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Have sensors really passed the resolution of lenses? NO!

 
 
Ray Fischer
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      02-26-2012
Wolfgang Weisselberg <(E-Mail Removed)> wrote:
>Ray Fischer <(E-Mail Removed)> wrote:
>> Kenneth Scharf <(E-Mail Removed)> wrote:
>>>On 02/11/2012 01:34 PM, David J Taylor wrote:

>
>>>> To get the full resolution out of a 36 MP sensor will likely require the
>>>> best of lenses, and some of the best of photographic techniques (such as
>>>> tripod, best focusing, reducing mirror-slap etc. etc.)
>>>Consider that many consider the resolution of Kodachrome 35mm film to be
>>>around 100-250 mega pixels.

>
>> According to the modulation-transfer diagram for Kodachrome 25 a good
>> estimate for an equivalent digital image is about 20 megapixels.

>
>> http://www.kodak.com/global/en/profe...bs/e55/e55.pdf

>
>> (70 cycles/mm * 2 pixels/cycle * 35mm) * (70 * 2 * 25)

>
>You want more than 2 pixels/cycle[1]. You want to seriously
>oversample[1]. You want 0% response, not 10%[2].


Absolute accuracy is neither the goal nor is it possible.
This is a useful guideline.

If you want to indulge yourself in pointless pedantry then go ahead.

--
Ray Fischer | None are more hopelessly enslaved than those who falsely believe they are free.
http://www.velocityreviews.com/forums/(E-Mail Removed) | Goethe

 
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Joe Kotroczo
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      02-26-2012
On 25/02/2012 20:49, Alan Browne wrote:
> On 2012-02-25 15:33 , Joe Kotroczo wrote:
>> On 25/02/2012 19:55, Alan Browne wrote:
>> (...)
>>> If a phase sensor could be done - and very small, I guess resolutions
>>> could be magnitudes higher and no lens at all would be needed.

>>
>> If you record phase as well as modulation in essence you'd have digital
>> holography, no?

>
> You certainly have the information to focus at any distance and resolve
> for any desired DOF. Not sure about holography.


Plenoptic camera maybe?

> The article in wikipedia is a good start, and it it:
>
>
> [1] A hologram represents a recording of information regarding the light
> that came from the original scene as scattered in a range of directions
> rather than from only one direction, as in a photograph. This allows the
> scene to be viewed from a range of different angles, as if it were still
> present.
>
> [3] A holographic recording requires a second light beam (the reference
> beam) to be directed onto the recording medium.
>
> and much more, of course.


Yes, but that's for analog holograms, with a physical recording medium
such as photopolymers or photorefractives.

If you look at <http://en.wikipedia.org/wiki/Digital_holography> you'll
find "The phase-shifting digital holography process entails capturing
multiple interferograms that each indicate the optical phase
relationships between light returned from all sampled points on the
illuminated surface and a controlled reference beam of light that is
collinear to the object beam"

Not that I truly understand any of it...

--
Illegitimi non carborundum
 
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Martin Brown
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      02-26-2012
On 26/02/2012 18:38, Alfred Molon wrote:
> In article<pqOdncpYH_IpodTSnZ2dnUVZ_tednZ2d@giganews. com>, Alan Browne
> says...
>> What set me off was when you said "lenses do not have a resolution".
>> (They do).

>
> Lenses have an MTF, but for sure not a resolution in megapixel.


For a specified zone of illumination they do. The choice of matching CCD
pixel size in microns to a given optical assembly on a telescope is
something astronomers think about much more than photographers.

The standard Rayleigh criterion for the angular resolution of a lens of
finite aperture is close enough for most purposes as 1.22*lambda/D
(radians).

Multiply that by the distance to the CCD and you have a bound based on
the sampling criterion that allows you to compute the pixel size and
resolution the lens can support. It is traditional to oversample
slightly by 1.5-2x to avoid jaggies in the raw data.

Advanced tricks can get you a factor of 3 better than classical
resolution performance in a very restricted set of circumstances.

--
Regards,
Martin Brown
 
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Martin Brown
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      02-26-2012
On 26/02/2012 22:19, Alfred Molon wrote:
> In article<g3x2r.18667$(E-Mail Removed)>, Martin Brown says...
>> On 26/02/2012 18:38, Alfred Molon wrote:
>>> In article<pqOdncpYH_IpodTSnZ2dnUVZ_tednZ2d@giganews. com>, Alan Browne
>>> says...
>>>> What set me off was when you said "lenses do not have a resolution".
>>>> (They do).
>>>
>>> Lenses have an MTF, but for sure not a resolution in megapixel.

