Diffraction limit and D2x

Bill Janes · Mar 18, 2005

cpw said:
Bill Janes said:

However, this MTF refers only to the lens. For the MTF50 of the
printed picture, we need also to take into account the MTF of the
sensor and printing device, and this is not done in Mike's
analysis. The MTF50 of the total system would be lower than 49
lp/mm and his print might not have the required resolution of 4
lp/mm. To take these factors into account would require some pretty
complex math involving Fourier transformation and convolutions as
explained by Norman Koren on his web site.

http://www.normankoren.com/Tutorials/MTF.html

Click to expand...

Hello Bill,

I agree with your doing the experimental approach to getting a good
MTF of your system, but let's get your math straight here. If you
have the MTFs of each independent component in your system (lens,
sensor, printer), then the total system MTF is just the product of
each individual MTF. This is the advantage of working in frequency
space; now, getting the individual MTFs of each component is
another story!

Take care,

Chris

Chris,

You raise a very valid point. The Imitest experiments give an effective MTF50 for the combination of lens and camera, but I have no way of determining the MTF of the printing device. Udoubtedly, even the best printer would degrade the MTF to some degree, but I would hope that a reasonably high end digital printer such as the 400 ppi Noritsu QSS-2901 would maintain most of the image detail.

As Norman explains on his web site, before you do the multiplication you have to use the Fourier transform to convert the data into the frequency domain, multiply the components, and then convert back to the spatial domain with an inverse transform.

This is completely beyond my capability, but the point of my post was that one must consider MTF beyond the lens component.

Apparently, the old formula 1/R = 1/Rl + 1/Rf + 1/Rp, where R is total system resolution Rl lens resolution, Rf film resulution, and Rp printer resolution only works for MTF of 10% or less.

As the Leica expert Erwin Puts points out on his web site 4 lp/mm at 15x magnification on the paper is difficult to achieve even with a Leica, the best Leitz lenses, and high resolution black and white film. His site contains some interesting insights, but he still uses the old formulas!

http://www.imx.nl/photosite/technical/highres.html

For field work focusing error,camera shake and subject movement probably are the most limiting factors and it remains to be seen how much our photos would be improved if we switched from a D70 to a D2X.

--
Bill Janes

cpw · Mar 18, 2005

Bill Janes said:
... one must consider MTF beyond the lens component.

Hi Bill,

So true!

How is that Imatest program? I have a D70 myself and was thinking of using it.

Chris

Bill Janes · Mar 18, 2005

cpw said:
cpw said:

... one must consider MTF beyond the lens component.

Click to expand...

Hi Bill,

So true!

How is that Imatest program? I have a D70 myself and was thinking
of using it.

Chris,

It is a great program if you like to experiment and measure things. It also is a great learning tool. To measure MTF you use a simple chart you print out yourself. If you want to make use of the color checking component, you need a Greytag-MacBeth color checker chart, which is also useful in calibrating Adobe Camera Raw if you use that program.

There is a free demo for download and I would try it. Norman told me he thinks he underpriced it. He thought the main audience would be photoenthusiasts, but it turns out that corporate types who wouldn't mind paying much more are getting it.

Norman supports it very well and now is constantly updating it.

cpw said:
Chris

--
Bill Janes

Ross · Mar 18, 2005

priya said:
The diffraction limit is reached when wavelength/pixel_size is less
than f ratio. Wavelength of yellow light (center of spectrum) is
about o.55 micron. So if pixel size is 6 micron, then we
diffraction effct sets in at f/11.

--priya

priya,
Your formula works better inverted or pixel size/wavelength.
--
Bob Ross
http://www.pbase.com/rossrtx

Mike Davis · Mar 19, 2005

Hi Bill,

Bill Janes said:
Mike,

I read your post with great interest but have one question. You
speak of resolution in lp/mm but do not state what MTF is
associated with that resolution. Are you using the Rayleigh limit
( 9% MTF), Dawes Limit (0%) or the MTF at 50% contrast, which is
better correlated with percieved image sharpness?

MTF was not considered in my calculation for when diffraction will become "visible".

