Відео

The Excel formula in Cell I22 for CTR is =-F12*F15*(E10+F10-(E18*F10*E10))/I7 The mathematical expression for CTR is found at 3:38.

It is a lot of surfaces isn't it. With 58 surfaces each having a 1% AR coating the back-of-the-envelope transmission is at best 0.99^58=56%. Not zero, but still a lot of loss. I'm not really sure if anything else is typically done to reduce loss with so many elements. The stray light from these reflections is probably no trivial matter either.

@@stephenremillard1 Yes, scattered internal light will also lead to a significant decrease in contrast, halos and blur at the borders. And this is the most necessary thing in lithography, a clear/sharp separation from dark to light. It would be interesting to see an image from this lens on a matrix camera. Creating one copy of a lens is an expensive pleasure. Only an enthusiast who risks time and expense will be able to do this.

​@@stephenremillard1 Maybe look at this with lens blocks immersion glued on. You may end up with universally changeable options.

Yes, and I think it would be really great if other viewers could chime in here with their own recommendations. I got a lot out of the treatment in Saleh and Teich, and also from Pedrotti, Pedrotti, and Pedrotti. But there was also plenty that I learned from experimentation, and I admit I have lost track of specifically what I have learned from reading and what I learned from experimentation. And when I say experimentation I am referring to both hard and soft experiments (hardware and software), but mostly soft experiment for me on this topic. And actually it's the soft experiments where you learn a lot and learn it quickly because it's versatile and quick, and you can immediately understand your results. For me there was the use of MatLab and also the use of nonsequential ray tracing, both of which I demoed a little bit in the video. But of course you're right in hitting the books first to get the foundational knowledge that you can then knowledgeably use in those experiments.

Yes sir. I think so.

I have always avoided critical coupling for my purposes, so maybe let's see what someone else says. But the insertion loss, Li would only ever be exactly zero at critical coupling if there were actually no dissipation, hence infinite unloaded Q. So, I don't find it alarming that Q blows up for zero Li at critical coupling. What is not baked into the equation is what the loaded Q should be at critical coupling if there were no dissipation, since in that specific case loaded Q is the product of zero and infinity. I am not familiar with the argument that the loaded Q is half the unloaded Q at critical coupling. In my experience, those two numbers diverge as critical coupling is approached from below.

@@stephenremillard1 But wouldn't he dissipation be dependent on the linewidth of the resonance and not the amplitude? I question myself what would happen at high coupling that approaches critical coupling. From theory I know that the linewidth of the resonance would increase. I found in literature that at critical couplings the linewidth is twice the unloaded linewidth. This comes from the definition of critical coupling: "A condition in a resonator system where the rate of energy transfer to an external load matches the intrinsic loss rate within the resonator, leading to maximum energy transfer and minimal reflection at resonance.". And in the very weak coupling regime one would measure the approximately the unloaded linewidth. But I believe there should be an expression for the calculation of the unloaded linewidth of a system with strong coupling, by knowing the amplitude at resonance. Do you know something that could help me?

The quantity that is computed in the loop from lines 136 to 138 isn't the complete total track length. It is only the distance up to the last glass vertex. The back image distance still needs to be added to it in order to have the quantity that is typically referred to as total track length. The effective focal length is calculated on Line 152 (@15:37).

@@stephenremillard1 Thank You for your expaining， The problem has been solved。 But have a new problem , when I using your all code for learning MATLAB and Optical fundamental konwledge , the speed for program is to slowly .I let this code run for 15 minutes, and the code still has not jumped out of 44 lines of code to find the loop of the chief ray.（@8:23)

@manarc-cs6uv It sounds like the loop termination condition is either being missed, or not reached. You might try changing the size of the variable incr. Increase it a factor of 10 or decrease it a factor of 10. Then try factors of 100 or more if that doesn't work. If it is taking too long to complete the loop, then incr is too small. If it never completes, then incr might be too large - which is why I had to give it a small value.

I'll also add that a better programmer than I can come up with a better way to execute that operation in the red box @8:23. A while loop isn't very efficient, and is perhaps evidence of my Fortran 77 upbringing.

​@@stephenremillard1 I tried the variables of reduction and enlarging INCR again, but it still seemed to be unable to meet the condition in the WHILE loop Done == 1.

Thanks for the nice compliment. If you can see the matrix operations, then you're the real deal, because that's exactly right.

