Is the Ideal Lens Calculation Patentable?

[kokes]

I have patent pending on this. I am publishing despite that. Patents last 20 years at the most, plus it can be researched already.

I choose a focal point. I dirve rays from it in a fan fashion toward lens, which for now is in shape of mere segment. The segment has its position in horizontal plane and its given height. For each ray I can calculate what inclination will our now straight curve have in point of intersection so that refracted ray leads horizontally. I move points of the curve so that it corresponds with calculated derivatives.

But this moves the points into different distance from focal point and originally calculated derivatives will no longer refract the rays horizontally. I repeat the process with more and more exact curve, until I reach required tolerance.

The script can be downloaded from github: https://github.com/jankokes/idealLens

EDIT: This thread should have been called "recalculated". To me it was extremely dificult to find any info on the ideal lens, so I thought it would be nice if I explained where I was wrong and which post shows correct answer right at the top. Jan

 

[berkeman]

I have patent pending on this. I am publishing despite that. Patents last 20 years at the most, plus it can be researched already.

I choose a focal point. I dirve rays from it in a fan fashion toward lens, which for now is in shape of mere segment. The segment has its position in horizontal plane and its given height. For each ray I can calculate what inclination will our now straight curve have in point of intersection so that refracted ray leads horizontally. I move points of the curve so that it corresponds with calculated derivatives.

View attachment 221342

But this moves the points into different distance from focal point and originally calculated derivatives will no longer refract the rays horizontally. I repeat the process with more and more exact curve, until I reach required tolerance.

The script can be downloaded from github: https://github.com/jankokes/idealLens

Welcome back to the PF.

Your description is not very clear, and I don't think folks are going to want to run any scripts on their PCs to try to figure out what you are saying. What is different in your procedure compared to standard ray tracing techniques of lens design?

Also, your figure is not very helpful -- do you have any PDFs of your technique? And how are you dealing with chromatic and spherical aberrations in your technique?

 

[Dale]

I have patent pending on this. I am publishing despite that.

If you already filed then that is fine.

 

[kokes]

Well, the main difference is that I calculated it. I firmly believe that someone wanting to understand this will not mind html as it is generally considered safe. Especially with no libraries included. Well, there is javascript, there is jquery... Programmers will understand.

Yes, I filed already.

This may sound little too selfish, but I think one day people will be amazed how this thread got moved from advanced to intermediate. I mean the lens has been known since old Babylonia and no one calculated the ideal lens before. Not even Newton. But that's an off topic. If you need it, you know where to find it. I have no pdfs yet and I am working on lens making machine since I need the lens for 3D printer, and I found it didn't exist. It exists now.

 

[Drakkith]

I choose a focal point. I dirve rays from it in a fan fashion toward lens, which for now is in shape of mere segment. The segment has its position in horizontal plane and its given height. For each ray I can calculate what inclination will our now straight curve have in point of intersection so that refracted ray leads horizontally. I move points of the curve so that it corresponds with calculated derivatives.

I'm guessing your lens ends up being shaped like a parabola?

This may sound little too selfish, but I think one day people will be amazed how this thread got moved from advanced to intermediate. I mean the lens has been known since old Babylonia and no one calculated the ideal lens before. Not even Newton. But that's an off topic. If you need it, you know where to find it. I have no pdfs yet and I am working on lens making machine since I need the lens for 3D printer, and I found it didn't exist. It exists now.

How does your lens correct chromatic aberration, coma, and astigmatism?

Well, the main difference is that I calculated it. I firmly believe that someone wanting to understand this will not mind html as it is generally considered safe. Especially with no libraries included. Well, there is javascript, there is jquery... Programmers will understand.

I'm not really a programmer, so I have no idea how to view your design, but I am an optical engineering student, the kind of person you'd want to easily be able to describe your lens to. Making it so that programmers are likely to be the only ones who will see your lens design runs counter to the idea of wanting to put your design out to the people who actually understand it. If you're not willing to put your design into a format that can be easily viewed by those people best able to evaluate and use it, why put it out at all? It's just lazy and it makes you look extremely bad.

