Interference of two differently polarised beams I would like to ask how the result of interference changes with the change of polarisation angle difference? Obviously we get the best results for 2 parallel polarised beams, and no intensity interference effects for orthogonally polarised, however what happens in the middle? I heard it's changing with $cos\Delta\phi$, so the result would be $I=I_1+I_2+\sqrt (I_1 I_2)cos \Delta \phi$ . Is it true and why? I tried to calculate it on my own, but I got something like $\sqrt(90°-\Delta\phi)/90°$ which gives the same results for 0, 45 and 90 degrees, but that's it :P
 A: Ok. The formula $I=I_1+I_2+\sqrt{I_1 I_2}\cos(\Delta\phi)$ is correct, but only if we consider that the two beams always have the same phase and $\Delta\phi$ is the angle between their unit polarisation vectors. However, this formula normally reads a little bit different. Normally $\Delta\phi$ denotes the phase shift of your two electromagnetic waves (which light is) and still it is a special case, in which the polarisation is parallel (like you said, you get the biggest contribution then).
Let's now look at the general case:
Let our two beams be $\vec E_1(\vec r,t)=E_1(\vec r)\hat e_1 e^{-i(\omega t-\phi_1)}$ and $\vec E_2(\vec r,t)=E_2(\vec r)\hat e_2 e^{-i(\omega t-\phi_2)}$, where $E_i(\vec r)$ is the amplitude, $\hat e_i$ the unit polarisation vector and $\phi_i$ the phase shift respectively. Then we got for our intensities (without interference) (omitting $\frac{1}{2}c\epsilon_0\epsilon$): $I_i(\vec r)=\langle|\vec E_i(\vec r,t)|^2\rangle$. Now with interference:
$$\vec E(\vec r,t)=\vec E_1(\vec r,t)+\vec E_2(\vec r,t)=[E_1(\vec r)\hat e_1e^{i\phi_1}+E_2(\vec r)\hat e_2 e^{i\phi_2}]e^{-i\omega t} \,.$$
Thus 
$$\begin{array}{}I(\vec r,t) &=\langle|\vec E(\vec r,t)|^2\rangle\\
&=[\langle|\vec E_1|^2\rangle+\langle|\vec E_2|^2\rangle+2\langle\operatorname{Re}[(\vec E_1\cdot\vec E_2)]\rangle]\\
&=I_1(\vec r)+I_2(\vec r)+2\langle\operatorname{Re}[E_1(\vec r)E_2(\vec r)(\hat e_1\cdot\hat e_2)e^{i(\phi_1+\phi_2)}]\rangle\\
&=I_1(\vec r)+I_2(\vec r)+2\sqrt{I_1(\vec r)I_2(\vec r)}\cos(\phi_1-\phi_2)(\hat e_1\cdot\hat e_2)\,.
\end{array}$$ 
Now we just call $\Delta\phi=\phi_1-\phi_2$. So as you thought, calling the angle between $\hat e_1$ and $\hat e_2$ $\theta$, we have with $(\hat e_1\cdot\hat e_2)=|\hat e_1||\hat e_2|\cos(\theta)=\cos(\theta)$ a cosine dependency.
Now to the different cases. Let's say $\hat e_1=\left(\begin{array}{c} 1 \\ 0 \end{array}\right)$. If we have parallel polarisation, then $\hat e_2=\left(\begin{array}{c} 1 \\ 0 \end{array}\right)$. This gives us $(\hat e_1\cdot\hat e_2)=1$, so we get the maximum interference. If we have orthogonal polarisation, $\hat e_2=\left(\begin{array}{c} 0 \\ 1 \end{array}\right)$. Then $(\hat e_1\cdot\hat e_2)=0$, so we have no interence after all. Now to the case that their angle is $45°$:
Naively, we would then assume that $\hat e_2=\left(\begin{array}{c} 1 \\ 1 \end{array}\right)$. However this is no unit vector. So, we have to make it one by dividing by its magnitude, which is $\sqrt2$. So then we get, $(\hat e_1\cdot\hat e_2)=\frac{1}{\sqrt2}$, which is the same as $\cos(45°)$.
