TRANSMISSION AND APERTURE
February 5, 2021
[This is a slightly rewritten response to a colleague with whom I was collaborating on a camera profiling project. He was trying to figure out what camera aperture would yield what he called "100% transmission," by which he meant that the light energy reflecting off the photographed object was equal in intensity to the light energy arriving at the film plane. I was not able to find the answer merely by looking it up but was able to (seemingly, I think?) figure out the answer through research and deduction. I think it's of interest more for the journey than for the destination -- in my process of getting to the answer, I learned some interesting stuff.]
Hi, J----, I need to check my work but I think I may have figured out what you’re after. I *think* but am not sure that the answer is that what you’re calling 100% transmission is indeed your first intuitive guess:
t/1.0
I have not found this formulation anywhere, so I'm not totally sure it’s right — I’ve come to the conclusion myself by putting together several disparate optical principles that I have found. So let me explain how I arrived at it and you can help check my homework and figure out if it’s right.
I went through a really long roundabout way to get to something simple: that an image of an object with magnification of 1.0 and 100% transmittance has the same light-energy averaged over the same area as the original object. And that, by the convention of f-numbers (or t-numbers) is what is meant by t/1.0.
Here's the long roundabout way I arrived at it. I'd imagine there must be a much more direct way to confidently arrive at "you get unity from unity" but this is all kinda new to me so I'm not sure what the shorter version is, but I do offer a guess down below.
Here's the long road I took to get there:
The first thing I had to understand is that we know from all the available information from relevant websites/papers about optics and apertures that ANY t/stop gives you 100% optical transmission — that’s literally the defintion of a t/stop: to compensate for loss of transmission due to real-world inefficiencies and restore 100% transmission. That means that our own definition of "100% transmission" that we're after here is unquestionably a bit different from the technical definition of "100% transmission" in the field of optics -- that's because EVERY t/stop yields 100% optical transmission in their sense of the term, but we know that only one specific t/stop yields our own definition of 100% transmission. It's okay that there are two different definitions, but we just need to keep in mind which we're talking about at any given time.
The next concept is that, as you can see from our previous exchange, our intended formulation of transmission only makes sense if we can average it over a uniform area. It has no meaning at an infinitesimal point. We need to average the illumination over some area, even if small (cd/m^2 is a unit that’s an average over area).
Next: if an object is reproduced at the film plane with magnification 1.0 (meaning that its image is actual-size at the film plane), then its ALSO actual size at the entrance pupil. Why? Because the equation for magnification is:
M = f / (f - d)
Where M is magnification, f is focal length and d is distance from the entrance pupil. So, if we set magnification to unity and the distance to 2*f, we get M = -1 which means the image has arrived at the image plane at actual-size (and upside down) when the object itself is at a distance from the lens equal to two times the focal length. And in that scenario, the image is also passing through the entrance pupil at actual size (but not upside down) since we can also solve the equation (getting M=1 instead of M=-1) by setting d to zero.
So, for the entrance pupil to collect the spreading rays of the object at the object's original size (neither brightening it by collecting a wider spread nor darkening by collecting a narrower spread), the aperture must equal the focal length.
The equation for f-number is
N = f / D
Where N is the f-number, f is the focal length, and D is the real/physical diameter of the entrance pupil.
So, in our target scenario wherein the aperture's physical diameter equals the focal length, we have an f-stop of f/1.0.
But we know that real world lenses are not perfect transmitters, so the purpose of using t/stops instead of f/stops is to correct that difference: to transmit in real life the amount that we'd transmit with an ideal/perfect lens. So t/1.0 transmits the amount that an f/1.0 would transmit if it were perfect lens with no real world inefficiencies.
So our own definition of "100% transmission" is indeed your original guess: t/1.0.
Doest this logic seem right? If so, the guess that I promised above for a shorter way to formulate it is something like: "all t-stops compensate for real-world inefficiencies and restore us to the optics defintion of 100% transmission. And, to have the same cd/m^2 as the original object -- not just a fixed proportion of that light energy (which is our own definition of "100% transmission") -- we need to not merely transmit 100% of the original overall light energy, but to spread that same overall light energy OVER THE SAME AREA. So we need not just 100% transmission but also 1.0 magnification."
This short version may sound like I'm saying that objects at different distances are different brightnesses, but if you track the logic further, you'll see that that's not the case but rather that I merely used 100% magnification as an easy reference point and the solution tracks for different distances. Here's why:
The distance where the object is at 100% magnification is where you get transmission unity by dividing one by one. At other distances you still get unity but, for example, by dividing 100 by 100. Or by dividing .1638 by .1638, and so forth.
See: let’s say you back the object up so that it’s half the surface area on the image plane: not double the distance, but just move it whatever distance it needs to go, given your lens’s angle of view, to become half the surface area. Now only half of the object’s light is passing through the entrance pupil (because the rays have spread farther now by the time they arrive at the entrance pupil and we know that exactly half of them now still contained in the entrance pupil’s scope). So, there’s half the light energy from that object entering the entrance pupil, but it’s also half the size on the image plane. Half the light over half the area averages back to the same cd/m^2.
Another way to think of it (if I’m doing this right) is that t/1.0 with the object at 2x the focal length is sort of all around unity and everything tracks coherently from there. Like, you can close the aperture by a stop (to t/1.4) and each position on the film plane will receive half the light of unity (it’s not like objects at different distances darken a different amount). Or, as in the example above, if instead of closing the stop, you back up the object, then it gets half as big with half as much light energy (half as much total, not have as much average) — and averaging the smaller amount of energy over the smaller surface area is why things don’t get brighter and darker as you move them around (if we ignore atmospheric effects).
Does that seem right?
-Steve