 

[kokes]

Unfortunately it is not parabola. It is ultimately unknown curve. I ruled out all the goniometric and exponential functions. I figured it is because of how it is created. If I had mathematical formula I would know the points of the curve directly, and I would approximate the derivatives. As is the case I calculate the derivatives first and approximate the curve from there. I create it. I made a set of functions and was testing which one it is. It is none. That is why I publish the program. It is html, you can view the curve in real time as you change the parameters, and in console you will find svg that can be imported into CAD or other software.

 

[berkeman]

Unfortunately it is not parabola. It is ultimately unknown curve. I ruled out all the goniometric and exponential functions. I figured it is because of how it is created. If I had mathematical formula I would know the points of the curve directly, and I would approximate the derivatives. As is the case I calculate the derivatives first and approximate the curve from there. I create it. I made a set of functions and was testing which one it is. It is none. That is why I publish the program. It is html, you can view the curve in real time as you change the parameters, and in console you will find svg that can be imported into CAD or other software.

You have not addressed our aberration questions yet. Have you used your process on multi-element lenses and been able to correct for aberrations?

 

[kokes]

Sorry. I made single interface of two materials, ie air and glass. This permits plano-covex or plano-concave lenses only. I can take it further, or anyone else, because first example is provided. I want to test it now. I imported it into Blender (3D software) and ran parallel rays through to see what Blender's internal engine thinks about the curve. I found the result satisfactory, so I am making the real lens now.

 

[Drakkith]

Unfortunately it is not parabola. It is ultimately unknown curve.

I'm nearly certain that the shape should be a parabola. I know for a fact that, for a mirror, the only shape which works is a parabola. Since the only difference between a lens and a mirror is that the index of refraction of the mirror is negative, the shape should remain the same, just stretched differently. This is identical to how two lenses with the same focal length but different refractive indices must have different strengths for their curvatures, but both can remain spherical. Just changing the refractive index does not mean you must change the shape.

Sorry. I made single interface of two materials, ie air and glass. This permits plano-covex or plano-concave lenses only. I can take it further, or anyone else, because first example is provided. I want to test it now. I imported it into Blender (3D software) and ran parallel rays through to see what Blender's internal engine thinks about the curve. I found the result satisfactory, so I am making the real lens now.

What about aberrations? You still haven't addressed these.

 

[kokes]

Blender confired the geometrical aberrations are taken care of. With monochromatic light, of course. Chromatic aberration is still present. The best I know that works to mitigate it is a focal point far away.

 

[Drakkith]

Blender confired the geometrical aberrations are taken care of. With monochromatic light, of course.

Did you test for points not at the focal point?

 

[kokes]

That is so. I started at focal point. It made debugging much easier.

 

[sophiecentaur]

That is so. I started at focal point. It made debugging much easier.

An 'ideal' lens has to function over a range of object and image positions. You have written some code which produces a result but what is its significance in terms of a real lens? I cannot / will not try to run this code and nor will many readers of your posts. Is that a surprise to you? I am not sure what your aim is in this.

 

[Drakkith]

That is so. I started at focal point. It made debugging much easier.

A lens takes light emitted from different points in 'object space' and maps them to different points in 'image space'. The problem is that a single lens cannot do this perfectly for all points in object space and for all wavelengths. You will have aberrations. To minimize aberrations, lens designers have to optimize their lens designs. It is trivial to optimize a lens to function near-perfectly for a single point and a single wavelength, but lenses like these are not usually useful by themselves in the real world except in very specific, simplified cases.

Also, please note that we already know an enormous amount about how lenses work and what shapes they should and shouldn't be. We've been working with lenses for more than 400 years, after all. If you're interested in optics and lens design, I recommend that you pick up a book on the subject. Here's one example, though there are many others.

 

[Khashishi]

So, what exactly is novel here? We've had aspheric lenses for some time now.
https://en.wikipedia.org/wiki/Aspheric_lens

 

[kokes]

Thank you for your replies and for feedback. I just wanted to let you know that this can be done. I am fine with the idea no one will try the script. It represents ideal lens, which took long time to calculate, and I am glad I managed to do so.

This lens can concentrate rays into a single point. I am well aware of the fact that there are many lenses that try to do the same.

 

[Drakkith]

I just wanted to let you know that this can be done.

That what can be done? We already have lenses capable of bringing rays of a single wavelength emitted from a single point in object space to focus at a single point in image space (or as close as possible, limited by diffraction and lens imperfections).

