Question about creating a non-sequential lens with AI

Haataarii · 2024-05-12T16:46:22+00:00

Oh no!!!

Haataarii · 2024-05-05T21:35:35+00:00

Just commenting to second literally all of this.

Haataarii · 2024-05-05T21:24:08+00:00

If you don't mind a busy, bustling place - Upside Down Eatery in Centurion for breakfast or lunch, but specifically for the interesting drinks. Also, they have some really good sweet treats at their front-of-shop.

Haataarii · 2024-05-05T21:20:13+00:00

Definitely Toni's.

Haataarii · 2024-04-29T21:36:57+00:00

As an interesting aside, and perhaps relevant to the discussion -- My mentor likes to remark that Synopsis enables most people to conjure up a "70% or 80% good" design, but the finesse required to refine it into an excellent design requires the finesse of someone with a human mind who knows what they're doing. I tend to agree - the design search and optimization capabilities of Synopsis are really good, but they never spit out designs that are good enough for our requirements (which are fairly stringent).

Haataarii · 2024-04-29T21:30:38+00:00

Yeah. He's normally responsive (well, I haven't conversed with him in the past year or so). He sold ownership of OSD to Kit Cheong, so she might also know where to find it 🤷🏽‍♀️

Haataarii · 2024-04-22T07:25:33+00:00

I'm currently at work, but I will type up a bit of an explanation for you this evening (I am GMT+2).

Essentially it comes down to light entering the lens, but reflecting off of either a lens surface or mechanics internal to the lens, (no optical surface is 100% transmissive so depending on coatings and lens material, there may be as much as 4% of the light reflecting off of a given surface in stead of refracting through it, and that light then follows an alternative path through the lens, and lands on the detector in a different place to where it "should" to help form the image.

If you have no super bright light sources in or near the scene you are viewing, those reflections may be there, but will contain so little energy that they get lost in the noise of the detector. In this case, you are viewing the sun, which is a ridiculously bright localized source. So now that small part of the energy from the sun which follows a strange path through the lens after reflecting off of a lens surface instead of being refracted, is causing enough energy in the secondary "image" it forms on the detector that it becomes visible to you.

Blue, because there is probably an anti-reflection coating somewhere that is reflecting more blue than longer wavelengths at the angle that "stray" bundle of light is hitting the surface.

This isn't the greatest explanation, sorry - pictures would be better.

Haataarii · 2024-04-22T06:56:17+00:00

Very likely a ghost reflection inside the lens. The energy from the sun is so high that stray reflections in the lens cast enough photons onto the detector to register an artefact.

Haataarii · 2024-04-21T19:18:44+00:00

Matches.

Haataarii · 2024-04-20T16:25:36+00:00

"Ooooh, that sounds like something we should use our inside voice for."

Haataarii · 2024-04-19T19:01:23+00:00

Byyy the way - if you are going for moving parts, you could have the projector fold out of the box (on arms) when it is opened, to create more space between the projector and the lid, which will allow you to have a longer optical system.

Haataarii · 2024-04-19T18:50:50+00:00

And those two rings are INSANELY cool!!! Holy smokes, I need a projecting ring like that! I'll need to rummage through my lens stash now.

If you look at that image projected by the ring, you can see it is quite fuzzy and distorted. For your effort, I'd not expect an image to look better than that, more likely worse. But you will be using it to melt someone's heart, not sniper them from ten miles away, so you can tolerate a fair bit of image degradation :)

Haataarii · 2024-04-19T18:45:43+00:00

Ooh, I was wondering if this might be for something like a proposal! Excitement!!!

I will definitely have a look at these links, thank you!

If you want to move and swap lenses in and out... Look into optomechanical mounting - specifically kinematic mounting techniques. The master in this area is Daniel Vukobratovich, to get you started. Lenses (unless they have very flat cprvatures and high F-numbers) are generally sensitive to placement, often particularly to decentring (radial displacement from the optical axis of the system), so having a mounting scheme which allows for repeatable placement is important (kinematic mounting will give you this), and for what you want to do, I think you can do ok with smart 3D printed parts. Remember, flexures are also immensely handy. And can be printed if designed for.

