The Schoenberg Lens
Frequently Asked (and Unasked) Questions
Q. What is the chromatic scale that Schoenberg refers to?
A. The chromatic scale comprises twelve notes across a full octave, where an octave is a doubling of the frequency. The frequency ratio between notes is constant. Going from C to the C of the next higher octave represents a doubling of frequency, so the ratio of the frequency of a note to that of the preceding note is . On a piano, this corresponds to the keys of a full octave, for example starting with C and ending with B. There are 7 white keys and 5 black keys: C, C♯, D, D♯, E, F, F♯, G, G♯, A, A♯, and B, where the sharps (♯) are the black keys and the un-sharped (or natural) notes are the white keys.
Q. How was the twelve tone technique applied in practice in musical composition?
A. A starting pattern of a particular sequence of the 12 notes, called the tone row, would be established using all 12 notes. All successive groups of 12 notes in the composition would then need to use this tone row either as is, inverted, backwards (retrograde), shifted, or as a combination of these. This allowed up to 48 variations of the original tone row that could be played in successive groups of 12 notes.
Q. Does the Schoenberg Lens use those same rules for the 10 digits?
A. No. Rather than forcing one particular sequence of the 10 digits to be used as is, reversed, or shifted, the Schoenberg lens allows the digits to be used in any order for the four separate lens parameters of each lens.
Q. Is there a limitation to the number of digits that can be used for any given parameter?
A. No. The 10 digits can be distributed however you want among the four parameters. The value of any given parameter can have anywhere from one to seven significant digits. All the significant digits before and after the decimal point count. If you do use seven digits for a given parameter, then the other three parameters can have only single digit integer values for a total of 10 digits.
Q. Does a single digit before or after the decimal count as using only one digit?
A. Yes and no. A single digit before the decimal (e.g., a thickness of 1 mm) counts as using only one digit. However, a single digit after the decimal (e.g., a thickness of 0.5 mm) counts as using two digits, the leading zero and the 5 digit.
Q. How do I use the digit 0?
A. The digit 0 must have significance. It can be used in three different ways. The first way is as a value of 0, for example a lens thickness of 0, an air spacing of 0, or the radius of a plano surface, which is conventionally specified as 0 in most lens design programs. The second way is as a leading 0 of a number less than 1, e.g., 0.5. The third way is imbedded in other numbers such as 10, 105, 1.05, etc.
Q. What if, for example, I specify a thickness to be 2.3 but the resulting lens file output by the design program lists it as 2.299999 or 2.3000001?
A. Since lens design programs do not have infinite precision, this often occurs. The evaluators will simply round the number as the actual number intended would be obvious.
Q. What happens if I accidentally use more than ten digits, repeat a digit, or omit a digit for a given lens?
A. If there is time (i.e., you did not submit your entry at the last minute), the evaluators will contact you about it and give you a chance to correct your design. If you do not correct it, your entry will be disqualified.
Q. Do I have to use all ten digits for each lens?
A. Yes.
Q. Can I omit a digit from one lens and then use it twice in another lens? Overall, this still uses all the digits the same number of times.
A. No. All ten digits must be used for each lens.
Q. Why is the focal length specified as 98.7 ± 0.1 and the overall length specified as ≤ 523.46 mm? These seem to be unusual specifications.
A. The focal length specification and overall length specification use all 10 digits once (9870152346). This is an application of the Schoenberg Rule to the specifications. Your focal length and overall length values do not need to meet the Schoenberg Rule (although they must meet the specifications)
Q. The wavelength of 587.6 nm is the standard d-line. Why is the index specified as 1.49032? I do not know of any real glass or material that meets this.
A. The wavelength and index value also use all 10 digits once (5876149032). This is another application of the Schoenberg Rule to the specifications.
Q. Why is the RMS wavefront error specified as ≤ 70.1 milliwaves rather than ≤ 0.0701 wave? And why the unusual distortion specification of ≤ 9.865432%?
