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Olympus M.ZUIKO 60mm 1:2.8 macro - minimum field size?
Eric Stevens wrote:
On Sun, 07 Aug 2016 15:52:02 -0800, (Floyd L. Davidson) wrote: Eric Stevens wrote: On Sun, 07 Aug 2016 02:02:36 -0800, (Floyd L. Davidson) wrote: Eric Stevens wrote: On Sat, 06 Aug 2016 23:06:37 -0800, (Floyd L. Davidson) wrote: Eric Stevens wrote: On Sat, 06 Aug 2016 03:07:47 -0800, (Floyd L. Davidson) wrote: David Taylor wrote: Could anyone please confirm what the field seen is for the Olympus M.ZUIKO 60mm 1:2.8 macro lens at maximum macro setting (closest focus): https://www.amazon.co.uk/gp/product/B009C742Y2 Is it the MFT frame size (17.3 mm Ã--- 13.0 mm) or twice that (i.e. 34.6 x 26 mm)? At 1:1 the field of view is exactly the same as the sensor size. So if you put it on a 17.3x13.0mm sensor, that is what the field of view will be. That will occur only when the lens is racked (screwed?) out to twice its nominal focal length. Before the size of the seen field can be determined it is necessary to now how far the lense can extended. Nonsense Eric. One thing, see that last paragraph below, that has not been stated at all is important to your comment, but really the only thing that counts is the "At 1:1" factoid. With a lens designed specifically for a minimum focus distance that produces 1:1 magnification ... Nobody has stated what minimum focus distance or its accompanying magnification actually is. Because it clearly does not make any difference what is actually is. That's a silly remark. The magnification is at the heart of the question. Your question was the silly remark! We know what the magnification is, that was specified. We don't care what the MFD actuall is because, for this discussion, it does not make any difference. That's not silly, that's a fact. Regardless of all of your obfuscation, at 1:1 magnification the size of the field of view will be *exactly* the size of the sensor. There can be no argument with that. My original response arose from the fact that (1) I did not know the near limit of the focussing range for the lens and (2) whether or not its focussing enabled it to give a 1:1 image/object ratio. None of this would have arisen if I had written 'focussed' rather than 'extended'. By definition. If the lens is specified to have 1:1 magnification at the MFD, we don't need to know the mfd we jus need to set focus the the minimum distance. The question was not what the MFD is, it was what is the field of view at MFD for that one specific lens. It turns out that lens is specified to have 1:1 magnification at MFD. ... the entirety of your comment is meaningless. Go back to basic optics. Good idea Eric. Do that! Just set the lens at minimum focus distance, and 1) it is at 1:1 magnification which brings about 2) the sensor size is exactly the field of view. But is that correct? Where do you get the information about minimum focus distance from? Try using Google to check the specifications for that specific lens, and then actually *read* what it says... Did you check this information before you first replied? If you did, why didn't you state the fact and give a source? Or did you just assume that the lens had 1:1 capability? Of course I looked at the URL that David provided. I also read it, and I even understood what it said!:-) You should have done the same... What I stated is correct. A simple lens will give 1:1 reproduction when both the object distance and the image distance = twice the focal You didn't say "A simple lens will", you said that *this* lens will. *t won't.* length. This is measured from the centre of the lens plane. With a complex compound lens such as the lens under discussion there is no single physical lens plane but there is a virtual lens plane. Eh? Making things up again huh. That has nothing to do with anything. What does make a difference, for this part of your error, is that this lens uses what is called an "Internal Focus" design. It doesn't move the lens as a unit away from the sensor to focus closer. Instead the sensor to lens distance is fixed, and internal element groups are moved in relation to each other. (Yes I know what wanted to say, and why. It makes not difference though. Even if you'd said it correctly.) Unfortunately this moves around and can even lie outside either end of the lens assembly. And then with built in telephoto lens units the image distance in particular can be telescoped. Nevertheless, my point stands: at 1:1 the object is 12cm from the virtual lens. You are talking about what is called the "principal focal plane". Obviously we don't really need to be concerned with that for this discussion. The principal focal plane is a different thing altogether. Well that is certainly what you described. Actually there are two "principal focal planes" and that refers to the rear principal focal plane. Maybe you were thinking of the exit pupil, or maybe the nodal points. Different. David gave us the needed information: "the Olympus M.ZUIKO 60mm 1:2.8 macro lens at maximum macro setting (closest focus): https://www.amazon.co.uk/gp/product/B009C742Y2" Otherwise, what your comment above missed entirely is that the lens in question uses an "Internal Focus" design. The sensor to lens distance does not change as it is focused. But what does the magnification do? You don't actually know. Maybe you don't know, but I do! At the MFD the magnification is specified as 1:1. I guarantee the focal length has changed, the effective aperture has changed, but the sensor to lens distance is exactly the same. But with focussing relying on three sets of independently driven moving elements, the sensor to virtual lens distance *will* have changed. Who cares? That still has nothing to do with this discussion. We can't see where the principal focal plane is, and never need to know while using a lens. We need to know where the sensor plane is, where the front of the lens is, and how those two relate to the object. http://www.bhphotovideo.com/c/produc...0mm_f_2_8.html or http://tinyurl.com/bm5l28o says: "You can focus as close as 7.4" (18.8 cm) with the maximum reproduction ratio of 1:1. Other minimum focusing distances Ha ha, that is funny! "Other minimum focusing distances"??? They aren't. Only the minimum focusing distance is a minimum focusing distance. For that lens it is 18.8cm. Measured from where? It is measured from the sensor. There is only one minimum focusing distance. Any other distance at which the lens can focus is not the minimum. Note that there is also a "minimum working distance", measured from the front element of the lens to the subject when focused at the MFD. available include 7.9" (20 cm) at a reproduction ratio of 1:1.3, 9.1" (23 cm) at 1:2, and 13.4" (34 cm) at 1:4." ... so 1:1 does seem to be the limit. And that is all we need to know! That is what the question was about. That is what the discussion was initially about. And note that it says nothing at all about changing the lens extension (because it doesn't change). But it changes the effective optical centre. So what? You can't see it and you can't use it for any purpose while making pictures with the lens. -- Floyd L. Davidson http://www.apaflo.com/ Ukpeagvik (Barrow, Alaska) |
#13
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Olympus M.ZUIKO 60mm 1:2.8 macro - minimum field size?
On 06/08/2016 12:07, Floyd L. Davidson wrote:
David Taylor wrote: Could anyone please confirm what the field seen is for the Olympus M.ZUIKO 60mm 1:2.8 macro lens at maximum macro setting (closest focus): https://www.amazon.co.uk/gp/product/B009C742Y2 Is it the MFT frame size (17.3 mm Ã--- 13.0 mm) or twice that (i.e. 34.6 x 26 mm)? At 1:1 the field of view is exactly the same as the sensor size. So if you put it on a 17.3x13.0mm sensor, that is what the field of view will be. Having now purchased the lens I can confirm that its minimum field of view is indeed 17.3 x 13.0 mm (or at least, very near to that). Thanks for all the help here. (The confusion arises as they also quote 2:1 magnification for a "35 mm equivalent". Yes MFT is providing a rather nice macro capability, and the working distance isn't too bad either. -- Cheers, David Web: http://www.satsignal.eu |
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