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The lenses we use
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ray_parkhurst



Joined: 20 Nov 2010
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PostPosted: Thu Aug 02, 2018 8:47 am    Post subject: Reply with quote

ChrisR wrote:
(Ray what was your issue with the Cannon 5DSR? I searched through and didn't find much criticism in your posts?)


I did not publish my results and impressions, so here they are...I had fairly high expectations from the 5DSR. In particular, the "Fine Detail" mode (or whatever it is called) seemed to offer perhaps an improved processing of the image at pixel level, better than my old HRT2i. The pixels are similar size vs the T2i, so what I expected from the 5DSR was a combination of improved 100% pixel detail and a FF sensor. My initial impressions were favorable, and indeed the 100% detail was good. What intrigued me was the new revision of demosaicing/editing software that came with the camera, which included a new sharpness function that offered better fine detail adjustments. I tried using the same settings on an image from my HRT2i, and got virtually the same output. Looks like all they did was to improve the sharpening algorithms, calling this "Fine Detail" mode. I can do this with my existing camera! So I'm still working with trying to get the best performance from the HRT2i...
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ray_parkhurst



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PostPosted: Thu Aug 02, 2018 8:52 am    Post subject: Reply with quote

mjkzz wrote:
Ray wrote:
0.7x-1.2x: Nikon 105mm Printing-Nikkor. I often need to shoot true reference shots of Cents to simultaneously document a die variety and all its details, and there is no better lens in this mag range for this work.


I think there is a Rayfact version of the printing lens, do you think it could be better? I am itchy to get it, just need some "justification" Very Happy


Far as I can tell, the Rayfact 105mm is identical to the "A" version of the Nikon 105PN. Some ratings are a bit different but that is probably all specmanship. So personally, I would not hesitate to pick up a Rayfact version for good price, but would not spend more for it.
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Lou Jost



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PostPosted: Thu Aug 02, 2018 10:35 am    Post subject: Reply with quote

"I think there is a Rayfact version of the printing lens, do you think it could be better?"

I think the "A" version is the best that has been tested.
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Lou Jost



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PostPosted: Thu Aug 02, 2018 10:40 am    Post subject: Reply with quote

Mike, if Nikon comes out with a medium format sensor at an affordable price, it would be a game-changer. However, even today we can get medium-format quality from Nikon cameras using shift adapters with medium-format lenses. I just began my first experiments with such a set-up, and they are encouraging.
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mawyatt



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PostPosted: Thu Aug 02, 2018 10:54 am    Post subject: Reply with quote

Lou Jost wrote:
Mike, if Nikon comes out with a medium format sensor at an affordable price, it would be a game-changer. However, even today we can get medium-format quality from Nikon cameras using shift adapters with medium-format lenses. I just began my first experiments with such a set-up, and they are encouraging.


Lou,

I'm betting (stockwise) on it. Of course I thought the same for some well known semiconductor companies Shocked

Only time will tell Rolling Eyes

Best,
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ray_parkhurst



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PostPosted: Thu Aug 02, 2018 11:11 am    Post subject: Reply with quote

Please help me out guys...for macro work, what is the lure of large format sensors? What I can see is that if you can quadruple the pixel count, then you can downsize the image by 2x and end up with a still-large image with much improved resolution and color fidelity due to elimination of whatever demosaicing/anti-aliasing/filtering/etc shenanigans are going on (ie same goal as the 4-image pixel-shift methods only better). But the price you pay (assuming same pixel dimensions) is double the magnification, and the associated optical penalties, difficulty in finding lenses, etc. For sensors with larger pixels, you can get a benefit of fewer diffraction effects on-sensor, but an even bigger required magnification and optical penalties. So, what gives? Where am I wrong in thinking there is an "optimum" sensor size/pitch versus magnification, and bigger is not necessarily better? If I am right, then maybe it would be worth discussing what does that optimum sensor size/pitch vs magnification curve look like...
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mawyatt



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PostPosted: Thu Aug 02, 2018 11:29 am    Post subject: Reply with quote

Lou Jost wrote:
Mike, if Nikon comes out with a medium format sensor at an affordable price, it would be a game-changer. However, even today we can get medium-format quality from Nikon cameras using shift adapters with medium-format lenses. I just began my first experiments with such a set-up, and they are encouraging.


Lou,

It would seem Nikon is already evaluating this medium format with a expanded version of the new full frame mirrorless sensor chip, as I'm sure they started working on this new Z mount many years ago and also started a team working on the medium format. They could expand the pixel size of the high resolution sensor to fit the medium format size (would have same number of pixels but bigger) and utilize the overall camera architecture, hardware & firmware without significant changes.

