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by evan-pak

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This question already has an answer here:

Zoom meaning the ability to make something distant appear closer. Consumer cameras are sold with a maximum zoom which is supposed to define how much closer. For example a camera might be 40x zoom, 4x optical, 10x digital.

Can someone explain that in simple English? Is a 2x zoom the same as taking a picture at half the distance?

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marked as duplicate by mattdm, Paul Cezanne, Itai, AJ Henderson, MikeW May 21 '13 at 18:37

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

The other questions really do cover this, but just to be clear, the "times zoom" notation tells you how much the lens can change, not how "powerful" it is. The focal length is the most useful number for what you're looking for, and it might be helpful to read about angle of view to see why. – mattdm May 21 '13 at 11:33
up vote 2 down vote accepted

As others have stated, "zoom" is a ratio, not a measure of focal length or angle of view. What the zoom ratio does tell you is how much narrower the field of view (FoV) will be at the longest focal length setting when compared to the shortest focal length setting. Another thing it normally tells you is the degree of unwanted things like distortion, chromatic aberration, and poor resolution you can expect from a particular zoom lens. The larger the zoom ratio is, the more optical compromises have to be made to allow the lens to operate over such a wide range or the more corrective elements must be included in the lens' design (and weight, size, and price) to counteract them. This is especially applicable in the case of lenses designed for larger sized sensors such as those found in DSLRs and other interchangeable lens cameras.

Digital zoom is just another way of taking a photo and trimming parts of it to leave what is in the middle of the image. If you have a 12MP sensor and use 2X digital zoom, what you have done is use only the 3MP at the center of the sensor and thrown the rest away. Remember Megapixels are computed using area, not linear measurements. If you use only one half of each side of a sensor, the area is one fourth as large.

You are correct that a focal length twice as long (2X zoom) as the original one will make objects appear twice as tall and wide in a photo taken from the same location. The perspective of the picture taken from twice as far with 2X zoom will be different than the one taken from half the distance and the same focal length, though. The longer the focal length you use, the more the distance between closer and further objects will be compressed, making it appear there is less space between them than there really is. On the other hand, a wide angle lens will expand the apparent distance between objects different distances from the camera.

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The term "zoom" is unfortunately absolutely meaningless - it is just the ratio of the longest and shortest focal length.


A 17mm to 50mm lens would be roughly 3x zoom. (50/17 = 2.94) However a 70-200mm lens would also be a roughly 3x zoom. (200/70 = 2.86)

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Does focal length affect the perceived distance to an object? If I found the longest focal length, could I use it to calculate perceived "zoom"? – just.another.programmer May 21 '13 at 9:29
A long focal length will make things "far away" seem closer. However you cannot related that to "zoom" which is just a ratio of focal lengths. (Property of the lens you are using) You could define 50mm as you "baseline" and then user zoom as a relative term to 50mm, BUT you would need to state what you use as your baseline. What do you want to achieve? (PS: I suggest you avoid the term zoom and *x magnification - they cause more problems than they solve.) – DetlevCM May 21 '13 at 10:16
A longer focal length can never be the same as being closer to an object, because you can't change perspective by doing anything other than moving. (Or, by taking a 3D image and re-rendering, but we'll leave that aside for a decade or so.) The focal length of a lens can be used to determine how much of the picture a distant object will fill, though. See the existing answers linked above, and particularly… – mattdm May 21 '13 at 11:30

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