Diffraction Limit Discussion Continuation

Started 8 months ago | Discussions
Jonny Boyd
Forum MemberPosts: 89
Like?
Re: Duck, duck, goose.
In reply to Ulric, 8 months ago

Ulric wrote:

Jonny Boyd wrote:

I get the impression that some people see the words 'diffraction' and 'limit(ed)' in close proximity and freak out without reading what I've written.

Johnny, for the benefit of everyone: just what is your definition of diffraction limited? A single, short sentence that anyone can understand. Mine is "Resolving power decreases when the lens is stopped down."

My definition comes at the end of this post. I wanted to put in some brief explanation first.

Diffraction always limits the resolving power of a lens, even wide open, and the effect gets stronger as you stop down. However lens aberations also limit resolution at all apertures but they're effect gets stronger as you open up the aperture. Hence all those curves we see of lens resolution at different apertures.

We could say that a meaningful definition of diffraction limited would be, not simply where diffraction limits resolution, because it always soes that, rather the point at which changing the aperture causes a drop in resolution due to diffraction effects. So if you take the peak aperture, then all smaller apaertures are diffraction limited while larger apertures are aberation limited.

What I was discussing though was a slightly modified definition which concerns when a lens is visibly diffraction limited. This would be when stopping down causes a visible decrease in resolution due to diffraction effects.

N.B. When a camera becomes dffraction limited at a larger aperture than another one and is therefore diffraction limited at a wider range of apertures,  that doesn't mean it has less overall resolution.

Reply   Reply with quote   Complain
Ulric
Senior MemberPosts: 2,623Gear list
Like?
Re: Thing is...
In reply to Great Bustard, 8 months ago

Great Bustard wrote:

Ulric wrote:

Great Bustard wrote:

Ulric wrote:

Great Bustard wrote:

Ulric wrote:

Jonny Boyd wrote:

I get the impression that some people see the words 'diffraction' and 'limit(ed)' in close proximity and freak out without reading what I've written.

Johnny, for the benefit of everyone: just what is your definition of diffraction limited? A single, short sentence that anyone can understand. Mine is "Resolving power decreases when the lens is stopped down."

...that's true of all lenses. It's simply a matter of what aperture the increasing diffraction softening outweighs lessening lens aberrations or, according to Johnny Boyd, what aperture you notice that diffraction is resulting in softer photos (which, of course, also depends tremendously, even primarily, on how large we display the photos, how closely we view them, and our visual acuity).

It is certainly not true of all lenses that resolving power is at a maximum wide open and decreases as it is stopped down.

You didn't initially say "is at a maximum wide open". So, if you are defining a "diffraction limited lens" as a lens that resolves best wide open, then we are in agreement.

I was just noting that "Resolving power decreases when the lens is stopped down" is true of all lenses once past their peak aperture. But, for sure, if we add "right from wide open" to the end of that statement, as I said, I agree.

Then we don't agree. Or we are talking past each other. It seems that my request to Johnny was unreasonable, there is no way to get one person to explain to another what he means by 'diffraction limit'. I'm sure there will be a third thread to follow up on this one.

Let me have another go. The effects of diffraction exist right from wide open, and only get worse as you stop down. However, two other effects typically offset this increasing diffraction at the wide end of the aperture: lessening lens aberrations and increasing DOF.

Good so far.

If the entire scene is within the DOF and the diffraction softening outweighs the lens aberrations from wide open, the lens is "diffraction limited".

I assume you mean that the effect of increasing diffraction is greater than the effect of lens aberrations decreasing.

Interesting, I think we may have different definitions. Mine does not require that to happen from wide open, just from the peak.

On the other hand, if either lessening lens aberrations or increasing DOF outweigh the effects of diffraction softening, then there will be an aperture that results in "optimal resolution", and the lens will be "diffraction limited" past this optimal aperture.

That is my definition. The peak can be wide open (the 7-14, or the Olympus 12-50 at the long end) but doesn't have to.

 Ulric's gear list:Ulric's gear list
Panasonic Lumix DMC-GF3 Olympus OM-D E-M5 Panasonic Lumix DMC-GM1 Panasonic Lumix G 20mm F1.7 ASPH Panasonic Lumix G 14mm F2.5 ASPH +6 more
Reply   Reply with quote   Complain
Steen Bay
Veteran MemberPosts: 6,237
Like?
Re: Cutting to the chase.
In reply to bobn2, 8 months ago

bobn2 wrote:

Life is too short to bother more with this pointless exchange. The nub of it is simple, you set out to 'prove' a mundane and obvious result using extensive calculations based on made-up numbers. That itself was an exercise in futility. My objection to that kind of bogusly quantitative exercise is that the might convince gullible people that there is something of substance in them, because on the surface they look deep and complex, when all they are is an exercise in numerbation (not even measurebation, because there are no genuine measurements involved). As for the result, yes if the pixel count is so small (we discover, about 200k) then you wont see the effects of diffraction or anything else. Every lens becomes perfect. What a great idea.

-- hide signature --

Bob

Also visible on cameras with more than 200k pixels. If someone for example upgraded from 5D to D800 (from 12.7mp to 36mp), then I'm almost sure that he/she would notice that diffraction would start to visibly reduce the resolution (of a good lens, at 100% view) at a larger aperture on D800 than it did on 5D. The higher the peak resolution is to start with, the more you have to lose to things like diffraction and camera shake. Nothing controversial about that.

