Diffraction limitation in 35mm lenses.

[crspe]

(Offtopic)

Furthermore, even with B/W film, getting 80 lp/mm in the print is not easy at all as discussed by the Leica authorty Erwin Puts (the enlarger or scanner also limits MTF):

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

I was just reading these interesting articles on very high resolution film and the technology behind it ... The real limits of resolution are still far away as proven by the semiconductor industry. Silicon chips are created using photo-sensative materials and optical exposure - the chips are simply prints made from the the negative (mask).   So, what resolutions are achievable at the limits of todays optic / photographical technology ...

Smallest technology in production today can produce structures on the chip of 0.065um which corresponds to 7700 lp/mm. Thats a good camera (you would hope so for $1M)!

Amazing, huh!


The structures on the 5D are made in a 0.5um technology - only 1000lp/mm needed to make those sensors.

http://www.chipworks.com/WebReports/ShowRe...ds=CAR-0601-805

(They also have a nice pic of the 20D sensor:
http://www.chipworks.com/WebReports/ShowRe...ds=PPR-0501-001 )

[bjanes]

Bill,

Yes you did. But both the D70 and the D200 are smaller format than the 5D and try to make up for it with greater pixel density and higher resolution.

My guess is that the 5D has no meaningful resolution beyond 45 lp/mm. If the 25-105 IS lens is optimised for good performance at f16, there is no resolution advantage to be gained by opening up the aperture, there's only a slight increase in MTF at resolutions up to 45 lp/mm, and with this lens such MTF increases appear to be negligible. Simply switching from ACR to Raw Shooter probably has a greater effect on image detail, at the plane of focus, than opening up from f16 to f8.

Maybe I'll shoot some Norman Koren test charts when I have the time.
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Although the 5D has more pixels than the D70, the pixel sizes are comparable. The 5D pixel size is 8.2 microns and the D70's is 7.8 microns. Diffraction should affect both cameras to about the same degree and other factors being equal I would expect the resolution in LP/mm to be about the same. Of course, the D70's crop factor would require that you move back to fill the field and the image details would be smaller.

Of course, the Imitest results would be more definitive. Why do you think that your lens is optimized for f/16? I would think that the "sweet spot" would occur at a somewhat larger aperture if it is well corrected for aberrations. You mention Raw Shooter vs ACR (I use the latter). Is there a significant difference in the resolution that the converters can extract from the file?

[Ray]

The 5D pixel size is 8.2 microns and the D70's is 7.8 microns. Diffraction should affect both cameras to about the same degree and other factors being equal I would expect the resolution in LP/mm to be about the same.

I didn't realise the pixel pitch was so close. I guess this is explained by the fact the 5D actually has 12.8MP and the D70 sensor is actually slightly bigger than the Canon D60 sensor, which was my first DSLR. You've got a crop factor of 1.5 there instead of 1.6.

I just checked the resolution charts at dpreview and it seems the D70 does have marginally greater resolution than the 5D. Eg. D70 lines per picture height (LPH) vertically = 1450. D70 sensor vertically = 15.6mm. 1450/15.6 = 93 lines.
93 lines = 46.5 line pairs.

A similar calculation for the 5D gives a result  of 42 lp/mm. These are described as absolute resolution figures relatively free of aliasing and moire effects.

One could argue that resolution in the centre of the image could be greater. However, the edge of the 5D frame vertically is only 12mm from the centre. The Photodo MTF charts for the Canon 50/1.4 at f8 show a remarkably flat response at 40 lp/mm all the way to 18mm from the centre, so I think the 43 lp/mm maximum, practical resolution for the 5D is likely to be quite accurate.

What we're left with, it seems, is the practical effect of an increase in MTF across all resolutions below 42 lp/mm as we open up the aperture from f16. This is obviously going to vary with the quality of the lens and one would expect prime lenses to produce a contrastier result.

