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let's consider a simple lens without diaphragm.

enter image description here

For a lens like this, the book Science for Curious Photographer states that:

The important fact is that the illumination of the image is proportional to the brightness B of the object and the square of sin θ' [that is f/2r = F Number squared]. Here the diameter 2r refers to the effective aperture of the lens. Th is is not the diameter of the front lens element or even the iris diaphragm. The effective aperture or entrance pupil is the image of the aperture stop as seen from the front of the lens.

Well, there is nothing intuitive for me. Let's start from the basics.

  1. The luminous subject has a luminance (you may express it as Watt or Lumen per square meter, or per steradians, as you prefer). This luminance is represented by its emitted rays, which the lens make converge to produce another luminance value on the sensor plane.

  2. Not all the rays enter the lens. I'd say that a lens limits the possible input rays by its diameter. I'd say that 2r = Lens diameter. If the lens were infinite, all the input luminance would be directed towards the sensor.

  3. The limited rays acceptance circle is just due to the lens diameter. I'd say that the focal length is considered because, if you increase it, some portion of image goes outside the sensor. So, it is not a problem of the lens. It is just the sensor that is too small. But, if that's true, shouldn't it be embodied into luminance definition? Luminance is Watt or Lumen per square meter. Big sensor -> more exposure because luminance is multiplied by a bigger area.

  4. Now let's introduce the diaphragm. Why should I consider for D the diameter of the entrance pupil? Why should the luminance be attenuated by the diameter of what I see in front of the lens? What does it have to do with the accepted power of the lens?

Every guide says the "brightness attenuation scales as (f/D)^2",... but I've not found an intuitive explanation for my points.

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3 Answers 3

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Consider this simplified diagram.

Without a diaphragm all rays that hit the front lens are used:

enter image description here

With a diaphragm rays that went though the rim of the lens are stopped so the effective diameter of the lens is reduced:

enter image description here

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    \$\begingroup\$ One more sentence about your diagram. Yes, the entrance pupil size at the front of the lens (the light that will enter the aperture) is of course larger diameter than the physical aperture diameter. The top two and the bottom two lines do not enter, but all the others do. \$\endgroup\$
    – WayneF
    Oct 15, 2022 at 17:04
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Optic rays are reversible. When you look into a camera lens from some vantage point that is in focus, every ray emitting from that point falls into two classes: those that get through the optics and fall on the respective sensor spot, and those that do not. The ratio determines the brightness of the lens. At the same time, any image of the aperture that gets out of the lens will reach your eyes through some path ultimately exiting the front lens in a straight line. Where the image changes from inside the aperture and outside the aperture determines the border between light reaching the sensor and not when taken from the other side.

This equivalence of "viewing rays" delivering a virtual image of the aperture as seen through the front lens and rays of light entering the front lens and reaching the sensor makes for the importance of the "entrance pupil" as a photographically significant measure that remains valid even as the single lens is replaced with a complex contraption of lens groups with an aperture somewhere in the middle.

For this equivalence to hold, the lens has to be built in a manner that once the aperture is passed by a light ray from the focused distance, it will reach the sensor in the right spot. This is not necessarily true for wide-aperture lenses yielding "cat-eye bokeh". But for most lenses, this is significantly true.

When you write: "The limited rays acceptance circle is just due to the lens diameter.", that is a pathological case, not the rule, and it would imply that there is no effective aperture in place.

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Your statements 1-3 are correct... without an aperture diaphragm behind the objective the objective element's diameter is the entrance pupil.

The aperture diaphragm is then (potentially) an optical/physical restriction to the ray's paths after the objective element; and it's diameter as seen from (magnified by) the objective element is then the entrance pupil (smallest optical restriction). Thus the entrance pupil cannot be larger than the diameter of the objective, but it can be smaller.

