Details from a telephoto lens?

D Cox · Mar 6, 2014

Mark Scott Abeln said:
But longer lenses don’t crop.

Long lenses don't crop but small sensors do. Take a 150mm lens designed for use with 5x4 film and put it on an APS-C camera, and there is extreme cropping. But within the cropped area, the image falling on the sensor has exactly the same definition and light level (at the same f number) as it does on the same area of the film.

Here a normal lens for one format has become a long lens for another. There is no change apart from the cropping.

Mark S Abeln said:
They have a proportionally larger light collecting area producing the same size of image circle at the sensor.

This area can be seen when you view lenses from the front: large lenses, opened up all the way, will have a large pupil diameter, while short lenses will have a small diameter.

This precisely cancels out the inverse law. That’s why we use the same exposure for any given f/number, no matter what the focal length happens to be.

--
http://therefractedlight.blogspot.com

D Cox · Mar 6, 2014

Pedagydusz said:
But suppose that you want to photograph this butterfly eye:

Which was photographed at very close quarters with a macro arrangement. To Imagine that you could do it at 12 km is ludicrous. You would need a huge FL, and of course also huge aperture, and such a monstrous equipment that it doesn't bear to think about!

We had a query on a dpReview forum a few months ago from somebody who wanted to do high power macro with a long vertical subject distance.

It worked out that he needed about a 200mm lens, and a camera stand about six feet tall, if I remember correctly. It was about the limit of what's practical.

mosswings · Mar 6, 2014

Lord metroid said:
As of lately I have been thinking about the effects that a telephoto lenses, particularly the super zoom lenses will have on the detail and sharpness of the photograph.

My reasoning tells me that details and sharpness of a super zoom lens should be relative to more normal lenses severely crippled.

Because the area that the zoom lens frames will be incredibly small and far away it follows that the amount of light will be incredibly minute and the intensity will have greatly decreased due to radiation decreasing with the square of the distance. Hence resulting in a low resolution and dark photograph(which can be compensated by higher f-stop).

Are my logical reasoning correct or am I missing some important details?

Other commenters have noted other problems with your reasoning, but I just want to touch on your use of the inverse square law - it's only appropriate for point sources, from which light spreads out radially. The light from a very distant object illuminated uniformly (the typical situation) behaves more like a plane wave, and decreases in intensity less rapidly:

You can use the inverse-square law to predict the rate of illumination fall-off for bare-bulb sources, speedlights, and studio strobes. Umbrellas and softboxes will follow the law quite closely for distances greater than twice their diameter. So, what are the exceptions?

Sources that use optical or physical means to collimate light will not follow the inverse-square relationship, at least when based on their physical location. Such devices include, optical spots (elliptical, Fresnel) and sources whose beam has been modified by a grid. When working very close to large, diffuse sources, such as softboxes and scrims, the fall-off rate will be somewhat less than predicted by the inverse-square law.

http://www.portraitlighting.net/inversesquare_law.htm

Mike Davis · Mar 6, 2014

Its RKM said:
MirekE said:

Mark Scott Abeln said:

Because of this, the relative amount of diffraction will reduce with increasing focal length, since a larger percentage of the photons won’t travel close to the iris or other edges in the lenses.

Inevitably, the sharpest lenses will be long lenses, and the shortest lenses will be the softest lenses.

Click to expand...

Mark, I believe the diffraction is also dependent on the distance from lens to the sensor, which increases with focal length and (for simple lenses) the effects of aperture size and focal length cancel each other.

The formula I have been using to calculate diffraction is d = 1.22 * lambda * N, where lambda is wavelength of the light and N is the f-stop.

Click to expand...

Well, almost... that formula gives the radius, r, of the first zero on the Airy disk projected on the focal plane. The diameter, d, is 2.44 * lambda * N.

Nevertheless, images of point sources separated by the radius, r, meet the Rayleigh criteria as "just resolvable".

xpatUSA · Mar 6, 2014

Lord metroid said:
Because a lesser number of photons are available and they have less intensity, the maximum detail of the subject will decrease.

