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  @linwang90552 күн бұрын

  I couldn't find the expression for CTR in the later part of the video, did i miss any part....?

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  @stephenremillard12 күн бұрын

  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.

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  @hakankabagac30544 күн бұрын

  Dear Stephen, Thank you for great explanation. I couldn't find related excel. Is there any way to download it? Regards

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  @myasterr6 күн бұрын

  Fantastic explanation. Many thanks!

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  @marmosetman7 күн бұрын

  I am a bit confused about Idsat. What if instead of evaluating it at x=L , we evaluate it at x=0. Idsat should be the same regardless of x. Then Idsat= wCox(Vgt-Vcs(0))vsat, which is completely different?

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  @stanley660217 күн бұрын

  Very important video for laser diode application. Thank you.

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  @sinecurve999919 күн бұрын

  Fantastic explanation! Cheers!

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  @Dstonephoto25 күн бұрын

  Could we offset some of these exagerated aberrations by utilizing curved first surface mirrors to zero out said distortions? I assume this wont work with mustache distortion . Could one further reduce these distortions by subsequently switching to a monochrome approach using monochrome light and complimentary optical filters on the lens?

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  @GoatzAreEpic28 күн бұрын

  I was making the exact same mistake as the viewer haha. I even asked chat gpt and it told me that it should indeed be 2d/cos(theta). However I was sceptical and luckily found your explanation. Thank you sir!

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  @Fujik1966Ай бұрын

  Hello Stephen. It seems to me that such a large number of glass-air-glass transitions will change all theoretical calculations to Zero due to re-reflections of glass surfaces. I'm a photographer, but I'm interested in microphotography. Eduard.

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  @stephenremillard1Ай бұрын

  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.

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  @Fujik1966Ай бұрын

  @@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.

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  @Fujik1966Ай бұрын

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

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  @GoatzAreEpicАй бұрын

  Thank you so much for the video. I was wondering if there are any books you can recommend on this subject? Thanks in advance. I really appreciate your work.

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  @stephenremillard1Ай бұрын

  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.

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  @svitloorsvetloАй бұрын

  Thanks you!!!

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  @TurboLoveTrainАй бұрын

  Thankyou for making this. Very fun to watch. The fine structure constant isn't as constant as advertised. I'm glad you commented that it doesn't matter what [a] constant is, it matters that it is "constant."

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  @thomasolson7447Ай бұрын

  It's a little bit different from what I taught myself using quadratics. That one in the middle (9:46) is a Second Kind type, or Fibonacci-like discrete homogenous sequence, even though it has that plus sign. That would make the magnitude equal to one, but that can be manipulated to r^((t-1)/2). I don't know how that changes, given the outside term. Is that a cubic? Are they triangle waves in 3d? The 'e' on the outside is vector angle addition. The magnitude is 1. That one is easier, r^t. There should be another function that pairs with this. Ψ(n+1)+Ψ(n-1)+f(n)=0 (I'm too lazy to do notation correctly). I might be wrong though, given that it is cubic. Anyway, that's wrong. You can't do the 2cos(dwt-dkx) thing. ChatGPT always simplifies that function, but it's wrong, I checked. It will work if time and displacement is an integer. It becomes a complex number when they are rational (fractional). 2cos(dwt-dkx) doesn't appear to become a complex number. Standup Maths: "Complex Fibonacci Numbers" kind of addresses it. Ψ(n+1)/Ψ(n) where n = -2 -i*2.. 2+i*2 should be a magnetic field. Three poles, I'm guessing, project it on a sphere. You might need to customize the tool you use to graph it because it's cubic.

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  @TurboLoveTrainАй бұрын

  Thankyou for the video. I can't see Lambda and not think, λ = (h/p), I like their version: λ /NA, just use 193nm/1.55. That made me giggle.

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  @hamiltonqrn2734Ай бұрын

  I can't find the double like button. Thanks for video!!

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  @rostyslav96Ай бұрын

  Thank you very much for the Video. For the same case: Would it be reasonable to use the method from the paper "Characterization of High- Resonators for Microwave-Filter Applications" by Raymond S. Kwok and Ji-Fuh Liang (1997)?

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  @stephenremillard1Ай бұрын

  Yes sir. I think so.

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  @rostyslav96Ай бұрын

  So this formula (Insertion loss method) breaks down for the case of a critically coupled resonator, because the when Li approaches 0, Qu should be double of Ql and not go to infinity. Is there a relation taking into account strong coupling near critical coupling or am I missing something obvious?

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  @stephenremillard1Ай бұрын

  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.

