On this datasheet of the Kowa LM25HC lens I have access to, there is a number called exit pupil, which equals -57.9 mm.

What does this number represent?

This is not a question about what the exit pupil is, I know this. I'd just like to understand what exactly does the given value represent. Position of the exit pupil? Relative to what? Size at a specific f-number and focus point? Then why is it negative? Things like that.

If it helps, I was able to figure out pupil magnification from the provided position of the entry pupil (E.P. in the drawing) and the front principal point (H1):

m_p = f / (f - s_ep) = 2.342

However, I can't seem to link this number to -57.9 in any way.

  • \$\begingroup\$ en.wikipedia.org/wiki/Exit_pupil \$\endgroup\$
    – db9dreamer
    Jul 17, 2018 at 14:17
  • \$\begingroup\$ @dav1dsm1th Like I said, I know what the exit pupil is. What I don't understand is which aspect does this number in the datasheet represent and why is it negative? \$\endgroup\$ Jul 17, 2018 at 14:35

3 Answers 3


In this case, you might be confusing the exit pupil size that is typically stated for binoculars, for the exit pupil coordinates given for this single lens.

In this case the '-57.9 mm' is the distance from the image to the exit pupil, the negative symbol means image object is being displayed to the left of the exit pupil. (The exit pupil is negative when the exit pupil is to the right of the apparent image).

For more, check out this detailed explanation: http://www.telescope-optics.net/terms_and_conventions.htm

  • \$\begingroup\$ I tried all the possible combinations of distances to the left or right of all other points and nothing made sense. I have no idea how it never occurred to me to try and position the pupil the specified amount from the image. Thanks. \$\endgroup\$ Jul 17, 2018 at 19:39
  • \$\begingroup\$ Just a small correction: in this case, negative seems to mean to the left of the image at infinite focus (exit pupil is closer to the object-side). Then all the numbers agree; calculating pupil magnification from either the entrance or exit pupil produces approximately the same number. \$\endgroup\$ Jul 17, 2018 at 19:42
  • \$\begingroup\$ Doesn't left or right depend on which side of the diagram the lens is in relation to the image? \$\endgroup\$
    – Michael C
    Jul 18, 2018 at 2:41
  • \$\begingroup\$ For some of the calculation, it seems so, yes. But for this one, it is the position of the exit pupil to the image created by the system: "distance from the image to the exit pupil is negative for exit pupil to the right, postive for exit pupil to the left of the image" \$\endgroup\$
    – cmason
    Jul 18, 2018 at 12:42

A lens has two principal points. From these points we make our measurements. If the lens was super simple in design, all measurements would be taken from the center of its center thickness. Camera lenses are a complex blending of several glass lens elements, each with different powers (degree of magnification). Because of this complexity, distances from lens to object to be images are measured from a point called the front nodal. Distances from lens to image are measured from the rear nodal. The locations of these principle points are determined by use of an optical bench.

Lens makers want to know the locations of these points. In a simple lens, 25mm focal length, the lens will be spaced 25mm forward of the film or sensors when imaging a far distance subjects. For technical reasons, 25mm back from the lens might be to0 close to accommodate mechanisms like focusing and shutter and reflex mirrors. It is sometimes desirable to lengthen the space between lens and film/sensor. The lens designer then uses a backwards telephoto (Retrofocus like looking through binoculars backward). This design lengthens the spacing called “back focus” distance.

Now the rear nodal will be key measuring point that tells the tale as to focal length and back focus distance. I think rear pupil location is sonorous with rear nodal. I think this principle point is located 58.8mm rearward from the image of an object that is a far distance away. I think this design the otherwise distance which would have been 25mm for the 25mm focal length lens.

Such a design allows the resulting image to have good definition with lowered vignetting and provides room for camera mechanisms between lens and sensor. I gave this my best shot. Be aware my best guess might be gobbledygook.


The camera lens, in many respects resembles the Human eye. Now the pupil of the human eye is that black circle in the center that expands and contracts involuntarily in response to the ambient light level. The larger the entry way diameter, the larger the surface area, the more light the eye can gather. The average human eye expands to about 7mm diameter under feeble light conditions.

All optical systems are plagued with defects called aberrations. To mitigate this the lens maker is forced to use multiple lenses constructed using different densities of glass each with a different shape (figure). Thus when we look into the lens from the front we get a view of aperture blades. This circular orifice is called the entrance pupil. Contrariwise, when peering into the lens from the rear we are looking at the exit pupil.

In both views, we are looking at the circular opening at the aperture stop, however they likely appear different in size because we are looking at this iris through lenses that likely magnify. The significance: It is the center of the front entrance pupil that defines the viewpoint of the lens. From the center of the rear pupil we trace the path of the un-deviated light rays that project an image on film or digital sensor.

Additionally, the diameter and distance from the exit pupil to film or digital sensor determines the size of the circle of good definition of the lens. In other words, will the amount of vignetting be tolerable?

  • \$\begingroup\$ So what does the -57.9 mm from the datasheet represent then? \$\endgroup\$ Jul 17, 2018 at 15:18
  • \$\begingroup\$ I don't think this data is of much value. The exit pupil is a variable based on the f-number setting. We can use the diameter to preform a ray trace to determine the field of view. We can trace to find limits of depth of focus. Diameter impacts aberration corrections and set the diffraction-limited resolution (see diffraction limitation based on f-number). \$\endgroup\$ Jul 17, 2018 at 16:39
  • \$\begingroup\$ What I mean is: what physical characteristic of the exit pupil (or what relationship to the other parts of the lens) does this number describe? \$\endgroup\$ Jul 17, 2018 at 18:18
  • \$\begingroup\$ I'm afraid that Alan's response does not answer your question because we cannot answer your question. The information given is ambiguous. The measurement -57.9 could be the distance of the exit pupil from the entrance pupil, distance from front vertex, distance from image plane, or any other arcane measurement. This question can only be answered definitively by experimentation or consultation with the manufacturer. \$\endgroup\$ Jul 17, 2018 at 18:26
  • \$\begingroup\$ @PhotoScientist Thanks. I hoped that it was a commonly understood value that I just don't understand because I'm new at this. \$\endgroup\$ Jul 17, 2018 at 18:34

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