Reply   Reply with quote   Complain
bobn2
Forum ProPosts: 30,584
Like?
Re: Cutting to the chase.
In reply to Steen Bay, 8 months ago

Steen Bay wrote:

bobn2 wrote:

Life is too short to bother more with this pointless exchange. The nub of it is simple, you set out to 'prove' a mundane and obvious result using extensive calculations based on made-up numbers. That itself was an exercise in futility. My objection to that kind of bogusly quantitative exercise is that the might convince gullible people that there is something of substance in them, because on the surface they look deep and complex, when all they are is an exercise in numerbation (not even measurebation, because there are no genuine measurements involved). As for the result, yes if the pixel count is so small (we discover, about 200k) then you wont see the effects of diffraction or anything else. Every lens becomes perfect. What a great idea.

-- hide signature --

Bob

Also visible on cameras with more than 200k pixels. If someone for example upgraded from 5D to D800 (from 12.7mp to 36mp), then I'm almost sure that he/she would notice that diffraction would start to visibly reduce the resolution (of a good lens, at 100% view) at a larger aperture on D800 than it did on 5D.

Are you sure? How are you sure?

I love the way that you haven't the confidence in your own views just to say you're sure, so you're 'almost sure'. Would you like to quantify your lack of surety? 5%, 10%, 20%? In which units?

The higher the peak resolution is to start with, the more you have to lose to things like diffraction and camera shake. Nothing controversial about that.

That's not how it's put, though is it? And is just as true about excellent lenses. We have regular threads on this forum saying how mFT lenses are the best. Well the downside of that is that they have more to lose - or is it?

-- hide signature --

Bob

Reply   Reply with quote   Complain
bobn2
Forum ProPosts: 30,584
Like?
Re: Thing is...
In reply to Ulric, 8 months ago

Ulric wrote:

That is my definition. The peak can be wide open (the 7-14, or the Olympus 12-50 at the long end) but doesn't have to.

The 7-14 is a bit of an odd example, since it's a faster lens that Olympus has stopped off at f/4 (true of most of the SHG's - the reason they are sharp 'wide open' is that 'wide open' isn't 'fully open'). Possibly also true of the 12-50. The advantage of a small format is that the lenses might be relatively humongous without being as big as people expect. The 7-14 is doing the same job as an f/8 on FF, while the 12-50 is doing the same job as a f/7-9, a range when many first class zooms are into diffraction limited mode (and if you think not, just think, you could make such a lens by taking a 24-100/7-9 and putting a 0.5x focal reducer behind it - it would probably end up much the same size).

-- hide signature --

Bob

Reply   Reply with quote   Complain
Steen Bay
Veteran MemberPosts: 6,237
Like?
Re: Cutting to the chase.
In reply to bobn2, 8 months ago

bobn2 wrote:

Steen Bay wrote:

bobn2 wrote:

Life is too short to bother more with this pointless exchange. The nub of it is simple, you set out to 'prove' a mundane and obvious result using extensive calculations based on made-up numbers. That itself was an exercise in futility. My objection to that kind of bogusly quantitative exercise is that the might convince gullible people that there is something of substance in them, because on the surface they look deep and complex, when all they are is an exercise in numerbation (not even measurebation, because there are no genuine measurements involved). As for the result, yes if the pixel count is so small (we discover, about 200k) then you wont see the effects of diffraction or anything else. Every lens becomes perfect. What a great idea.

-- hide signature --

Bob

Also visible on cameras with more than 200k pixels. If someone for example upgraded from 5D to D800 (from 12.7mp to 36mp), then I'm almost sure that he/she would notice that diffraction would start to visibly reduce the resolution (of a good lens, at 100% view) at a larger aperture on D800 than it did on 5D.

Are you sure? How are you sure?

I love the way that you haven't the confidence in your own views just to say you're sure, so you're 'almost sure'. Would you like to quantify your lack of surety? 5%, 10%, 20%? In which units?

OK, then my guess would be that something like a 20% drop in MTF-50 resolution/sharpness would be 'noticeable' or 'clearly visible' at at least 100% view.

The higher the peak resolution is to start with, the more you have to lose to things like diffraction and camera shake. Nothing controversial about that.

That's not how it's put, though is it? And is just as true about excellent lenses. We have regular threads on this forum saying how mFT lenses are the best. Well the downside of that is that they have more to lose - or is it?

-- hide signature --

Bob

Wouldn't call it a downside, but you'll have to be a bit more careful and think a bit more about what you are doing if you want to take full advantage of more MPs or a good lens.

Reply   Reply with quote   Complain
Jonny Boyd
Forum MemberPosts: 89
Like?
Re: Um...
In reply to Great Bustard, 8 months ago

Great Bustard wrote:

Jonny Boyd wrote:

Great Bustard wrote:

Jonny Boyd wrote:Diffraction causes a decrease in resolution, agreed?

Agreed.

When resolution drops due to stopping down from the peak aperture, that is due to diffraction, agreed?

Agreed.

At the aperture at which diffraction is reducing resolution, you can say that diffraction is limiting the resolution of the final image, agreed?

Agreed.

