the intensity change after passing through any ONE polariser. Can you set up a way to solve the problem now?
  6. Dec 6, 2013 #5
    I just can't seem to put it all together. I've tried countless different values for theta using the S(n+1) = S(n)cos^2 equation. I guess I just don't get how the formulas relate to give me what I'm looking for.
  7. Dec 6, 2013 #6
    Ok. Every polariser is rotated at the same angle relative to the previous one so,

    (S(n+1)/Sn) = (cos(θ))^2 = constant = c

    because theta is the same for them all.

    So S(6) = c S(5) = c c S(4) = c^2 cS(3) ....

    Does that make sense? Can you now find an expression for S(6) in terms of S(0)?
  8. Dec 6, 2013 #7
    Okay I think I'm getting it. So would the relationship between them would be: S6=(costheta^2)^6*S0? And if my thinking is correct, S6 should be equivalent to S0? Leaving me with (costheta^2)^6=1?
  9. Dec 6, 2013 #8
    This is right: S6=(costheta^2)^6*S0

    But S6 only equals S0 if costheta^2 = 1 i.e the angle between them is 0.

    But you have chosen that S6 = 1/6 * S0
    Last edited: Dec 6, 2013
  10. Dec 6, 2013 #9
    Got it thank you
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