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@mwerensteijn 7 күн бұрын
Thank you so much for your time to create this video!
@maheshnamboodiri4097 20 күн бұрын
Thank you for this.
@alihassan9690 24 күн бұрын
profassional presentation i have seen most of your videos, i hope if i could achive a copy of MATLAB code of ray tracing
@yuhui8109 24 күн бұрын
Hi Dr Remillard, i have simulated a cylindrical cavity resonator with a TE 011 mode at 8.5 GHz, when i placed a DR inside, the resonant frequency is increasing to 9 GHz, i am expecting to be lower than 8.5 GHz, do you know what had happened sir ?
@stephenremillard1 24 күн бұрын
You might be expecting the resonant frequency to be lower since it does decrease as dielectric constant increases. But this rule is for resonators of the same size and same field distribution. When the DR is placed in the cavity, the dielectric constant certainly goes up. But the TE011 mode is redefined and its shape is altered. My guess is that your result is because now a lot of the electric field energy is being constrained to a smaller volume (the dielectric), and the resonator configuration is not the same.
@shilahosseinzadeh9440 24 күн бұрын
🙏🙏
@shilahosseinzadeh9440 24 күн бұрын
thanks
@captainspire9094 26 күн бұрын
I have been looking for this for a year! I remake old toys using those 5v green laser diodes and sometimes it's night er have a slightly thicker laser beam. But I just couldn't figure out which type of lenses to get! Now I do. Thank you.
@masonmault979 28 күн бұрын
Great video
@cylosgarage Ай бұрын
Stephen, as a precision engineer who has no formal optics training but is having to learn it for both work and hobby related projects, I find your videos immensely enjoyable and educational. Thank you for making them
@stephenremillard1 Ай бұрын
Thanks, and I'm also one of your 18K+ subscribers. I especially like your video about diamond turning a spherical mirror (Part 2). Your Bath interferometer gave beautiful MSFR wavey fringes.
@stephenremillard1 Ай бұрын
Typo at 7:00 - There should be a factor of t-squared multiplying the ratio of cosines. At 7:20 the next line appears correctly with a factor of t-squared multiplying each cosine ratio.
@LTC_dingzhen-zhenzhu Ай бұрын
Hi,MR! Now i am wondering about fermi pinning effect(fermi level pinning), would you please explain some about it ?😁
@yuhui8109 Ай бұрын
Dear Dr. Remillard, I have a question, may I know is there a function to slice/half cut the geometrical design, so that Im able to observe the signal propagation inside, I'm simulating a cylindrical cavity resonator ? Thank you in advance.
@stephenremillard1 Ай бұрын
The short answer is yes. You can plot fields and currents on the Global xy, xz, and yz planes, and you can also define your own planes for observing the fields and currents. You can ask HFSS to animate the fields, which might give the representation of signal propagation that you are looking for. I don't have an active license right now, so I can't take you through doing that, but hopefully the content starting at 26:41 is at least a little bit helpful.
@yuhui8109 24 күн бұрын
Thank you sir for the reply, it is helpful. @@stephenremillard1
@ArsenedeBienne Ай бұрын
thanks a lot!
@Saullavado44 Ай бұрын
That is so interesting. Thank you!
@SIAT2023 Ай бұрын
thanks for this video , how can i calculate QL that you mentioned in last equation to calculate Qu ? because all parameters (s11,s22) can be extracted from simulation but QL i dont know how , could you please how can i get these value in this method
@stephenremillard1 Ай бұрын
Take a look at this video. Hopefully your question is answered here. kzbin.info/www/bejne/qJOZoK1phc5nnaM
@刘宇熙-o5u Ай бұрын
First of all, thank you for your video; it’s very beginner-friendly. Recently, I tried to replicate the lens design based on your video and the patent mentioned in it. All the surface parameters match the table exactly, and I kept five decimal places in the simulation, just like you did. However, I noticed that my layout is somewhat different from the one in the patent and your layout. Specifically, lenses 1 and 2 are very close to each other, with their edges overlapping, which results in the edge light not converging on the image plane. The shapes of the other lenses appear to be consistent. I carefully checked the parameters for the four surfaces of lenses 1 and 2 and didn’t find any issues. Additionally, I used EFLY to check the focal length of each lens, and they are essentially the same as in the table. I personally believe that the shape of an aspherical lens is determined by its radius, thickness, and aspherical coefficients. Since these data match the patent data, why do the results differ? Could you please advise if there’s something I might have overlooked that’s causing my results to differ?
