TITLE: Conversion of polarized light to unpolarized light
QUESTION [17 upvotes]: Unpolarized light can be converted into polarized light. However, can polarized light be converted into unpolarized light? If so, how can this be done?

REPLY [4 votes]: The answers above are probably already more than you were looking for. The standard techniques are adequate for most applications, though, as WetSavannaAnimal aka Rod Vance pointed out, they don't produce a truly mixed state of the incident photon.
However, I disagree with him that they cannot produce a truly mixed state of polarization. In fact, it is quite easy to entangle the internal degrees of freedom of a photon. For example, one can entangle arrival time and polarization by sending a photon through a sufficiently thick slab of birefringent material (the slab must be thick enough that the two polarization states end up in completely distinguishable time bins, i.e., one polarization should be delayed relative to the other by more than the coherence time of the light source). Certainly the photon is still in a pure state, but if frequency is traced over (i.e., ignored), its polarization is completely mixed.
Similarly, one can produce a spatially-varying pure polarization: here, the polarization is entangled with the spatial mode, and if the spatial mode is traced over, the output is completely unpolarized.
For any such technique to work in a specific application, the key point is that the extra degree of freedom needs to be traced over. For example, if we filter out a very narrow frequency bandwidth before measuring our photon's polarization, the first depolarizer I described will cease to work. Similarly, the second scheme won't work if you only collect light from a small portion of the beam where the polarization doesn't vary much.


TITLE: Why can't we solve this question by using conservation of energy?
QUESTION [0 upvotes]: A block of mass $m$ is attached to one end of a light string which is wrapped on a disc of mass $2\,m$ and radius $R$. The total length of the slack portion of the string is $l$. The block is released from rest. The angular velocity of the disc just after the string becomes taut is:

Solution:
Using the principle of conservation of angular momentum, $$m\sqrt{2gl}\times R=\left[\frac{(2m)R^2}{2}+mR^2\right]\times\omega\\
\implies \omega=\sqrt{\frac{gl}{2R^2}}$$

I am getting a different answer when I tried solve it by using conservation of energy.

REPLY [3 votes]: You can try to solve this problem using law of conservation of energy and using law of conservation of angular momentum and you will get different answers.
The thing is that you can't use conservation of energy law in this case. At the moment when the string became taut some kind of inelastic impact would happen and some portion of energy would  lost. Imagine, that the string is made of elastic rubber so that no energy would be lost. But in this case there will be some oscillations. In case the string is rigid these oscillations will fade out quickly, but some energy will be lost in the process.
Use the law of conservation of angular momentum.
Update.
Since it's a homework I will not provide a full solution. Just main steps.
In the beginning the block just falls. You can find it's speed $V_1$ at the moment just before the string gets taut.
Total angular momentum of the system at that moment would be $-m V_1 R$ - we calculate the momentum around the center of disk.
After the string is taut the speed of the block will be some some $V_2$, the angular velocity of disk $\omega$, and $V_2 = \omega * R$.
Now you calculate the total momentum of disk and block.
These two calculated angular momentum should be equal. And this would be the first equation on your picture.