How does optical zoom relate to subject size in the image?
Asked 1/10/2018
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If a subject appears 100 pixels tall at the widest setting, can I make it about 150 pixels tall by setting the camera to 1.5x zoom? My camera’s optical zoom range is 4.3–129mm.
I’m trying to understand how subject size in the frame relates to focal length, zoom ratio, and angle of view. Is the relationship approximately linear, and when does it stop being exact? I’ve also seen angle-of-view formulas that use atan(), which makes it seem non-linear. How should I think about this in practice?
Originally by Photography Stack Exchange contributor. Source · Licensed CC BY-SA 4.0
Photography Stack Exchange contributor
8y ago
2 Answers
1
The focal length of a lens is a measurement taken when the lens is imaging an object at infinity. Light rays from such an object arrive parallel. The lens then focuses this image and we take a measurement from a point called the rear nodal to the image plane. This distance is inscribed as the focal length.
Now the lens has limited ability to refract (bend inward) light rays. If the object is closer than infinity, the distance downstream from lens elongates. Thus we must rack the lens forward to obtain focus. We can approximately calculate the back focus if the focal length is known and the distance lens to object.
Assume 30mm mounted and an object 250mm forward of the lens. We approximately calculate the back focus distance by converting both focal length and distance to diopter units.
For the 30mm = 1/30 X 1000 = 33.333d
For the 250mm = 1/250 X 1000 = 4d
We change the sign of the object distance diopter power and add.
33.333 + -4 = 29.33d
We convert back to millimeters
1/29.33 X 1000 = 34.09mm back focus distance
An easy approach: Draw imaginary lines from the top and bottom of the object to the center of the lens. This traces out a triangle. Say the object is 250mm forward of the lens and the object is 10mm in height. The ratio height to distance is 10/250 = 0.04.
Inside the camera the image forming rays trace out a similar triangle, the image triangle has the same angles as the object triangle however the sides and height are different but proportional.
The image triangle back focus distance is the height of the image triangle. The height of image triangle is 34.09mm. An object 10mm in height will image 34.09 X 0.04 = 0.96mm in height.
The magnification is 0.96/10 = 0.96X
As to angle of view:
As to angle of view: Assume compact digital format size 16mm height by 24mm length. The diagonal of this rectangle is 28.8444. Assume 30mm lens
Height angle of view = 29.9°
Length angle of view = 43.6°
Diagonal angle of view = 51.4°
With the object at 250mm distance the back focus of the 30mm lens is 34.09mm
Height angle of view = 26.4°
Length angle of view = 38.8°
Diagonal angle of view = 45.9°
Originally by user44949. Source · Licensed CC BY-SA 4.0
user44949
8y ago
0
Generated from our catalog & community — verify before relying on it.
Yes—approximately. For the same camera position and the same subject distance, image size on the sensor is roughly proportional to focal length. So if a subject is 100 pixels tall at 4.3mm, increasing focal length by 1.5× should make it about 150 pixels tall.
In your example, 1.5× from the wide end means about 6.45mm, and that should give about 1.5× the subject size in the image.
Why this works: zooming changes focal length, which narrows the angle of view. A narrower angle of view makes the subject occupy more of the frame. Over ordinary shooting distances, this behaves close to linearly with focal length.
Why formulas can look non-linear: angle of view is given by a trig formula such as 2·atan(d/(2f)), where d is sensor size and f is focal length. Angle of view itself is not linear with focal length, but subject magnification in the image is still approximately proportional to focal length when distance stays fixed.
The main caveat is close focus. Focal length is defined for focus at infinity; when focusing on nearer subjects, lens extension/focus breathing can make the relationship less exact. At typical distances, the approximation is usually good enough.
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