>>
>> For a specified zone of illumination they do. The choice of matching CCD
>> pixel size in microns to a given optical assembly on a telescope is
>> something astronomers think about much more than photographers.
>>
>> The standard Rayleigh criterion for the angular resolution of a lens of
>> finite aperture is close enough for most purposes as 1.22*lambda/D
>> (radians).
>>
>> Multiply that by the distance to the CCD and you have a bound based on
>> the sampling criterion that allows you to compute the pixel size and
>> resolution the lens can support. It is traditional to oversample
>> slightly by 1.5-2x to avoid jaggies in the raw data.
>>
>> Advanced tricks can get you a factor of 3 better than classical
>> resolution performance in a very restricted set of circumstances.

>
> Once again, lenses have a response curve. The higher the spatial
> frequency, the lower the contrast. But you can't say that a lens will
> only let pass through 80 lp/mm and not 120lp/mm, although at a certain
> spatial frequency the MTF will be too low to be usable.


Actually you *can* to a very good approximation. Once the Rayleigh
criterion is exceeded you cannot sensibly distinguish two adjacent dots.

If you haven't sampled a spatial frequency in the image then it is lost.
The highest spatial frequency you can measure is determined by the ratio
of the wavelength of light to the diameter of the lens. This is all
classical wave optics and well understood physics.

> By the way, the formula you are quoting has nothing to do with the
> quality of the lens (and the ability to resolve detail depends on the
> lens quality). The only parameters are the wavelength of light and the
> diameter of the aperture.


That formula assumes diffraction limited optics typical of telescopes
but a condition that is seldom met in consumer level photographic lenses
at apertures wider than f5.6 (though there are exceptions).

The Rayleigh criterion assumes near perfect lens performance with
overall rms errors in the optical surfaces of lambda/4 or less.

--
Regards,
Martin Brown
 
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Wolfgang Weisselberg
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      03-02-2012
Alan Browne <(E-Mail Removed)> wrote:

> There is also the "Kodak equation" which used say something like the
> resolution of the system is that of the lens and that of the film
> combined - it's not simply the worst of the two. I can't find the
> equation offhand, it may have been: sysres = 1/(1/filmres+1/lensres)


Isn't it
MTF_{system} = MTF_{lens} * MTF_{sensor}
?

That would make sense --- MTF is 0--1, the whole system is weaker
than all (imperfect) components, the weakest component limits the
system most (and increasing the better component(s) only results
in slight improvements overall), ...

-Wolfgang
 
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Wolfgang Weisselberg
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      03-03-2012
Ray Fischer <(E-Mail Removed)> wrote:
> Wolfgang Weisselberg <(E-Mail Removed)> wrote:
>>Ray Fischer <(E-Mail Removed)> wrote:
>>> Kenneth Scharf <(E-Mail Removed)> wrote:
>>>>On 02/11/2012 01:34 PM, David J Taylor wrote:


>>>>> To get the full resolution out of a 36 MP sensor will likely require the
>>>>> best of lenses, and some of the best of photographic techniques (such as
>>>>> tripod, best focusing, reducing mirror-slap etc. etc.)
>>>>Consider that many consider the resolution of Kodachrome 35mm film to be
>>>>around 100-250 mega pixels.


>>> According to the modulation-transfer diagram for Kodachrome 25 a good
>>> estimate for an equivalent digital image is about 20 megapixels.


>>> http://www.kodak.com/global/en/profe...bs/e55/e55.pdf


>>> (70 cycles/mm * 2 pixels/cycle * 35mm) * (70 * 2 * 25)


>>You want more than 2 pixels/cycle[1]. You want to seriously
>>oversample[1]. You want 0% response, not 10%[2].


> Absolute accuracy is neither the goal nor is it possible.
> This is a useful guideline.


> If you want to indulge yourself in pointless pedantry then go ahead.


Ah, that's how you react to facts. Should have known.

-Wolfgang
 
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Ray Fischer
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      03-03-2012
Wolfgang Weisselberg <(E-Mail Removed)> wrote:
>Ray Fischer <(E-Mail Removed)> wrote:
>> Wolfgang Weisselberg <(E-Mail Removed)> wrote:
>>>Ray Fischer <(E-Mail Removed)> wrote:
>>>> Kenneth Scharf <(E-Mail Removed)> wrote:
>>>>>On 02/11/2012 01:34 PM, David J Taylor wrote:

>
>>>>>> To get the full resolution out of a 36 MP sensor will likely require the
>>>>>> best of lenses, and some of the best of photographic techniques (such as
>>>>>> tripod, best focusing, reducing mirror-slap etc. etc.)
>>>>>Consider that many consider the resolution of Kodachrome 35mm film to be
>>>>>around 100-250 mega pixels.

>
>>>> According to the modulation-transfer diagram for Kodachrome 25 a good
>>>> estimate for an equivalent digital image is about 20 megapixels.