One fundamental law, not mentioned yet, is that a photon has an energy determined only by it's frequency (1/wavelength) and therefore it's energy can not vary and neither could it's "intensity" - not that intensity is a property of a photon in the first place.

So, I'm wondering if something else was meant. I'm also wondering if photons have a place at all in a side discussion about diffraction. Isn't diffraction a wave thing?

--
Cheers,
Ted

Its RKM · Mar 6, 2014

xpatUSA said:
I'm also wondering if photons have a place at all in a side discussion about diffraction. Isn't diffraction a wave thing?

Well, as things usually are with photons, that's one way of looking at it. ;-)

Another is that diffraction is merely a manifestation of Hysenberg's Uncertainty Principle. The position of the photon is constrained in the plane of the aperture by the aperture, resulting in uncertainty in its momentum in that plane. Consequently the path of the photon has a probability of spreading out into the well known diffraction pattern.

Yes, I know that diffraction occurs in apparently pure wave media, such as water/air interfaces and sound, but there are also phonons and HUP equations apply equally to them, resulting in the same diffraction as classical wave analyses.

So it isn't so much that diffraction is a consequence of HUP, as it is that HUP is a consequence of diffraction. Nevertheless, both views are equivalent.

Its RKM · Mar 6, 2014

Mike Davis said:
Mike Davis said:

If we assume that lambda is yellow-green light at a wavelength of 555nm (nanometers), in the middle of the visible spectrum, setting our units to millimeters, we get:

Click to expand...

The diameter of the Airy disk, d = 2.44 * 0.000555 * N

And thus, for f/22.6, we would have d = 2.44 * 0.000555 * 22.6 = 0.031 mm

Which is really close to the 0.03mm "standard" maximum permissible Circle of Confusion diameter used for fullframe (35mm format) DoF calculations.

Which suggests that if we adhere to DoF calculations calculated against a maximum permissible CoC diameter of 0.03mm, there's no need to stop down below f/22.6, because doing so would cause diffraction's Airy disk diameters to become larger throughout the entire image than our largest CoC's at the near and far extremes of DoF.

Not quite.

The CoC from typical optical blur assumes that a point source is rendered as a circle of even intensity. However a point source is rendered by diffraction as the Airy disc, where intensity falls from the central peak to its first zero at 2.44 * lambda * N.

For example, 90% of the intensity of the Airy disc within that central zero is contained within approximately 60% of its diameter (although only 86% of the total intensity falls within the first zero), while 90% of the intensity of the CoC falls within 90% of its area, or 80% of its diameter.

In other words, diffraction blur and CoC blur are not the same effective size.

D Cox · Mar 7, 2014

xpatUSA said:
Lord metroid said:

Because a lesser number of photons are available and they have less intensity, the maximum detail of the subject will decrease.

Click to expand...

One fundamental law, not mentioned yet, is that a photon has an energy determined only by it's frequency (1/wavelength) and therefore it's energy can not vary and neither could it's "intensity" - not that intensity is a property of a photon in the first place.

You could say it is a measure of how many friends a photon has.

If we stick to one wavelength of light, "intensity" here is the number of photons. Usually the number that arrive during the exposure (sampling) time.

The number of photons arriving affects the certainty with which you can say that one photodetector is receiving more photons than its neighbour, which is what you need to know if you want to detect details in an image. (Or to detect an image at all.)

xpatUSA said:
So, I'm wondering if something else was meant. I'm also wondering if photons have a place at all in a side discussion about diffraction. Isn't diffraction a wave thing?

A photon is wave shaped.

mosswings · Mar 7, 2014

D Cox said:
xpatUSA said:

Lord metroid said:

Because a lesser number of photons are available and they have less intensity, the maximum detail of the subject will decrease.

Click to expand...