This sounds like a job for either a cylindrical lens or an aspheric lens, both of which can be purchase from Thorlabs (www.thorlabs.com/navigation.cfm?guide_id=2087). You can specify an AR coating when ordering. Not sure about AS coatings though. I would probably go with a cylindrical lens element with a fairly large radius of curvature (www.thorlabs.com/newgrouppage9.cfm?objectgroup_id=2803). This way you introduce a small anamorphic effect. That's my spur-of-the-moment thought on the matter.

Small range of refractive index, high coefficient of thermal expansion and poor optical properties I have seen plastic lenses on aliexpress, they have poor quality but low price

It is - canon has got quite good at moulding elements at that size, so their cheap lenses tend to include some very extreme plastic aspherics (e.g. the RF 28mm f/2.8)

I'm in the same boat as Google. Not really something I know how to do.

@@stephenremillard1 ok, thank you.

@@j.w.8663 I wish I could be more helpful. Your question did stir up a memory for for me, though. When I was in grad school I had some microwave measurement questions, and I contacted someone at NRAO (National Radio Astronomy Observatory) who was very helpful. I even went to visit since I was close. It impressed me how eager these guys were to be a resource for a student. I'm going to guess that there is well-established know-how for your specific question that resides squarely in the radio astronomy community. I don't work in that field, but it does have a wealth of knowledge about microwaves, the sky, noise, etc.

Using two glass types helps to keep the outgoing angle and height the same for all wavelengths. But as I found, it isn't perfect. You certainly can use two prisms of the same glass type, which just means a little more variation with wavelength in the outgoing angle. Using two different glasses doesn't affect the beam quality for monochromatic light, but it does help to keep all colors on the same path if the spectrum is broad.

You are right. What is written is correct. Thanks for pointing that out.

This is a really good question. Wavefront aberration coefficients can be computed analytically through fourth order aspheric. But for higher order aspherics, ray tracing is exclusively used to understand the final image locations. A fourth order aspheric will cause a small departure from a spherical surface, unlike the higher order terms in the surfaces used here. As you noted, the distortion is hybrid, meaning that there is a change in sign moving from the center to the image edge, and ray tracing, rather than a single number such as a Seidel coefficient, is the only way to look at it. Zemax, and all other programs, do compute a table of Seidel coefficients. But when higher order aberrations dominate, and high order aspherics are used, I really don't know what meaning, if any, they have. I'm sure they are meaningful, and maybe someone can help us out here. By the way, you can get hybrid distortion without using aspherics. Some lenses balance out the third order distortion leaving higher order field dependent magnifications that can result in a wavey distortion versus field plot. Bear in mind that a distortion plot is not the result of only the third order polynomial term, but of all polynomial terms that describe a displacement in the chief ray.

​ @stephenremillard1 thank for great explaination, I did some experiments in cooke triplet lens structure and realize that distortion's magnitude is really hybrid through lenses but in another lens structure, it not really working that way, when the system get signigiciant magnitude on barrel distortion, I add into it another got barrel distortion lens and the final result make the distortion decreased but image smaller, this is quite interesting, the image is really compressed to expand FOV with lower distortion. But I still dont know the way it works.

As much as I would really like to say that you can master this craft by watching my UA-cam videos, that probably won't do it. So, you are right in looking for resources. I think it's critical to engage, use software, follow guidance from knowledgeable teachers, and collaborate with a cohort. After all, expertise only comes with a lot of practice. There are courses at UDEMY, CourseEra, UC Irvine DCE, and others, and which is best depends on your background (science, art, etc) and where you want to go with this knowledge (lens design, photography, etc.).

@@stephenremillard1 Thanks for the response I will check them out. While I do have a stem (biochem) background did cover maths and physics, we didn't touch on optics a whole lot. I am a hobby photographer and microscopist, lens design is something that interests me and understanding it surely will come in handy as well. I subscribed to you, I'm sure I can at least pick up some things listening to someone so knowledgeable!

Take a look at 3:46. This is the equation at the top of the screen, (n'u'=nu-y*phi). In Line 49, yt(i) is the chief ray height at surface i. uprime is the chief ray angle after the designated surface. So, uprime(i-1) is the chief ray angle incident at surface i, since it is the chief ray angle after surface i-1. Also, the angles are replaced with tangent of the angles to improve the accuracy.