Creating a Manufacturable Lens Using an Extended Polynomial Surface

  • 16 September 2022
  • 5 replies

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          I am currently exploring different ways to create manufacturable lenses (circular, rectangular, plano-convex/concave) using different surfaces.  I am wondering what the “most efficient” surface is to use to create these types of basic lenses?  I created a plano-convex lens using the extended polynomial surface using 30 terms and am playing around with more and less terms to see the effects.  Is there a rule of thumb for the coefficients of the polynomials that they should be some order of magnitude  so the lens is realistically manufacturable?

Thanks in advance for any advise.


Best answer by Mark.Nicholson 17 September 2022, 20:32

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Wow, this is a great question. It goes right back to the foundations of optical design theory.

Remember that the definition of the optical power of a surface is


where delta_n is the change of index across the surface and curvature is

[from Local curvatures and its measurements of an optical surface or a wavefront: a review (]

The curvature therefore depends on the first and second derivatives of the surface sag. A flat surface has no power; a tilted surface has only prismatic power; and a spherical surface has power 1/R where R is the radius of curvature. The sphere has no higher-order derivatives than second, and so the power of a spherical surface is the same everywhere. If you add higher order (aspheric) terms then you get non-zero higher order derivatives and so power becomes a function of position on the surface too. Only the sphere has the property that the power is everywhere the same, as its first and second derivatives are everywhere the same.

Now when you approximate the sphere with polynomials you need to keep your fingers crossed that the polynomial series converges quickly. If you have terms going as fast as r^30, then you’ll have derivatives going as fast as r^29 and second derivatives going at r^28. Keeping everything balanced when you add manufacturing tolerances will be very hard. This is why most surface sag expressions tend to use a base sphere (strictly, a base conic asphere) plus other terms which are expected to converge quickly. But the manufacturing difficulty of a surface will go as the derivatives of the fastest significant derivatives.

Does that help? The paper I reference above just came out in the latest issue of Optical Engineering, and it’s a really super review of the whole area of curvature in arbitrary continuous surfaces. Highly recommended.

I should add that within OpticStudio there are soem excellent features to help you understand the curvature of a surface. Look on the Analysis tab, under Surface:



  • Mark
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Hi Mark,

               Thank you for the feedback.  The paper was very helpful.

I understand the features of analysis for OptisStudio and have been looking over the sag curvature and departure over the aperture of the lenses I have created.  From my perspective everything looks acceptable to a manufacturer in that the departure does not have steep drops and the slope is not so steep as to necessitate high precision manufacturing.

I know the Extended polynomial is “more accepted” by manufacturers and the “Q-Type Asphere” is less so.  With the extended polynomial you can’t tell which term is affecting performance, so you have to optimize carefully, and the Q-Type you can tell which term is affecting performance.

This is just from reading the OpticStudio documentation and some knowledge about manufacturers.  I was hoping to get information on the surface types from people who have experience in designing manufacturable lenses.  Is the choice of surface just manufacturability (i.e. the manufacture can physically use the surface specifications to create the lens) or optical designer preference?

Again thank you for the response.

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Let’s take the sphere as the gold standard for manufacturability, and say that however hard it is to make a sphere, test sphere and other test equipment, that the sphere represents some accepted standard of difficulty.

Plane surfaces are generally easier to make, and tilted planes (prisms) harder than planes but easier than spheres. This is because the first and second derivatives of the sag of a plane are zero, and the tilted surface has a first derivative but no second. Hence the sphere, with its curvature of 1/R everywhere is the standard.

As we add higher order derivatives to the sag, the derivatives of curvature (which depends on the ratio of second to first derivatives) becomes higher order too. For a rotationally symmetric surface, this tends to be a series of spatial frequencies going as p(rho)^2, p^4, p^6 etc.

Now by all means you can group these terms into an expression which is orthogonal, but they are still the same deviations. The deviations don’t become more or less important for being described by an orthogonal polynomial than a non-orthogonal polynomial. The insights gleaned, if any, accrue to the designer, not the manufacturer.

As you create ever more aspheric, but still rotationally symmetric lenses, we can assess manufacturability by treating the sag as the intended base sphere, plus an additional function that represents the deviations. This is traditionally taken as a Zernike expression, and so is orthogonal, although for lower spatial frequencies something like adding spherical and astigmatism can also be used. So the more complex a surface is, the more likely we are to use an orthogonal description of the deviations, for both the designer and manufacturer. Most wavefront analysis software describes the waveform produced as a base sphere plus Zernike terms, especially as we go to more complex surfaces.

Does this help? On re-reading it, I can’t help but feel it’s a long-winded way of saying planes are easier than spheres, spheres are easier than aspheres, and mild aspheres are easier than extreme ones. 🤓

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          I can’t properly express my gratitude in this medium for your description and the time you took to write it all out.  What you wrote is very clear to me and very helpful.

Thank you!

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My pleasure! I had to go back to square one for this question...haven’t been there fort a long time!