One fundamental law, not mentioned yet, is that a photon has an energy determined only by it's frequency (1/wavelength) and therefore it's energy can not vary and neither could it's "intensity" - not that intensity is a property of a photon in the first place.

Click to expand...

You could say it is a measure of how many friends a photon has.

If we stick to one wavelength of light, "intensity" here is the number of photons. Usually the number that arrive during the exposure (sampling) time.

The number of photons arriving affects the certainty with which you can say that one photodetector is receiving more photons than its neighbour, which is what you need to know if you want to detect details in an image. (Or to detect an image at all.)

D Cox said:

So, I'm wondering if something else was meant. I'm also wondering if photons have a place at all in a side discussion about diffraction. Isn't diffraction a wave thing?

Click to expand...

A photon is wave shaped.

Does anyone remember what the OP's original question was?

Mike Davis · Mar 7, 2014

Then you disagree with David Jacobson - just one author out there, who's Lens Tutorial I referenced earlier.

Can you recommend a different constant, than the aforementioned 0.00135383, with which to determine the f-Number at which diffraction's Airy disks would have the same diameter as the maximum permissible CoC diameter? I would also appreciate an explanation of how you derived it.

If not f-Number "limit" = max. CoC / 0.00135383, then what?

And by the way, I very much appreciate that there are several ways to boil this egg. I'd just like to know how you approach it.

Thanks!

xpatUSA · Mar 7, 2014

I suppose CoC does have a place in this thread, but I'm failing to see the relationship between CoC and diffraction. My puzzlement comes from two claims, as follows:

I can double the CoC of my GH1 by viewing it on my monitor as opposed to a standard 8x10 print.

I can double the diffraction through my Sigma 50mm macro lens by extending it to 1:1 magnification from infinity.

So both those parameters appear to be quite variable, irrespective of aperture setting, focal length and the direction of the wind ;-)

--
Cheers,
Ted

Chris Malcolm · Mar 8, 2014

Another factor affecting detail resolution from long lenses is dirt in the air, slight amounts of mist, particulate pollution, etc., each little particle causing a tiny amount of diffraction. Long lenses are often used to photograph distant things, which means the light travels through more air. The effect is very obvious with distant mountains tens of miles away, especially if you look through binoculars. But when you're talking about ultimate sha!rpness in an image the difference between clean and dirty air is noticeable over hundreds of yards.

Another effect already mentioned is thermal turbulence causing refraction, such as the wriggling effect you see when looking over the top of a bonfire. When the sun is out this can have a noticeably degrading effect over distances as short as a hundred yards.

Its RKM · Mar 8, 2014

Mike Davis said:
Then you disagree with David Jacobson - just one author out there, who's Lens Tutorial I referenced earlier.

Can you recommend a different constant, than the aforementioned 0.00135383, with which to determine the f-Number at which diffraction's Airy disks would have the same diameter as the maximum permissible CoC diameter?

As I explained in the previous message, they are different forms of blur so, although there is a specific f/# at which the Airy disc has the same diameter as the CoC, it does not have the same effect. As for the f/# at which it has the same effect that, again, depends on your definition of the effect, since the form of the blur is different. Is it the 50% encircled energy you want to be the same? Or the 90% point, as I gave the example of earlier? At this stage I should point out an error at the end of that example - 90% of the area of a uniform circle corresponds to 95% of the diameter. So the comparison for 90% of the encircled energy within the blur to be equal size would require an Airy disc some 95/60 larger than the CoC. If the CoC is 30um, then the Airy disc should be 47.5um which, from the previous equations would occur at a relative aperture of f/35 - just over a stop higher. However that is only for one comparison point.

Mike Davis said:
And by the way, I very much appreciate that there are several ways to boil this egg. I'd just like to know how you approach it.

I'd approach it a bit more openly since, as I mentioned, the different forms of blur mean there isn't a fixed relationship between the two. However the more gradual blur of the diffraction spot compared to the hard blur of the out of focus spot means that you can be a lot more liberal with diffraction than you can with defocus.

