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If an electromagnetic wave is linearly polarized, the intensity of the light that goes through a polaroid is proportional to the square of the cosine of the angle between the polarization plane and the axis of the polaroid (Malus'law). This is because we consider only the component along the axis of the polaroid. If the wave is circularly polarized, the electric field moves on a circle, but on average, if we decompose it in a component along and perpendicular to the axis of the polaroid, these two components will be equal, so that the intensity will be $I_0/2$, where $I_0$ is the intensity of the wave before going through the polaroid.

Now, if the electric field varies on an arbitrary curve and we know the parametric equation of this curve (for example, it rotates around an ellipse), given a particular direction of the axis of the polaroid, how can we compute the intensity of the resulting wave?

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    $\begingroup$ On an arbitrary curve? That would be impossible to start with. An ellipse is rather readily decomposed into a superposition of phase-shifted linear waves: for every ellipse in the $(x,y)$ plane, there exists an orthonormal basis $(x',y')$ such that the ellipse is parameterized by $(\tfrac{x'}{r_{x'}})^2+(\tfrac{y'}{r_{y'}})^2=1$. In that basis, the light wave is simply a superposition of a linear wave polarized in $x'$-derection at magnitude $r_{x'}$ and one in $y'$ direction with magnitude $r_{y'}$. $\endgroup$
    – leftaroundabout
    Commented Feb 23, 2012 at 0:13
  • $\begingroup$ Ok, it's the ellipse I am mostly interested in. Can I treat the two linear waves independently, use Malus'law for each of them and simply add the resulting intensities? Another question: suppose to plot in a polar graph the intensity of an elliptically polarized wave (as a function of the angle, you plot the intensity measured by an oscilloscope). In this case, will I get again an ellipse? The same ellipse of the electric field? (apart from rotations and dilations) $\endgroup$
    – quark1245
    Commented Feb 23, 2012 at 8:53
  • $\begingroup$ Yes, the two waves can be treated independently, And since they're orthogonal the intensities will indeed add. I don't understand that other question of yours, what angle do you mean? $\endgroup$
    – leftaroundabout
    Commented Feb 23, 2012 at 10:57
  • $\begingroup$ I'll try to explain myself better. Suppose you have some elliptically polarized light. You also have a polaroid, that is used as analyzer. You vary the angle of the polaroid with respect to the plane of polarization of the incident light. Behind the polaroid the intensity of the light going through is measured by a photodiode. I suppose that, since the electric field goes around an ellipse, when the polaroid is oriented along the major axis the intensity is big; when it is oriented along the minor axis it is less. $\endgroup$
    – quark1245
    Commented Feb 23, 2012 at 14:27
  • $\begingroup$ So, I suspect that if you plot the intensity as a polar plot of the angle you get again an ellipse: I mean you plot data of this type $(r,\phi)$=(intensity, angle of orientation of the polaroid). Is this true? And what relation is there between this ellipse and the one of the electric field? $\endgroup$
    – quark1245
    Commented Feb 23, 2012 at 14:28

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A standart way of computation is to use Jones calculus. In this formalism complex amplitude of the electric field is expressed as a two-component vector: $$\left|\mathcal{E}\right>=\left( \begin{array}{c} \mathcal{E_x}\\ \mathcal{E_y}\\ \end{array} \right)= \left|\mathcal{E}\right|\left( \begin{array}{c} \mathcal{a}\\ \mathcal{\sqrt{1-a^2}e^{i\varphi}}\\ \end{array} \right),$$ whith $0\leq a\leq1$ (usually one uses normalized Jones vectors, but let's leave it as is). Any polarization-sensitive optical element is described by some operator acting on Jones vectors. A polarizer, rotated at angle $\alpha$ with respect to $x$ axis is essintially a projector on $\left|P_\alpha\right>=\left(\cos{\alpha},\sin{\alpha}\right)^T$, and the intensity is modulus squared of the inner product $I=\left|\left<P_\alpha|\mathcal{E}\right>\right|^2$. Simple calculation leads to the following result: $$I(\alpha)=\left|\mathcal{E}\right|^2\left(a^2\cos^2\alpha+(1-a^2)\sin^2\alpha+a\sqrt{1-a^2}\sin 2\alpha \cos \varphi\right),$$ which is in general not an ellipse. As an example, that is how it looks like for $a=1/\sqrt{2},\varphi=\pi/3$: enter image description here

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  • $\begingroup$ Thank you! In the mean time I had figured that out, but nonetheless good answer. $\endgroup$
    – quark1245
    Commented Jul 20, 2012 at 22:32

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