So utilize a bigger pixel size version of the high resolution sensor to be offered in the upcoming Z mount mirrorless. This would allow test mules to be developed and evaluated very quickly and without significant cost other than the expanded chip design/layout/fabrication and necessary larger chip mounting. In the meantime a separate design team could be developing the high resolution medium format chip with many more pixels for the future medium format version offered.

Anyway, this is all speculation and only time will tell.

Best
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mawyatt



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PostPosted: Thu Aug 02, 2018 11:59 am    Post subject: Reply with quote

Ray, Lou,

I think this might have a lot to do with what lenses you can use. Newly designed lens may be able to take advantage of the medium format better than our present lens. However some lens may work well and give full coverage like your superb Nikkor PN105 you both have Very Happy
I'll leave the optical/sensor optimal details to the experts, but I'll listen carefully.

Another interesting aspect of this new Z mount is the 16mm sensor to flange distance. With an additional 30.5mm extension adapter Nikon's F mount lenses will be placed at the optimal point and should resolve images indistinguishable from our revered D800E, D810 and D850 other than the pixel size/number (OK maybe some noise issues also). Nikon has hinted somewhere that they were going to try and make some of the Nikon DSLR FX lenses work as well on the new mirrorless as they do on the high end FX bodies. My guess is these will be the newer and expensive Nikon lenses.

With such a short 16mm distance, most of the 3rd party lenses should be able to work well manually also, even ones that haven't performed well because of the F mount 46.5mm flange distance to sensor. So I would expect to see new and older Canon and Sony lenses working nicely manually. AF with 3rd party or "other" lenses is another story though.

Anyway, interesting time for the new Nikon. Hopefully we won't be disappointed, I know I wasn't, and haven't been, with the superb D850 Very Happy

!

Best,
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rjlittlefield
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PostPosted: Thu Aug 02, 2018 12:18 pm    Post subject: Reply with quote

ray_parkhurst wrote:
Please help me out guys...for macro work, what is the lure of large format sensors?

Well, there's an emotional aspect and a physical one.

I can't speak to emotions, but from a physical standpoint the key issue is just pixel counts.

Suppose you want to shoot a subject field of 24x36 mm.

Also suppose you have optics that can cover that field with effective f/5.6 on the subject side. (On the image side, the effective f-number will vary in proportion to sensor size, assuming you adjust the magnification to maintain constant FOV.)

There's a handy law of physics that the cutoff limit imposed by diffraction is just nu_0 = (2*NA)/lambda = 1/(lambda*f-number).

Assuming lambda = 0.00055 mm (green light), this gives nu_0 = 325 cycles/mm.

Across the 36 mm subject field, that means 36*325 = 11,000 cycles per field width.

By the Shannon sampling theorem, you need at least two samples per cycle, so to fully capture that f/5.6 optical image you need 2*11000 = 22,000 pixels across the field.

Of course that also means 22,000 pixels across the sensor, and with a 2:3 aspect ratio, that's 15,600 pixels in height.

Then just multiply those last two numbers to get 365 megapixels as the minimum count necessary to capture the optical image from subject-side f/5.6 over a 24mm x 36mm field.

There are good arguments that this number is actually too small, but in any case it should be clear that with large macro fields there's plenty of detail in the optical image to require huge pixel counts if you actually want to capture all of it. Huge pixel counts require large sensors, and that's one physical reason for the lure of large format sensors.

Whether you actually do want to capture all that detail is an entirely different question. But since that one leans toward the emotional side, I'll not speak to that.

--Rik
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ray_parkhurst



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PostPosted: Thu Aug 02, 2018 1:42 pm    Post subject: Reply with quote

rjlittlefield wrote:
ray_parkhurst wrote:
Please help me out guys...for macro work, what is the lure of large format sensors?

Well, there's an emotional aspect and a physical one.

I can't speak to emotions, but from a physical standpoint the key issue is just pixel counts.

Suppose you want to shoot a subject field of 24x36 mm.

Also suppose you have optics that can cover that field with effective f/5.6 on the subject side. (On the image side, the effective f-number will vary in proportion to sensor size, assuming you adjust the magnification to maintain constant FOV.)

There's a handy law of physics that the cutoff limit imposed by diffraction is just nu_0 = (2*NA)/lambda = 1/(lambda*f-number).

Assuming lambda = 0.00055 mm (green light), this gives nu_0 = 325 cycles/mm.

