Doug Kerr
Well-known member
Why do we use "telephoto" lenses - lenses with "greater" focal length?
The layman may say, "to bring distant objects closer".
But we know that this doesn't really say anything.
A more meaningful notion is "to make a distant object fill a more substantial portion of the frame than we could otherwise".
But why use a costly and bulky long focal-length lens to do that? We can just crop the region surrounding the object from the total image.
So that is not really the object either.
What is the object is to make get the greatest possible sharpness of the small object field we want in our final frame (the object and however much surround we want in our composition).
Qualitatively, we recognize that if we have a camera with a high resolution, we can successfully crop a smaller fraction of the taken frame and still attain some desired resolution in the delivered image (as related to the object). That is, we can perhaps meet our objective without such a great focal length.
Perhaps, an APS-C sensor camera with a 28 mm lens (such as a Ricoh GR) can do as good a job on a delivered image embracing a certain distant object as a Canon PowerShot G16 with its pitiful "1/1.7 inch" sensor and its zoom lens at its maximum focal length (140 mm ff35 mm equivalent).
But how can we quantify this concept - how can we make actual numerical comparisons?
One way is to speak of a parameter some call the "reach" of the camera, since that term immediately suggests its significance to us. But the actual parameter is the angular resolution of the camera.
By that I mean the number of cycles of resolution per unit of angle of the view of the camera
For example, if, with a certain lens focal length in effect, the angular resolution of the camera is 5 cycles per milliradian (a handy-sized unit), it means that as visualized on the object, for an object at a distance of 100 feet, the camera's resolution is 4.16 cycles per inch.
As a quick approximation, I will assume that the resolution of the camera at the sensor (in cycles/mm) is 1/2P, where P is the pixel pitch in mm. (It will usually be only about 75% of that - Kell factor and all that), but that's OK).
Taking the units into account, the angular resolution in cycles per milliradian is f/2p, where f is the focal length in mm and p is the pixel pitch in µm.
So for a pixel pitch of 10 µm, and a focal length of 100 mm, the angular resolution would be 5 cycles per milliradian.
Now lets look at what this comes out to for some cases that have recently been discussed in connection with comparisons of "compact" cameras.
Case 1: Canon PowerShot G16
Pixel pitch: 1.86 µm
Maximum focal length: 30.5 mm
Thus the reach at maximum focal length is 8.19 cycles per milliradian.
Case 2: Ricoh GR
Pixel pitch: 4.81 µm
Focal length: 28 mm.
Thus the reach is 2.91 cycles per milliradian.
Thus, compared to the GR16 at maximum zoom, the GR is at a 2.8:1 disadvantage with respect to the capture of an object an any given distance.
I will tabulate the "reach" values for other cameras under current discussion shortly.
Best regards,
Doug
The layman may say, "to bring distant objects closer".
But we know that this doesn't really say anything.
A more meaningful notion is "to make a distant object fill a more substantial portion of the frame than we could otherwise".
But why use a costly and bulky long focal-length lens to do that? We can just crop the region surrounding the object from the total image.
So that is not really the object either.
What is the object is to make get the greatest possible sharpness of the small object field we want in our final frame (the object and however much surround we want in our composition).
Qualitatively, we recognize that if we have a camera with a high resolution, we can successfully crop a smaller fraction of the taken frame and still attain some desired resolution in the delivered image (as related to the object). That is, we can perhaps meet our objective without such a great focal length.
Perhaps, an APS-C sensor camera with a 28 mm lens (such as a Ricoh GR) can do as good a job on a delivered image embracing a certain distant object as a Canon PowerShot G16 with its pitiful "1/1.7 inch" sensor and its zoom lens at its maximum focal length (140 mm ff35 mm equivalent).
But how can we quantify this concept - how can we make actual numerical comparisons?
One way is to speak of a parameter some call the "reach" of the camera, since that term immediately suggests its significance to us. But the actual parameter is the angular resolution of the camera.
By that I mean the number of cycles of resolution per unit of angle of the view of the camera
For example, if, with a certain lens focal length in effect, the angular resolution of the camera is 5 cycles per milliradian (a handy-sized unit), it means that as visualized on the object, for an object at a distance of 100 feet, the camera's resolution is 4.16 cycles per inch.
As a quick approximation, I will assume that the resolution of the camera at the sensor (in cycles/mm) is 1/2P, where P is the pixel pitch in mm. (It will usually be only about 75% of that - Kell factor and all that), but that's OK).
Taking the units into account, the angular resolution in cycles per milliradian is f/2p, where f is the focal length in mm and p is the pixel pitch in µm.
So for a pixel pitch of 10 µm, and a focal length of 100 mm, the angular resolution would be 5 cycles per milliradian.
Now lets look at what this comes out to for some cases that have recently been discussed in connection with comparisons of "compact" cameras.
Case 1: Canon PowerShot G16
Pixel pitch: 1.86 µm
Maximum focal length: 30.5 mm
Thus the reach at maximum focal length is 8.19 cycles per milliradian.
Case 2: Ricoh GR
Pixel pitch: 4.81 µm
Focal length: 28 mm.
Thus the reach is 2.91 cycles per milliradian.
Thus, compared to the GR16 at maximum zoom, the GR is at a 2.8:1 disadvantage with respect to the capture of an object an any given distance.
I will tabulate the "reach" values for other cameras under current discussion shortly.
Best regards,
Doug