Its RKM · Mar 8, 2014

xpatUSA said:
I suppose CoC does have a place in this thread, but I'm failing to see the relationship between CoC and diffraction. My puzzlement comes from two claims, as follows:

I can double the CoC of my GH1 by viewing it on my monitor as opposed to a standard 8x10 print.

Two issues here. First is that CoC is specific to a final image size. CoC for 4x6 is twice that of 8x10.

Secondly, the concept of CoC depends on resolution not being limited by the medium. Your monitor almost certainly has much lower resolution than the print. Your example suggests that the monitor resolution is around half of the print CoC, assuming you are comparing similar image sizes as per the first point.

xpatUSA said:
I can double the diffraction through my Sigma 50mm macro lens by extending it to 1:1 magnification from infinity.

At 1:1 magnification you have moved the lens pupil to twice the distance from the focal plane as it would be at infinity. Consequently the effective relative aperture has doubled, or increased by two stops. The effect of diffraction due to the aperture has therefore doubled.

xpatUSA said:
So both those parameters appear to be quite variable, irrespective of aperture setting, focal length and the direction of the wind ;-)

Not if you are precise about the definitions. ;-)

xpatUSA · Mar 8, 2014

Its RKM said:
xpatUSA said:

I suppose CoC does have a place in this thread, but I'm failing to see the relationship between CoC and diffraction. My puzzlement comes from two claims, as follows:

I can double the CoC of my GH1 by viewing it on my monitor as opposed to a standard 8x10 print.

Click to expand...

Two issues here. First is that CoC is specific to a final image size. CoC for 4x6 is twice that of 8x10.

Secondly, the concept of CoC depends on resolution not being limited by the medium. Your monitor almost certainly has much lower resolution than the print. Your example suggests that the monitor resolution is around half of the print CoC, assuming you are comparing similar image sizes as per the first point.

Its RKM said:

I can double the diffraction through my Sigma 50mm macro lens by extending it to 1:1 magnification from infinity.

Click to expand...

At 1:1 magnification you have moved the lens pupil to twice the distance from the focal plane as it would be at infinity. Consequently the effective relative aperture has doubled, or increased by two stops. The effect of diffraction due to the aperture has therefore doubled.

Its RKM said:

So both those parameters appear to be quite variable, irrespective of aperture setting, focal length and the direction of the wind ;-)

Click to expand...

Not if you are precise about the definitions. ;-)

Thanks for the info, all of which I already knew. I admit to PWI (posting while intoxicated) and was really trying, as you say above, to point out the lack of precision when folks mention just "the CoC" or "the diffraction".

Notes to self:

Only post when sober

Do not try to be so subtle ;-)

Mike Davis · Mar 8, 2014

Thanks for that! I understand what you're saying.

I'm usually careful to incorporate the phrase "begin to inhibit a desired print resolution" when posting about diffraction's impact, because it does indeed come on slowly.

Example from a post I made three months ago: http://www.dpreview.com/forums/post/40100813

Its RKM said:
Its RKM said:

Here's a similar formula for determining the aperture, f/N, at which diffraction's Airy disks will begin to inhibit your desired final image resolution (in lp/mm) at the anticipated enlargement factor for a 25cm viewing distance:

N = 1 / desired print resolution (lp/mm) / anticipated enlargement factor / 0.00135383

Notice that it's just the calculated CoC value divided by the constant 0.00135383.

Click to expand...

So, yes, it's safe to stop down a little further than the f-Number calculated by my formula, but how much further? A whole stop?

It's a breath of fresh air having an opportunity to discuss this topic with someone who actually understands the need to select CoC and Airy disk maximums with consideration of enlargement factor and a desired final image resolution (which itself requires consideration of anticipated viewing distance - as discussed in the Wikipedia article on Circle of Confusion).