If resolution appears to be the same at an aperture smaller than the peak aperture then diffraction doesn't become the dominant factor in limiting resolution until later than the peak aperture, agreed?

Agreed.

Therefore, for practical purposes, as far as the eye can see, a system where resolution visibly drops immediately after peak aperture is more limited by diffraction than a system where the visible drop happens later. Agreed?

Not agreed, and am surprised you do not understand this. For example, let's say for a particular display size, viewing distance, and visual acuity, I can resolve 1000 lw/ph. All else equal, the photo from the lower MP sensor will dip below that threshold before the higher MP sensor.

What you're saying is that the lower resolution sensor produces lower resolution images. No kidding, that's what I've always said.

But how does this make the higher MP sensor more "diffraction limited" than the lower MP sensor?

As I've repeatedly explained, when diffraction produces a visible drop in resolution in situations where a lower resolution sensor wouldn't exhibit a visible drop in resolution, then diffraction is more a limiting factor for the image produced by the high res sensor. The high res sensor will still produce a higher resolution image, it's just that the resolution of that image is determined largely by diffraction effects, where as the lower resolution sensor's performance is dominated by the impact of its sensor and diffraction effects are relatively negligible.

My argument is that for a sufficiently low resolution sensor, an image taken at peak aperture and an image taken at a smaller aperture will have a difference in resolution that is indistinguishable to the naked eye because it is so minor. For practical purposes therefore the perceived resolution is not being limited by diffraction until an even smaller aperture than the actual peak.

But you can't see that the higher MP sensor had greater resolution at the peak aperture, so the higher MP sensor isn't being "visibly limited" by diffraction, either, until a much smaller aperture.

Of course it has higher resolution.

You're bringing up an interesting point though about whether the drop in resolution will be noticed for a high res sensor. To be honest, I hadn't really thought about things from that perspective, but that would be worth looking into, whether it is possible for the drop in resolution for a high resolution sensor and a high resolution lens being stopped down, is not visible.

In contrast, a higher resolution sensor will exhibit a drop in resolution immediately after the peak aperture which will be greater in relative and absolute terms and be more noticeable...

But it won't be noticeable -- that's the whole point. The higher MP sensor is resolving better than you can see, so you don't notice the drop in resolution.

Well the phase 'here MP sensor' is a relative one, comparing the resolution of two sensors, rather than comparing to the human eye, so a higher resolution sensor may not actually resolve better than you can see.

But I take your point that conceivably there could be a lens/camera combination where the changes could be undetectable because the resolution is so high. So Sensor resolution could have an impact at both the high end and the low end.

I suppose in this situation what you're really talking about is a system where the sensor effectively isn't limiting the resolution at all, just diffraction effects, so the question is then really whether the resolution of the lens at the next selectable aperture after the peak is sufficiently lower to be distinguished by the human eye. In essence you're taking the sensor out of the picture and asking whether the lens has a plateau instead of peak in terms of visible resolution changes. But even that would be determined by factors such as the physical size of the image, viewing, distance etc., so like with the low res sensor situation, in principle, under the right conditions, what you're saying is true. It's not trivial though to say when this would happen.

...therefore the system is limited by diffraction at an earlier aperture, while (as I have said on numerous occasions) having greater resolution than the lower resolution image.

No. Both the high and low MP sensors reached their peak at the same aperture,

Agreed.

and the higher MP sensor had higher resolution at every stop,

Agreed.

which was beyond what you could see.

Not necessarily. That would depend on the factors I mentioned above. To be fair, that's also true of the low res sensor situation. A difference that is imperceptible at a certain print size under certain viewing conditions would be visible in another situation.

Thus, if anything, the photo from the higher MP sensor dropped below that visible threshold at a smaller aperture than the photo from the lower MP sensor.

Under certain conditions, yes. Under other conditions, no.

I get the impression that some people see the words 'diffraction' and 'limit(ed)' in close proximity and freak out without reading what I've written.

At best what you're trying to say is that the higher MP sensor might go from, say, 3000 lw/ph at it's peak to 2000 lw/ph stopped down to some point, whereas the lower MP sensor might go from 2200 lw/ph to 1900 lw/ph at the same stopped down aperture.

That's hardly the best case scenario. It's entirely possible that the difference could be 3000–2000 vs. 200–199.9999

You are then arguing that we would notice a drop in resolution from 3000 lw/ph to 2000 lw/ph, but we wouldn't notice a drop in resolution from 2200 lw/ph to 1900 lw/ph. Thus, you conclude that the lens on the lower MP sensor is "diffraction limited" at a more narrow aperture than the lens on a higher MP sensor.

So, are you trying to define "diffraction limited" as when the lens resolution falls to a particular resolution of its peak value?

It's when the drop in resolution is visible. Which will depend on image size, viewing distance, etc.

I've done a few calculations to try and get more definitive numbers here, making use of the formula 1/r^2 = 1/l^2 + 1/s^2 where r is image resolution, l is lens resolution, and s is sensor resolution

If we take a lens with resolution at peak aperture of l_p, the width of the line pair in an image produced by a sensor will be w_p.

Similarly, at another aperture the resolution will be l_a and the width of a line pair in an image produced from the same sensor will be w_a.

