opticspolarization
I know that when circularly polarized light passes through a linear polarizer, the resulting intensity is half the initial intensity. This is because we can decompose the components of $\mathbf{E}$ on the axis of the polarizer and the axis perpendicular to the polarizer, and on average the field on both axes will be equal.
Does the same effect occur with elliptical polarization? Can we say that $I=I_0/2$ after passage through a linear polarizer?
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 $a=1/\sqrt{2},\varphi=\pi/3$:
I also found this surprising when first introduced.
The light has no memory. When it passes through the second polariser, there is no information whatsoever of its previous polarisation. It could have been whatever!
When light goes through a polariser at $30ยบ$, it gets tilted at the cost of some lost intensity. In other words, it gets projected on the polariser axis, that is rotated a bit with respect to the light polarisation vector. This new light then goes on to the next one, and gets tilted again.
You get the picture. You can use this idea to arbitrarily change the polarisation angle of light by putting many of them in series. The more steps you do, the more polarisers you use, and thus, the smaller the steps, more light will come out on the other side (in reality, you always loose some light because they are not completely transparent, so you don't want to use many of them).
Best Answer
You should again split up $\vec{E}$ in its components and work out the resulting field. You will notice that the resulting intensity, after passing through the polariser, is generally not half the original intensity.