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The bellows extension formula takes the focal length and draw as input and produces an exposure compensation as an output.

A Nikon 4T magnifier is a -2.9 diopter lens. Sticking it on the front would seem to produce, then, a ~184mm lens. However, using that as the focal length for bellows factor, or, indeed, in the lens equation for distance-to-subject and draw, doesn't seem to give me working results.

Am I just holding it wrong, or am I thinking about it wrong?

bmargulies
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2 Answers2

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Bellows Factor: As you know, modern cameras are adjustable and allow focusing over a wide range of subject distance. However, many camera models prevent extreme close focusing. We are talking about subject distances closer than 5 focal lengths. At distances greater than 5 focal lengths, the exposure error induced by bellows factor is trivial thus ignored.

As we close focus, the f-numbers engraved on the lens barrel become gradually invalid. At magnification 1 (unity – life size) the subject distance is twice the focal length and the error induced in the f-numbers, as denoted. This error is 2 f-stops, which is a factor of 4. We use the factor value as a multiplier. As an example, if the normal shutter speed is 1/100 of a second, at magnification 1 the exposure factor is 4 thus to compensate, we multiply 1/100 X 4 = 4/100 = 1/25 of a second.

One solution is to mount a close-up lens. This is a simple magnifier lens akin to the lenses used in reading eye glasses. Normally we classify lenses by expressing their power in terms of focal length in the millimeter scale.

Opticians often switch to a unit of measure called the diopter. A lens with a focal length of 1000mm is classified as 1 diopter written as 1d. A 500mm lens converts to 2d and a 250mm lens is 4d. The conversion formula is 1/focal length in millimeters times 1000. Thus, a 50mm lens converted to diopter is 1/50 X 1000 =20d. Opticians do this to simplify the math used when combining lens elements.

When a camera’s design does not permit extreme close-up imaging, we can resort to several remedies. The simplest is to mount a close-up supplementary lens. The close-up lens, when doing close up work, forces the light rays to enter the camera lens as nearly parallel rays. This allows close imaging while eliminating any need to adjust the exposure (no bellows factor compensation required).

If we seek different remedies such as extension tuber bellows attachments, we now must recognize that an exposure error has been induced. The formula is bellows factor = magnification + 1 squared.

Suppose we work at magnification 1 (life-size). Magnification =1 = 1 + 1 squared = 4. This is the bellows factor for unity. We compensate for this elevated magnification by multiplying the unadjusted shutter speed by multiplying it by 4.

With a +2.9 supplementary close-up lens mounted, set the camera for infinity focus, place the object to be imaged 1/2.9 X 1000 = 345mm forward of the first lens of the camera barrel.

Alan Marcus
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  • In your final paragraph, what base focal length are you assuming? FWIW, I'm working in large format. – bmargulies Aug 02 '23 at 03:01
  • The focal length of the camera lens is moot in this instance. The positive power supplemental close-up lens has a finite focal length. If the object being imaged is place at a distance from the camera equal to this value, the light rays from this object arrive at the camera as parallel rays. The camera, set to infinity focus, accepts these rays as if they originated at infinity (about 3000 meters distant). – Alan Marcus Aug 02 '23 at 03:14
  • Since the camera can focus closer than infinity, I would expect to be able to use the lens equation some how for the combo of the lens and the magnifier to calculate alternative positions? – bmargulies Aug 02 '23 at 18:56
  • You set the focus scale to 3 feet = 914.4mm. Convert this distance to diopter power as if it was the focal length of a lens. Thus 1/914.4 X 1000 = 1.096d. The supplemental close-up to be used has a power of 2.9d. Given the 3-foot focus scale position, the total power of this lash-up is 2.9 + 1.06 = 3.96d. Now convert 3.96d to millimeters. Thus 1/3.96 X 1000 = 252.5mm. This is the distance object to center of the supplement close-up lens. The bellows factor is not applicable, the light rays are refracted by the close-up lens and enter the prime lens as parallel rays no bellows factor needed. – Alan Marcus Aug 02 '23 at 19:31
  • Above has math error: – Alan Marcus Aug 03 '23 at 18:14
  • Above has math error: 3 feet = 914.4mm in diopter power 1/914.4 X 1000 = 1.0936d add supplementary lens power 2.9 = 3.9936d total power covert to millimeters 1/3.9936 = 250.39mm distance object to center of supplemental lens camera focused at 3 feet. – Alan Marcus Aug 03 '23 at 18:20
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Bellows factor is about how extension changes the exit pupil of the lens and thus the light intensity at the image plane. Basically, if you move the lens farther away, the size of the exit pupil decreases, the image circle expands, some of the light collected does not fall on the image area; and therefore the exposure is reduced.

The Nikon 4T is an add-on diopter and affects the entrance pupil of the lens. Which is about the amount/area of light which enters the lens. But because it changes both the apparent size of the aperture (entrance pupil) AND the apparent size of the subject simultaneously/equally the etendue of the system does not change... the light intensity/size of area seen does not change as it would if you change distance to the subject or image plane.

That's just a long way of saying bellows factor does not apply to an add-on diopter lens (objective end).

Steven Kersting
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