rehess Loc: South Bend, Indiana, USA

Isn't there also another function, which just happens to partially offset (3)
(4) limit light passing through outer part of lens {area furthest from axis} - what I guess you could call the "unsweet part"

A perfect lens would have parabolic surfaces, but our machines produce spherical surfaces. As you make the aperture smaller, the remaining surface looks more like a parabola, so the light behaves "better". This is related to the "sweet spot" of a lens that people some times talk about. The smaller the aperture, the more you are depending on this "sweet spot".

Added: but regardless of how many functions we identify, the fact is that finding one modified f-stop which preserves all these is a problem.

rmalarz Loc: Tempe, Arizona

Would you kindly like to explain, in some detail, the statement you made regarding parabolic and spherical surfaces.
--Bob

rmalarz Loc: Tempe, Arizona

Focal lengths and apertures do not mystically change when you place an FX lens on a DX body. The only thing that changes is the amount of the image circle that the sensor uses. This gives the appearance of multiplying the focal length of the FX lens when placed on a crop sensor body.
--Bob

Gene51 Loc: Yonkers, NY, now in LSD (LowerSlowerDelaware)

I think this may address your point very well:

http://www.cambridgeincolour.com/tutorials/diffraction-photography.htm

rehess Loc: South Bend, Indiana, USA

My initial comment came from an optics course I took in college - roughly fifty years ago - so I had to do some on-line research {actually I do still have the textbook, but copying the drawings and simplifying the explanations sounded like real work to me}. This source is the best I've found so far.
https://en.wikipedia.org/wiki/Spherical_aberration
We were told back then that having parabolas as boundaries of the cross-section will give the upper drawing, while having circles as boundaries will give the lower drawing, and that lenses like the lower drawing are much easier to manufacture.

whitewolfowner

The f stop is not affected. All that is going on is that the circle the lens is producing on the sensor is larger than it needs to be, thus the multiplication factor on a smaller sensor. The amount of light entering to the sensor is still the same.

SS319

A quick look at the ray diagram makes this an easy question.The outer blue lines are the primary rays for a given lens at a given distance from the red arrow forming the image inverted on the left. With the same lens, focused at the same distance, the focal point (where the rays cross) remains the same. Now, you can see how the crop size of the snsor creates the multiplication factor of the lens focal length, and you can see how the outer portions of the lens are not effectively used and the light coming through that part of the lens is lost to the image, thus the equivalent of a smaller aperture.

THIS ONLY AFFECTS THE WIDE OPEN APERTURE! when you impose an iris and close the lens down one stop, and from there to f/36, there is no further change in the f stop.

So, on a f/2.8 FF lens, on a crop sensor, your f stops would look like this:

Full Frame F/2.8 F/4 F/5.6
Crop Sensor F/4 F/4 F/5.6

blackest Loc: Ireland

Doesn't seem right to me

The f-stop of the lens does not change when you mount it on a crop sensor camera. All you are doing is cropping the image that the lens produces. The brightness of the image obviously doesn't change when you crop it, so cropping has no effect on image brightness. Since the brightness is related to the f-stop, the f-stop obviously doesn't change either.

The key here is that you don't change the focal length of the lens when you mount it on a crop sensor body. What you change is the angle of view. Since we're used to thinking of angle of view in terms of 35mm focal length, the "digital multiplier" is just a quck way to get an impression of the field of view of the lens. So on a full frame camera a 28mm lens is a wideangle, but a 24mm lens on a crop sensor is still a 28mm lens, it just gives a narrower angle of view. The "digital multiplier" is just a quick way to see that, so you can say that the lens woul have the same FOV as a 45mm lens on a crop sensor camera (28 x 1.6 = 45) EVEN THOUGH IT'S STILL A 28mm LENS.

TriX Loc: Raleigh, NC

Interesting. I'm familiar with using parabolic surfaces as reflectors (both optically in mirrors and as microwave antennas for RF), but had not considered their use as photographic lenses. Do high quality lenses typically have parabolic elements?

SS319

If the Focal length does not change (and it doesn't), then, for the same distance to subject, the focal point remains the same - as shown in the diagram. ALL light from the subject going through the lens goes through the focal point. For the crop sensor, the rays reaching the edge of the sensor and going through the focal point cannot come through the extreme diameter of the lens. The rays from the extreme diameter of the lens and going to the focal point fall beyond the perimeter of the crop sensor - but within the perimeter of the full frame sensor. This is the same effect as closing the iris on the lens - i.e. changing the f stop.

The light passing through the lens and not landing on the sensor, do not contribute to the image in the same way that closing the iris eliminates light from the outer portion of the lens - effectively, you are only using the central portion of the lens to generate your image.

Brucej67 Loc: Cary, NC

I am not knowledgeable on this subject and would love to be enlightened. I have heard that two different lenses with the same focal length and aperture (let's say a Nikon 50mm f1.4 and a Sigma 50mm f1.4) when shot at any aperture on both lenses) will not necessarily have the same amount of light transmission ( let say bot lenses are used at f2.8 and everything else on the settings are the same) and this effect is called T/Stop versus F/Stop. Here is a link to what I am talking about and he explains it better: https://www.picturecorrect.com/tips/f-stop-vs-t-stop-whats-the-difference/

TriX Loc: Raleigh, NC

But doesn't the luminance per area cd/m2 remain the same? If that were not the case, why does a FF f2.8 lens produce the same exposure on both a FF and crop body camera? According to your theory, the exposure from the crop body would be a stop less than the FF, but actual practice shows otherwise I believe.

rehess Loc: South Bend, Indiana, USA

My physics goes back fifty years ago; the theory remains, but practice has changed.

We have to recognize that reflectors are much easier to work with. My brother, who worked for an astronomer at one time, tells me that they make reflecting mirrors by rotating molten glass - and physics will form the parabola for them; my understanding is that lenses are formed by numerically controlled grinders - improved computers improve that process, but I don't know for certain what they're able to do these days. I do know they work with some kind of "aspherical" lenses these days. Another issue is that lens designers have to deal with the fact that different colors behave differently in refraction {the prism effect}, while all colors reflect together.

My earlier comment was based on some personal testing I did. I found lens performance improving at apertures where I would have expected diffraction to be pushing the other way, and my conclusion was that improving "sweet spot" was still dominant - but I don't have an actual theoretical justification for that conclusion.

Gene51 Loc: Yonkers, NY, now in LSD (LowerSlowerDelaware)

"This is the same effect as closing the iris on the lens - i.e. changing the f stop."

Not at all. Using a crop sensor is more like cropping the image in Photoshop. The reason the light doesn't change even though you are using the "center" of the lens is that the angle of view is narrower, so you are not using the edges and corners, the image itself is narrower and shorter. If you were to actually close the iris to just use the center of the lens, physically blocking any light rays from the wider perimeter of the lens from being focused at the sensor plane, of course you would get less light. But in your example, such is not the case.

Try this experiment. Get a flashlight, turn off the lights int the room, get two pieces of paper - 8 1/2" b 11" and 11" by 17" The big piece is 4x the area of the

Stick the big piece on the wall, and adjust the flashlight so the circle of light fully covers the corners of the paper. Use your camera in spot meter mode to measure an exposure for the paper.
Substitute the small piece of paper for the bigger one placing it approximately in the center. The light from the flashlight is much wider than the small paper - completely overshooting it all the way around. Read the light reflecting from the smaller paper. Can you see or measure any difference in light?

BebuLamar

Your ray diagram is wrong and thus all what you said about the effective aperture is wrong. Even Tony Northrup wouldn't agree with you.