It seems the majority of "diffraction-aware" shooters want to believe that, for each camera sensor, there exists a single f-Number at which diffraction will become a problem - for any and all enlargement factors and desired resolutions. Not true. And even if they understand that there is an f-Number at which diffraction can become visible in a 100% crop viewed on their monitor, some shooters then confine themselves to always shooting at wider apertures, even though their enlargement factors and desired print resolutions might allow the use of much larger f-Numbers (and the greater DoF and slower shutter speeds that come with stopping down further).

--

Mike Davis

Home

MirekE · Mar 8, 2014

Mike Davis said:
It's a breath of fresh air having an opportunity to discuss this topic with someone who actually understands the need to select CoC and Airy disk maximums with consideration of enlargement factor and a desired final image resolution (which itself requires consideration of anticipated viewing distance - as discussed in the Wikipedia article on Circle of Confusion).

These considerations need to be made at the time of shooting and it is usually difficult to predict the enlargement at that time. I don't know the level of enlargement, but I know that the resolution of my image won't be higher than resolution of my camera. So I use my camera resolution as the factor to determine the CoC.

Mike Davis · Mar 8, 2014

MirekE said:
MirekE said:

It's a breath of fresh air having an opportunity to discuss this topic with someone who actually understands the need to select CoC and Airy disk maximums with consideration of enlargement factor and a desired final image resolution (which itself requires consideration of anticipated viewing distance - as discussed in the Wikipedia article on Circle of Confusion).

Click to expand...

These considerations need to be made at the time of shooting and it is usually difficult to predict the enlargement at that time. I don't know the level of enlargement, but I know that the resolution of my image won't be higher than resolution of my camera. So I use my camera resolution as the factor to determine the CoC.

That covers all possibilities, for sure, but sometimes with compromised DoF and/or shutter speeds - when you stopped down more than necessary to secure sufficient DoF for a smaller print than the largest dimensions supported by your pixel count.

Still working under the pixel-limit ceiling, I make the enlargement factor decision for a given scene at the time of exposure and then stick to it later - not tough to do when shooting static subjects like landscapes, architecture, interiors, etc. Smaller prints viewed at similar distances allow sufficient DoF to achieve a desired print resolution using smaller f-Numbers - yielding faster shutter speeds. High-depth scenes that can't be shot at all due to subject motion not covered by 1/15th (at f/16 on a tripod, for example), might be possible with 1/60th (at f/8), on the decision to cut the print dimensions in half. If a given print size is mandatory, then I consider changing my camera position (getting farther away from the nearest subject), or changing my focal length (going shorter), or simply not shooting. I go out of my way to avoid compromising my desired print resolution goal or going to higher ISOs. I'd rather secure an image that has both acuity and resolution - with lots of fine detail that can reward scrutiny even at a viewing distance of 10 inches - than a high enlargement factor. Personal choice. Neither right nor wrong.

Thanks again for your insights.

Roland Karlsson · Mar 10, 2014

MirekE said:
MirekE wrote:
Super teles are usually the best resolving lenses from a given manufacturer.

Hmmmm ... that is not my experience. Even the most expensive long teles have less resolution than even the cheapest normal lenses.

You can look here http://www.photozone.de

What I think gives you the impression that long teles are sharp is that you get better resolution for far away subjects. But ... that is only the magnifying effect.

It is the same effect as macro photo. You think it is very sharp as you normally cannot see that detail.

Erik Magnuson · Mar 10, 2014

Roland Karlsson said:
You can look here http://www.photozone.de

Where are the super-teles tested on that site? I think the Nikkor 200mm f/2 is the only one and it doesn't support your argument.

Roger C. at Lens Rentals has tested the Canon 300mm f/2.8 II and the 200-400L. At f2.8 he gets similar results from the 300mm f/2.8 as any 50mm lens at f/2.8 in the center and slightly better averaged. The 50mm's may do a tiny bit better at f/4 than the 200-400 @ 400mm.

Details from a telephoto lens?

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