A. These two specifications as written also use all 10 digits only once (7019865432). This is yet another application of the Schoenberg Rule to the specifications. Your RMS wavefront error and distortion values do not need to meet the Schoenberg Rule (although they must meet the specifications).
Q. Why are intermediate images not allowed?
A. First of all, it would probably be difficult to include an intermediate image in the 523.46 mm overall length restriction. Secondly, and mainly, it is to make all the entries have similar design forms so as to not favor one design form over another.
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. Why is an extra dummy surface allowed before the image surface?
A. This is to give extra flexibility, if needed, to meet the RMS wavefront requirement. Not every requirement is designed to make the problem harder!
Q. Can the length of the dummy surface inserted before the image surface be negative?
A. No.
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 near diffraction-limited, 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. Can we use TIR within any optical element?
A. No. All non-blocked rays 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. Why are cemented doublets not allowed??
A. In a cemented doublet, the back radius of the first lens and the front radius of the following lens are the same and there is no airspace. Since both glasses are the same, this effectively makes a single fatter lens with the cemented surface meaningless. Per the Schoenberg Rule for each lens, however, this would allow the digits of the front surface of the first lens to be used again at the back surface of the following lens. This violates the spirit of the Schoenberg Rule for the overall fat lens. Thus, cemented doublets are not allowed.
Q. Can I have a “cemented” doublet but with an airspace of 0 between the two lenses?
A. No. Assuming the airspace does not cause total internal reflection, this also would be equivalent to an overall fatter single lens since both glasses are the same. This would allow the digits of the front surface to be used again at the back surface, violating the spirit of the Schoenberg Rule. Thus, this is not allowed.
Note, however, that air spaces with zero thickness are allowed. In such a case, not only do both lenses need to follow the Schoenberg Rule, the back radius of the first lens and the front radius of the following lens must have different values and also assure non-negative edge thickness of the air gap. If both radii are identical, the design would be considered as a cemented element and not be allowed.
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. Where must the aperture stop be located?
A. The aperture stop must be on a plano dummy surface between two lenses. The aperture stop is then treated as an element in the overall lens. For the lens preceding the stop, the 10 digits are allocated among that lens’s two radii, center thickness, and the distance to the stop. The stop itself is plano, so its radius is 0, using that digit. The other 9 digits must then all be used in the value of the spacing to the next lens. Curved dummy stops are not allowed.
Q. Is vignetting allowed?
A. The requirement 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 on the stop surface. For any point in the field of view, all the rays 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.
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 portion of 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 the lens must meet the distortion specification, these lateral shifts will all be within the distortion specification. Therefore, tilt can be removed from the wavefront to provide a better fit to the wavefront and improve the RMS wavefront error.
Removing residual focus error from a wavefront is accomplished by shifting the center of the reference sphere along the chief ray, either forward or aft of the image surface to obtain a better fit to the wavefront. 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. Therefore, any focus error in the wavefront is not removed.
Q. How is the RMS wavefront value calculated?
A. A uniform, rectangular grid of rays is traced across the entrance pupil for each field point. The density of this grid is adjusted as needed to ensure accurate calculation of the wavefront error. 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. 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 £ 0.0701 wave requirement.
Q. What happens if you calculate that my lens has an RMS wavefront error at some point in the field of view greater than 0.0701 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 0.0701 wave at some point in the field of view, the evaluators will reduce the field of view and/or the entrance pupil diameter until the wavefront requirement is satisfied. This will be tried to be done in a way which minimizes 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 evaluators 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 and/or field of view 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. Unlike the previous seven IODCs where Dave Shafer came up with the problem idea, this problem is the brainchild of Thomas Nobis. The Lens Design Problem committee consisted of Julie Bentley (University of Rochester – chair), Thomas Nobis (Zeiss), Richard Juergens (retired), Scott Sparrold (Keysight/CODE V), and Erin Elliott. (Ansys/Zemax).
Q. Isthat Juergens idiot going to run the problem again? Isn’t he getting too old for this?
A. No. Juergens has finally retired. Julie Bentley has assumed chairing the problem. It is about time we had someone running the IODC Lens Design Problem who actually knows something about optics!
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