Details from a telephoto lens?

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?

Same F-stop, same scene illumination = same light on the sensor.

F-stop is focal length divided by aperture size.

See:

http://en.wikipedia.org/wiki/F-number

So an f/2.8 on a telephoto is much larger than F/2.8 on a 55mm - so the telephoto gathers more light to compensate for the narrower field of view.

The early photographers invented f-stop so they had a number for exposure that was invariant with focal length.

Okay, forget about my F-stop comment!

What about the resolution and details of the photograph?

Lord metroid said:

Okay, forget about my F-stop comment!

What about the resolution and details of the photograph?

Click to expand...

Some lens aberrations are harder to correct in a longer lens, while others are easier.

It is harder to avoid camera movement with a long lens, in proportion to the angle of view. This is probably the main limit on definition.

Are we talking about super zoom or super telephoto?

Under super zooms I understand lenses that have very large zoom range, from wide angle to tele.

Under super teles I understand lenses over 300mm in focal length. These are the white lenses used by wildlife and sport photographers.

Super zooms are constructed with some compromises and details may suffer. They are not very fast and usually used handheld, so camera shake can further compromise image quality, especially at the long end.

Super teles are usually the best resolving lenses from a given manufacturer. While the optical quality is excellent, there are some rules the photographer needs to follow to ensure maximum sharpness and detail. Most important one is that these lenses need to be very steady during exposure (or the exposure needs to be very short).

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.

Click to expand...

In practice it appears that if your area of interest fits in the angle of view at the 'super zoom' focal length, detail is typically better when captured at the 'super zoom' focal length than, say, at half of it - all other things being equal.

Lord metroid said:

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).

Click to expand...

If you think about it, the light will have traveled exactly the same distance, arriving at a far lens as the same plane wave both at the 'super zoom' focal length and at one half of it. Luminance from your area of interest hitting the lens will also be the same in both situations. So if shutter speed, f/number, lens transmission and all else are the same the number of photons reaching the sensor from your area of interest will ideally be the same at both focal lengths. With equal processing both captures should appear equally bright.

However keep in mind that f/number can be simplified as the ratio f/D (with f focal length and D aperture diameter). So for the same f/number, twice the focal length means twice the diameter of the lens iris. And usually longer focal lengths require more lens elements so transmission may tend to penalize them, making the relative captures a little darker aotbe.

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?

Click to expand...

The inverse square law does not hold here.

Diffraction is a major source of blurriness of a lens — diffraction around the camera’s iris and other internal edges. By definition, the aperture setting of a camera is defined as the focal length divided by the front pupil width (which is the apparent diameter of the hole of light coming from the back of the lens, when viewed from the front), and so a longer focal length will aways have a larger apparent pupil width for any given aperture setting. Take a telephoto lens, and the front pupil width will be quite large, almost as wide as the lens itself, but if you take a super wide-angle lens, the front pupil width will be tiny.

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.

Mark Scott Abeln said:

Diffraction is a major source of blurriness of a lens — diffraction around the camera’s iris and other internal edges. By definition, the aperture setting of a camera is defined as the focal length divided by the front pupil width (which is the apparent diameter of the hole of light coming from the back of the lens, when viewed from the front), and so a longer focal length will aways have a larger apparent pupil width for any given aperture setting. Take a telephoto lens, and the front pupil width will be quite large, almost as wide as the lens itself, but if you take a super wide-angle lens, the front pupil width will be tiny.

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...

Are you talking about the size of detail that can be resolved on the subject, or the number of line-pairs per image height ?

In the first case, the longer lens has an obvious advantage.

In the second case, as a general rule the sharpest lenses are mid-range or longish; lenses with extreme angles of view, wide or narrow, tend to be less good.

To start with lets review some fundamental physical laws of electromagnetic radiation which visible light is.

D Cox said:

Light and other electromagnetic radiation @ Wikipedia:

The intensity (or illuminance or irradiance) of light or other linear waves radiating from a point source (energy per unit of area perpendicular to the source) is inversely proportional to the square of the distance from the source; so an object (of the same size) twice as far away, receives only one-quarter the energy (in the same time period).