Why do you think that your lens is optimized for f/16? I would think that the "sweet spot" would occur at a somewhat larger aperture if it is well corrected for aberrations

Just my eyes. That's what my sample images posted earlier in the thread are telling me. The 'sweet spot' is probably still f8, but the differences between f8 and f16 are so small as not to matter.

I must get myself one of those Canon 50/1.4 lenses. I'd forgotten it has such a flat response at 40 lp/mm and f8.

You mention Raw Shooter vs ACR (I use the latter). Is there a significant difference in the resolution that the converters can extract from the file?

Yes. There appears to be; far greater than the 24-105 differences at f8 and f16.
However, perhaps it would be more accurate to say, if one restricts oneself to using sharpening options only available from within the RAW converter, then Raw Shooter does a much better job than ACR. If one is a real whiz at using other sharpening programs and one has a perfected technique of post conversion sharpening, then who knows?

[jani]

Offtopic

Okay! I've learned that b in brackets means something else in this program  :D .

You can turn off emoticons by unchecking "Enable emoticons?" just below the text box.

See: (B)

But unfortunately, there appears to be no way to disable ( C ) -> ©

[jani]

(Offtopic)
I was just reading these interesting articles on very high resolution film and the technology behind it ... The real limits of resolution are still far away as proven by the semiconductor industry. Silicon chips are created using photo-sensative materials and optical exposure - the chips are simply prints made from the the negative (mask).   So, what resolutions are achievable at the limits of todays optic / photographical technology ...

Smallest technology in production today can produce structures on the chip of 0.065um which corresponds to 7700 lp/mm. Thats a good camera (you would hope so for $1M)!

Amazing, huh!
The structures on the 5D are made in a 0.5um technology - only 1000lp/mm needed to make those sensors.

You should read up a bit on how they're having problems with photolitography because of the wavelengths of the extreme ultraviolet they're using today (the mentioned 65 nm structures; 0.065 µm = 65 nm) ...

That they're having problems doesn't mean that they're confident about solving them; Intel have just recently started mass production of 65 nm technology, and several actors in the market are working on 45 nm prototypes.

But the usefulness for visible light, which is one order of magnitude greater in wavelength, is ... limited.

I'm sure Bjørn Rørslett would be delighted, should he ever get the lenses and sensor filters that would transmit UV light at these wavelengths, but your hopes are not based in reality.

[Slough]

Having got out my college physics book, I will now prove that focal length does not affect resolution loss due to diffraction. *drum roll*

Now, we know that diffraction creates these 'Airy disks' which look like bulls-eyes - a central dot surrounded by concentric rings. The smaller the aperture, the bigger this central dot becomes - this is as described in http://www.cambridgeincolour.com/tutorials...photography.htm, to wit:
Some images may help, hopefully he won't mind me linking them here:

Airy disk in 2D


Airy disk in 3D


In order to figure out the maximum resolution of a lens, we figure out the angular separation of two points which are barely resolved - which for the purposes of this exercise will be two points whose central dots are just touching edge-to-edge - this will be the same as the angular size of either central peak.

The equation to figure this out is:

sinθ = 1.22(λ/D)

Where θ is the angular separation, λ is the wavelength of light (considered a constant in this exercise, and set at 500nm) and D is the aperture diameter.

For simplicity's sake, we'll make sinθ just plain old θ, as θ is so small in any real world scenario to make the difference negligible. That gives us:

θ = 1.22(λ/D)

Now, the experiment: We take two lenses, of 50mm and 100mm focal lengths, and set their apertures to f/2. Take a picture of an object 10 meters away with the 100mm lens, and one of an object 5 meters away with the 50mm lens. We'll take the wavelength of light to be 500nm as a constant.

We do this because for this experiment to be valid, the size of the image on the negative/sensor must be the same for both lenses. Since a 100mm lens gives 2x the magnification of a 50mm lens, we stand back twice the distance with the 100mm lens.

