I have recently acquired a Meopta Belar 75mm/f4.5, to use with my Durst M605 enlarger. I was remiss in not checking the mount of this lens (M23.5) before I bought it, and have therefore ended up with a lens much too small for any of the Durst lens boards (M39) I currently own. For this reason I intend to 3D print a new lens board, but I am stuck with the following:

As you can tell from the image below, lens boards mount lenses recessed at various 'depths'. What determines how deeply recessed a lens should be mounted? I intend to make prints of 30x40cm from 6x6 negatives.

enter image description here


3 Answers 3


Years ago, I fitted enlarger lenses to high-speed photofinishing printers I using this this math. It proved accurate enough.

Now for the gobbledygook:

The negative carrier masks the negative plus we desire some over-spill at the easel. I use an over spill of 1.5% otherwise easel placement to avoid shabby borders is laborious.

We now figure magnification based on the cropping of the negative by the negative carrier and factoring in the over-spill.

A 35mm negative measures 24mm height by 36mm length. The negative carrier crops these dimensions, , we will use 23.5mm height by 35.5mm length. Since we desire an 8x10 inch print, we will use, for the paper size 203mm height by 254mm length.

To figure the needed magnification we divide paper dimension by negative dimension. The 35mm frame is an elongated rectangle, the 8x10 paper size is more square. We must calculate the magnification requirement for both height and width and then use the greater value.

For height: 203 ÷ 23.5 = 8.64X For length: 254 ÷ 35.5 = 7.15X

We choose 8.64X and apply a 1.5% factor to allow some over-spill.
Thus 8.64 X 1.015 = 8.77 we will use 8.77 as the desired magnification.

Now we compute lens to negative distance base on focal length lens and magnification required. This task places the lens further away from the negative than the focal length. In other words, to achieve this magnification, we compute the back-focus distance. Say we mount a 75mm enlarging lens.

Next we compute the lens to negative distance based on published lens focal length and operating magnification.

Back-focus distance: focal length multiplied by magnification plus 1 divided by magnification. This is the lens to negative distance.

For a 75mm lens magnification 8.77X: 75 X (8.77 +1) ÷ 8.77 75 X 9.77 ÷ 8.77 = 83.6mm

This answer i.e. 83.6mm is approximate but good enough for what you want to accomplish

  • \$\begingroup\$ Thanks Alan, this is a great answer as it allows me to accurately calculate what I need to know! Is the outcome from these calculations a minimal or maximal distance-from-negative value? \$\endgroup\$
    – timvrhn
    Commented Jan 25, 2020 at 0:17
  • \$\begingroup\$ I'd still love to hear from you \$\endgroup\$
    – timvrhn
    Commented Jan 27, 2020 at 7:29
  • \$\begingroup\$ @ timvhn - A lens has two cardinal points. We measure object distance from one of these and image distance from the other. We don’t know where they fall, so best make our measure from about the center of the lens barrel (length). This formula computes lens-to-negative distance for a specified magnification. \$\endgroup\$ Commented Jan 27, 2020 at 14:57
  • \$\begingroup\$ Alan from what I gather the formula returns the distance for one single magnification. I want to make enlargements at smaller magnifications with the same lens board, can I do that with this distance (and by lowering the enlarger head)? \$\endgroup\$
    – timvrhn
    Commented Jan 27, 2020 at 16:11
  • 1
    \$\begingroup\$ I showed you how to compute magnification given negative size and projected print size. You must do this for your typical enlarger usage. Once you know the magnification for a specific setup, use that data plus focal length of the lens used. Now use the formula to compute negative-to-lens distance. \$\endgroup\$ Commented Jan 28, 2020 at 2:31

The recessed board helps in focusing a very short focal lenght enlarging lens (say 30mm, the kind used for enlarging 110 film). A short lens requires such a short distance from the film plane that the bellows extended to minimum would still not suffice, and had to be recessed.

This was common issue only for lenses for smaller format that 35mm, which was obsolete by the 70's or so... These can still be theoretically used for enlarging large details from 35mm film, but today are more of a curiosity item.

On the Durst 605 model neither 50 nor 75 or 80mm focal lens requires a recessed board, the regular is just fine.

