I used my jig and projected a test slide on the wall from a fixed distance with each of my five lenses. I marked (and then later measured) the the four corner points of a 50x50mm square and the vertical and horizontal distances across the edges of the square.
|Lens No.||Serial No.||Condition||Diag.1/Diag.2||Sum Diags||Horiz./Vert.||Sum H+V||Relative FLs (diag)||Relative FLs (h+v)||Distortion*|
|1||35915||EX, f3.2 viewer||619.0/618.0 mm||1237.0||436.5/433.0||869.5||Reference lens (0.0%)||Reference lens (0.0%)||0.60%|
|2||29432||EX, f3.2 viewer||622.0/621.5 mm||1243.5||438.0/434.5||872.5||0.53%||0.34%||0.78%|
|3||21316||EX, f3.5 viewer||621.5/620.0 mm||1241.5||437.5/435.5||873.0||0.36%||0.23%||0.56%|
|4||14469||rear element separation||626.0/625.0 mm||1251.0||440.5/438.0||878.5||1.13%||0.92%||0.69%|
|5||26372||front coating scratched||620.5/619.5 mm||1240.0||not avail./435.5||435.5 (V only)||0.24%||0.58% (based on Vert. only)||0.67% (based on Vert. only)|
* Distortion is the amount the diagonal is longer than you would expect from Diag^2 = H^2 + V^2.
Marking and measuring the distances is probably accurate to with 1mm (+/- 0.5 mm). Thus diagonal measurements are accurate to 0.08% and comparisons between diagonals are accurate to 0.2% or so. Thus I am fairly confident that the top 3 lenses are close matches. The measurements of both diagonals and the horizontal and verticals increases the confidence since they agree to with the 0.2% error.
The distortion measurments are most error-prone since you are basically subtracting two measurements. Thus their error rate is likely higher.
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