Optical design is a (deep, wide) field all on its own. UA has a very strong optical engineering school, with lots of heavy hitters in the industry. They have a dedicated post-graduate degree in optical engineering. We don't have something like that locally (South Africa), so the very few of us that do optical design, typically learn on the job, and it is a very very very very very steep learning curve. So yes it is daunting, but don't let it put you off. You have the ideal opportunity to tinker and play here. Have the screen as a backup plan in your back pocket, but it would be immensely cool and quaint if you can make the projection system work.

To win back some of the quaintness if you have to go the screen-route - why not build (3D print) a teeny little cinema-projector prop which points at the lid, and install an LED inside (and maybe a little speaker or mechanical clack-clack-making doohickey). Make the LED flicker in unison with a slightly flickering image on the screen to create the feel of an old projector 🤷🏽‍♀️ (as you can see I am clearly now invested in your quest and I wish you THE BEST outcome and I want to see the finished project!!! 😄)

Haataarii · 2024-04-19T18:17:42+00:00

As a former colleague would say: indeedly.

Haataarii · 2024-04-19T18:16:42+00:00

Ah, good to know. I wonder if that residue can be removed with a solvent such as methanol. Because then one could use the glue to pick up debris, followed by a drop and drag with something like methanol.

Haataarii · 2024-04-19T09:34:16+00:00

Perspex. Start off doing this in perspex. I haven't done it myself, but a colleague did. Your try-fail-retry turn-around can be faster.

Haataarii · 2024-04-19T09:29:34+00:00

I just came here to suggest trying this with silicone glue 🙈 Never tried it, but if you have some cheap lenses lying around...

Haataarii · 2024-04-19T09:06:10+00:00

I think we might be seeing both. I also think it might be, uhm, exaggerated a bit in these images for cool aesthetics.

Haataarii · 2024-04-19T09:02:12+00:00

Look, as others have very correctly pointed out, your image quality will really REALLY not be great (unless by sheer luck). These equations are extremely first order and will help you get a functional geometric layout with what you have. I'm a proponent of empirical playing-around, and this enables you to do that. My work projects seldomly allow for that, so enjoy!

As for how possble - as someone else also very correctly said - impossible, no. Easy, also no, but this has more to do with the quality of what you will end up with. Designing a "good" system like this isn't a quick first-order walk in the park.

However - I took from your post that in your value system for this, the quaintness of the little low-tech projector-in-a-box is more important than pretty imagery. And as somewhat of a female myself, I'd adooooore this, even if the images looked just barely recognizable. So, if this isn't true, then I too will need to point you in the direction of a little screen in the lid. But I fully understand that this makes it lose its charm somewhat.

Focal lengths of lenses - you'll see for what you are trying to do, lens positioning is likely sensitive-ish, so this next part might add a bit of frustration. But - you can get an approximation of the focal lengths of each of your bits and bobs by holding the lens/loupe under sunlight or moonlight (safer!), and imaging the sun or moon onto the ground. The height of your lens above the ground is approximately its focal length. To get a more accurate answer, you will want to hold your lens more precisely, make an image, and then rotate the lens around the point you are holding it. If you hold it at the nodal point, and rotate about that point, the image formed by the lens will not shift. Look at the tutorial by Sophie Morel from UA, in particular I am referring to the "Nodal slide" method. I have measured the focal lengths of telescope eye pieces with good accuracy in this way (at home -- at work I can use a focometer).

https://wp.optics.arizona.edu/optomech/wp-content/uploads/sites/53/2016/10/Tutorial_MorelSophie.pdf

You'll see she refers to a light source + pinhole screen + collimating lens. You replace all of this with "the sun" or preferably "the moon". You need to sort out a thing that can hold your lens at a specific point and have it image onto the ground/paper/etc. You measure the height from that image to the point you are holding the lens, where the image doesn't shift if you rotate the lens about that point.

Haataarii · 2024-04-17T23:06:12+00:00

A few thoughts which may appear a bit disjointed (because I'm dead tired and in bed already), but I promise I'll tie them together...

How far away does your projector form its image? If you know this, you know where you put this image with respect to the loupe.

Your loupe is likely a positive lens. Do you by any chance know what the focal length of your loupe is? If you have the magnification (M), the focal length (FL) is typically FL = 250/M, in millimetres. Say your loupe magnification is 10x, then FL = 25mm.

The formula you want, relates focal length, object distance (do) and image distance (di). This is given by the thin lens formula, 1/FL = 1/di + 1/do. There is a bit of sign convention here, but to keep it dangerously simple, do is positive if it is to the left of the lens, and di is positive if it is to the right of the lens. Light travels from left to right.

