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Unpolarized Light

  1. Apr 4, 2009 #1
    PART 1: Unpolarized light has an intensity of 100 W/m^2. It passes though a polarizer whose transmission axis is rotated 18 degrees with respect to the vertical. What is the intensity of the light that emerges from the polarizer?

    I = (1/2)(Io)(cos angle)^2
    I = (1/2)(100 W/m^2)(cos 18)^2
    I = 45 W/m^2 -- I am not sure if this is correct.

    PART 2: This light passes through a second polarizer and the intensity of the light that emerges is 40 W/m^2. What angel does the transmission axis of the second polarizer make with the vertical?

    I don't understand this part. Would I use the same equation above and plug in 40 W/m^2 for I, 100 W/m^2 for Io, and solve for cos?

    Thank you!
     
  2. jcsd
  3. Apr 4, 2009 #2

    rl.bhat

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    PART 1: From a polarizer 50% of light will be transmitted irrespective of the orientation of the transmission axis.
    For part 2, use Malus's law.
     
  4. Apr 5, 2009 #3
    Thank you!
    For PART 1, would it just be I = (1/2)(100 W/m^2) then? So, I = 50 W/m^2 ?
    I still don't understand PART 2. What do I plug into Malus' Law?
     
  5. Apr 5, 2009 #4

    rl.bhat

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    I = Imax*cos^2(theta)
     
  6. Apr 5, 2009 #5
    Part 1 is okay then? The answer is 50?

    Part 2: 40 W/m^2 = 100 W/m^2 cos^2(theta)
    and then solve for cos?
     
  7. Apr 5, 2009 #6

    rl.bhat

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    No.
    For second polarizer Imax is 50 w/m^2
     
  8. Apr 5, 2009 #7
    Oh! Thank you so much for helping me. I'm sorry I'm a little confused.
    Would the answer be 27 degrees?

    I am not sure if I solved for the angle correctly. I'm forgetting my math skills here. haha
     
  9. Apr 5, 2009 #8

    rl.bhat

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    Part 2: 40 W/m^2 = 50 W/m^2 cos^2(theta)
    cos^2(theta) = 0.8
    Find cos theta. Add 18 degrees to this angle to get the result.
     
  10. Apr 5, 2009 #9
    cos^2(theta) = 0.8
    then take the sqrt of both sides?
    cos(theta) = 0.9
    cos-1(theta) = 26 degrees

    Angle: 26 + 18 = 44 degrees ... I hope. :redface:
     
  11. Apr 5, 2009 #10

    rl.bhat

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    Yes. Correct. You can take 26.6 as well.
     
  12. Apr 5, 2009 #11
    That was very helpful. Thanks, again!!
     
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