Dumb question regarding mFT lenses

Djarum

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Ok, so I got to thinking. Dangerous, I know.

One of the things about the mFT format vs the FT format is the flange distance. This supposedly allows lenses to be smaller. I noticed that the lens diameter for equivalent mFT in comparison to FT is smaller. How is this? For a given f stop and focal length, the size of the diameter of the lens should be the same if the sensor size is the same. Or does this only apply to simple lenses or single focal length lenses.

I guess what I am saying/asking, is that for a focal length of 14mm and f stop of 3.5, the diameter of the lens would need to be 39.2mm.

Why does the lens diameter need to be larger for a larger flange distance, if the above situation is not the case?
 

sebastel

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hi.

the optically effective aperture diameter (the entrance pupil) for a 14mm lens at aperture 3.5 is 4mm
(not 39.2mm).

you can see it by looking into the lens from the front. this opening diameter is not identical to the lens diameter. especially with wide angle lenses, the required lens diameter can be much larger than the computed aperture diameter.

this is especially true for retrofocus constructions. if the focus length is bigger than the flange distance, the optical construction gets much more voluminous than what your simplifying division 14mm/3.5 seems to suggest.

http://en.wikipedia.org/wiki/Aperture
http://en.wikipedia.org/wiki/Entrance_pupil
http://en.wikipedia.org/wiki/Retrofocus
 

Djarum

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hi.

the optically effective aperture diameter (the entrance pupil) for a 14mm lens at aperture 3.5 is 4mm
(not 39.2mm).

you can see it by looking into the lens from the front. this opening diameter is not identical to the lens diameter. especially with wide angle lenses, the required lens diameter can be much larger than the computed aperture diameter.

this is especially true for retrofocus constructions. if the focus length is bigger than the flange distance, the optical construction gets much more voluminous than what your simplifying division 14mm/3.5 seems to suggest.

http://en.wikipedia.org/wiki/Aperture
http://en.wikipedia.org/wiki/Entrance_pupil
http://en.wikipedia.org/wiki/Retrofocus
I suppose I'm coming from telescope terminology, where aperture = objective size. If I hooked an f 5 telescope to the front of a camera, wouldn't the f-stop be f5? If not, what would the f-stop be?

Secondly, wouldn't a lens design allow the iris to be open as much as possible to utilize the entire diameter of the the lens so that the light path is not impeded before hitting the sensor?

I hate being in two hobbies that are closely related, have similar vernacular, but mean different things, LOL.

Dj
 

Djarum

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Ok, I think I understand. Telephoto lenses use positive front element while using a negative lens behind it to increase the focal length. This is equavalent to using a barlow lens in telescopes. This allows the lenses to be shorter. Makes sense. This is why a short tube f/6 refractor can be turned into an f/10 refractor without adding length to the tube. Wide angle lenses uses negative front elements while using positive lenses in the back.

Now I see why lenses are larger in diameter in the four thirds format due to the longer flange distance.
 

sebastel

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hi dj,
yes, you got it.
for long focus length, the aperture is (close to) identical to the lens diameter.
(telescopes usually have a very long focus length!)
but, as soon as your optical construction is more complicated than a single convex lens or a two lens element, the deviation from this simplistic rule becomes apparent.

even with symmetric constructions like the biogon, you will find much bigger lens diameters. retrofocus design increases this difference even more, and the biggest clumsiness in glass element is required by retrofocus wide angle zoom lens.

cheers,
sebastian
 

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