Unpolarized Light Suppose I had a ray of unpolarized light, and I was sitting inside the beam and looking at the electric fields oscillating, then , if I am looking at a point how would the oscillations look like? I cannot seem to understand it.
 A: I guess you have this image in mind:

This hods just for a single photon, or elementary packet of light. An unpolarized beam of light contains a bunch of photons with many different polarizations. The total electric field will randomly jump all around, still the interactions with matter typically involve a single photon at a time. This means that if you ride the beam, you are allowed to look at the photons one by one, each time seeing a different, but consistent with itself, oscillating field.
A: They will look all over the place. If you take a particular photon, you will see it in a particular polarisation state, but it will have nothing to do with the photon next to it, or the one before.
Actually, getting a perfectly unpolarised source of light is quite difficult. One of the best cheap options are the sodium discharge lamps, used extensively in undergrad optics labs. They work by passing an electric current through some sodium vapour, so that each individual atom gets excited, and then releases the energy in light. They are very hot, so the state when the light is emitted is quite random; but still you will see a faint polarisation in the direction of the discharge. That is, if you measure the intensity of the light through a very good polariser at 90º, you will see a small difference.
A: A full classical answer to your question is actually rather subtle and tricky, but its summary is pretty much as in Davidmh and DarioP's answer, both good answers: the electric field vector's direction is changing wildly. The classical description of the nature of depolarised light is bound up with decoherence and partially coherence, a topic which Born and Wolf in "Principles of Optics" give a whole chapter to. The following is roughly analogous to Born and Wolf's desctipyion: if the transverse (normal to propagation) plane is the $x,y$ plane, then we represent the electric field at a point as:
$$\mathbf{E} = \left(\begin{array}{cc}E_x(t) \cos(\omega t + \phi_x(t))\\E_y(t) \cos(\omega t + \phi_y(t))\end{array}\right)\tag{1}$$
where $\omega$ is the centre frequency and the phases $\phi_x(t)$, $\phi_y(t)$ and envelopes $E_x(t)$, $E_y(t)$ are stochastic processes, which can be as complicated as you like. The formulas I cite above just assume that:


*

*$E_x$, $E_y$ and $\phi$ behave like independent random variables, and

*They vary with time swiftly compared to your observation interval (the time interval whereover you gather light in a sensor to come up with an "intensity" measurement) but not so swiftly that the light's spectrum broadened so much that we cannot still think of the light as roughly monochromatic.


Although it cannot really be "visualised", I actually find the quantum description of depolarised light is a great deal clearer and easier: certainly it involves a great deal less of the heavy duty stochastic process theory expounded on in Born and Wolf. We think of a lone photon; it propagates following Maxwell's equations, which are the first quantised propagation equations for the photon (only here the $\vec{E},\,\vec{B}$ represent a quantum state, not measurable electric and magnetic fields). Then we think of an ensemble of pure quantum states, exactly analogous to classical, fully polarised light states. A partially depolarised state is nothing more than a classical mixture of pure polarisation states: what this means is illustrated by the Wigner's Friend Thought Experiment. For a large ensemble of photons, you simply do any calculations you need for all the different pure polarisation states present in the classical mixture, then sum the squared amplitudes weighted by the classical probabilities. The density matrix formalism is invaluable for such calculations, and I talk about these calcualtions for light in this answer here.