For the 50mm lens, D, the aperture diameter is 50/2 = 25mm.
This gives us:

θ = 1.22(500x10⁻⁹/25x10⁻³)
so...
θ = 2.44x10⁻⁵ rad

For the 100mm lens, D, the aperture diameter is 100/2 = 50mm.
This gives us:

θ = 1.22(500x10⁻⁹/50x10⁻³)
so...
θ = 1.22x10⁻⁵ rad

So, what does that tell us? It tells us that the Airy disk of the 100mm lens is half the angular size of that of the 50mm lens. This is critical to understanding why focal length doesn't matter in terms of diffraction. Longer lenses produce a narrower 'beam' of diffracted light - remember, the number above is in 'rad', so it's an angle, not a distance.

Even though the diffracted light has to travel further to reach the film/sensor in the 100mm lens than the 50mm lens, it's spreading out much more slowly. By the time it reaches the film/sensor it will produce a spot of light exactly as large as the 50mm lens does.

Now to prove this concretely, we want to measure what that means in terms of resolution. To do this, we want to measure the minimum distance between two points on an object that can be resolved with both lenses at the distances and aperture given.

From basic optics:
y/s = y'/s'

Where y is the separation of the points on the object, y' is the separation of the corresponding points on the film/sensor, s is the distance from lens to object and s' is the distance from lens to film/sensor.

Thus the angular separations of the object points and the corresponding image points are both equal to θ, so we get:

y/s = θ
and
y'/s' = θ

Because s is greater than the focal length of the lens, the image distance s' is approximately equal to the focal length.

For the 50mm lens, s = 5m, s' = 50mm and θ = 2.44x10⁻⁵ rad:

y/5m = 2.44x10⁻⁵
so...
y = 1.22x10⁻⁴m = 0.122mm
and
y'/50mm = 2.44x10⁻⁵
so...
y' = 1.22x10⁻³mm = 0.00122mm

For the 100mm lens, s = 10m, s' = 100mm and θ = 1.22x10⁻⁵ rad:

y/10m = 1.22x10⁻⁵
so...
y = 1.22x10⁻⁴m = 0.122mm
and
y'/100mm = 1.22x10⁻⁵
so...
y' = 1.2x10⁻³mm = 0.00122mm

Phew! So, where does that leave us? Both lenses can resolve a minimum distance of .122mm on an object, which translates to a minimum resolvable dot on the film/sensor of .00122mm at those distances and an aperture of f/2. If we change the aperture, we will get different numbers, but the important thing is that they are the same for both lenses - so we've just proved that for the same image size and the same aperture, but different focal lengths, resolution is identical.

Again, the reason for this is that the longer lens produces a narrower cone of diffraction, so even though the light has to travel further to get to the film/sensor, it ends up producing the same size spot of light in both cases.

Thus, focal length does not contribute to resolution loss from diffraction.

Cheers,
Peter
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Peter: An interesting post.

You say that: "so we've just proved that for the same image size and the same aperture, but different focal lengths, resolution is identical."

I understand what you are saying, and I agree with the point, but I'm not happy with the use of the term resolution, as it is very confusing. In optics the resolution of a lens is a well defined quanity, and it does depend on the focal length as the equations you posted show. I think that what you mean is the sharpness of the image on the sensor (whether that be film, or a solid state device), though I'm sure there is a more precise phrase than "sharpness on the sensor".

Leif

[Slough]

You should read up a bit on how they're having problems with photolitography because of the wavelengths of the extreme ultraviolet they're using today (the mentioned 65 nm structures; 0.065 µm = 65 nm) ...

That they're having problems doesn't mean that they're confident about solving them; Intel have just recently started mass production of 65 nm technology, and several actors in the market are working on 45 nm prototypes.

But the usefulness for visible light, which is one order of magnitude greater in wavelength, is ... limited.

I'm sure Bjørn Rørslett would be delighted, should he ever get the lenses and sensor filters that would transmit UV light at these wavelengths, but your hopes are not based in reality.
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You are right that according to conventional optics that we are approaching the limit of what can be produced on a chip.

