I wondered if there may be different thicknesses of foil due to focal length but at .003" it appears to be the same for 900mm as 710mm.
I'm not sure how the math works for achromatic elements, my understanding is the 4 surfaces (1&2 crown,3&4 flint) usually have surfaces 2&3 that define the air space between crown/flint as almost identical radii, with surface 1&4 determining the focal length? maybe someone who knows how element design works could confirm or correct?
if i'm correct then surfaces 2&3 could be the same across many focal lengths? just guessing i know too little about the design:)
Edmund used .005" air space on the 4 inch objectives they imported from Japan as noted in Sam Browns book All about telescopes on page 177. When they started designing their own in house with Dr. Rank they changed to .004 air space. This is the specs on the in house objective originally dated 8-10-78 when designed and updated with Eng changes through 4-24-81. I find it very interesting that 2/3 down the page they annotate the following: "All surfaces figured to 1/8 wave + or - 15 rings.
What does "15 rings" mean? Newton rings? Maybe one of our resident experts can tell us.