 

[kokes]

I don't think it is possible to get "as close as possible" without the actual knowledge of the curve. I will appreciate valid links.

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[sophiecentaur]

I just wanted to let you know that this can be done.

We know it can be done because every decent lens you use will use ray tracing in its design (or some short cuts, no doubt). If you want more information about methods for this why no google "ray tracing lens design"? I did that and there were dozens of hits which would be much more relevant than I got with other search terms. Find one that is useful to you.

 

[Drakkith]

I don't think it is possible to get "as close as possible" without the actual knowledge of the curve. I will appreciate valid links.

My apologies, it appears the correct shape for a plano-convex lens is a hyperbolic shape, not a parabolic.
Here's a link: http://httprover2.blogspot.com/2012/03/plano-convex-hyperbolic-lens.html
https://wp.optics.arizona.edu/jgreivenkamp/wp-content/uploads/sites/11/2017/07/Perfect-Plano-Convex-Lens.pdf an example problem with solution by one of my professors where he requires that the shape of a perfect plano-convex lens be derived using both Fermat's principle and Snell's law. Both give you an exact shape in the form of a hyperbola. (if you can't access this link, let me know)

 

[kokes]

Thank you both. I looked at mathematical derivation. It looks impressive.

Section of hyperbola near peak could approximate the curve fairly well. However I must insist ideal lens is not hyperbic, because I must use only small section, as close to appex as possible, which disqualifies hyperbola in my eyes, as whole curve would have to produce the result. With hyperbola, the further I go, the more off I am. Which is the same as with sphere, only not as bad. Not an ideal lens.

I downloaded pyOptTool and will see if I can import my lens in it. I will also try hyperbola and see how they compare.

 

[jbriggs444]

However I must insist ideal lens is not hyperbic, because I must use only small section, as close to appex as possible

Perhaps you need to rephrase. You are agreeing that a hyperboloid is optimal but complaining that a portion of a hyperboloid is not?

 

[kokes]

Yes, this is the case. See my screencast below.
One cannot use whole hyperbola, only infinately small part of it, such as

Lim x -> 1 (y = 1/x)

as shown here:


As for the blog link I am not certain what the gentleman calculated and how he did it. Other people have already asked questions to no avail. The curve isn't too close to mine:

As for the mathematical paper, I am not all that good at diferential equations. Let me assume it is correct.

I tested more recent OpticalRayTracer where hyperbolic lens can be simply chosen. The curve is much closer to what I have. The focal point is not ideal, but close:

I failed to install some libs for pyoptools yesterday (no internet at home), but I downloaded a showcase video about it which shows set of three lenses to bring rays into common points (yet it never shows the focus clearly).

If I understand correcly, the situation up till today has been that someone theoretically calculated the curve and the calculation was obviously correct. The only problem was that whoever tried running it through ray tracers found out it wouldn't focus rays into a common point. The curve they got was close to what I get after my first approximation. I suspect I would end up with similar result if I didn't realize I need to repeat the processes recursively.

So if I am correct we were all correct, except I was much more correct since I did actually have the curve and the algorithm. In addition I am presenting to your attention a mathematical formula that works. (However I don't think it is plausible to use it since it requires infinately small limits stretched out later on along with heavy usage of differential equations which I am yet to see used correctly, while my algorithm utilizes nothing but Snell's law.)

 

[sophiecentaur]

So if I am correct we were all correct, except I was much more correct since I did actually have the curve and the algorithm.

You have 'an' algorithm and not necessarily 'the' algorithm. in post #20 there is a link which describes an algorithm that doe not agree with your result. That algorithm shows that the focus is 'very near' a point.
Ray tracing methods are all subject to error so, perhaps you could re examine your algorithm and check before assuming you are 'right'.

 

[jbriggs444]

I tested more recent OpticalRayTracer where hyperbolic lens can be simply chosen.

The lens shown in the diagram following this passage showed a biconvex lens. The hyperbolic lens that is proven optimal is a plano-convex lens.

There is no conflict between claims of optimality for both designs.

