One can scan the intensity $I$ of incoming linear polarized light by a linear polarizer, the outcome is the well known Malus' law.

$$I = I_0 \cos^2 \Theta$$

Deriving the dependence $I(\Theta)$ is rather easy, when going one step back to the electric field component and the relationship $I \sim |E|^2$.

Now, I was wondering, what happens if I scan circular or even elliptically polarized light by this method. The former should bring $I(\Theta) = const.$, while the latter should bring a more complicated expression. Especially the latter is of interest because linear and circular polarizations are just a special case of generally elliptical polarization. Now, if one could find a dependence $I(\Theta,\phi)$, with $\phi$ being the phasedifference between two orthogonal polarized beams (for simplicity (surely depending on the reader), but I'm thinking of linear x- and y-polarized light), one could easily alter Malus' law for the more general case of elliptical polarization.

Finally I'm interested in $I(\Theta,\phi)$ with $\phi$ being the phase difference between two orthogonal polarized beams and $\Theta$ being the scan angle. Can anyone help me to figure out the right derivation?


1 Answer 1


This isn't enough information to fully specify the state of the light, since you have yet to specify the relative strength between the two components. It seems, however, that you're thinking of light composed of a superposition of $x$- and $y$-polarized light, and that you want to have the ellipse's axes line up with the coordinate frame, but that constrains the relative phase between the $x$ and $y$ components to be $\pi/2$.

(If you don't do this, then you will have an ellipse with axes at some arbitrary angle, which introduces some gratuitous complications to the geometry, for essentially zero gain. If you do have two arbitrary orthogonal polarizations at some arbitrary relative phase, then the first thing to do is to extract a frame where the same light is decomposed as two orthogonal linear polarizations at a relative phase of $\pi/2$ $-$ a frame which always exists, and which can be found via the method in this answer of mine $-$ and then carry on with the analysis on that frame.)

Thus, the simplest nontrivial elliptical polarization has the form $$ \mathbf E(t) = \frac{E_0}{\sqrt{1+\varepsilon^2}}\begin{pmatrix}\cos(\omega t) \\ \varepsilon \sin(\omega t)\end{pmatrix}, $$ where $\varepsilon$ is the (signed) ellipticity of the light. If you take this form, it is then easy to calculate the equivalent for Malus' law, by taking the inner product with the unit vector $\hat{\mathbf u}=(\cos(\theta),\sin(\theta))$, squaring, and time-averaging the result: \begin{align} I(\theta) &= \left<(\mathbf E(t) \cdot \hat{\mathbf u})^2\right> \\&= \frac{E_0^2}{1+\varepsilon^2}\left<\left(\cos(\theta)\cos(\omega t) + \varepsilon \sin(\theta)\sin(\omega t)\right)^2\right> \\&= \frac{E_0^2}{1+\varepsilon^2}\left<\left( \cos^2(\theta)\cos^2(\omega t) + 2\varepsilon \sin(\theta)\cos(\theta)\sin(\omega t)\cos(\omega t) + \varepsilon^2 \sin^2(\theta)\sin^2(\omega t) \right)\right> \\&= \frac12 \frac{E_0^2}{1+\varepsilon^2}\left(\cos^2(\theta) + \varepsilon^2 \sin^2(\theta)\right). \end{align} That's it, pretty much. You can rephrase it in a number of ways, and you can choose several other representations of the initial elliptical light, but they're all equivalent.


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