In this equation:

Maximum N =
1 / desired print resolution / enlargement factor / 0.00135383

The constant 0.00135383 is the diameter of the first zero of the diffraction pattern, as calculated using the Rayleigh function:

Diameter of first zero = 2.43934 * N * lambda

where lamda is the wavelength of light (I use 555 mm for this wavelength, in the yellow-green portion of the visible spectrum.)

The Rayleigh critierion for a circular aperture says that diffraction's first zero is at 1.21967 lambda/(2 R), where R is the radius of that aperture and lamda is the wavelength of light in nm.

The constant I'm using, 2.43934, is simply 2 x 1.21967 because I'm calculating the diameter of the first zero, instead of the radius.

So... this formula (again):

Maximum N =
1 / desired print resolution / enlargement factor / 0.00135383

Makes the assumption that the achievable resolution (in lp/mm) in the final print will be limited by diffraction to the reciprocal of the diameter of diffraction's Airy disk, after enlargment.

In other words, if the diameter of your Airy disks is 0.25 mm after enlargmenet, for example, taking the reciprocal of that diameter gives us a maximum possible on-print resolution of:

1 / 0.25mm = 4 lp/mm

This reveals an assumption that a line pair will not be distinguishable as independent lines if the Airy disk diameter is anything less than the width of that line pair.

Primitive, I admit, but my personal experience shooting with this approach has convinced me that it works very well. (And it's so much easier than performing Fourier transforms. Simple algebra is about as far as my math skills go.)

For anyone interested in more information on the Rayleigh criterion, see:

http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/raylei.html

http://www.photo.net/photo/optics/lensTutorial

Thanks,

Mike Davis

---
http://www.accessz.com

Mike Davis · Mar 19, 2005

Hi Kathy,
[snip]

Kathy_in_Maryland said:
Could you put "4 lp/mm" into a dpi for me. Meaning, I normally
send an image to the printer at 240dpi or 300dpi depending on the
printer. How does dpi relate to line pairs per mm?

Good question!

Just multiply your desired lp/mm value by 50!

Pixels/inch = lp/mm * 50

The assumption here is that the diameter of one pixel equates to the width of one line in a line pair. If your goal is 4 line pairs per millimeter in the final print, then your goal is 8 lines per millimeter in the final print -OR- 8 pixels per millimeter. There are 25.4 mm in an inch, so if we want 8 pixels per millimeter, we'll need 8 * 25.4 pixels per inch, which equals 203.

Rather than multipling lines/mm by 25.4, just multiply line pairs/mm by 50 to get pixels/inch.

So... If you want to achieve 4 lp/mm in the final print (at the soft end of what the human eye can resolve at a viewing distance of 10 inches and more than enough for viewing at 20 inches), the Data Density of your print must be:

50 * 4 lp/mm = 200 pixels/inch

Here's a short table of equivalents:

2 lp/mm = 100 pixels/inch
3 lp/mm = 150 pixels/inch
4 lp/mm = 200 pixels/inch
5 lp/mm = 250 pixels/inch
6 lp/mm = 300 pixels/inch
7 lp/mm = 350 pixels/inch
8 lp/mm = 400 pixels/inch

In my opinion, there's never any need to set your data density any higher than 300 pixels per inch, not to mention that you'll need a very expensive camera (lots of megapixels) to avoid the tiny prints had with higher data densities. Most people simply can't appreciate resolutions higher than 6 lp/mm at a viewing distance of 10 inches. Consider this: A color lab I use routinely produces digital prints on their LightJet and Lamda printers using image files with a data density of only 150 pixels per inch. That's the density they go for when scanning transparencies for these printers. They have lots of satisfied customers. Remember that at a viewing distance of 20 inches, a 3 lp/mm (150 dpi) print looks just as sharp as a 6 lp/mm (300 dpi) print viewed at 10 inches.

How do you adjust this thing I'm calling Data Density?

In Photoshop, for example, if you do an Image> Resize> Image Size, in the "Document Size" frame, you can specify a "Resolution" in "pixels/inch". Think of this as how dense the data is in the image file to be stored on your hard drive and sent to the printer. This Data Density remains constant whether you tell your printer to use a "Print Quality" setting of 360, 720, 1440, or even 2880 dpi.

By the way, as a rule of thumb, this "Print Quality" setting you specify for the Printer should always be at least three (3) times the image file's Data Density. For a 4 lp/mm print (from an image saved at a Resolution of 200 dpi), the printer's Print Quality should be set to at least 3 x 200 = 600, so 720 would be sufficient.