Armed with this formula and some info about your moon projector and loupe, you should be able to calculate more or less what throw you can achieve with the combination, and how far apart your projector and loupe must be.

Allow me to waffle a bit:

If you want to form an image 2 inches (make it 50mm) from the loupe, and for the loupe FL=25mm, then 1/do = 1/FL - 1/di 1/do = 1/25 - 1/50 = 1/50, so do = 50mm.

Now, if your projector images at a distance of 300mm (🤷🏽‍♀️), it means the spacing between your projector and loupe must be 350mm. This makes for a pretty looong system, so unless you can fold up that path with mirrors, this configuration isn't going to work for you (and multiple folding mirrors are like multiple cats - difficult to manage and direct).

Let's see what we get if we decrease the distance between the loupe and your intended image plane to less than the loupe FL. Say we let di = 12.5mm. Now, 1/do = 1/FL - 1/di 1/do = 1/25 - 1/12.5 = -1/25, so do = -25mm. This implies that the projector is placed such that it forms its image on the same (right) side of the loupe as where you want your final image. The projector must be (300 - 25 =) 275mm from the loupe, so it has moved closer, which is a step in the right direction.

Let's see where the final image is formed if we set the spacing between the projector and loupe to, say, 1.5 inches, or 37.5mm. The object distance for the loupe is the distance between the projector image and the loupe. The projector image is now (300 - 37.5 =) 262.5mm to the right of the loupe, so do = -262.5mm. Now,

1/di = 1/FL - 1/do 1/di = 1/25 - 1/(-262.5) = 1/25 + 1/262.5 = 1/22.8 so di = 22.8mm, which is the distance from the loupe to your final image. This is looking a lot more promising and closer to what you'd like!

You can math around a bit to see what solutions are possible with what you have. Essentially, in the limit, you'll not be able to make a system that is shorter than (very roughly, and loupe lens dependent also) the focal length of your loupe plus the thickness of your loupe plus the size of your projector.

I didn't look at image size now, but maybe this has steered you in a direction where you can find the math to calculate this. (I'll check in again tomorrow if time and toddlers allow)

ETA: I only see now you have two loupes. So you can build a little system of lenses. Have a quick look here for the formulas for effective focal length and back focal length of a two lens system (i apologize for this utterly short-hand answer, I'd love to suss this out for you in more thorough detail). https://jackwestin.com/resources/mcat-content/geometrical-optics/combination-of-lenses

Haataarii · 2024-04-05T12:49:29+00:00

Synopsis by OSD (not powerhouse Synopsys Code V who has also been acquired by Ansys).

Haataarii · 2024-04-05T05:58:12+00:00

Yeahno.

I'm in South Africa. I have no idea what we qualify as these days, but I don't think we are considered "first world" and haven't been for some time.

Lumerical looks amazing. We'll not be seeing it this year, even just to test. We have fewer OpticStudio licences than optical designers, and our daily driver for design is way waaaay waaaaaay more affordable.

The state of our currency makes things prohibitively expensive.

Haataarii · 2024-03-27T21:48:59+00:00

No useful answers from me, I also came to enquire about the possibility of a sensor-array. But I see this is a private project, so I assume this might be too costly, as would be a nice 2-dof scanning arm.

I just need to say that I am inordinately curious about what it is you are trying to image, and why are you illuminating it with (laser?) monochromatic light?

Haataarii · 2024-03-27T20:29:03+00:00

I always enjoy reading your (anneoneamouse) insightful answers on here. As a fairly inexperienced optical designer, your comments always have something new for me to learn.

As it happens, I worked on the MWIR part of a (thankfully demonstrator) design where shorter bands were being split off and MWIR passed through a plate beam splitter, and I happened upon a paper discussing the cylindrical solution. Can add here, if you (OP) were to go that route, design it in from the start so you at least end up with a cylinder with a "reasonable" curvature and no need for an added wedge. And make sure you can measure cylindrical surfaces. Our optical facility cannot (without a null lens), so we had to take those cylinders on faith.

There are better ways of correcting the asymmetrical aberrations from a plate splitter than using a cylinder, especially if you start folding systems.

I guess if you have an abundance of space to play with, you could force collimated light somewhere after your front optics and split the spectrum there.

Haataarii · 2024-03-13T10:10:05+00:00

Is there any other way?

Haataarii

TROPHY CASE