Talk:Degree of polarization

Question

DOP=(max power-min power)/max power*100%, is it right? thanks,—Preceding unsigned comment added by Jyqiao (talk • contribs) 08:55, March 10, 2008

Example

Your measurement of DOP does not take into account that light can be circularly polarized, what is quite different than unpolarized.

In fact, the expriment you describe permits to Degree Of Linear Polarization

Look at the Stokes Parameter page for further details —Preceding unsigned comment added by 192.93.158.26 (talk) 11:47, 23 May 2008 (UTC)[reply]

I've removed the simple example below, because of the same concern raised above: light can be completely circularly polarized (DOP=100%), and the transmission through the polarizer will then be independent of its rotation. A linear polarizer cannot by itself distinguish between circular and unpolarized light. --Srleffler (talk) 23:50, 12 June 2008 (UTC)[reply]

A simple example for measuring DOP is to place a perfect polarizer in the path of the wave, and measure how much energy is transmitted through the polarizer, while the polarizer is rotated 180 degrees. If the wave is perfectly polarized (DOP=100), there will be a maximum signal transmitted in one orientation which is equal to the incident energy, and a minimum signal equal exactly to zero (no power is transmitted when the polarization of the incident wave is perpendicular to the polarizer direction). On the other hand, if the wave is completely unpolarized (DOP=0), equal power will be transmitted through the polarizer no matter how the polarizer is rotated. For signals with partial polarization, we can determine the DOP by comparing the maximum and minimum power transmitted through the polarizer during the 180 degree rotation. This method is complicated by the fact that polarizers are not perfect (i.e. have a finite extinction ratio). There are numerous other ways to measure DOP.