The 2023 IODC Shafer Cup Competition – FAQ

The Yin Yang Lens

Frequently Asked (and Unasked) Questions

Q.    Why did you select the name Yin-Yang Lens?

A.    The concept of yin and yang as positive and negative aspects of the same object is similar to the goodness and badness of color correction at different wavelengths. Plus, we needed a catchy name, and the other choices we considered were not as good. For the record, the name was suggested by Rich Pfisterer of Photon Engineering (he was not told the actual problem, only that it had to do with goodness and badness in lenses).

Q.    Why are the paraxial image distances for the two wavelengths required to be the same? In most achromat designs these are allowed to differ as long as the color correction is achieved to the desired level.

A.    We want the increase in wavefront error at 540 nm relative to 550 nm not to be due to simple axial color but to some actual designed-in image aberration, such as spherochromatism. There are no freebies in this problem!

Q.    Does the common image distance used for the two wavelengths need to be at the paraxial focus?

A.    No.

Q.    Why are cemented surfaces not allowed?

A.    Cemented surfaces can allow high angles of incidence, especially at steeply curved buried surfaces, which can produce high spherochromatism. Without the cement (i.e., with air between the surfaces), these high angles of incidence may cause TIR, forcing the surfaces to be less steeply curved. So basically, the answer is to make you work harder at making the 540 nm performance bad.

Q.    Why are intermediate images not allowed?

A.    First of all, it would probably be difficult to include an intermediate image in the 500 mm overall length restriction. Secondly, and mainly, the spherochromatism is added to by each intermediate image. If we allowed intermediate images, it might be possible to design a system with multiple intermediate images where the spherochromatism is added by each subsystem to the previous spherochromatism sum. Thirdly (of course), it is to make you work harder to introduce the spherochromatism!

Q.    Why is there a length restriction?

A.    To avoid entries that are so long that it is not possible to adequately see the lens details. In past IODC lens design problems, some entries, especially many of the higher-scoring ones, were very long (kilometers or longer) and it was often very difficult, or impossible, to see the actual lens structure. The problem committee simply decided to limit the overall length to avoid this (and, of course, to make the problem harder).

Q.    Are semi-fields of view greater than 90° allowed?

A.    No. The requirement that the object be flat at infinity limits the maximum semi-field angle to 90°.

Q.    Are very fast f/numbers (e.g., faster than f/0.5) allowed?

A.    There is no restriction. However, the lens is required to be diffraction-limited at 550 nm, which usually results in the lens obeying the sine condition. This would normally limit the f/number to be f/0.5 or slower. However, if you can design a lens faster than f/0.5 that meets the RMS wavefront requirement over the field of view, more power to you (pun intended).

Q.    Why the restriction to use just N-BK7 and N-SF6 glasses?

A.    These two glasses have significantly different refractive indices. If all glasses were allowed, then the lens could be designed monochromatically for high etendue at 550 nm (especially with a high index glass), and then lens splitting could be done with a glass with similar refractive index but with different dispersion. This would introduce the needed spherochromatism without significantly upsetting the monochromatic aberration balance. This makes the problem too easy. With significantly different refractive indices, adding the second glass to introduce the spherochromatism at 540 nm also upsets the monochromatic aberrations at 550 nm and thus forces the designer to rebalance all the aberrations. We want you to work to win the Shafer Cup!

Q.    Can we use TIR within any optical element?

A.    The problem description prohibits any TIR action. All non-blocked rays at both wavelengths must refract at every surface. Also, all rays must refract only through the surface’s front and back surfaces and not touch any lens’s edge.

Q.    Does lateral color need to be corrected?

A.    No. Only the on-axis image quality (or lack of it) at 540 nm is considered.

Q.    Do we need to allow for extra lens diameter beyond the clear apertures for mounting?

A.    No. Lenses are allowed to go to zero edge thickness at the maximum clear aperture, which would not leave any extra diameter for mounting. Lens thicknesses are also allowed to go to zero thickness, and lens spacings are allowed to go to zero at the axis and at the clear apertures. Of course, none of this is realistic from a manufacturing or mounting standpoint, but the Lens Design Problem has never been practical or realistic!

Q.    Are there any restrictions on the aperture stop location?

A.    The aperture stop can be anywhere between the object and the image. If the aperture stop is not on a lens’s front or rear surface, then a plano dummy surface must be used to define the stop surface (note that this can be inside an optical element). The stop surface cannot be virtual (negative thickness from the stop to the first optical surface), which, although it ray traces correctly, is not physically possible.