Also: When using Photoshop to Resize your image for the purpose of setting the "Resolution" to 200 pixels/inch, you should start by disabling the "Resample Image" option, so that the Width and Height values will show you the largest print dimensions possible with the amount of real, uninterpolated data you've captured with your camera. If those print dimensions are larger than you need, feel free to Resample to smaller print dimensions, but avoid resampling to larger diminsions.

Mike Davis

--
http://www.accessz.com

Mike Davis · Mar 19, 2005

Bill,

You've covered a lot of material I have no experience with. Thanks for covering it so well and providing the links for more information. I've got some homework to do.

I am very excited about how closely your Imatest results correlate to my method for determining the aperture at which diffraction will become "visible".

Bill Janes wrote:
[snip]

Bill Janes said:
http://www.imatest.com/

I performed an Imitest analysis of the central image using my Nikon
D70 and 50mm f/1.8 lens and the results are shown graphically at
this link.

[snip]

Bill Janes said:
http://bjanes.smugmug.com/gallery/438435

[snip]

Bill Janes said:
In summary, I would not advise stopping down to f/16 (or f/15) if
maximal resolution is required for 8 by 10 inch prints with this
camera/lens combination, but would be a little more consevative. Of
course, if you are making 4 by 6 inch prints, you have more
lattitude. Nonetheless, I congratulate Mike for an excellent
analysis.

So, for 8x10 prints of "maximal resolution", you have advised Nikon D70 users to use apertures wider than f/15.

The D70's 3008 x 2000 pixels will have to be cropped to produce an 8x10 print because of the aspect ratio of its sensor, so we would end up using 2500 x 2000 pixels in the final print.

2500 pixels divided by 10 inches equals a pixel density of 250 pixels/inch. This is equivalent to 5 lp/mm, assuming one pixel per line.

The enlargement factor to 8x10 for the D70 sensor is 13.0 (dividing 8 inches, converted to mm, by the sensor height of 15.6mm).

Using the formula provided in my first post:

Maximum N =
1 / desired print resolution / enlargement factor / 0.00135383

For an 8x10 print from the D70...

Maximum N =
1 / 5 / 13.0 / 0.00135383 = 11.36 (or f/11)

So, my method says we can stop down to f/11 and no farther if we want to prevent diffraction from inhibiting 5 lp/mm in the 8x10 print. Imatest results indicated that one should not stop down to f/15. Somewhere between Mike saying it's OK to use f/11.36 and Bill saying it's not OK to use f/15 lies the deviation between the two methods, but we can certainly say this: Both methods conclude that f/11 is acceptable, and f/16 is not.

I feel validated! Joy!

Mike Davis

--
http://www.accessz.com

Jay Turberville · Mar 19, 2005

Mike Davis said:
The Rayleigh critierion for a circular aperture says that
diffraction's first zero is at 1.21967 lambda/(2 R), where R is the
radius of that aperture and lamda is the wavelength of light in nm.

And the Rayleigh criterion is about equal to MTF 9%.

Mike Davis said:
The constant I'm using, 2.43934, is simply 2 x 1.21967 because I'm
calculating the diameter of the first zero, instead of the radius.

So... this formula (again):

Maximum N =
1 / desired print resolution / enlargement factor / 0.00135383

Makes the assumption that the achievable resolution (in lp/mm) in
the final print will be limited by diffraction to the reciprocal of
the diameter of diffraction's Airy disk, after enlargment.

So this doubling of the spacing of the Rayleigh criteria should yield an MTF of about 50% - which jibes with Bill's deduction.

Mike Davis said:
In other words, if the diameter of your Airy disks is 0.25 mm after
enlargmenet, for example, taking the reciprocal of that diameter
gives us a maximum possible on-print resolution of:

1 / 0.25mm = 4 lp/mm

This reveals an assumption that a line pair will not be
distinguishable as independent lines if the Airy disk diameter is
anything less than the width of that line pair.

But the reality is somewhat different. What your assumption actually yields and why it is a good formula for your purposes is the criteria where the lines should be very distinctly distinguishable. The Rayleigh criterion tries to determine where you might barely be able to make the distinction. I made a little illustration to show the point for those not familiar and who don't want to get too deep into the technical aspects.