We'll call the difference in these widths delta_w = w_a - w_p

At a distance of 1m, the acuity of a human eye allows it to resolve line pairs with separation of approximately w_0 = 0.3mm. In more general terms therefore, w_0 = 3 * 10^-4 * d.

If we took two images with a camera at peak aperture and another aperture, but which are otherwise identical, and viewed at a distance d, then at what point does it become impossible to tell that there has been drop in resolution? If the difference in the width of the line pairs at the two apertures in lower than the minimum line pair width that the eye can distinguish then the resolutions will appear to be the same.

If delta_w = w_a - w_p then we will perceive no difference in resolution when delta_w < w_0.

delta_w = w_a - w_p

= 1/r_a - 1/w_p

= sqrt (1/l_a^2 + 1/s^2) - sqrt (1/l_p^2) + 1/s^2)

If you choose appropriate units so that l_p =1, then you can plot graphs for how various sensor resolution (scaled relative to l_p) perform at apertures with various resolutions (scaled relative to l_p). Having chosen units so that l_p = 1, such graphs will give you delta_w * l_p and look like this:

I've chosen sensor resolutions that cover a useful range, from s = l_p x 10^-3 to s = l_p * 10^1. As you can see from the chart, this covers pretty much the entire range. Real world sensors will fall somewhere in between these values.

So where does delta_w become too small to notice a change in resolution? When delta_w < w_0. But we've charted delta_w / w_p, so we need to find:

delta_w < w_0

delta_w * l_p < w_0 * l_p

delta_w * l_p < 3 * 10^-4 * d * l_p

This so far assumes that the image is the same size as the sensor, wo now also need to include an enlargement factor e = h_i / h_s where h_i is the height of the image and h_s is the height of the sensor.

So what we need now is to find:

delta_w * l_p * e < 3 * 10^-4 * d * l_p

delta_w * l_p < 3 * 10^-4 * d / e * l_p

So now we need a chart of results for different values of d/e and l_p.

This shows w_0 * l_p for lenses of different resolutions (in lp/mm ) for different values of d/e (in mm)

So for example, an enlargement factor of 50, viewing distance of 50 mm, and a lens with resolution 1,000 lp/mm, w_0 * l_p = 0.3.

Looking at the first chart, this means that when the lens set to an aperture when the resolution is 0.7 times the peak, then delta_w < w_o for cameras with sensors of lower resolution than l_p.

So when the lens resolution is 1,000 lp/mm, the sensor resolution is lower than this, the viewing distance is 5cm and the enlargement factor is 50, then you have to stop down to an aperture with resolution less than 0.7 of the peak aperture to notice any drop in resolution due to diffraction. That's just one example.

How does that look?

Reply   Reply with quote   Complain
bobn2
Forum ProPosts: 30,584
Like?
Re: Cutting to the chase.
In reply to Steen Bay, 8 months ago

Steen Bay wrote:

bobn2 wrote:

Steen Bay wrote:

bobn2 wrote:

Life is too short to bother more with this pointless exchange. The nub of it is simple, you set out to 'prove' a mundane and obvious result using extensive calculations based on made-up numbers. That itself was an exercise in futility. My objection to that kind of bogusly quantitative exercise is that the might convince gullible people that there is something of substance in them, because on the surface they look deep and complex, when all they are is an exercise in numerbation (not even measurebation, because there are no genuine measurements involved). As for the result, yes if the pixel count is so small (we discover, about 200k) then you wont see the effects of diffraction or anything else. Every lens becomes perfect. What a great idea.

-- hide signature --

Bob

Also visible on cameras with more than 200k pixels. If someone for example upgraded from 5D to D800 (from 12.7mp to 36mp), then I'm almost sure that he/she would notice that diffraction would start to visibly reduce the resolution (of a good lens, at 100% view) at a larger aperture on D800 than it did on 5D.

Are you sure? How are you sure?

I love the way that you haven't the confidence in your own views just to say you're sure, so you're 'almost sure'. Would you like to quantify your lack of surety? 5%, 10%, 20%? In which units?

OK, then my guess would be that something like a 20% drop in MTF-50 resolution/sharpness would be 'noticeable' or 'clearly visible' at at least 100% view.

Actually I was asking you to quantify your unsurety, not make a spurious estimate of where this border of noticeability is. I had this discussion with someone else about using made-up figures. What is your '20%' figure based on? Does it matter what it's 20% of? Is it 20% regardless of size of viewing size? Is it subject dependent? How does acuity come in? How does image contrast come in? Have you a guess to make about any of those?

The higher the peak resolution is to start with, the more you have to lose to things like diffraction and camera shake. Nothing controversial about that.

That's not how it's put, though is it? And is just as true about excellent lenses. We have regular threads on this forum saying how mFT lenses are the best. Well the downside of that is that they have more to lose - or is it?

-- hide signature --

Bob

Wouldn't call it a downside, but you'll have to be a bit more careful and think a bit more about what you are doing if you want to take full advantage of more MPs or a good lens.

I would have thought that goes without saying. Somehow thought, people feel moved to say it about pixel count but not about good lenses.

-- hide signature --

Bob

Reply   Reply with quote   Complain
bobn2
Forum ProPosts: 30,584
Like?
Re: Um...
In reply to Jonny Boyd, 8 months ago

Jonny Boyd wrote:

Great Bustard wrote:

Jonny Boyd wrote:

Great Bustard wrote:

Jonny Boyd wrote:Diffraction causes a decrease in resolution, agreed?