Across the 36 mm subject field, that means 36*325 = 11,000 cycles per field width.

By the Shannon sampling theorem, you need at least two samples per cycle, so to fully capture that f/5.6 optical image you need 2*11000 = 22,000 pixels across the field.

Of course that also means 22,000 pixels across the sensor, and with a 2:3 aspect ratio, that's 15,600 pixels in height.

Then just multiply those last two numbers to get 365 megapixels as the minimum count necessary to capture the optical image from subject-side f/5.6 over a 24mm x 36mm field.

There are good arguments that this number is actually too small, but in any case it should be clear that with large macro fields there's plenty of detail in the optical image to require huge pixel counts if you actually want to capture all of it. Huge pixel counts require large sensors, and that's one physical reason for the lure of large format sensors.

Whether you actually do want to capture all that detail is an entirely different question. But since that one leans toward the emotional side, I'll not speak to that.

--Rik


What MTF would that correspond to, and is that really a valid goal? What would the pixels look like at 100%? I can understand the desire to capture 100% of the information available, but again, is that really a valid (let alone achievable) goal?

The subject size you give as an example is arbitrary, but I believe the resolution of 1.64um per "subject pixel" scales. Then the bottom line of the discussion is achieving 1.64um on-subject, assuming this is actually a valid goal. So for instance, it would not make sense to shoot at higher than 1:1 on a sensor with 1.64um pixels, or 2:1 with higher than 3.28um, etc, correct?

You morphed from object to image space without stating the 1:1 mag assumption inherent in your numbers. At 1:1, not many lenses (maybe zero?) are diffraction-limited at f5.6 effective across a FF sensor, and the availability diminishes at higher mags. Scaling to large format makes the situation worse.

So I suppose the "optimum" needs to include limits of availability.
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Lou Jost



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PostPosted: Thu Aug 02, 2018 3:46 pm    Post subject: Reply with quote

If we are talking about infinity-corrected objectives, it's just a matter of using a longer tube lens, which however does have its practical limitations...

For example, to go from MFT's 20Mp to FF 36MP or 42MP, I can replace the Raynox 125mm lens for a Precision Optics 250mm lens, and get exactly the same aerial image (though twice as big) with about twice the pixels under it. Replace that with the Canon 500mm close-up lens and I can project the same aerial image onto a medium-format 6x7 sensor (or scanning back).

Alternatively I'd just use the MFT sensor + pixel-shifting to give me the equivalent of the full-frame image with the Raynox 125mm lens.
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mjkzz



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PostPosted: Thu Aug 02, 2018 4:49 pm    Post subject: Reply with quote

rjlittlefield wrote:
ray_parkhurst wrote:
Please help me out guys...for macro work, what is the lure of large format sensors?

Well, there's an emotional aspect and a physical one.

I can't speak to emotions, but from a physical standpoint the key issue is just pixel counts.

Suppose you want to shoot a subject field of 24x36 mm.

Also suppose you have optics that can cover that field with effective f/5.6 on the subject side. (On the image side, the effective f-number will vary in proportion to sensor size, assuming you adjust the magnification to maintain constant FOV.)

There's a handy law of physics that the cutoff limit imposed by diffraction is just nu_0 = (2*NA)/lambda = 1/(lambda*f-number).

Assuming lambda = 0.00055 mm (green light), this gives nu_0 = 325 cycles/mm.

Across the 36 mm subject field, that means 36*325 = 11,000 cycles per field width.

By the Shannon sampling theorem, you need at least two samples per cycle, so to fully capture that f/5.6 optical image you need 2*11000 = 22,000 pixels across the field.

Of course that also means 22,000 pixels across the sensor, and with a 2:3 aspect ratio, that's 15,600 pixels in height.

Then just multiply those last two numbers to get 365 megapixels as the minimum count necessary to capture the optical image from subject-side f/5.6 over a 24mm x 36mm field.

There are good arguments that this number is actually too small, but in any case it should be clear that with large macro fields there's plenty of detail in the optical image to require huge pixel counts if you actually want to capture all of it. Huge pixel counts require large sensors, and that's one physical reason for the lure of large format sensors.

Whether you actually do want to capture all that detail is an entirely different question. But since that one leans toward the emotional side, I'll not speak to that.

--Rik


Sure you can over sampling it and have a minimum resolution, but the "true" picture (signal) is still 91.25MP, so you still have to down sample it by the same factor because pixels pulled from over sampling are diffraction limited.

To get diffraction free pixels, maximum MP is 91.25MP where each pixel is diffraction free. without down sampling, and I believe this is what most photographers mean.