The distribution of "light" in the disks is not technically accurate, but it should be close enough to make the point that detail can be distinguished when separated only by the radius of the Airy disk and it can be very readily distinguished when separated by the diameter of the Airy disk.

Mike Davis said:
Primitive, I admit, but my personal experience shooting with this
approach has convinced me that it works very well. (And it's so
much easier than performing Fourier transforms. Simple algebra is
about as far as my math skills go.)

I've found the similar results when calculating limits of diffraction for digiscoping rigs. I target an MTF of 50% and hope to get 20-30% (digiscoping components are not optimized for each other). The results usually respond well to moderate sharpening.

I'd say your formula is a bit stringent and that there is more detail in the image than what it implies. But its an excellent and simple way to get in the ballpark and people can certainly fudge it one way or the other to their personal taste by simply using a different lp/mm number.

--
Jay Turberville
http://www.jayandwanda.com

Bill Janes · Mar 19, 2005

Mike Davis said:
Bill Janes said:

Mike,

I read your post with great interest but have one question. You
speak of resolution in lp/mm but do not state what MTF is
associated with that resolution. Are you using the Rayleigh limit
( 9% MTF), Dawes Limit (0%) or the MTF at 50% contrast, which is
better correlated with percieved image sharpness?

Click to expand...

MTF was not considered in my calculation for when diffraction will
become "visible".

In this equation:

Maximum N =
1 / desired print resolution / enlargement factor / 0.00135383

Mike,

It's nice to know that our two methods are in agreement. With regard to MTF, see Norman Koren's link below. The formula for the size of the Airy disk and Rayliegh limit are similar but with a different constant. Norman says a MTF of 9% is implicit in the definition of that limit. By changing the constant you can get the MTF50 value. If you go through your math, I think that your constant is pretty close to that of the MTF50, which explains why the two methods are in agreement. The lens tutorial and Roger Clark's web site cover the same material from a different viewpoint.

http://www.normankoren.com/Tutorials/MTF6.html#Diffraction

I agree that Fourier transforms are too cumbersome for practical photography and I wouldn't have the slightest idea on how to apply them. Fortunately, we can leave the heavy lifting to the sceintists

--
Bill Janes

Bill Janes · Mar 19, 2005

Jay Turberville said:
And the Rayleigh criterion is about equal to MTF 9%.

Jay Turberville said:
So this doubling of the spacing of the Rayleigh criteria should
yield an MTF of about 50% - which jibes with Bill's deduction.

Jay Turberville said:
But the reality is somewhat different. What your assumption
actually yields and why it is a good formula for your purposes is
the criteria where the lines should be very distinctly
distinguishable. The Rayleigh criterion tries to determine where
you might barely be able to make the distinction. I made a little
illustration to show the point for those not familiar and who don't
want to get too deep into the technical aspects.

The distribution of "light" in the disks is not technically
accurate, but it should be close enough to make the point that
detail can be distinguished when separated only by the radius of
the Airy disk and it can be very readily distinguished when
separated by the diameter of the Airy disk.

Jay Turberville said:
--
Jay Turberville
http://www.jayandwanda.com

Jay,

A very nice visual demonstration.

It seems intuitive to me that if the Airy disc is appreciably larger than the pixel spacing of the sensor (7.8 microns for the D70), the discs will overlap and the picture will be blurred.

For green light, at f/5.6 the disc is 6.3 microns, 8.9 microns for f/8, and 12 microns for f/11.

These values are in the critical range we are discussing here, but I can't quite visualize the relationship. Your visualization is helpful, but the pixel scale seems a bit off to me.
--
Bill Janes

Jay Turberville · Mar 19, 2005

Bill Janes said:
Bill Janes said:

Click to expand...

Bill Janes said:
A very nice visual demonstration.

It seems intuitive to me that if the Airy disc is appreciably
larger than the pixel spacing of the sensor (7.8 microns for the
D70), the discs will overlap and the picture will be blurred.

I think this is where Nyquist becomes important and the exact position of the Airy disc comes into play. But in general, as the Airy disc gets larger than the sensor pitch, the MTF should drop.

