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The reproduction ratio formula is:

object size in image = object size  * focal length
                        ------------------------
                             object distance

It has been widely discussed there, but I've found no discussions about my specific doubts.

  1. Is it valid for close-up photography?

From basic optics, for a lens with focal length f it results:

object size in image = object size  * focal length
                      -----------------------------
                     (object distance - focal length)

Of course, if the object distance is much higher than the focal length, this formula results in the first one. But what if you were using a 60mm lens at 50mm from a small object, such as in a macro?

  1. Let's image to neglect the focal length term compared to the object distance. The actual magnification still does depend on the object distance from the camera. This means that further objects will have less magnification than closer objects. Is this the cause of wide angle lens perspective warping?
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3 Answers 3

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Your first formula is correct, it is just similar triangles, simulating a Thin Lens model (single lens element).

Object size in image / object size is obviously always the exact magnification.

And the first formula equating that magnification to focal length / object distance is generally correct, except that we don't know focal length. The marked focal length applies to an infinity focus, and at macro focus, it is longer. In fact, at 1:1 focal length is generally 2x the marked number (but modern macro lenses use internal focusing which shifts things a small bit).

This first formula could compute focal length, knowing magnification and object image size (in mm). And then, knowing focal length, the first formula is valid (except object distance is still measured to the unknown internal node).

FWIW, for general lenses, Field of View and Depth of Field calculations also use the Thin Lens (similar triangles) model, using marked focal length and subject distance actually to an lens node, which is generally internal except in telephoto lenses, it is at the front of the lens (definition of telephoto), and in wide angle lenses in SLR, it is well behind the lens, to move lens forward to leave room for the SLR mirror to rise. In a 60 mm macro lens, it is somewhere probably around the middle of the lens.

Otherwise, subject distance in regular photography is measured to the sensor plane (or film plane if you prefer). But the similar triangles in the first formula is instead measured to a node inside or near the lens (Thin Lens properties).

The second formula just tries to state that difference, referring to the focal length node.

There is a macro property called Working Distance, which is 1:1 object distance measured from the front of macro lenses, determined for specific lenses by amateurs measuring it instead of the factory. http://www.jeffree.co.uk/pages/macro-lens-calcs.html is a good sample. But the Thin Lens similar triangles are measured from internal nodes (probably internal) instead of the front of the lens.

Further distance is less magnification. Magnification at infinity is zero (too small to see, unless bright like stars, where we can see the bright, but not the size). Conversely, general lenses (as opposed to macro lenses) don't focus much closer than about 0.1x magnification, because the focal length and f/number become awkwardly larger at closer focus.

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    \$\begingroup\$ I was taught that the focal length of a lens was the distance of focus when the source was at infinity. If the object is close enough that the reproduction is 1:1 the focal length does not change, but the distance to the image does. \$\endgroup\$ Sep 15, 2022 at 3:26
  • \$\begingroup\$ A thin lens is not a single lens, it is a theoretical refracting lens with zero thickness. It should be fairly obvious that as the ratio of the lens' actual thickness to subject distance changes as the subject distance is reduced, the thin lens equation will be more and more inaccurate. \$\endgroup\$
    – Michael C
    Sep 15, 2022 at 7:39
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    \$\begingroup\$ Field of View is just the similar triangles so I don't see how macro could be wrong, it can only work that way (if we knew the numbers). But the focal length is generally unknown, and the subject distance is so short and imprecise, so it could clearly be mistaken. The formula measures that distance to an internal node inside the lens somewhere (and it moves with focus). So macro people really don't use field of view (due to being so short on precision), but instead they use magnification for size. The resulting object size on the sensor or film vs. the real object size is pretty clear. \$\endgroup\$
    – WayneF
    Sep 20, 2022 at 23:36
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    \$\begingroup\$ @Kinka-Byo The approximation begins to show significant divergence at a magnification factor of around 0.10X-0.12X. That is, when the subject's actual size is 10X-8X the size of the subject in the focused image projected by the lens. This is because that is the point at which the distance between the lens' rear nodal point and the image plane start becoming significantly longer than the lens' focal length at infinity focus. \$\endgroup\$
    – Michael C
    Sep 21, 2022 at 4:09
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    \$\begingroup\$ @Kinka-Byo At unity (1:1 reproduction), for a theoretical thin lens (zero thickness) the distance from subject to image plane is 4X the lens' focal length at infinity, with the lens halfway between the subject and image plane. With a simple single lens having real thickness, the lens is still exactly halfway in between, but the distance between the lens' front and rear nodal points must be added to the total distance. \$\endgroup\$
    – Michael C
    Sep 27, 2022 at 11:21
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Your expression is only valid when the object is far enough away that one can regard the distance from lens to image as the focal length. When you get closer the distance from lens to image increases. The correct expression, valid for all distances is

object size in image=object size * distance from lens to image/distance from lens to object

This comes from similar triangles in the thin lens approximation. With real lenses it is hard to know where to measure the distances from lens to image and object from. It also assumes that the light rays are close to the axis, which is not the case with wide angle lenses.

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The focal length of a lens is only valid when the lens is imaging a object at infinity. Light rays from an object at infinity arrive as parallel rays. The job of the lens is to bend these rays inward (refract) causing them to take on a conical shape. The distance, apex of the cone to the rear nodal is the focal length.

If the lens is imaging an object closer than infinity, the light rays from this object arrive as diverging rays. The lens has limited power to refract thus these diverging rays come to an apex further downstream.

We are talking "focus travel". This is distance we must rack the lens forward to achieve focus.

For your purposes, we can approximately calculate the focus travel substitute this value for the focal length and then do the math.

A 50mm lens is close focusing. The subject distance is 100mm. I convert both values to diopter units.

1/50 X 1 000 = 20d

1/100 X 1000 = 10d

Now we subtract 20 – 10 = 10d

Covert 10d to millimeters 1/10 X 1000 = 100mm

This is the revised distance lens to focused image when a 50mm lens is working a object 100mm distant.

Now, in your math replace 50mm focal length with 100mm and work the formula.

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  • \$\begingroup\$ 1/50 X 2000 should be 1/50 X 1000. \$\endgroup\$
    – qrk
    Sep 15, 2022 at 20:31
  • \$\begingroup\$ Thanks 1/50 X 2000 is a typo. To convert 50mm lens to diopter power 1/50 X 1000 = 20d -- now corrected. \$\endgroup\$ Sep 15, 2022 at 23:18

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