Recent content by Ornit

Yes, people usually speak in term of fraction of lambda (aka "wave") to describe surface figure or TWD (Transmitted Wavefront Distortion). My interpretation is that this is the "A" defined in the picture. For example, if the peak-to-valley in the picture is lambda/4, A=1/4 and my scaling is...

Lambda is indeed the wavelength. The industry uses units of "waves" to measure Transmitted Wavefront Distortion (the deviation from a perfect wavefront, either plane or sphere). It is quite common to talk in units of waves (or fringes), therefore it seemed natural to me to assume those are the...

Looking at the picture. I interpret A as PV (peak to valley) in waves therefore multiplied the sine expression by A*λ. The 0.5 factor comes from the (sin(y)+1): this expression will vary between 0 and 2, therefore I multiplied by 0.5. Have I misunderstood your question?

True, but it doesn't help. Let me write it down fully and properly (well, I hope not - if you find a mistake it will actually solve the problem :smile:): w(x)=0.5*A*λ*(sin(π+2*π*x*n/D)+1) (0.5 and +1 are for the wavefront to swing from 0 to +1 PV as you mentioned; π is just for phase shift...

Thanks soarce, Thanks for your help. You actually made me find the reason for the pi... At first I approximated the wavefront by saw-tooth, missing the (clearly noted) sine wave mention, hence no π in my initial derivation. But I still can't prove the book is right. I think the author meant...

Thanks soarce, but I still can't understand the textbook version. Derivative with respect to what? I assumed with respect to unit length, this is why I tried dividing by the typical length of half a bump. I completely don't understand how did pi got into the numerator of the textbook equation...

This is no homework; but I cannot work it out so I assume one of my assumption is off... no matter how much I looked at it I can't get to the textbook answer. I tried searching (Google is my friend); I got to Prof. Burge's wonderful lectures slides (Arizona.edu) but he follows a different path...

I found a Wiki list with index of refraction as high as ~4 (Germanium): http://en.wikipedia.org/wiki/List_of_refractive_indices I don't think there is a theoretical maximum, just a practical one.

3 correct... Honestly, I was not expecting more.

(almost) final answer is no, we cannot remove the AR coating without ruining the waveplate. I am still interested in gathering this forum's experience to help figuring out the mystery of the flipped HWP.

Thanks. Yes, they have AR coating on both sides, for the wavelength we use.

Thanks. It would certainly be helpful to remove the coating from a "bad" HWP and see if it behaves as expected without the coating, or not. Worth a try. I asked our vendor, but will have an answer only in two days.

Thanks! I am confident the plates are parallel (have no wedge). I used a camera instead of the sensor, and the bit of structure I see there comes from my source (removing the HWP does not change the structure, just the intensity). Also the HWP is perpendicular to the beam, by checking the back...

Thanks M Quack and mfb, I tried modeling the input polarizer and the analyzer angle misalignment, together with random errors of the HWP, using Zemax. Still got the same minimum extinction when flipped. I was thinking of reflection after a colleague of mine suggested it on Friday. This means the...

Hi all, We are measuring our custom-made Half-Wave-Plates (HWPs) for controlling retardation error as part of incoming QC. We don't have a device for measuring retardation nor do we have access to X-ray analyzer. We have experimental result with no model supporting it; the actual question is...