I am struggling with this "resolving power" of the lens, say at 7um for a modest lens, what effect does it have in terms of resolution?
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Lou Jost



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PostPosted: Thu Aug 02, 2018 5:03 pm    Post subject: Reply with quote

Diffraction is not an all-or-none thing, and there is no such thing as diffraction-free pixels. Diffraction is linear with respect to aperture setting, and there is some diffraction blurring even at apertures that are normally considered to be diffraction-free. The more pixels you can put under a subject feature, the better the image, though we do soon reach a point of diminishing returns.
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Last edited by Lou Jost on Thu Aug 02, 2018 5:19 pm; edited 1 time in total
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rjlittlefield
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PostPosted: Thu Aug 02, 2018 5:06 pm    Post subject: Reply with quote

ray_parkhurst wrote:
You morphed from object to image space without stating the 1:1 mag assumption inherent in your numbers.

There is no 1:1 mag assumption in anything I wrote.

I specified optics that are effective f/5.6 on the subject side, and that the subject field is 24x36 mm.

IF you have mag=1, then you will need a 24x36 mm sensor and the optics will be f/5.6 on the sensor side.

But IF you have mag=0.5, then you will need a 12x18 mm sensor and the optics will be f/2.8 on the sensor side.

Given the same FOV at the subject and the same effective f-number on the subject side, when you change the sensor size by some factor, you also change the magnification and the sensor-side effective f-number by the same factor. This is basic optics, but it is a simple truth that is easy to miss.

Regarding pixel count, I'm hoping it's obvious that if you need N pixels on the subject (or on a sensor at 1:1 magnification), then you also need N pixels on any other sensor that is going to capture the same information.

So, again, there is no assumption of mag=1 in the analysis. The pixel count is determined by FOV and subject-side effective f# (or NA, if you like to think in NA). It is not at all affected by sensor size.

Quote:
What MTF would that correspond to

By definition nu_0 is the frequency at which MTF drops clear to zero. So by putting 2 pixels per cycle, I've just barely met the Nyquist–Shannon requirement, at a frequency where all detail disappears.

This is the same approach that is used on Nikon's website for configuring microscope cameras. At first I thought it was completely crazy. But then I realized that the same specification puts 4 pixels per cycle at a frequency where the MTF is 39%. I know from other work that ~4 pixels per cycle is a good number to avoid sampling artifacts, so I decided that the approach was not at all crazy and I'd go along with it for simplicity.

Quote:
What would the pixels look like at 100%?

They will not be "sharp". That is, there will be no cases where you see single-pixel edges; no cases where you see 0/255/0/255 on four consecutive pixels. There is an unavoidable tradeoff that if you have those sorts of sharp edges, then you have failed to capture all the information in the optical image. See the discussion "On the resolution and sharpness of digital images..."

Quote:
I can understand the desire to capture 100% of the information available, but again, is that really a valid (let alone achievable) goal?

I have no idea how you're using the word "valid".

As for achievable, that depends on how large an FOV and how much resolution you need. Obviously 24mm x 36mm by 325 cycles/mm is pretty challenging. But is it out of reach? My guess is that if you used a microscope to look at the optical image of your PN105 wide open, you'd see a level of finest detail right in line with these numbers.

Quote:
So for instance, it would not make sense to shoot at higher than 1:1 on a sensor with 1.64um pixels, or 2:1 with higher than 3.28um, etc, correct?

That's a pretty fair statement, still assuming effective f/5.6 on the subject side. Again, f/5.6 on the subject side at 1:1 implies f/5.6 on the sensor side, so then for green light nu_0=325 lines/mm which implies 1.54 microns at the Nyquist/Shannon limit. The exact numbers change depending on how many pixels per cycle you want, at what MTF, at what wavelength, but what you've said has the right pattern.

Quote:
So I suppose the "optimum" needs to include limits of availability.

That word always scares me unless it comes with well specified criteria attached. Some people consider 4 pixels per cycle to be overkill because it means that their images never look sharp when pixel-peeped. Other people consider 4 pixels per cycle to be perfect because it means they won't lose any information that was present in the optical image.

--Rik
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rjlittlefield
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PostPosted: Thu Aug 02, 2018 5:08 pm    Post subject: Reply with quote

mjkzz wrote:
To get diffraction free pixels, maximum MP is 91.25MP where each pixel is diffraction free. without down sampling, and I believe this is what most photographers mean.

I would be interested to know where that number comes from.

Can you state the assumptions and show the calculation?

--Rik
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