Q.    Is vignetting allowed?

A.    The intent is to have no vignetting, or ray clipping, by any surface other than the stop surface. The stop surface is a physical aperture somewhere in the lens system. The clear aperture radius of the stop surface is the height of the on-axis real marginal ray at 550 nm at the stop surface. Note that this means the stop diameter is the same for both wavelengths, and also means the entrance pupil diameter (which is sized to meet the stop diameter) may end up different for the two wavelengths. For any point in the field of view, all the rays at both wavelengths that hit the stop surface within this clear aperture must not be blocked (i.e., must make it to the image plane). All the rays that hit the stop surface outside this clear aperture are blocked; this would normally occur only if you have pupil aberration in your system between the entrance pupil and the stop surface. If you have pupil aberration in your system, depending on your lens design program you may require vignetting factors to shape the entrance pupil for off-axis field points to match the stop aperture. Note that since for any field point the chief ray goes through the center of the stop surface, the chief ray can never be blocked.

Q.    Why are piston and tilt removed prior to the RMS wavefront error calculation, but focus is not removed?

A.    The RMS wavefront error is computed by comparing the actual wavefront to a reference sphere centered at the chief ray intersection at the image surface. The radius of the reference sphere is the distance from the chief ray intersection on the image surface to the axial location of the real exit pupil for that field point. If the radius of the reference sphere is not an exact multiple of the wavelength, then there will be a constant term in the wavefront. This is piston, and can be large relative to the wavefront error due to aberrations. This piston term is removed so only the residual wavefront error due to aberrations is included in the calculation.

Removing tilt from the wavefront is accomplished by shifting the center of the reference sphere laterally from the chief ray location so as to get a better fit to the wavefront. It effectively tilts the reference sphere relative to the wavefront to get a better fit. This lateral image shift is equivalent to a distortion. Since distortion is not specified for this problem, the tilt component of the wavefront is also removed.

Removing residual focus error from a wavefront is accomplished by shifting the center of the reference sphere axially. This is the same as adding a focus shift to the image, and the amount of the focus shift may vary across the field of view due to curvature of field or other causes. We want to evaluate the wavefront error across the field of view at a fixed image location, namely, the image location specified in the lens prescription. So any focus error in the wavefront is not removed.

Q.    How is the RMS wavefront value calculated?

A.    A uniform, rectangular 50×50 grid of rays is traced across the entrance pupil for each field point. These rays are traced to the exit pupil and the RMS of their OPD errors computed. All the rays which make it to the exit pupil without being blocked by the aperture stop are included in the RMS calculation. For the 550 nm wavelength, this is computed for 25 field angles uniformly spaced across the specified semi-field of view, and each of these field points must meet the £ 0.070 wave requirement. For the 540 nm wavelength, the calculation is only performed on-axis.

Q.    What happens if you calculate that my lens has an RMS wavefront error at some point in the field of view greater than 0.070 wave, even if I think it meets the requirement?

A.    This may happen because different lens design programs may use different numbers of rays or use different algorithms to compute the RMS wavefront error. To eliminate these differences, all the entries will be converted to CODE V format and evaluated using CODE V. This ensures that all the entries will be evaluated equally.

If the RMS wavefront error calculated by CODE V is greater than 0.070 wave at some point in the field of view, the evaluator will reduce the field of view and/or the entrance pupil diameter until the wavefront requirement is satisfied. This is tried to be done so as to minimize the overall impact on the merit function – but this is not guaranteed! Note that any entrance pupil diameter and/or field of view adjustments made by the evaluator are final and no appeal is allowed. (Of course, you will not know that your parameters have been changed until the results are presented at the IODC, and by that time it is too late to complain!)

Q.    What if my values for entrance pupil diameter, field of view, or the 540 nm RMS wavefront error differ from the evaluator’s values for my lens?

A.    The evaluator’s values win, of course. After all, the evaluator is the judge of which values are the right ones!

Q.    Who thought up this crazy lens design problem?

A.    Dave Shafer, of course. That is why we call the contest the Shafer Cup Competition!

Q.    Isthat Juergens idiot going to run the problem again? Isn’t he getting too old for this?

A.    Unfortunately, yes, he is going to both chair the problem and give the presentation at the IODC. However, do not despair – after all, he is getting old and is bound to retire someday!

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