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  @rostyslav96Ай бұрын

  @@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?

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  @manarc-cs6uvАй бұрын

  @Stephen Remillard. Hi Stephen, I am a student in optics.Could you explain the equation at line and 142 of your code (@15:11); In this section I don't konw the value EFL why not mentioned here?Also,I couldn't find the new variable EFL you used in line 159 (@15:49) in the previous code ? Please tell me the reason.

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  @stephenremillard1Ай бұрын

  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).

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  @manarc-cs6uvАй бұрын

  @@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)

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  @stephenremillard1Ай бұрын

  @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.

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  @stephenremillard1Ай бұрын

  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.

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  @manarc-cs6uvАй бұрын

  ​@@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.

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  @zacherychen484Ай бұрын

  Thank you for this video!

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  @TurboLoveTrainАй бұрын

  Beautiful work. Thankyou! You are an absolute treasure. As a computer scientist all I see are chains of matrix operations :)

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  @stephenremillard1Ай бұрын

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

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  @TurboLoveTrainАй бұрын

  Thankyou Your videos are incredible

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  @Berk-lf6geАй бұрын

  Thank you so much. You are a fantastic teacher

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  @capionstudioАй бұрын

  Stephen this is great information. Thank you for explaining this. Oddly, I am looking to introduce sagittal astigmatism preferably with AR/AS coatings. Do you know what type of optic I should be looking at (specific name / keywords) and where to source this? I've used these terms with no results. I am looking for an 82MM diameter optical filter that can effectively swirl the edges of my frame with preexisting lenses. Thanks for your help!

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  @stephenremillard1Ай бұрын

  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.

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  @cordi-fm9tbАй бұрын

  Thanks! I came out confused from my quantum class and this was very helpful!

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  @pushprajprasad1902Ай бұрын

  Wow! The electron tunneling equation looks cool!

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  @javieranavarro5462Ай бұрын

  Hi stephen! Is there a way in contacting you? Im doing the same experiment in uni but im having problems getting the info of the laser/bullseye on the arduino camera. How did you get the data?

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  @-eduarth_ab6693Ай бұрын

  What is your opinion of Leica Co working with Xiaomi?

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  @user-wb4rl9jm7yАй бұрын

  Stephen you always much such great and clear videos! please keep up the great work.

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  @julianprzybysawski8543Ай бұрын

  I wonder why plastic is not used for photograph lenses in larger formats.

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  @user-wt8vq6ce8kАй бұрын

  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

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  @maxhammick948Ай бұрын

  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)

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  @danieleppelsheimer9273Ай бұрын

  Please continue Your helping me and Many others unravel The process of lens Design in the digital image capture.

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  @mgpetertАй бұрын

  Great presentation! Thank you for showing the simulation results. Much easier to understand the discussion by seeing pictures

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  @zircron452 ай бұрын

  Very good descriptions

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  @j.w.86632 ай бұрын

  I will be using pyramidal horn antennas with an LNA and a spectrum analyzer to scan in a 360 degree azimuth to look for possible interference in the X and S bands (2GHz and 8GHz). One of the requirements is to "Calculate cold sky noise temperature at 10 degree elevation." Would you (or anyone) perhaps know how one does this? Is this simply measurable somehow? Google doesn't get me too far...😕

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  @stephenremillard12 ай бұрын

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

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  @j.w.86632 ай бұрын

  @@stephenremillard1 ok, thank you.

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  @stephenremillard12 ай бұрын

  @@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.

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  @user-nv3rw6sw4f2 ай бұрын

  The video you made is really good! I just started to learn to use ZEMAX software. In front of you, I feel like I am like a boat that has just been launched, and there is a long way to go. Thanks you !

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  @ShopperPlug2 ай бұрын

  Excellent video on such important topic for laser diodes. Is it possible to use the Achromatic Anamorphic Prisms pairs of the same glass type and get great beam spot size? Why two different glass types are needed?

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  @stephenremillard12 ай бұрын

  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.

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  @rayjones94182 ай бұрын

  Thanks for these videos. I wish I had seen these 30 years ago!

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  @nikunjbheda77662 ай бұрын

  How I can get this excel file?

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  @nguyensontung59232 ай бұрын

  20 minutes and it solves my 2-day problem. Big thanks!

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  @testboga59912 ай бұрын

  Excellent!

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  @pawetrznadel77702 ай бұрын

  Great video. Thank you for sharing.I have to watch it again to understand more than 10%😂

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  @chevestong3 ай бұрын

  16:09: Dr. Remillard says "the group velocity is LESS than the phase velocity", which I believe was a mistake, since it's written that the group velocity is GREATER than the phase velocity, which is true for d v_p / d omega > 0.