With a Bayer mask sensor, things get a bit more difficult. This is the point where I punt and just consider the sensor to be a "black box" that yields (for DX sized sensors) a resolution of around 80% of the implied pixel count. So your Airy disc can be a bit larger than the sensor spacing would suggest.

The other point that I hope my illustration shows is that the diameter of the disc is indicated by the "zero" point where there is almost no light. The apparent size of the disk is smaller because its intensity diminishes towards it limits. This is why we can resolve detail smaller than its diameter.

Bill Janes said:
For green light, at f/5.6 the disc is 6.3 microns, 8.9 microns for
f/8, and 12 microns for f/11.

Are you sure about these numbers? It looks like you calculated with a wavelength of 460nm instead of 560 nm?

I have a test at f/5.1 online that I did a few years ago with my CP995 that has a 3.5 micron sensor pitch.
http://www.jayandwanda.com/digiscope/reztest/reztest.html
(go to the "CP995 lens only at 31mm f5.1" section).

It resolved detail (barely) to a bit better than 100 lp/mm. Thats a detail size of 5 microns. The Airy disc (using green light as 560 nm) for f/5.1 is almost 7 microns. But we are still resolving detail slightly smaller than 5 microns - although at a very low MTF.

Bill Janes said:
These values are in the critical range we are discussing here, but
I can't quite visualize the relationship. Your visualization is
helpful, but the pixel scale seems a bit off to me.

Perhaps the problem is that the discs look smaller than 100 pixels? That is because the edge of the disc becomes very faint with very little energy near the edges. The seem smaller than 100 pixels. Of course, the falloff show is not necessarily correct and the discs should be accompanied by their diffraction rings to be more accurate.

Maybe this page will help.
http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/raylei.html#c1

--
Jay Turberville
http://www.jayandwanda.com

Bill Janes · Mar 19, 2005

Jay Turberville said:
Jay Turberville said:

For green light, at f/5.6 the disc is 6.3 microns, 8.9 microns for
f/8, and 12 microns for f/11.

Click to expand...

Are you sure about these numbers? It looks like you calculated
with a wavelength of 460nm instead of 560 nm?

mea culpa! I copied the figures from a chart on Roger Clark's web site, supposedly for 500 nm, but the math doesn't work out. For 530 nm the figures should be 7.2, 10.3, and 14.2

Jay Turberville said:
Perhaps the problem is that the discs look smaller than 100 pixels?
That is because the edge of the disc becomes very faint with very
little energy near the edges. The seem smaller than 100 pixels.
Of course, the falloff show is not necessarily correct and the
discs should be accompanied by their diffraction rings to be more
accurate.

I was expecting the disc to be about the size of the pixel, not 100 pixels. What disc size and pixel size are you using for illustration?

Jay Turberville said:
--
Jay Turberville
http://www.jayandwanda.com

--
Bill Janes

Jay Turberville · Mar 19, 2005

Bill Janes said:
I was expecting the disc to be about the size of the pixel, not 100
pixels. What disc size and pixel size are you using for
illustration?

Well these are not "real" Airy discs, but are approximations to illustrate the point that you can resolve detail smaller than the Airy disc's diameter. I think that concept is counterintuitive. The illustration shows how this can occur. The way I think of it is that the apparent size of the Airy disc is actually significantly smaller than the diameter of the first zero point.

The disc diameter is set to 100 pixels and hence the spacing is 100 pixels or 50 pixels. If you measure the pixels in Photoshop you will find that the center never quite goes to fully black - though its close.

--
Jay Turberville
http://www.jayandwanda.com

Ronny Poelmans · Mar 20, 2005

Thom Hogan said:
On paper, for example, the Kodak Pro SLR/n ought to be a winner. It
doesn't have an antialising filter or microlenses and it's got 14mp
at full frame. Unfortunately, it doesn't resolve to expectations:
the noise reduction clips detail visibly, it has signficant
side-to-side issues that are resolved with software adjustments,
and don't even get me started on the JPEG engine. Most of the
Nikkors I work with are a little softer in the corners (a much more
visible problem than diffraction).

Hi Thom,

Can I react to your kodak paragraph... (contains some dx vs ff content)

The kodak ought to be a winner, but it isn't.. but I keep asking myself what is wrong ?