On a side note: the Belar is a triplet, and was considered a budget item in the Meopta lineup. It is worse than worthless now, and does not deserve the kind of attention you are considering. You will be better off by throwing it out and getting a decent 4 element lens (Anaret in Meopta lineup) or even better a 6 element lens (Meogon in Meopta lineup, or one of the Germans - Rodagon, Componon, or Neonon by Durst).

  • \$\begingroup\$ Thank you very much! Just the answer I was looking for. \$\endgroup\$
    – timvrhn
    Commented Jan 24, 2020 at 9:29
  • \$\begingroup\$ Jindra, what is your take on the E-Ikor 75mm/4.5? \$\endgroup\$
    – timvrhn
    Commented Jan 24, 2020 at 9:40
  • 1
    \$\begingroup\$ @timvrhn - to be honest, I never heard about it. Given the current prices I would stick to reputable brands - Rodenstock, Schneider, Nikon; I am a Meopta fan mainly out of my Czech national pride :) If you are Europe based and looking for a purchase I suggest ffordes in the UK: the prices (and GBP exchange rate) seem right, and quality is good: ffordes.com/c/99/enlarging-lenses \$\endgroup\$ Commented Jan 24, 2020 at 12:07
  • \$\begingroup\$ Even the Apo-Rodagon-N, which is Rodenstock's most superior enlargement lens and quite a bit better than the 'plain' Rodagon lens, can be found for a fair price if you have a bit of patience. \$\endgroup\$
    – jarnbjo
    Commented Jan 24, 2020 at 18:38

Lenses are fitted to cameras and projectors (enlargers are subset of projectors) based on the diagonal measure of the film format. As an example, for 35mm film, the format is 24mm by 36mm with a diagonal of 43.3mm. This corner-to-corner measure is somewhat unconventional, talking about lens focal length, thus it is industry standard to round this value up to 50mm. Now longer is better, because, if the fitted lens is too short, the resulting image will display a vignette. A vignette is a reduction in image brightness at the edges.

Now all lenses vignette. This is due to the apparent shape of the exit pupil of the lens. Pretend you have been shrunk to microscopic size and you are walking about on the photo paper looking back at the enlarger lens. If you are directly centered, looking up, you will see a round illuminated aperture. As you walk away from center, looking up, you will see that aperture (iris) is no longer circular; it takes on an elliptical shape. The further off center, the more eccentric the ellipse appears. The elliptical shape of the iris when view obliquely, passes less light; thus the edges of the projected image are more feeble. Stated differently, all lenses project a circular image, and only the central portion is photography useful. The usable portion is called the circle of good definition.

We mitigate the vignette by choosing lenses with a focal length that is equal to or longer than the diagonal measure of the film format. The hitch is, shorter lenses yield higher magnifications. We can raise the enlarger to get more magnification, but likely we run out of room, and thus we must resort to mounting a shorter lens. Now a shorter lens must be positioned closer to the negative than a longer lens. This fact and the mechanics of the enlarger dictate whether or not the lens board that mounts the lens must be flat or recessed. In other words, shorter focal length lenses yield higher magnifications. At risk is the vignette, and the necessity to recess the lens to achieve focus. Additionally, the walls of the recessed lens cone must not encroach into the optical path as this would contribute to a vignette.

  • 2
    \$\begingroup\$ Thank you for your lengthy reply on why we recess lenses Alan, but it did not answer my question: "What determines how deeply recessed a lens should be mounted?" I need to the amount of recession (or lack thereof) in order to produce the right lens board. \$\endgroup\$
    – timvrhn
    Commented Jan 23, 2020 at 17:04
  • \$\begingroup\$ Why would any reasonable mounting recession matter at all on an enlarger? Enlarger focusing moves the lens up and down with respect to the film. Recession differences would simply involve very slightly different up and down focusing amounts. There seems very little possibility that you would be near any maximum movement limit. \$\endgroup\$
    – WayneF
    Commented Jan 23, 2020 at 18:28
  • 3
    \$\begingroup\$ @WayneF Recession is needed for shorter lenses- usually 50mm or shorter- because the minimum bellows travel on the enlarger doesn't allow the lens to get close enough to the film for focus at normal enlargement distances. It's debatable whether you would need a recessed lens board for a 75mm lens under normal circumstances but depending on the enlargement and the enlarger it may make sense. \$\endgroup\$
    – BobT
    Commented Jan 24, 2020 at 4:43

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