Focal length is also a factor beyond your point in #3; because as you increase magnification of the subject by increasing the power at the objective end it (can) simultaneously increases the magnification of the aperture restriction. Thus increasing the effective size of the entrance pupil. This is essentially how a constant aperture zoom lens functions... the physical size of the aperture restriction doesn't change, but it's apparent size does.

And all of this relates to the property of Etendue; how spread out the light is within a system. Etendue is defined by the apparent/relative size of the source, and simultaneously the apparent/relative size of receiving area (entrance pupil). It is essentially the volume of the area defined by their fields of view... anytime there is an increase in one the etendue decreases.

The first drawing shows how this works. In A both the subject and aperture are being increased in apparent size by increased magnification (constant Ap zoom). In B the lens is moved closer to the subject increasing both the entrance pupil and subject relative/apparent sizes. C shows exposure increasing because one's relative size increased (increased FoV), and D shows a decrease in exposure due to one's decrease in relative size (decreased FoV).

enter image description here

The second diagram shows mathematically how the etendue/spread of light decreases (volume of area increases) in scenario A where the source and entrance pupil are both increased in relative/apparent size by increased magnification. E.g. a constant aperture zoom lens w/ the green FOV representing a shorter FL zoom position. (Both FOV cones go to a point at the sensor).

It secondarily attempts to show an additional increase in volume/decrease in etendue that would occur with a physical increase in objective area/entrance pupil (but that is not carried to a point/resolution). And it correlates the relative size of the source within the recorded fields of view. I.e. there is less light received from the source with the wider focal length, but it is also recorded smaller within the wider FoV, and so it's exposure remains constant.

enter image description here

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  • \$\begingroup\$ A wide angle lens can have a larger entrance pupil than front element diameter when the focus distance is significantly shorter than infinity, particularly in the case of certain compound lens designs. Please see here and here. \$\endgroup\$
    – Michael C
    Oct 16, 2022 at 0:21
  • \$\begingroup\$ @MichaelC, The entrance pupil is the apparent size of the objective as seen by the source through the objective element; how can it be larger than the objective? Your first example does not show it as being larger than the objective, but it is larger than the physical aperture (which is typical). Your second example makes little sense, it has the exit pupil at the object side and the entrance pupil at the image side, and both are larger than the objectives involved. Almost all wide angle lenses have front elements larger than their focal length; yet none have extremely fast apertures. \$\endgroup\$ Oct 16, 2022 at 12:55
  • \$\begingroup\$ The entrance pupil is measured from a point on the center of the lens' optical axis. The entrance pupil can be larger than the lens' objective diameter in the same way that one can look through a keyhole and see more of the room on the other side than an area the size of the keyhole. If the entrance pupil is more distant than the front of the lens, then the cone formed by drawing two rays from the point of focus on the center of the lens' optical axis can continue to diverge after passing through the front of the lens. \$\endgroup\$
    – Michael C
    Oct 17, 2022 at 6:40
  • \$\begingroup\$ @MichaelC, For what you are describing to work the entrance pupil would have to be (nearly) co-located with the objective element, or be the objective element; and there would have to be a secondary lens of larger diameter to collect and redirect those divergent rays. Otherwise, the rays hit a physical obstruction (lens barrel) and do not contribute to image formation (i.e. do not pass the entrance pupil). Even fisheye lenses do not have entrance pupils larger than the objective element. \$\endgroup\$ Oct 17, 2022 at 13:15
  • \$\begingroup\$ Quite the opposite. The further behind the objective the entrance pupil is, the greater the possibility that the e.p. can be larger than the objective. As you yourself said, the magnification of the optics behind the objective allow the e.p. to be larger than the lens barrel, just as they allow an e.p. to be significantly larger than the physical aperture diaphragm on many lenses. Most WA lenses have very large objectives in order to reduce peripheral illumination falloff (often inaccurately called "vignetting"), but it's not a requirement to design every lens that way. \$\endgroup\$
    – Michael C
    Oct 17, 2022 at 17:34

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