# Focal length, zoom-x, magnification, and their relationship to image size projected on sensor

I have been doing remedial research on zoom and magnification. Here are some of my conclusions ("sensor" here assumes a digital camera):

1. The zoom-x number of a zoom lens is the maximum focal length ÷ the minimum focal length.
2. If I am a fair distance away from an object (say 40 to 300 feet,) the size of the image projected on the sensor is nearly in direct proportion to the focal length.
3. EDIT: If I am a fair distance away from an object (say 40 to 300 feet,) the size of the image projected on the sensor when focused is a function of the focal length and the distance of the object. Thus, for a given distance of the object from the lens, the in-focus size of the image projected on to the sensor is in (nearly) direct proportion to the focal length.
4. Thus, if I am a fair distance away from an object (say 40 to 300 feet) using a zoom lens, the size of the image projected on the sensor at maximum zoom ÷ the size of the image at minimum zoom will equal the zoom factor of the lens.

Ie, if I have a 70mm/210mm zoom lens, the zoom factor is 3x. So if I am viewing an object, the image projected on the sensor will be 3x larger when zoomed all the way in than when zoomed all the way out.

Say I have a 3x zoom lens on a digital camera. I take a photo of an object which is 40 feet away at minimum zoom, then take a photo of the same object from the same position at maximum zoom. Then I open both images in Photoshop. Using my select tool I carefully measure the pixel height of the object in each image. I would assume that if I take the pixel height of the object in the max zoomed image, and divide it by the pixel height of the object in the min zoomed image, the ratio would be nearly equal to the zoom factor of the lens.

Does this all seem correct?

I am asking this question because I have purchased a surveillance camera which advertises 5x zoom, however when I test it the way I described above (pixel height of object in max zoomed image ÷ pixel height of object in min zoomed image,) it comes out to only about 3x. When questioning the manufacturer about this, they replied "The zoom number is not equal to the magnification," with a link to this SO question: How do zoom, magnification, and focal length relate?. However I find nothing in the answers to that question which contradict what I have described above. I don't see an error in your basic understanding.

However, a lens' stated focal length is only accurate when focused to infinity; which is another way of saying, when the lens is focused so that everything beyond a particular distance is in reasonable focus (i.e. maximum depth of field). This distance is called the hyperfocal distance and it varies with the lens' focal length, aperture setting, and sensor size. In this case it would need to be set for the longer zoom position and (resulting) aperture... the hyperfocal distance can be found with most DoF calculators.

If the lens is focused to less than infinity it can introduce focus breathing; which is an effective focal length (magnification) less than indicated... that could be part of what you are seeing. But usually this is not as significant and requires the lens to be focused far short of infinity.

And manufacturers tend to round numbers; some out of convenience (e.g. 59.5mm is labeled 60mm), and some because they think it sounds better (e.g. long effective focal lengths due to crop factor). I would guess that at least some of what you are seeing is because of something like that.

• If the found discrepancy was only 10%, say 4.5x versus the advertised 5x, I could accept that. But it is more like 40% - 3x versus 5x. That just seems a bit too far off to me to be attributable to the kinds of errors you mention here. Feb 15 at 16:49
• @KevinHJ, I didn't want to necessarily call "rounding" outright "lying/misrepresentation"... and it very well might not be. The effect of focus breathing can be quite significant; the Nikon 70-200 @ 200mm and minimum distance is only about 120mm effective (~ 40% loss). Feb 15 at 17:11
• Okay, I'm starting to see there are more factors to this than my naive conclusions. Feb 15 at 17:51
• Just watching this short video about focus breathing, and especially the comparison at the end, shows me how much an effect it can have. So maybe the camera I have just doesn't have good correction for that. dpreview.com/videos/8010417625/… Feb 15 at 18:20

You are correct that for objects far away the image on the sensor for a given object scales with the focal length. A 5x zoom ought to be a 5x change in focal length which corresponds to a 5x change in image size. For this purpose far away is more than 20 or so (some might say 100) times the longer focal length. Your scene seems to satisfy this easily. The reason for the far away is that the lens needs to be one focal length from the sensor and we don't want to worry about the distance from the lens to the object changing so the difference in distance to the subject should be negligible.

Simple math discovers image heigh of the object being imaged: Symbols used when working lens problems m = magnification (power) v = Image distance from lens u = object distance forward of the camera

When working problems, all measurement unit to be the same, best if all are initially expresses using millimeters. Image distance is that distance measured from a point called the rear nodal to the focused image. Unless doing macro or close-up imaging, OK to use focal length for v. When doing close focusing, the lens to film/sensor distance increases. In this case we must know this revised value and use it for v.

Problem: Find image of object height which is unknown Object to be imaged is 1/2 meter (500mm) tall Lens focal length is zoomed to 200mm = v Object is 10 meters distant v = 10 X 1000 = 10,000mm First calculate m (magnification) Formula m = u ÷ v m = 200 ÷ 10000 = 0.02 (written as 0.02X)

Note: m is magnification factor, in this case it is a reduction. This is because modern cameras are of the miniature type meaning they are small portable devices.

Now that we know the value of m, we can easily find the resulting height of the image of the object.

Image height of object = linear height of object multiplied by m Image height = 500 X 0.02 = 10 mm

(This was going to be a comment)

Size is a slippery idea here.

A 100mm focal length is 2x as long as a 50mm focal length.

But the objects in the scene cover 4x the area at 100mm versus 50mm because both height and width are doubled.

The inverse square law appears in many places within photography.

• It is advisable to stay away from area magnification. Area magnification is a gimmick used in advertising magnification. Say binoculars magnify 10X. We can advertise it makes objects 100 times bigger. In actuality the object is seen to be 10 times larger. Feb 16 at 1:54
• @AlanMarcus There is 100x the information at 10x. Likewise digital storage requires 4x the space. An 8x10 negative has 4x the film as 4x5. To me it is better to not be sloppy with math. YMMV Feb 16 at 7:10
• In terms of optical magnification. while mathematically true, using area magnification in place of linier magnification is unwise. It adds sensationalism / hype. I don't like it! Feb 16 at 14:53
• @AlanMarcus I don’t like or dislike it inherently. Sometimes it is useful to talk about it one way. Others useful to talk the other. Sometimes both. 2x can be useful when describing focal lengths. 4x can be useful for describing the effects of focal length. Both can be when the inverse square law is at play. That’s always and why I discuss both in my answer and provide a link. Feb 16 at 19:11