Q1. Suppose the laser beam has a certain (i.e. arbitrary) polarization state but contains only TEM00. Also suppose the PSB is perfect (reflect all S and transmit all P). What results do you expect from your expereiment?
Q2. Suppose the above condition but the PBS is not perfect (i.e. reflects most of S but also small leakage of P to the reflection port.) How are the expected results modified?
Q3. In reality, the laser may also contain some thing dirty (e.g. deporarization in the laser Xtal, higher order modes in a certain polarization but different from the TEM00's one, etc). What actually is the cause of 170mW rejection from the PBS? Can we improve the transmitted power through the PBS?
Q4. Why is the visibility for the lambda/4 with 330deg better than the one with 326deg? Yes, as I already explained to Kevin, I suppose it was caused by the lack of the data points in the wider angle ranges.
I measured the reflected power from the PBS as a function of half wave plate rotation for five different quarter wave plate rotations.
The optimum angles that minimize the reflected power are 330° for the quarter wave plate and 268° for the half wave plate.
The following data was taken with 2.102 A laser current and 32.25° C crystal temperature.
For each of five quarter wave plate settings around the optimum value, I measured the reflected power from the PBS with an Ophir power meter. I measured the power as a function of half wave plate angle five times for each angle and averaged these values to calculate the mean and uncertainty for each of these angles. The Ophir started to drift when trying to measure relatively large amounts of power. (With approximately 1W reflected from the PBS, the power reading rapidly increased by several hundred mW.) So I could only take data near the minimum reflection accurately.
The data was fit to P = P0 + P1*sin^2(2pi/180*(t-t0)) with the angle t measured in degrees with the following results:
where V is the visibility V = 1- P_max/P_min. These fits are shown in attachment 1. We would like to understand better why we can only reduce the reflected light to ~150 mW. Ideally, we would have V = 1. I will redo these measurements with a different power meter that can measure up to 2 W and take data over a full period of the reflected power.