@stephenremillard1 Ай бұрын
The first thing I would check are the clear apertures on those four surfaces. After I optimized, the first lens element had a front-side clear semi-diameter of 0.724mm and a back-side of 0.792mm. If all of the surface geometry is entered without error, then the next thing I would check is how the field and the aperture are defined. I used Real Image Height for the field and set the max field to 2.801mm. I used Image Space F/# for the aperture, and I set it to 2.35. Hope this helps.
@dilipdas5777 2 ай бұрын
Optics is a complicated Physics
@linwang9055 2 ай бұрын
I couldn't find the expression for CTR in the later part of the video, did i miss any part....?
@stephenremillard1 2 ай 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.
@hakankabagac3054 2 ай бұрын
Dear Stephen, Thank you for great explanation. I couldn't find related excel. Is there any way to download it? Regards
@myasterr 2 ай бұрын
Fantastic explanation. Many thanks!
@marmosetman 2 ай 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?
@stanley6602 2 ай бұрын
Very important video for laser diode application. Thank you.
@sinecurve9999 2 ай бұрын
Fantastic explanation! Cheers!
@Dstonephoto 2 ай 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?
@GoatzAreEpic 3 ай 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!
@Fujik1966 3 ай 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.
@stephenremillard1 3 ай 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.
@Fujik1966 3 ай 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.
@Fujik1966 3 ай бұрын
@@stephenremillard1 Maybe look at this with lens blocks immersion glued on. You may end up with universally changeable options.
@GoatzAreEpic 3 ай 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.
@stephenremillard1 3 ай 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.
@svitloorsvetlo 3 ай бұрын
Thanks you!!!
@TurboLoveTrain 3 ай 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."
@thomasolson7447 3 ай 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.
@TurboLoveTrain 3 ай 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.
@TheGreatHammys 3 ай бұрын
I can't find the double like button. Thanks for video!!
@rostyslav96 3 ай 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)?
@stephenremillard1 3 ай бұрын
Yes sir. I think so.
@rostyslav96 3 ай 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?
@stephenremillard1 3 ай 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.
@rostyslav96 3 ай 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?
@manarc-cs6uv 3 ай 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.
@stephenremillard1 3 ай 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).
@manarc-cs6uv 3 ай 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)
@stephenremillard1 3 ай 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.
@stephenremillard1 3 ай 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.
@manarc-cs6uv 3 ай 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.
@zacherychen484 3 ай бұрын
Thank you for this video!
@TurboLoveTrain 3 ай бұрын
Beautiful work. Thankyou! You are an absolute treasure. As a computer scientist all I see are chains of matrix operations :)
@stephenremillard1 3 ай бұрын
Thanks for the nice compliment. If you can see the matrix operations, then you're the real deal, because that's exactly right.
@TurboLoveTrain 3 ай бұрын
Thankyou Your videos are incredible
@Berk-lf6ge 3 ай бұрын
Thank you so much. You are a fantastic teacher
@capionstudio 3 ай 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!
@stephenremillard1 3 ай 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.
@cordi-fm9tb 3 ай бұрын
Thanks! I came out confused from my quantum class and this was very helpful!
@pushprajprasad1902 3 ай бұрын
Wow! The electron tunneling equation looks cool!
@javieranavarro5462 3 ай 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?
@-eduarth_ab6693 3 ай бұрын
What is your opinion of Leica Co working with Xiaomi?
@תומרום 3 ай бұрын
Stephen you always much such great and clear videos! please keep up the great work.
@julianprzybysawski8543 4 ай бұрын
I wonder why plastic is not used for photograph lenses in larger formats.
@AleksandrCheplakov 4 ай 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
@maxhammick948 3 ай 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)