>
>>>> http://www.kodak.com/global/en/profe...bs/e55/e55.pdf

>
>>>> (70 cycles/mm * 2 pixels/cycle * 35mm) * (70 * 2 * 25)

>
>>>You want more than 2 pixels/cycle[1]. You want to seriously
>>>oversample[1]. You want 0% response, not 10%[2].

>
>> Absolute accuracy is neither the goal nor is it possible.
>> This is a useful guideline.

>
>> If you want to indulge yourself in pointless pedantry then go ahead.

>
>Ah, that's how you react to facts.


Ah, that's how you lie. By claiming that you have "facts".

--
Ray Fischer | None are more hopelessly enslaved than those who falsely believe they are free.
(E-Mail Removed) | Goethe

 
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David J Taylor
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      03-03-2012
"Wolfgang Weisselberg" <(E-Mail Removed)> wrote in message
news:(E-Mail Removed)...
> Alan Browne <(E-Mail Removed)> wrote:
>
>> There is also the "Kodak equation" which used say something like the
>> resolution of the system is that of the lens and that of the film
>> combined - it's not simply the worst of the two. I can't find the
>> equation offhand, it may have been: sysres = 1/(1/filmres+1/lensres)

>
> Isn't it
> MTF_{system} = MTF_{lens} * MTF_{sensor}
> ?
>
> That would make sense --- MTF is 0--1, the whole system is weaker
> than all (imperfect) components, the weakest component limits the
> system most (and increasing the better component(s) only results
> in slight improvements overall), ...
>
> -Wolfgang


Wolfgang, you also need to include the atmosphere, processing and display
to get the system MTF, but in terms of lens and sensor, you are correct.

Likely, if you convert MTF into a single figure (e.g. by integrating the
area under the MTF curve), the "Kodak equation" is equivalent.

Cheers,
David

 
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Wolfgang Weisselberg
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      03-14-2012
Ray Fischer <(E-Mail Removed)> wrote:
> Wolfgang Weisselberg <(E-Mail Removed)> wrote:
>>Ray Fischer <(E-Mail Removed)> wrote:
>>> Wolfgang Weisselberg <(E-Mail Removed)> wrote:
>>>>Ray Fischer <(E-Mail Removed)> wrote:
>>>>> Kenneth Scharf <(E-Mail Removed)> wrote:
>>>>>>On 02/11/2012 01:34 PM, David J Taylor wrote:


>>>>>>> To get the full resolution out of a 36 MP sensor will likely require the
>>>>>>> best of lenses, and some of the best of photographic techniques (such as
>>>>>>> tripod, best focusing, reducing mirror-slap etc. etc.)
>>>>>>Consider that many consider the resolution of Kodachrome 35mm film to be
>>>>>>around 100-250 mega pixels.


>>>>> According to the modulation-transfer diagram for Kodachrome 25 a good
>>>>> estimate for an equivalent digital image is about 20 megapixels.


>>>>> http://www.kodak.com/global/en/profe...bs/e55/e55.pdf


>>>>> (70 cycles/mm * 2 pixels/cycle * 35mm) * (70 * 2 * 25)


>>>>You want more than 2 pixels/cycle[1]. You want to seriously
>>>>oversample[1]. You want 0% response, not 10%[2].


>>> Absolute accuracy is neither the goal nor is it possible.
>>> This is a useful guideline.


>>> If you want to indulge yourself in pointless pedantry then go ahead.


>>Ah, that's how you react to facts.


> Ah, that's how you lie. By claiming that you have "facts".


You didn't see/remember the URLs? Could be a brain tumor.
Could be just you.

-Wolfgang
 
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Wolfgang Weisselberg
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      03-14-2012
David J Taylor <(E-Mail Removed)> wrote:
> "Wolfgang Weisselberg" <(E-Mail Removed)> wrote in message
>> Alan Browne <(E-Mail Removed)> wrote:


>>> There is also the "Kodak equation" which used say something like the
>>> resolution of the system is that of the lens and that of the film
>>> combined - it's not simply the worst of the two. I can't find the
>>> equation offhand, it may have been: sysres = 1/(1/filmres+1/lensres)


>> Isn't it
>> MTF_{system} = MTF_{lens} * MTF_{sensor}
>> ?


>> That would make sense --- MTF is 0--1, the whole system is weaker
>> than all (imperfect) components, the weakest component limits the
>> system most (and increasing the better component(s) only results
>> in slight improvements overall), ...


> Wolfgang, you also need to include the atmosphere, processing and display
> to get the system MTF, but in terms of lens and sensor, you are correct.


Depends on how the 'system' is defined, but yes, you're right.
If you include atmosphere again and the human eye, that is.

> Likely, if you convert MTF into a single figure (e.g. by integrating the
> area under the MTF curve), the "Kodak equation" is equivalent.


However, you need something else, you need to factor in how the
eye perceives the sharpness (which depends on print size and
viewing distance).

-Wolfgang
 
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