Interestingly there is a new field of optics called photonics, whereby light can image detail at scales less than the wavelength of the light, a seemingly nonsensical proposition. I seem to recall that this is based on using materials with negative dielectic constants, and until recently they were only a theoretical curiosity. But I seem to recall that someone, probably in America, has created such a device. Whether this ever escapes from the labs, and impacts on photography remains to be seen. (Well, I'm sure it will. But whether or not we will be alive to see it is another question.)

Also they are using shorter wavelengths of 'light'. Maybe they will end up using x-rays. But here the problem is that the 'wires' on the chips become so small that weird quantum effects (such as quantum tunnelling whereby current leaks between adjacent 'wires' when classical physics says it shouldn't) start to dominate, along with excessive heating and other problems.

Leif

[Ray]

Peter: An interesting post.

You say that: "so we've just proved that for the same image size and the same aperture, but different focal lengths, resolution is identical."

I understand what you are saying, and I agree with the point, but I'm not happy with the use of the term resolution, as it is very confusing. In optics the resolution of a lens is a well defined quanity, and it does depend on the focal length as the equations you posted show. I think that what you mean is the sharpness of the image on the sensor (whether that be film, or a solid state device), though I'm sure there is a more precise phrase than "sharpness on the sensor".

Leif
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Leif,
My understanding is the resolution per unit area (or unit length) at the focal plane will not change with a change of focal length provided the f stop remains the same. But this is partly due to the fact that f stop is dependent upon both focal length and physical aperture diameter. Another way of putting it would be to say, diffraction limited resolution will not change with focal length provided the physical aperture diameter changes in proportion to the change in focal length, ie. provided the f/stop remains the same.

But line pairs per mm are not all we are concerned about when making pictures. A camera like the 20D is capable of delivering substantially greater resolution in terms of lp/mm than the 5D, but the 5D can deliver more lines per picture height.

This is the distinction which Peter's proof does not address.

[BJL]

Leif and others,

    it might be nice to use the optical jargon which is something like angular resolution of the subject. Roughly this is the smallest angle between different parts of the subject, as measured from the camera, that the camera (lens plus sensor!) can resolve. For me, this gets to the heart of what I usually care about; how much detail we get of a particular subject. Sensor/film/lens resolution in lp/mm are only means to that end.

For example, I believe that matching the angular resolution of the human eye requires aperture diameters no smaller than about 3-4mm due to diffraction which is about f/16 with a 50mm lens.

Angular resolution is about the same as "line per picture height" or "lp/mm on pictures of the same size", once the pictures we use to compare cover the same angular field of view.

Both diffraction and out of focus effects have the same effect on angular resolution of a subject when the effective aperture diameter (focal length divided by f-stop) is the same, regardless of the focal length and format.

[bjanes]

Leif and others,

    it might be nice to use the optical jargon which is something like angular resolution of the subject. Roughly this is the smallest angle between different parts of the subject, as measured from the camera, that the camera (lens plus sensor!) can resolve.

Both diffraction and out of focus effects have the same effect on angular resolution of a subject when the effective aperture diameter (focal length divided by f-stop) is the same, regardless of the focal length and format.
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According to Widipedia, angular resolution (θ) is given by this formula:

sin (θ) = 1.22  λ / D, where λ is the wavelength of the light and D the diameter if the lens. The f/stop is not involved here and it is the diameter of the lens that matters. This is why astronomers rate a telescope by the diameter of the lens or mirror, not the focal length or f/number.

Spatial resolution ( Δl) is given by the formula Δl = 1.22  f λ/D, where f = the f focal length of the lens and lambda and D are as above.

Since f/stop = D/f, This simplifies to  Δl = λ * f/stop. In this case, the diameter of the lens is not involved.

[a href=\"http://en.wikipedia.org/wiki/Angular_resolution]http://en.wikipedia.org/wiki/Angular_resolution[/url]

I hope this helps  

Bill Janes