We all know the kodak body is derived from the F80 (has parts in common) and lacks the proffessional feeling of an F6 (or D2 series) with battery problems etc.. slow AF , pentaprism viewfinder etc.. so it's a $300 body for 10 times the price. with popup flash.. enough said.

But the imager CCD is a different story.. assumed 80 tot 125 iso (14n)

It surprises me how good and sometimes how restricted it is. When I put on the Nikon AF 85 1.4 it performs absolutely stunning (even wide open), unbelievable details (landscapes, portraits) compared to my D70/D2H.
But when there is a lightsource or even a reflection in the frame it fails.

The noise reduction artifacts is something you can get completely rid of when using the raw files in photoshop and setting NR to zero. Using jpg is not an option with this camera.

I am wondering where the problems are coming from

1. Is it the absense of an AA-filter or microlenses in front of the sensor which are giving the sharpness but also the color-shifts. Kodak software is applying lens-optimization algorithms in camera (thus not changable afterwards) to the RAW file for some reason. We also know the sensor has been modified from 14n to slr/n for these color shifts.

2. Is it a problem in lens design, maybe FF lenses for film and digital ought to be designed very differently. (especially wide angle)

If you compare a 50mm for Hasselblad MF and a 35mm Nikon, they have a different design and the distance between filmplane and lens is very different.

Maybe the way to go for fullframe 24x36 is to make a 17-35 2.8 with the same design as the 12-24 (telemetric).. different exit pupil...

When I put on the 12-24dx on the 14n it is also a stunning performer (3000x3000 pixels cropped at 12-15mm) or limited range 17-24mm

3. Is it the software.. or something wrong with the sensor.. why has it lots of noise compared to the 12mp (D2x) sensor from sony.. If sony manages to create a perfect 12mp dx sensor it should also be able to provide us the same sensor in a FF package with 12mp or more pixels at INCREASED sensitivity because it is larger, has larger photosites why is this not happening ? And why is the kodak sensor only good at base ISO (80/100/125) talking about 14n here. Kodak provides a FF sensor upgrade 14N to 14NX for $1400 which includes a FULLFRAME sensor so don't tell us manufacturing cost is the limiting factor..

The kodak is not a bad camera, but it is very limited and requires lots of attention (skill) : Use the right lenses, precize focussing, avoid strong backlight (overexposure), limit iso to 125..

Also using a D2H, I am tempted to replace the kodak for a D2X.. but the thought of a mature FF "F6" like digital camera in the future really excites me. And get rid of the need for having every focal length covered twice (dx-FF) therefore having to many lenses and having problems making up my mind which ones to bring with me.. this is of course a luxury problem...

The future will probably bring us the one camera (F6D) we are all waiting for.. make us concentrate completely on photography and stops us complaining about all the compromises we are stuck with in our equipment.. having too much equipment to cover all of our needs is also a large handicap.

Regards
Ronny (Belgium)

The kodak has some nice concepts which I would like to see in future nikons.

different raw-formats for example.. kodak kan make 3mp or 6mp raw files in addition to its 14mp.
click white balance.

it is a pitty there is not more cooperation between Nikon and Kodak, together they would shine.

Kathy_in_Maryland · Mar 20, 2005

Thanks for taking the time for the lengthy explanation, I appreciate it.

Kathy

Mike Davis said:
[snip]

Mike Davis said:

Could you put "4 lp/mm" into a dpi for me. Meaning, I normally
send an image to the printer at 240dpi or 300dpi depending on the
printer. How does dpi relate to line pairs per mm?

Click to expand...

Good question!

Just multiply your desired lp/mm value by 50!

Pixels/inch = lp/mm * 50

The assumption here is that the diameter of one pixel equates to
the width of one line in a line pair. If your goal is 4 line pairs
per millimeter in the final print, then your goal is 8 lines per
millimeter in the final print -OR- 8 pixels per millimeter. There
are 25.4 mm in an inch, so if we want 8 pixels per millimeter,
we'll need 8 * 25.4 pixels per inch, which equals 203.
Rather than multipling lines/mm by 25.4, just multiply line
pairs/mm by 50 to get pixels/inch.