More generally, the irradiance, i.e., the intensity (or power per unit area in the direction of propagation), of a spherical wavefront varies inversely with the square of the distance from the source (assuming there are no losses caused by absorption or scattering).

Click to expand...

The number of photons available to the sensor will also decrease with a greater focal lengths as the area being photographed decreases. Illustrated with the below image.



The maximum detail of a subject is hence determined by the physical properties of electromagnetic radiation and the area being photographed. It follows that:

1. The further away the subject is, the less detail can be recorded.
2. Using a greater focal length to capture a distant subject will limit the number of photons available and their intensity due to the distance and the area being photographed.
3. Because a lesser number of photons are available and they have less intensity, the maximum detail of the subject will decrease.

Mark Scott Abeln said:

[snip], since a larger percentage of the photons won’t travel close to the iris or other edges in the lenses. [snip]

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But every photon travels via every path through the iris with equal probability, so every photon "sees" the diffracting edge equally. It's like asking: which slit did the photon go through in the double-slit experiment? We have to work out the sum of the probability amplitudes for the photon passing through both slits to get an answer that agrees with the experiment...

Joe

Joe Pineapples said:

Mark Scott Abeln said:

[snip], since a larger percentage of the photons won’t travel close to the iris or other edges in the lenses. [snip]

Click to expand...

But every photon travels via every path through the iris with equal probability, so every photon "sees" the diffracting edge equally. It's like asking: which slit did the photon go through in the double-slit experiment? We have to work out the sum of the probability amplitudes for the photon passing through both slits to get an answer that agrees with the experiment...

Click to expand...

For the purposes of diffraction, you need to use wave theory. Then, the relatively smaller area of edge influence is seen to predominate in longer focal length lenses.

But longer lenses don’t crop. 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.

Mark Scott Abeln said:

But longer lenses don’t crop. 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.

Click to expand...

I would say it cancels out the angle of view rather than the inverse square law as such. But they are aspects of the same geometry.

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.

I think that your reasoning is flawed in a peculiar way: you seem to be concerned with a non-problem.

Yes, a subject at 100 m will be far more difficult to photograph than if it is at 10 m. You have to use an expensive tele lens (not a super zoom), which is bulkier and heavier, making the act of taking the photograph more difficult. The amount of photons that arrive at the sensor are the same, as explained above, in particular by Mark, because the lens has a very wide opening, but it is much more difficult to use. If the subject is at 10 km, it is much harder again, but doable. Take this photograph:



The fireworks display is at some 12 km away or so, photographed with an Olympus Pen with a Zuiko 50-200 mm lens. It is basically as good as if photographed at 100 m with a wider lens.

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!

To photograph details in the Moon, you need a lens (or mirror) with a diameter measured in meters, rather than millimeters!

But at large distances, even a few dozen meters, another problem starts to creep into photography: air turbulence. It can easily ruin tele lens photography, from the slightly soft to the totally distorted. This is at about 1 km:



BTW: the first image also suffers from turbulence (badly, but less noticeably)

Regards

--
Antonio
http://ferrer.smugmug.com/

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...

Yes! Thank you!

We should all take note that focal length is not a factor in determining the f-Number at which diffraction's Ariy disks will reach any given diameter at the sensor - possibly inhibiting a desired print resolution after enlargement.

See David Jacobson's Lens Tutorial and search for the word "diffraction."

--

Mike Davis

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...

MirekE,

You are correct. I was tired when I wrote that, and didn’t take that into account.

Mark Scott Abeln said:

Joe Pineapples said:

Mark Scott Abeln said:

[snip], since a larger percentage of the photons won’t travel close to the iris or other edges in the lenses. [snip]

Click to expand...

But every photon travels via every path through the iris with equal probability, so every photon "sees" the diffracting edge equally. It's like asking: which slit did the photon go through in the double-slit experiment? We have to work out the sum of the probability amplitudes for the photon passing through both slits to get an answer that agrees with the experiment...

Click to expand...

For the purposes of diffraction, you need to use wave theory. Then, the relatively smaller area of edge influence is seen to predominate in longer focal length lenses.

Click to expand...

For calculating diffraction through simple apertures you don't need to use wave theory: the classical and quantum theories both lead to exactly the same answer; but the original statement about photons was just plain wrong.

Joe

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".