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Optics - Three polarizers with 1 moving

  1. Jul 2, 2014 #1

    DataGG

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    1. The problem statement, all variables and given/known data
    Thee linear polarizers are in sequence. Let first and last polarizer be crossed (perpendicular to each-other) and the middle polarizer rotate with angular frequency ##\omega##. Show that under such circumstances, ##I## is given by: $$I = I_0 \frac{1}{16}(1-\cos (4\omega t))$$.

    The light that strikes the first polarizer is unpolarized.

    Please, see the following link (yes, my paint skills are horrible! I tried to make it in inkspace, but I don't know how to work with that program yet!): https://imgur.com/UwQShhx

    2. Relevant equations

    Malus Law, which is given by ##I(\theta) = I_0 \cos^2 (\theta)##

    3. The attempt at a solution

    If you saw the following link, you noticed two angles, ##\theta## and ##\alpha##.

    I think ##\theta=\omega t ## and ##\alpha = 90 - \omega t##

    So, after the first polarizer:

    $$I_1= \frac{I_0}{2}$$

    After the second polarizer:

    $$I_2= I_1 \cos^2 (\omega t)$$

    After the third polarizer:

    $$I_3 = I_2 \cos^2 (90-\omega t) \Leftrightarrow I_3 = I_1 \cos^2 (\omega t) \cos^2 (90-\omega t) $$

    I can play around with that but I never arrive at the ##I = I_0 \frac{1}{16}(1-\cos (4\omega t))##.

    Any ideas?
     
    Last edited: Jul 2, 2014
  2. jcsd
  3. Jul 2, 2014 #2

    Doc Al

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    You need to monkey around with some trig identities. Step 1: Get rid of that ##90 - \omega t## term. (Get everything in terms of just ##\omega t ##.) Then you are just two identities away from the answer.
     
  4. Jul 2, 2014 #3

    DataGG

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    Hey Doc, thx for answering.

    I did do that in paper, which I'll copy to latex below:

    $$ I_3 = I_1 \cos^2 (\omega t) \cos^2 (90-\omega t) = I_1 \cos^2 (\omega t)[\cos (90) \cos (\omega t) - \sin (90) \sin (\omega t)]^2 = I_1 cos^2 (\omega t) (-sin^2 (\omega t)) $$

    Oh, btw, I forgot to say that the light that strikes the first polarized is unpolarized. Will edit OP.

    EDIT: Sorry, that latex line is long. Kind of ruins the forum template for me. Don't know how to make it so the equal signs are aligned in different lines..
     
    Last edited: Jul 2, 2014
  5. Jul 2, 2014 #4

    Doc Al

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    OK, except for a sign error.

    Keep going.
     
  6. Jul 2, 2014 #5

    DataGG

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    Indeed, there is a sign error.

    $$I_3 = I_1 cos^2 (\omega t) (sin^2 (\omega t)) = \frac{I_0}{2} \cos^2 (\omega t) [1-cos^2(\omega t)] = \frac{I_0}{2} [\frac{1}{2}(1+ \cos (2 \omega t)][1-\frac{1}{2}(1+ \cos (2 \omega t)] $$

    Then comes

    $$I_3= \frac{I_0}{4}(1+ \cos (2 \omega t)) (\frac{1}{2}(1- \cos (2 \omega t)) = \frac{I_0}{8} (1 - cos^2 (2 \omega t))$$

    And now I'm not sure what to do.. Maybe expand the square? I'm not seeing anything..

    EDIT: Actually, expanding the square might probably work. I'll then have another co-sin term squared, and If I do the substitution I used above, a 1/16 will appear and.. ya, let me try it on paper..
     
    Last edited: Jul 2, 2014
  7. Jul 2, 2014 #6

    Doc Al

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    You have another sign error: When you express ##\cos^2 \omega t## in terms of ##\cos 2\omega t##. Fix that and you're almost home.
     
  8. Jul 2, 2014 #7

    DataGG

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    I made a sign mistake there, again! Will correct asap

    edit: damn, you were faster!

    edit2: Fixed the mistake, I think.
     
    Last edited: Jul 2, 2014
  9. Jul 2, 2014 #8

    Doc Al

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    Believe it or not, you still have a sign error. In your second line.
     
  10. Jul 2, 2014 #9

    DataGG

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    humm
     
  11. Jul 2, 2014 #10

    Doc Al

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    Consider the very last factor in your first line, which is correct. When you simplified that factor in the second line, you messed up a bit.
     
  12. Jul 2, 2014 #11

    DataGG

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    I'm an idiot! Alright, I now finally got it. Edited the post and below is the solution:

    $$I_3 = \frac {I_0}{8}[1- \cos^2 (2 \omega t)] = \frac {I_0}{8} (1 - \frac {1}{2} - \frac {\cos (4 \omega t}{2})$$

    and then finally,

    $$I_3 = \frac {I_0}{16} (1- \cos (4 \omega t))$$

    Thank you Doc Al!
     
  13. Jul 2, 2014 #12

    Doc Al

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    :thumbs: (And you are welcome!)
     
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