So... If you want to achieve 4 lp/mm in the final print (at the
soft end of what the human eye can resolve at a viewing distance of
10 inches and more than enough for viewing at 20 inches), the Data
Density of your print must be:

50 * 4 lp/mm = 200 pixels/inch

Here's a short table of equivalents:

2 lp/mm = 100 pixels/inch
3 lp/mm = 150 pixels/inch
4 lp/mm = 200 pixels/inch
5 lp/mm = 250 pixels/inch
6 lp/mm = 300 pixels/inch
7 lp/mm = 350 pixels/inch
8 lp/mm = 400 pixels/inch

In my opinion, there's never any need to set your data density any
higher than 300 pixels per inch, not to mention that you'll need a
very expensive camera (lots of megapixels) to avoid the tiny prints
had with higher data densities. Most people simply can't
appreciate resolutions higher than 6 lp/mm at a viewing distance of
10 inches. Consider this: A color lab I use routinely produces
digital prints on their LightJet and Lamda printers using image
files with a data density of only 150 pixels per inch. That's the
density they go for when scanning transparencies for these
printers. They have lots of satisfied customers. Remember that at
a viewing distance of 20 inches, a 3 lp/mm (150 dpi) print looks
just as sharp as a 6 lp/mm (300 dpi) print viewed at 10 inches.

How do you adjust this thing I'm calling Data Density?

In Photoshop, for example, if you do an Image> Resize> Image Size, in
the "Document Size" frame, you can specify a "Resolution" in
"pixels/inch". Think of this as how dense the data is in the image
file to be stored on your hard drive and sent to the printer. This
Data Density remains constant whether you tell your printer to use
a "Print Quality" setting of 360, 720, 1440, or even 2880 dpi.

By the way, as a rule of thumb, this "Print Quality" setting you
specify for the Printer should always be at least three (3) times
the image file's Data Density. For a 4 lp/mm print (from an image
saved at a Resolution of 200 dpi), the printer's Print Quality
should be set to at least 3 x 200 = 600, so 720 would be sufficient.

Also: When using Photoshop to Resize your image for the purpose of
setting the "Resolution" to 200 pixels/inch, you should start by
disabling the "Resample Image" option, so that the Width and Height
values will show you the largest print dimensions possible with the
amount of real, uninterpolated data you've captured with your
camera. If those print dimensions are larger than you need, feel
free to Resample to smaller print dimensions, but avoid resampling
to larger diminsions.

Mike Davis

--
http://www.accessz.com

Robin Casady · Mar 20, 2005

Your explanation only works if you are changing lenses to match field of view. A 200mm lens on 4x5 film will have the same DOF as a 200mm lens on a D100 sensor. It is more accurate to say that DOF is dependent on lens focal length. The sensor size has no actual effect on DOF.

Mark Burnham said:
Robin Casady said:

I don't understand this. What does sensor size have to do with DOF?

Click to expand...

At a given aperture, DOF is a function of distance from the
subject. The closer the subject, the less DOF; the farther away
from the subject, the greater the DOF. The crop factor of the APS
sensor requires greater distance from the subject for the same FOV
as a larger sensor, and therefore you end up with more DOF at the
same aperture than a larger sensor due to that increased distance
from the subject.

-Mark

Ron Parr · Mar 20, 2005

Robin Casady said:
Your explanation only works if you are changing lenses to match
field of view. A 200mm lens on 4x5 film will have the same DOF as a
200mm lens on a D100 sensor. It is more accurate to say that DOF is
dependent on lens focal length. The sensor size has no actual
effect on DOF.

You still need to consider output size. If you keep aperture, distance to subject, and focal length constant and apply the same amount of enlargement, then everything will be the same since the output from the smaller sensor will be the same size as a crop of the output from the larger sensor.

OTOH, if you keep output size constant so that you are looking at two prints of the same size, then the print from the smaller sensor will be enlarged more and will have less DOF.

--
Ron Parr
FAQ: http://www.cs.duke.edu/~parr/photography/faq.html
Gallery: http://www.pbase.com/parr/

Jay Turberville · Mar 21, 2005

Bill Janes said:
It is a great program if you like to experiment and measure things.
It also is a great learning tool. To measure MTF you use a simple
chart you print out yourself.

I downloaded the program Sunday, ran a few tests and also ran some of the DPreview tests through it.

