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Linearly polarized light homework problem

  1. Sep 14, 2005 #1
    You need to rotate the polarization direction of some linearly polarized light by an angle φ. That is, the light is initially linearly polarized along some direction (let's call it the y direction), and you need to change it so that the polarization direction is φ away from the y direction.
    In order to do this, you have two polarizers available. Prove that in order to get the maximum final intensity, you should choose to orient first polarizer at an angle φ=θ/2 from the initial polarization direction. (Note that to achieve the desired final polarization, you have no choice about how to orient the second polarizer.)

    Anyone care to take a stab at this? im unsure of how to go about proving this. Any help to point me in the right direction would be appriciated, thanks.
  2. jcsd
  3. Sep 14, 2005 #2
    See if you can find a formula somewhere (I won't give it away so soon!) that gives you the percentage of light transmitted through a polarizing filter as a function of its angle relative to the polarization direction of the incident light beam. That should prove quite useful in answering the question. If you need an extra hint, just ask. Good luck!
  4. Sep 14, 2005 #3
    yes the intensity I=I*cos^2(x), i still do not see the steps used to prove the given true.
  5. Sep 14, 2005 #4
    Right, this will involve some trickery with trigonometry formulas. Here goes.

    Suppose we call the deflection angle of the first filter [tex]\theta[/tex] and the deflection angle of the second filter relative to the first filter [tex]\varphi - \theta[/tex], so they add up to [tex]\varphi[/tex]. Now we can set up a formula describing the intensity of the light beam that has passed through both polarisation filters. The intensity of the beam as it has passed through the first filter is:

    [tex]I_{1} = I_{0} \cdot \cos^{2}(\theta)[/tex].

    For the second filter we get the equation:

    [tex]I_{2} = I_{1} \cdot \cos^{2}(\varphi - \theta)[/tex].

    Combining them yields:

    [tex]I_{2} = I_{0} \cdot \cos^{2}(\theta) \cdot \cos^{2}(\varphi - \theta)[/tex].

    We need [tex]I_{2}[/tex] to be a maximum by choosing a suitable [tex]\theta[/tex]. Because [tex]0 \le \varphi \le \pi / 2[/tex] (can you see why?), this amounts to maximising [tex]\cos^{2}(\theta) \cdot \cos^{2}(\varphi - \theta)[/tex] on the interval [tex]0 \le \varphi \le \pi / 2[/tex]. This maximum can be found in several different ways. I'll just differentiate the whole lot here and set the result to zero to get:

    [tex]-2 \cos(\theta)\sin(\theta)\cdot \cos^{2}(\varphi - \theta) + 2 \cos(\varphi - \theta) \sin(\varphi - \theta)\cdot \cos^{2}(\theta) = 0[/tex]

    After a lot of rewriting (I'll let you check it) we get:

    [tex]\tan(\theta) = \tan(\varphi - \theta)[/tex].

    The only possibility for these tangents to be equal is for their arguments to be equal. This means that [tex]\varphi = 2 \theta[/tex], and thus that both relative deviation angles used in the setup must be equal.

    Phew, I hope I didn't leave a typo in there.
    Last edited: Sep 14, 2005
  6. Sep 14, 2005 #5
    Thanks a bunch. yea that makes perfect sense.
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