Agreed.

When resolution drops due to stopping down from the peak aperture, that is due to diffraction, agreed?

Agreed.

At the aperture at which diffraction is reducing resolution, you can say that diffraction is limiting the resolution of the final image, agreed?

Agreed.

If resolution appears to be the same at an aperture smaller than the peak aperture then diffraction doesn't become the dominant factor in limiting resolution until later than the peak aperture, agreed?

Agreed.

Therefore, for practical purposes, as far as the eye can see, a system where resolution visibly drops immediately after peak aperture is more limited by diffraction than a system where the visible drop happens later. Agreed?

Not agreed, and am surprised you do not understand this. For example, let's say for a particular display size, viewing distance, and visual acuity, I can resolve 1000 lw/ph. All else equal, the photo from the lower MP sensor will dip below that threshold before the higher MP sensor.

What you're saying is that the lower resolution sensor produces lower resolution images. No kidding, that's what I've always said.

But how does this make the higher MP sensor more "diffraction limited" than the lower MP sensor?

As I've repeatedly explained, when diffraction produces a visible drop in resolution in situations where a lower resolution sensor wouldn't exhibit a visible drop in resolution, then diffraction is more a limiting factor for the image produced by the high res sensor. The high res sensor will still produce a higher resolution image, it's just that the resolution of that image is determined largely by diffraction effects, where as the lower resolution sensor's performance is dominated by the impact of its sensor and diffraction effects are relatively negligible.

My argument is that for a sufficiently low resolution sensor, an image taken at peak aperture and an image taken at a smaller aperture will have a difference in resolution that is indistinguishable to the naked eye because it is so minor. For practical purposes therefore the perceived resolution is not being limited by diffraction until an even smaller aperture than the actual peak.

But you can't see that the higher MP sensor had greater resolution at the peak aperture, so the higher MP sensor isn't being "visibly limited" by diffraction, either, until a much smaller aperture.

Of course it has higher resolution.

You're bringing up an interesting point though about whether the drop in resolution will be noticed for a high res sensor. To be honest, I hadn't really thought about things from that perspective, but that would be worth looking into, whether it is possible for the drop in resolution for a high resolution sensor and a high resolution lens being stopped down, is not visible.

In contrast, a higher resolution sensor will exhibit a drop in resolution immediately after the peak aperture which will be greater in relative and absolute terms and be more noticeable...

But it won't be noticeable -- that's the whole point. The higher MP sensor is resolving better than you can see, so you don't notice the drop in resolution.

Well the phase 'here MP sensor' is a relative one, comparing the resolution of two sensors, rather than comparing to the human eye, so a higher resolution sensor may not actually resolve better than you can see.

But I take your point that conceivably there could be a lens/camera combination where the changes could be undetectable because the resolution is so high. So Sensor resolution could have an impact at both the high end and the low end.

I suppose in this situation what you're really talking about is a system where the sensor effectively isn't limiting the resolution at all, just diffraction effects, so the question is then really whether the resolution of the lens at the next selectable aperture after the peak is sufficiently lower to be distinguished by the human eye. In essence you're taking the sensor out of the picture and asking whether the lens has a plateau instead of peak in terms of visible resolution changes. But even that would be determined by factors such as the physical size of the image, viewing, distance etc., so like with the low res sensor situation, in principle, under the right conditions, what you're saying is true. It's not trivial though to say when this would happen.

...therefore the system is limited by diffraction at an earlier aperture, while (as I have said on numerous occasions) having greater resolution than the lower resolution image.

No. Both the high and low MP sensors reached their peak at the same aperture,

Agreed.

and the higher MP sensor had higher resolution at every stop,

Agreed.

which was beyond what you could see.

Not necessarily. That would depend on the factors I mentioned above. To be fair, that's also true of the low res sensor situation. A difference that is imperceptible at a certain print size under certain viewing conditions would be visible in another situation.

Thus, if anything, the photo from the higher MP sensor dropped below that visible threshold at a smaller aperture than the photo from the lower MP sensor.

Under certain conditions, yes. Under other conditions, no.

I get the impression that some people see the words 'diffraction' and 'limit(ed)' in close proximity and freak out without reading what I've written.

At best what you're trying to say is that the higher MP sensor might go from, say, 3000 lw/ph at it's peak to 2000 lw/ph stopped down to some point, whereas the lower MP sensor might go from 2200 lw/ph to 1900 lw/ph at the same stopped down aperture.

That's hardly the best case scenario. It's entirely possible that the difference could be 3000–2000 vs. 200–199.9999

You are then arguing that we would notice a drop in resolution from 3000 lw/ph to 2000 lw/ph, but we wouldn't notice a drop in resolution from 2200 lw/ph to 1900 lw/ph. Thus, you conclude that the lens on the lower MP sensor is "diffraction limited" at a more narrow aperture than the lens on a higher MP sensor.

So, are you trying to define "diffraction limited" as when the lens resolution falls to a particular resolution of its peak value?

It's when the drop in resolution is visible. Which will depend on image size, viewing distance, etc.