This is a very simple way to do some solid testing of your cameras and lenses. This is great for the average photographer in that the setup for the test isn't very hard at all. Its quick, easy and give pretty objective results with very little of the subjective "finger in the wind" interpretation that is required when reading a standard resolution chart. It also puts an emphasis on MTF response which is a better way to try to understand resolution and image quality than standard lines per whatever ratings.

It would be nice if Phil would start including an Imatest evaluation summary along with his battery of test results.

Highly recommended. The current low price expires in a few days BTW.

--
Jay Turberville
http://www.jayandwanda.com

Bill Janes · Mar 21, 2005

Jay Turberville said:
Jay Turberville said:

It is a great program if you like to experiment and measure things.
It also is a great learning tool. To measure MTF you use a simple
chart you print out yourself.

Click to expand...

I downloaded the program Sunday, ran a few tests and also ran some
of the DPreview tests through it.

This is a very simple way to do some solid testing of your cameras
and lenses. This is great for the average photographer in that the
setup for the test isn't very hard at all. Its quick, easy and
give pretty objective results with very little of the subjective
"finger in the wind" interpretation that is required when reading a
standard resolution chart. It also puts an emphasis on MTF response
which is a better way to try to understand resolution and image
quality than standard lines per whatever ratings.

It would be nice if Phil would start including an Imatest
evaluation summary along with his battery of test results.

Highly recommended. The current low price expires in a few days BTW.

--
Jay Turberville
http://www.jayandwanda.com

Jay,

Glad you liked Imitest. Yes, I think Phil should use something more objective to supplement photos of the resolution chart. At least he publishes full resolution images of the test chart than you can further analyze.

Near the Nyquist limit, the resolution charts become very difficult to interpret because of alaising.

For those who do not want to download the full resolution chart, Norman's web site shows a good example of alaising near the Nyquist limit. At first glance, you want to go back to film! (Note to other readers: this resolution chart is not used by Imitest).

http://www.normankoren.com/EOS-10D_3.html#Res_vs_35mm
--
Bill Janes

Jay Turberville · Jun 5, 2005

Jay Turberville said:
Jay Turberville said:

First, as Bjorn points out further in the thread, all bets are off
with digital.

Click to expand...

OK. Here's my bet. The Nikon D2x will test out to have a
resolution of 72 lp/mm. This will equate to about 2200 LPH
(horizontally) in Phil's review. Based on past experiences, this
should be accurate to within about + or - 5% (I know of one result
that fell outside this range). I would be very surprised if the
actual tested resolution exceeded 2300 or dropped below 2100. Time
will tell.

So Thom, I'll bet $20 with you that all bets aren't really off.

Thom never took this bet and I would have lost it by too being specific and pinning the LPH number to the horizontal value alone and being very strict on the tolerances for accuracy. But the fact that my prediction falls right in the middle of the measured resolution for the vertical and horizontal resolutions makes my point that all bets really should not be off and that the resolution of this sensor was very predictable coming in with resolution number about what you'd expect.

76.4 lp/mm horizontally
63.7 lp/mm vertically
70.1 lp/mm on average.

I said it would be about 72 lp/mm

There was no resolution surprise here where the technology was pretty well understood and is not groundbreaking in respect to issues that would effect the resolution per pixel.

Bjorn's resolution (bark) test was and is flawed. The 1Ds-Mark II outresolves the D2x by pretty much what simple math predicts. The 1Ds-Mark II is soft at the edges compared to the D2X as experience and our understnding of 35mm lens optics would have us expect. The resolution differences were pretty predictable (and also largely academic - as expected).

Jay Turberville said:
Its nice to list all the potential variables, but we should also
remember that the various sensor and camera makers have not
demonstrated that they have any huge edge in any of these areas.
So it is very likely that we will get very predictable results.

I suspect the degree of deviance from expected resolution figures
is as much due to the subjective nature of interpretting a
resolution chart as they are from real differences in sharpness. It
is often hard to see differences unless special test images are
taken. So even if I'm wrong and the D2X is 10% better or worse
than expected, the visual impact of such a difference will be
small, representing only one fourth of the 40% improvement implied
by pixel count.

--
Jay Turberville
http://www.jayandwanda.com

Diffraction limit and D2x

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