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  @stephenremillard13 ай бұрын

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

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  @ToanNguyen-vf3hc3 ай бұрын

  Thanks for wonderful video but I am still a little confused that the magtify of distortion should change following the curve of lens shape but how can you get a number telling about distortion since lens 5 and 6 are not spherical lens so distortion might change from barrel to pincushion based on how chief ray is reflacted on curve, can you help me to get deeply in it? I am college student and trying to research about distortion, thank you alot

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  @stephenremillard13 ай бұрын

  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.

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  @ToanNguyen-vf3hc2 ай бұрын

  ​ @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.

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  @Not_Qwake3 ай бұрын

  Do you have resource recommendations to get started?

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  @stephenremillard13 ай бұрын

  As much as I would really like to say that you can master this craft by watching my KZread 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.).

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  @Not_Qwake3 ай бұрын

  @@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!

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  @danielchatrie66143 ай бұрын

  Hello I need and answer to this important question if a mosfet is rating at 94amp and 380 amp pulse current at 580watt power dissipation rating if the frequency is 250khz does the current rating remain the same or it reduces the current rating please explain to me thank you

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  @brisingreye52093 ай бұрын

  @Stephen Remillard. Could you explain the equation at line and 49 of your code (@7:18)? How did you come to this equation? thanks in advance!

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  @stephenremillard13 ай бұрын

  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.

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  @DianaCantini3 ай бұрын

  Thank you!!

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  @user-wt3ev1zg2l3 ай бұрын

  wonderful

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  @utube321piotr3 ай бұрын

  Love your channel. I would welcome our commentary or video coverage about adaptability of vintage lenses to various DSLR and mirrorless camera brands, bodies and sensor sizes. Specifically the effect of glass elements covering the digital sensor itself and their effect of thickness and optical properties upon the Marginal and Chief ray. Reason for my question is that some particular vintage lenses may be outstanding on some camera bodies and not so on others. For example SONY is known for using thicker glass over the sensor than Fuji on their x-trans. In essence, the glass over the sensor is yet another "lens" element unaccounted for in the original design of a vintage lens.

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  @stephenremillard13 ай бұрын

  Thanks for raising this question. It's definitely worth looking at if and how the presence of a microlens array can influence the performance of the lens. Although I'm sure that the know-how is out there, I have not looked into this. But you have made me curious about it.

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  @utube321piotr3 ай бұрын

  @@stephenremillard1 Actually, I didn't mean the "micro lens", but a piece of flat glass that sits on top and protects the micro lens array along with silicon sensor., and which in routine practice is cleaned when we want to "clean the sensor". This glass is around 0.8-1.2mm thick depending on camera brand.

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  @stephenremillard13 ай бұрын

  Ah, thanks for the clarification. Such a glass is usually included as an IR filter and its presence is also taken into account during the lens design. I covered it a little bit in my patent review of smartphone camera lenses kzread.info/dash/bejne/Y5V3uq97gJvFY6w.html . Out of curiosity, I just now opened the Zemax file for that patent study and removed the slab of glass over the sensor. I adjusted the sensor position a fraction of a mm to the new point of best focus and the focus was good, although degraded from the original design. The RMS spot size increased from diffraction limited with the glass present (about 1.7 microns) to 4 microns without the glass. The MTF dropped from over 50% at 100 LPMM to about 30%. So, imaging is still possible with or without the slab of glass. But if the lens is designed without assuming the slab, the performance might suffer a little by adding a slab.

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  @Dstonephoto25 күн бұрын

  Busy body chiming in here. When I was trying to research older medium format backs (pre 2010) i came across some interesting conversations whereby users observed differences with backs which had and didn’t with regards to microlenses. Fast forward to today and I’ve seen some people posit the notion that Fuji GFX sensors sacrificed sharpness for image fidelity by integrating microlenses. I find the entire domain of microlenses intriguing, yet there isnt much information available on this domain. We’re also seeing microlenses being utilised with lightin as well. Makes you wonder if microlenses could be integrated with things like film cameras , existing microlensed sensors, projectors, enlargers, and microscope cameras, and possibly enhance existing devices, possibly with the introduction of trqdeoffs, but with favorable results. Would you mind doing an analysis on focal reducers ? They’re interesting lenses. Now, back to microlenses, to compound things a bit, I wonder how electrowetting and liquidmlenses could come into play by manioukating focal length and polarisation. Alright 😅.