I spent most of the day today (Tuesday) looking at the Fabry-Perot effect which is easy to see in the temporal pulses. Take the following plot:
This is a plot of the (0,0) position on the Si wafer. Like I said in a recent post, there are three distinct pulses -- the initial pulse which has the greatest amplitude which is then followed by a pulse of slightly less amplitude and then yet another peak of slightly less amplitude. If we look at the horizontal scale, which is measured in picoseconds, we can see that the difference between peaks is approximately 12 ps. We also notice that the three peaks are equally spaced with this same increment.
The math for this calculation is pretty straight-forward. Basically, we have the Si wafer of width l and refractive index n. Most of the signal transmits directly through the wafer, and is seen in the primary pulse in the above scan. Then some of the signal reflects off the boundary between the back of the Si disc (call it b), then reflects onto the first boundary which it has already passed through between the front of the Si and the air (call this a), and finally back through b. This gives us that second pulse. The third pulse is explained in precisely the same fashion -- signal transmits through a, reflects off b, reflects off a, reflects off b, reflects off a, and is then transmitted through b. Since the thickness of the wafer determines the path length of each piece of the pulse, they each show up at a different position on the time scale.
I understand the basic physics behind this effect, but I am interested in determining a method to effectively remove it from the data.
I also spent some time in changing the lens used to focus the excitation beam. We found a lens which had a longer focal length so as to get a tighter focus of the excitation beam on the Si wafer.
After doing this, I took more lateral scans across the Si wafer to determine if the beam shape is still about the same as before. The data that I found are shown in the plots below. First is that of the x- and y-direction on the same set of axis.
The scan traced with the blue line and markers is of the x-direction, while the red line and markers display the scan for the y-direction. We can see from the above plot that the beam profile for the x-direction seems to be even more distorted than what we saw when using the other lens. The reason for this, as Antoine explained, might be due to a low-pass filter effect in which the size of the excitation beam effectively changes the profile of the THz beam. Basically, since the excitation beam is so much tighter and we are creating more accurate measurements than with the wider beam from before, we may expect some of this sort of over-distortion to come about. We also see the same sort of skewness in this scan as the x-scan that was done with the other lens.
The y-scan shows the same sort of Gaussian distribution from before. It is still rather symmetric and looks relatively normal. It is also interesting to note that the peak amplitudes for the two scans do not match up perfectly. I am not really sure what this is due to, since I took both scans within about 30 - 60 minutes of each other.
The distortion in the x-direction and the normal, Gaussian shape in the y-direction suggest that perhaps the x-incidence excitation beam is distorted in some way and thus we see such strange profiles for the x-scans.
To further investigate the matter, I took another scan in the x-direction but at about -HM of the y-direction. This plot is shown below.
It is clear that we see the same sort of distortion in this scan... and skewness. Again, I feel as though the reason for this might be the incidence of the x-beam on the wafer. It would be interesting from here to take even more lateral scans and try to actually build a beam profile from this. I might look into taking what one might consider to be circular contours about the xy-plane to see if I can create a better 3D image, though there may be more interesting things to do.