I've done a few calculations to try and get more definitive numbers here, making use of the formula 1/r^2 = 1/l^2 + 1/s^2 where r is image resolution, l is lens resolution, and s is sensor resolution

If we take a lens with resolution at peak aperture of l_p, the width of the line pair in an image produced by a sensor will be w_p.

Similarly, at another aperture the resolution will be l_a and the width of a line pair in an image produced from the same sensor will be w_a.

We'll call the difference in these widths delta_w = w_a - w_p

At a distance of 1m, the acuity of a human eye allows it to resolve line pairs with separation of approximately w_0 = 0.3mm. In more general terms therefore, w_0 = 3 * 10^-4 * d.

If we took two images with a camera at peak aperture and another aperture, but which are otherwise identical, and viewed at a distance d, then at what point does it become impossible to tell that there has been drop in resolution? If the difference in the width of the line pairs at the two apertures in lower than the minimum line pair width that the eye can distinguish then the resolutions will appear to be the same.

If delta_w = w_a - w_p then we will perceive no difference in resolution when delta_w < w_0.

delta_w = w_a - w_p

= 1/r_a - 1/w_p

= sqrt (1/l_a^2 + 1/s^2) - sqrt (1/l_p^2) + 1/s^2)

If you choose appropriate units so that l_p =1, then you can plot graphs for how various sensor resolution (scaled relative to l_p) perform at apertures with various resolutions (scaled relative to l_p). Having chosen units so that l_p = 1, such graphs will give you delta_w * l_p and look like this:

I've chosen sensor resolutions that cover a useful range, from s = l_p x 10^-3 to s = l_p * 10^1. As you can see from the chart, this covers pretty much the entire range. Real world sensors will fall somewhere in between these values.

So where does delta_w become too small to notice a change in resolution? When delta_w < w_0. But we've charted delta_w / w_p, so we need to find:

delta_w < w_0

delta_w * l_p < w_0 * l_p

delta_w * l_p < 3 * 10^-4 * d * l_p

This so far assumes that the image is the same size as the sensor, wo now also need to include an enlargement factor e = h_i / h_s where h_i is the height of the image and h_s is the height of the sensor.

So what we need now is to find:

delta_w * l_p * e < 3 * 10^-4 * d * l_p

delta_w * l_p < 3 * 10^-4 * d / e * l_p

So now we need a chart of results for different values of d/e and l_p.

This shows w_0 * l_p for lenses of different resolutions (in lp/mm ) for different values of d/e (in mm)

So for example, an enlargement factor of 50, viewing distance of 50 mm, and a lens with resolution 1,000 lp/mm, w_0 * l_p = 0.3.

Looking at the first chart, this means that when the lens set to an aperture when the resolution is 0.7 times the peak, then delta_w < w_o for cameras with sensors of lower resolution than l_p.

So when the lens resolution is 1,000 lp/mm, the sensor resolution is lower than this, the viewing distance is 5cm and the enlargement factor is 50, then you have to stop down to an aperture with resolution less than 0.7 of the peak aperture to notice any drop in resolution due to diffraction. That's just one example.

How does that look?

Very pretty, lots of coloured lines.

-- hide signature --

Bob

Reply   Reply with quote   Complain
Jonny Boyd
Forum MemberPosts: 89
Like?
Re: Um...
In reply to bobn2, 8 months ago

bobn2 wrote:

Jonny Boyd wrote:

How does that look?

Very pretty, lots of coloured lines.

And what value does that comment add to the discussion? You asked for a more detailed consideration of factors such as visual acuity, viewing distance, image size and so on. I've had a go at doing that. If there's an error somewhere, then I'd be happy to discuss that. But given how quickly you posted, I doubt you took the time to read over it and see what was worth commenting on.

Reply   Reply with quote   Complain
bobn2
Forum ProPosts: 30,584
Like?
Re: Um...
In reply to Jonny Boyd, 8 months ago

Jonny Boyd wrote:

bobn2 wrote:

Jonny Boyd wrote:

How does that look?

Very pretty, lots of coloured lines.

And what value does that comment add to the discussion?

About as much as all the pretty coloured lines.

-- hide signature --

Bob

Reply   Reply with quote   Complain
Just another Canon shooter
Senior MemberPosts: 3,237Gear list
Like?
Re: Um...
In reply to Jonny Boyd, 8 months ago

Jonny Boyd wrote:

So where does delta_w become too small to notice a change in resolution? When delta_w < w_0.

What if that happens when delta_w/w_p<w_0 (which is approximately d(log w)<w_0)? Imagine all the colors then!

 Just another Canon shooter's gear list:Just another Canon shooter's gear list
Canon EOS 5D Mark II Canon EF 15mm f/2.8 Fisheye Canon EF 35mm f/1.4L USM Canon EF 50mm f/1.2L USM Canon EF 135mm f/2.0L USM +3 more
Reply   Reply with quote   Complain
Ulric
Senior MemberPosts: 2,623Gear list
Like?
Re: Duck, duck, goose.
In reply to Jonny Boyd, 8 months ago

Jonny Boyd wrote:

Ulric wrote:

Jonny Boyd wrote:

I get the impression that some people see the words 'diffraction' and 'limit(ed)' in close proximity and freak out without reading what I've written.

Johnny, for the benefit of everyone: just what is your definition of diffraction limited? A single, short sentence that anyone can understand. Mine is "Resolving power decreases when the lens is stopped down."

