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Rotating Polarizer 90 Degrees With Only 10 Percent Loss

  1. Oct 27, 2011 #1
    1. The problem statement, all variables and given/known data
    Describe how to rotate the plane of polarization of a plane-polarized beam of light by $90^\circ$ and produce only a 10 percent loss in intensity using "perfect" polarizers.


    The solution says to use 24 polarizers, with each at an angle of $3.75^\circ$ with the next. How is this allowed by the problem statement which says you must rotate the plane of polarization?


    2. Relevant equations
    Polarization equation - [itex]\cos^2(\theta)[/itex].


    3. The attempt at a solution
    My solution was let the initial angle of polarization be $\theta$ so we have
    [itex]\frac{\cos^2(90+\theta)}{\cos^2(\theta)}=\frac{9}{10}[/itex]
    [itex]\frac{\cos(90-(-\theta))}{\cos(\theta)}=\frac{3}{\sqrt{10}}[/itex]
    [itex]\frac{\sin(-\theta)}{\cos(\theta)}=-\frac{\sin(\theta)}{\cos(\theta)}=\frac{3}{\sqrt{10}}[/itex]
    [itex]\tan(\theta)=\frac{\sqrt{3}}{10}[/itex]
    [itex]\theta\approx \boxed{43^\circ}[/itex]

    The solution says to use 24 polarizers, with each at an angle of [itex]3.75^\circ[/itex] with the next. How is this allowed by the problem statement which says you must rotate the plane of polarization? How should I interpret the problem?
     
  2. jcsd
  3. Oct 28, 2011 #2

    collinsmark

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    Polarizes rotate the plane of polarization. [Edit: in other words, if you have a light wave that's already polarized, you can rotate the plane of polarization of that light wave by using a polarizer. However, not all the light will make it past the polarizer, and the amount of light that does make it past depends on the relative difference between the light's polarization angle and the polarizer's polarization angle. My point is that ignoring intensity losses for the moment, you can use polarizers to change the polarization angle of the light.]

    Here are some examples. To start, assume that the beam light is linearly polarized at an angle of 0o, and you have as many linear polarizers as you wish. The goal is to rotate the polarization angle of the beam of light to 90o, after it passes through the polarizer.

    Let's start with 1 polarizer. Since it's the only polarizer, we must rotate the polarizer to an angle of 90o. The intensity ratio is
    [tex] \frac{I}{I_0} = \cos ^2 \left( 90 ^{\circ} \right) = 0. [/tex]
    Well, that didn't let any light through at all. Obviously we'll need more polarizers. Let's try two of them, each rotated at 45o. [Edit: by that I mean each is rotated at 45o with respect to the previous. So the first is at the absolute angle 45o, and the second at the absolute angle 90o.]

    After the first polarizer, half of the light gets through, per [itex] \cos ^2 \left( 45 \deg \right) = 0.5 [/itex], and the polarization of the light is 45o. But unfortunately, only half of that light gets through the second polarizer when it polarization is the 90o. The intensity ratio is now (after the last polarizer) only 0.25. But at least some light is getting through. Time to use more polarizers.

    Let's try three, each rotated by 30o [Edit: I mean each rotated by 30o with respect to the previous. I.e., 30o, 60o, 90o in absolute angles.] The intensity ratio after the first polarizer is 0.75. The final intensity ratio comes out to be (0.75)(0.75)(0.75) = 0.421875. 'Lookin better at least.

    Based on this, can you derive a formula which relates the final intensity ratio as a function of the number of polarizers N, with each polarizer at an angle of 90o/N with respect to the previous polarizer?

    If I'm not mistaken, the final formula is a transcendental function. You won't be able to solve for N algebraically. But you can keep plugging numbers in for N until the final intensity ratio rises above 0.9. (Try to find the smallest natural number N that causes the final intensity ratio to be above 0.9. Use trial and error.)
     
    Last edited: Oct 28, 2011
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