 

[Drakkith]

I tested more recent OpticalRayTracer where hyperbolic lens can be simply chosen. The curve is much closer to what I have. The focal point is not ideal, but close:

You have not chosen the correct shapes. I've attached a screenshot showing a design that focuses rays to a point all the way down to the limit of the program's ability to display it. If I zoom in any further the rays simply disappear. The magnification is ~33,700 and moving my mouse cursor from the left side of the screen to the right side does not change the X coordinate shown in the bottom left. It remains at 19.1681. This means that, if we take our units to be millimeters, the entire screen, from left to right, does not take up more than 100 nanometers. If we use 100 as the upper limit, then the grid lines are, at most, a little less than 20 nanometers apart (see note below screenshot), and the rays focus to a point with an error less than 1 nanometer. The smallest wavelength of violet light is roughly 380 nm, so this raytrace shows that the rays focus to a point with an error less than 1/380th the size of the smallest wavelength of light a person can see.

This is, to within the limit of the program's ability to show, an absolutely perfect diffraction limited lens.

Edit: The original screenshot I uploaded showed 5 gridlines, but I had to crop the image to make the details viewable and now it only shows 4. However you can use the values shown in the image and see for yourself that what I typed above is correct.

 

[kokes]

Sophiecentaur: I agree that proving the result is correct is the most difficult part. My algorith may be wrong. However my horizonal lines are horizontal. If you look inside the script, you will notice that I did not take points of the curve and move them horizonally, I took the points and orbited the focal point about the angle I calculated. And they came out perfectly parallel, which to me is proof, because this can be only achieved if both the algorithm and the ray tracer are without an error. However Drakkith has meanwhile presented another design that must have met the same criteria. In at least one instance two errors must have occurred that canceled each other out, and I need to dig now and find it.

jbriggs: Yes. I should add that in case I prove my design is indeed small section of hyperbola, it will send my own patent rights straight to garbage since I won't be able to prove which algorithm has been used. I only have patent pending on one of them.

Drakkith: Let me doublecheck first. In the worst case scenario I will find out what section of hyperbola must be used at the expense of me looking like an ignorat piece of mud. Thank you. I very much prefer that over not knowing the answer at all, since I desperately need a better 3D printer. Either way you helped me immensly.

horizontalLines.png in github

Check out y coordinates of points of curve vs. endpoints of horizontal lines. They are off by miniscule amount, which corresponds to chosen precision of 5 (meaning any ray must refract within 1/100.000 of 1° from horizontal). I hate suggesting this, but perhaps there is something wrong with the way Fermat's principle is used. I cannot prove it, but there are clues. When it was used during creation of ideal lens, its usage didn't result in definite shape. It was correct but incomplete. In all available raytracers Fermat's principle is used once again. This, along with notion that two errors must have occured, one during calculation and one during evaluation in raytracers, leads me to speculative conclusion that error is not in my reciever.

 

[sophiecentaur]

If you look inside the script,

Unless you are asking someone to check a small piece of code (and that would be on a different forum) you cannot expect us to reverse engineer what you have done and decide what your algorithm actually does. If you can't state what your algorithm actually does then it is a secret process as far as most people are concerned.
I suggest that, if you just built the code from scratch, it is a very unsafe approach and I would expect hidden errors to find their way in.

 

[jbriggs444]

jbriggs: Yes. I should add that in case I prove my design is indeed small section of hyperbola, it will send my own patent rights straight to garbage since I won't be able to prove which algorithm has been used. I only have patent pending on one of them.

If both designs were plano-convex (plane side facing the parallel rays at right angles, convex side facing the focal point) then correctness of the hyperbolic solution would automatically mean that any different plano-convex solution would automatically be invalid. (*)

However, as I understand it, that is not the case here. You have a symmetric, biconvex lens (convex with your special shape toward the parallel rays, convex with the same special shape toward the focal point). The shape of the required curve would be uniquely defined but would almost certainly be something other than hyperbolic.

More generally, one would expect there to be infinitely many optimal asymmetric (front and back faces not identically curved) designs.

(*) A piece-wise hyperbolic, Fresnel sort of variation could arguably be another optimal, shape for a lens with one planar face. A plano-convex lens canted with the plane face not at right angles to the incoming parallel rays might not result in a hyperbolic solution for the curved face.

 

[kokes]

The lenses seem identical. Hyperbolic factor is what determines what portion of hyperbola will be used. I basically calculated something using analytic geometry that has been calculated using algebra before. I will contact the administrator and ask to delete this thread as it brings nothing new. Thank you for opening my eyes, and thank you for your time. My deepest apologies.