My definition comes at the end of this post. I wanted to put in some brief explanation first.

Diffraction always limits the resolving power of a lens, even wide open, and the effect gets stronger as you stop down. However lens aberations also limit resolution at all apertures but they're effect gets stronger as you open up the aperture. Hence all those curves we see of lens resolution at different apertures.

I realize that you understand that, but so do everyone else and yet there are different opinions on what the implications are.

We could say that a meaningful definition of diffraction limited would be, not simply where diffraction limits resolution, because it always soes that, rather the point at which changing the aperture causes a drop in resolution due to diffraction effects. So if you take the peak aperture, then all smaller apaertures are diffraction limited while larger apertures are aberation limited.

What I was discussing though was a slightly modified definition which concerns when a lens is visibly diffraction limited. This would be when stopping down causes a visible decrease in resolution due to diffraction effects.

N.B. When a camera becomes dffraction limited at a larger aperture than another one and is therefore diffraction limited at a wider range of apertures, that doesn't mean it has less overall resolution.

So your single-sentence definition is something like "Resolving power decreases visibly when the lens is stopped down one stop"?

 Ulric's gear list:Ulric's gear list
Panasonic Lumix DMC-GF3 Olympus OM-D E-M5 Panasonic Lumix DMC-GM1 Panasonic Lumix G 20mm F1.7 ASPH Panasonic Lumix G 14mm F2.5 ASPH +6 more
Reply   Reply with quote   Complain
Jonny Boyd
Forum MemberPosts: 89
Like?
Check your maths
In reply to Just another Canon shooter, 8 months ago

Just another Canon shooter wrote:

Jonny Boyd wrote:

So where does delta_w become too small to notice a change in resolution? When delta_w < w_0.

What if that happens when delta_w/w_p<w_0 (which is approximately d(log w)<w_0)? Imagine all the colors then!

I don't have a maths degree, but I'm pretty sure you can't compare units with different dimensions, so asking if delta_w/w_p (dimensionless) is less than w_0 (length) is nonsensical.

Reply   Reply with quote   Complain
Jonny Boyd
Forum MemberPosts: 89
Like?
Re: Duck, duck, goose.
In reply to Ulric, 8 months ago

Ulric wrote:

Jonny Boyd wrote:

Ulric wrote:

Jonny Boyd wrote:

I get the impression that some people see the words 'diffraction' and 'limit(ed)' in close proximity and freak out without reading what I've written.

Johnny, for the benefit of everyone: just what is your definition of diffraction limited? A single, short sentence that anyone can understand. Mine is "Resolving power decreases when the lens is stopped down."

My definition comes at the end of this post. I wanted to put in some brief explanation first.

Diffraction always limits the resolving power of a lens, even wide open, and the effect gets stronger as you stop down. However lens aberations also limit resolution at all apertures but they're effect gets stronger as you open up the aperture. Hence all those curves we see of lens resolution at different apertures.

I realize that you understand that, but so do everyone else and yet there are different opinions on what the implications are.

Yes. i just wanted to be careful with the definition to make at absolutely clear that there is no disagreement over facts like this.

We could say that a meaningful definition of diffraction limited would be, not simply where diffraction limits resolution, because it always soes that, rather the point at which changing the aperture causes a drop in resolution due to diffraction effects. So if you take the peak aperture, then all smaller apaertures are diffraction limited while larger apertures are aberation limited.

What I was discussing though was a slightly modified definition which concerns when a lens is visibly diffraction limited. This would be when stopping down causes a visible decrease in resolution due to diffraction effects.

N.B. When a camera becomes dffraction limited at a larger aperture than another one and is therefore diffraction limited at a wider range of apertures, that doesn't mean it has less overall resolution.

So your single-sentence definition is something like "Resolving power decreases visibly when the lens is stopped down one stop"?

Close.

'The diffraction limit for a system occurs at the aperture at which any further reduction in settable aperture size will cause a visible decrease in image resolution.'

'for a system' clarifies that the sensor resolution and viewing method are factors.

'settable aperture size' allows for the fact that different lenses have different limitations on how precisely the aperture is controlled.

Reply   Reply with quote   Complain
Just another Canon shooter
Senior MemberPosts: 3,237Gear list
Like?
Re: Check your maths
In reply to Jonny Boyd, 8 months ago

Jonny Boyd wrote:

Just another Canon shooter wrote:

Jonny Boyd wrote:

So where does delta_w become too small to notice a change in resolution? When delta_w < w_0.

What if that happens when delta_w/w_p<w_0 (which is approximately d(log w)<w_0)? Imagine all the colors then!

I don't have a maths degree, but I'm pretty sure you can't compare units with different dimensions, so asking if delta_w/w_p (dimensionless) is less than w_0 (length) is nonsensical.

It is, if you put there a constant that changes with the units, like the gravitational constant in Newton's law.

Let me revise it then, if R is the "resolution" in fixed units, why not dR/R^2<0.05. That would be d(log R)/R<0.05, i.e., the rate of change of R on a log scale. And yes, 0.05 would change if you change the units, and whether this is the right law or not, can be decided by experiments with people. It is not such a strange low if you think about it, it reflects the expectation that when you start increasing R too much, you notice that less and less. If you do not like log, I can replace it with your favorite function with a decreasing derivative.

