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Light through 3 linear polarizers

  1. Jan 21, 2013 #1
    1. The problem statement, all variables and given/known data

    What is the most light that can be transmitted through 3 linear polarizers if you start with light polarized in the (i unit vector) direction and end with light polarized in the (j unit vector) direction for theoretically perfect polarizers

    2. Relevant equations and attempt at solution.

    Well I reason since the light is already polarized, I need to use the cosine squared law

    I=IoCos^2∅ Where I is the intensity after going through a polarizer and Io is the initial intensity.

    So the light in already polarized through the i direction, here in what I am thinking, If I send this light through 2 polarizers who's filters are parallel with the light in the i direction, then then angle is zero and all the light is retained.

    But when I send it through a third filter in the j direction, as required in the problem, that is perpendicular to i of course, then no light emerges.

    I know these are theoretically perfect polarizers but my result seems counterintuitive.

    Is this right?
    Last edited by a moderator: Jan 21, 2013
  2. jcsd
  3. Jan 21, 2013 #2
    You can also put the second polarizer at a 45° angle to the first polarizer, and then some light will emerge.
  4. Jan 21, 2013 #3


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    Once you decide on what the orientation of the first and last filters must be, then there is only one variable to adjust. Write the final intensity as a function of that variable and see if you can maximize the function.
  5. Jan 21, 2013 #4
    Yes that was the one thing I was wondering, if I put the second polarizer at 45 degrees.

    Could you please help me with maximizing the function? There is definitely only 1 variable, which is the second polarizer, but I am not sure how to maximize it.

    I am also having trouble visualizing if it at the maximum at 45 degrees, I would like to see it mathematically but I'm not sure now.
  6. Jan 21, 2013 #5
    To find the maximum of a function, take the derivative and set it to zero. So, write an equation for the light passing through the filters, leaving the angle variable only for the middle filter, take the derivative of that function, and solve for the unknown. If you're stuck, post the derivation as far as you can go.
  7. Jan 21, 2013 #6


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    Let θ be the angle between the transmission directions of the first and second filters. If Io is the original intensity before the first filter, then you know that all the light gets through the first filter so the intensity is still Io before entering the 2nd filter.

    How would you write the intensity just after the 2nd filter?

    What would be the direction of the polarization of the light after the 2nd filter?

    Just before the light strikes the 3rd filter, what is the angle between the polarization direction of the light and the transmission axis of the 3rd filter?
  8. Jan 21, 2013 #7
    See if I use the equation I=IoCos(∅), If I set the equation equal to zero (IoCos∅) and take the inverse cosine of both sides, it's saying that the angle is at a maximum at zero. That is where I get to in that derivation.

    I'm not sure how I would write the intensity after the second filter, I know that the incident light will be reduced, so I could write it as Io=Io2Cos(∅2), and similarly for the third filter after the light passes through.

    I understand this, it just seems that 45 degrees is the middle ground for the i and j directions of that middle filter the more I think about it, but I am having trouble representing it mathematically.

    Then if it is 45 degrees between the third filter and and the incident light, then the light is again reduced in intensity, but I am unsure of that direction as well, maybe (∅ - Pi/4).

    I think I understand the problem intuitively but I am obviously having trouble representing it mathematically.
  9. Jan 21, 2013 #8


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    This is what I get from what you wrote:
    $$I=I_0\cos\theta = 0$$
    Then you take the inverse cosine of both sides:
    $$\arccos(I_0\cos\theta) = \arccos 0$$
    Therefore, ##\theta=0##. It doesn't make sense. What did you actually do?

    By the way, your expression for the intensity is wrong. You had it right in your original post: ##I = I_0 \cos^2 \theta##.
  10. Jan 21, 2013 #9
    Yes, that's right, I forgot the square on the cosine,

    So that is exactly what I did, I'm not sure how to find the maxium of the cosine, I just, okay, you're right, what I did does make no sense. This is what I meant to write

    As TSny and opaka had said, I should have taken the derivative of the equation,


    So then setting IoCos^2(θ)=0

    And taking the derivative, I get -2Iocosθsinθ=0

    I think I can write this as -Iosin2θ=0

    That is about as far as I can get here.
  11. Jan 21, 2013 #10
    yeah thanks everybody for all the help so far, I think i understand the problem, I just need to see the math.
  12. Jan 21, 2013 #11


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    When finding the extrema, you don't set I=0. You set the derivative I' equal to 0. There's no reason to write that last line.

    That's right, and sine vanishes at 0 and 180 degree, right? That means the extrema occur at θ=0° and θ=90°. If you plug these values back into the original expression for I, you get I=I0 for θ=0°, which makes sense. If the angle between the polarization of the light and the axis of the polarizer is 0, all the light gets through. For θ=90°, you find I=0. When the two are perpendicular to each other, no light gets through.

    So you've derived what you already knew about how to maximize or minimize the light coming through one polarizer, but in this problem, you have two polarizers. (The first one doesn't matter because it lets all the light through.) So reread TSny's last post and figure out the answers to the questions asked. You want to find a function I(θ) that tells you how much light gets through to the end. Once you have that, maximize it.
  13. Jan 21, 2013 #12


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    Look at the attached figure which assumes the light is originally polarized horizontally. Go step by step. The green lines show the transmission axis direction for each filter. You have already figured out how the first and last filters must be oriented. So, you just need to find the angle θ for the middle filter such that you get maximum transmission overall.

    How would you express I1 in terms of Io?

    How would you express I2 in terms of I1 and θ?

    How would you express I3 in terms of I2 and θ? Here you will need to think about how to express the angle shown with the question mark in terms of θ.

    Attached Files:

  14. Jan 21, 2013 #13
    God thank you guys so much. I still am really unsure of how to formulate this mathematically, but I am going to read over polarizers and see if I can bring any new information to the discussion.
  15. Jan 22, 2013 #14
    I don't understand why this was moved here, this is a pretty hard problem in a 500 level optics course at my school, which is a 3rd year optics course. Perhaps because we are doing the easier stuff from the intro books first, although really fast, before we get into physical optics, eh.

    So anyways I thought about this for awhile and this is what i've got

    I = Io1Cos(0) = Io1

    Then the second polarizer must be positioned at a 45 degree angle relative to the first and the third, which is in the j direction, so

    Io2 = Io1Cos(45) = [(2)^1/2]/(2)

    But then the next angle is 90, so I think I have two parts of the problem represented properly, but of course the cosine of 90 is zero, so the math there says no light comes through, which doesn't make sense.

  16. Jan 22, 2013 #15


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    Don't forget to square the cosine functions.

    The last angle is not 90o. What is the direction of polarization of the light just before it reaches the last filter? What is the angle between that polarization direction and the transmission axis direction of the last filter?
  17. Jan 22, 2013 #16
    Last edited: Jan 23, 2013
  18. Jan 22, 2013 #17
    The angle between the polarization filter and the polarized light is 60 degrees. I'm not sure what my problem is.
  19. Jan 23, 2013 #18


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    Not sure how you got 60o. Remember, when light passes through a polarizing filter, the light comes out with its polarization direction the same as the direction of the transmission axis of the filter. So, if you set the middle filter at 45o from the horizontal, then after the light passes through the filter the light will be polarized at 45o to the horizontal. It then proceeds to the last filter (which has its transmission axis vertical). So, what is the angle between the transmission axis of the last filter and the polarization direction of the light that strikes the last filter?
  20. Jan 23, 2013 #19
    45 degrees, I think.
  21. Jan 23, 2013 #20
    I am going to do all the problems in the book until i figure this out, i'm really determined not to be this bad at physics.
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