BTW, I am still playing your game, in which I do not believe to begin with.

 Just another Canon shooter's gear list:Just another Canon shooter's gear list
Canon EOS 5D Mark II Canon EF 15mm f/2.8 Fisheye Canon EF 35mm f/1.4L USM Canon EF 50mm f/1.2L USM Canon EF 135mm f/2.0L USM +3 more
Reply   Reply with quote   Complain
bobn2
Forum ProPosts: 30,584
Like?
Re: Check your maths
In reply to Just another Canon shooter, 8 months ago

Just another Canon shooter wrote:

whether this is the right law or not, can be decided by experiments with people.

What a silly suggestion, introducing experiments and real data into the exercise. You should know the obvious problem with real data, that it doesn't always give the answer that you want.

-- hide signature --

Bob

Reply   Reply with quote   Complain
tt321
Senior MemberPosts: 3,802Gear list
Like?
Re: Duck, duck, goose.
In reply to Jonny Boyd, 8 months ago

Jonny Boyd wrote:

Close.

'The diffraction limit for a system occurs at the aperture at which any further reduction in settable aperture size will cause a visible decrease in image resolution.'

'for a system' clarifies that the sensor resolution and viewing method are factors.

'settable aperture size' allows for the fact that different lenses have different limitations on how precisely the aperture is controlled.

You do realize that for practical purpose, this definition of 'diffraction limit' has zero meaning for cross-system comparisons, esp. regarding systems with the same sensor size but different sensor pixel counts (i.e. different physical pixel sizes)?

And more specifically, the particular interpretation that sensors with higher pixel counts are somehow worse than those with lower pixel counts because one could notice the 'visible diffraction limit' earlier is pure nonsense.

Reply   Reply with quote   Complain
Jonny Boyd
Forum MemberPosts: 89
Like?
Re: Check your maths
In reply to Just another Canon shooter, 8 months ago

Just another Canon shooter wrote:

Jonny Boyd wrote:

Just another Canon shooter wrote:

Jonny Boyd wrote:

So where does delta_w become too small to notice a change in resolution? When delta_w < w_0.

What if that happens when delta_w/w_p<w_0 (which is approximately d(log w)<w_0)? Imagine all the colors then!

I don't have a maths degree, but I'm pretty sure you can't compare units with different dimensions, so asking if delta_w/w_p (dimensionless) is less than w_0 (length) is nonsensical.

It is, if you put there a constant that changes with the units, like the gravitational constant in Newton's law.

Let me revise it then, if R is the "resolution" in fixed units, why not dR/R^2<0.05. That would be d(log R)/R<0.05, i.e., the rate of change of R on a log scale. And yes, 0.05 would change if you change the units, and whether this is the right law or not, can be decided by experiments with people. It is not such a strange low if you think about it, it reflects the expectation that when you start increasing R too much, you notice that less and less. If you do not like log, I can replace it with your favorite function with a decreasing derivative.

To clarify, do you mean -d/dL ([log (R)] / R) where R = (1/L^2 + 1/S^2)^-1/2 ?

If so, then I think it looks like this (though my calculus is admittedly rusty):

If the threshold under certain viewing condition was 0.05, then a sensor in the range of l_p*10^-0.25 to l_p*10^0.5 would show a noticeable drop in resolution when the lens resolution drops to 0.95*l_p. Higher and lower resolution sensors wouldn't exhibit any noticeable change, presumably because every image from the higher res sensors would be too high to show change and the lower resolution ones wouldn't show enough change to be noticeable.

If under different viewing conditions the threshold was 1.5, then for sensors with resolution l_p or worse, you could stop down as far as 0.65*l_p while a camera with resolution roughly a third of the lens would never show any visible loss in resolution.

BTW, I am still playing your game, in which I do not believe to begin with.

I'm not sure what 'game' that's supposed to be, or why you don't 'believe' in it.

Reply   Reply with quote   Complain
Jonny Boyd
Forum MemberPosts: 89
Like?
Re: Duck, duck, goose.
In reply to tt321, 8 months ago

tt321 wrote:

Jonny Boyd wrote:

Close.

'The diffraction limit for a system occurs at the aperture at which any further reduction in settable aperture size will cause a visible decrease in image resolution.'

'for a system' clarifies that the sensor resolution and viewing method are factors.

'settable aperture size' allows for the fact that different lenses have different limitations on how precisely the aperture is controlled.

You do realize that for practical purpose, this definition of 'diffraction limit' has zero meaning for cross-system comparisons, esp. regarding systems with the same sensor size but different sensor pixel counts (i.e. different physical pixel sizes)?

Who ever mentioned cross-system comparisons? Though the work I've done looks at sensor resolution without specifying sensor size.

And more specifically, the particular interpretation that sensors with higher pixel counts are somehow worse than those with lower pixel counts because one could notice the 'visible diffraction limit' earlier is pure nonsense.

I'm baffled as to why you would need to say that when it has repeatedly been affirmed over and over and over again. I mean really, where did anyone ever suggest that a higher pixel count would produce a worse image? I'm sick of that being thrown out as if anyone had made the claim.

Reply   Reply with quote   Complain
Keyboard shortcuts:
FForum MMy threads