Like most of us, I thought it was yes. Until I really look into the Lens Equation:

(1/subject distance) + (1/image distance) = (1/focal length)


Please refer to this simlpe diagram:

enter image description here

As the focal length f increases, left size keeps unchanged. However the real image moves further away from the lens, and image distance increases.

By similar triangles, we know that the enlargement of the real image is proportional to the image distance, S2.

So let's get back to the Lens Equation and see what happens when focal length is increased by 2x. Let's simply plug in some logical values into the lens formula. And try seeing what happens.

(1/subject distance) + (1/image distance) = (1/focal length)

For a 50mm lens, having subject standing 5 metres away. That is:

(1/5000) + (1/image distance) = (1/50)

image distance = 50.5050505050505 mm

Same subject distance, 100mm lens (2x):

(1/5000) + (1/image distance) = (1/100)

image distance = 102.040816326531 mm (2.020408163 x)

Roughly. But not exact.

What if a 500mm lens (10x)?

(1/5000) + (1/image distance) = (1/500)

image distance = 555.555555555556 mm (11x)

Much more deviated.

With this chart:

enter image description here

We can see that, the larger multiplier the focal length, the more inaccurate the image magnification ratio.

On the other way, the closer the subject distance, the more inaccurate the image magnification ratio. (Is this count to the perspective distortion?)

High school classes are over. Here are my photographic concerns.

(1) Thick Lens Equation?

Afterall, that is called the Thin Lens Formula. Could it be used to model the thick lens (that we are using) correctly? (ref: How to use the Thin Lens Formula to model a thick lens)

(2) Crop factor?

I am quite sure that, the crop factor (e.g. DX is 1.5x) really means subject enlargement. As DX is 24x16 mm, 24 * 1.5 = 36 and 16 * 1.5 = 24. It all works on the 2D image plane on the sensor. Simple. Here comes the problem. For example, a 400mm lens using on DX, we will say its focal length is acting as a 600mm lens on FX. Suppose shooting a subject on 2m (2000mm) away. The image distance of a 400mm lens is 500mm. To have the image enlarged by 1.5x, we need a image distance of 750mm (remember the similar triangles). However, on an actual 600mm lens, the image distance is 857mm. Having a longer image distance means a larger image (similar triangles). So, the image of 600mm on FX is actually larger than 400mm on DX! Can we still say that, 400mm on a 1.5x crop factor body, act as a 600mm? (Or saying it acts like 550mm would be more close.)

(3) Marking of Zoom Ranges?

On most compact cameras, they mark 2x or 10x of zoom ranges. All of them I have seen are based on their max/min focal lengths. By the proofs above, it doesn't mean subject enlargement at all. Customers are getting more enlargements. Seems like a benefit. But not as accurate as they expected. The term "zoom range" is misleading, isn't it?

(Added on May 27)

One more weird behaviour i observed was, let's take two subject one located at 3000mm and one located at 5000mm away from the lens. When the lens zooms from 50mm to 400mm, the magnification of the 3000mm subject is 9.08x, while the magnification of the 5000mm subject is 8.61x. Notice that the closer the subject, the higher the magnification ratio. Assume a certain part of the far subject is obstructed by the near subject. Does it mean that some more parts/area of the far subject is obstructed by the near subject along the zoom?

Of course not! We all know that in real life experiences, no matter lens zooms or digital magnifications, this only affects the FOV or sizes, perspective will always be the same - as long as we are standing at the same point. We will never "see more" at different zooms. This is crazy.

After a further inspection into the ray diagrams, i noticed that, the further the subject, the closer the focused image forming (respect to the lens).

(refer only to a and b)

enter image description here

Which means that while focal length at 50mm, the image of the 3000mm subject and the 5000mm subject is forming at different distances (this is obvious). Bear in mind that we can only have one image plane at a time (sensor/film can't happen in different places at the same time), this comparison is not valid. Let's say we first focus on the near subject, and having the far subject partially obstructed. Focusing on the near subject means the far subject is out-of-focus. If taking the out-focused image of the far subject, at the near subject's image plane, its magnification ratio is the same as the near subject (by similar triangles). This also conforms to our real life perception - we will not "see more" at different zooms.

A point to note that, the above paragraphs still cannot explain why, for a certain subject as a certain distance, doubling the focal length does not equal to doubling the subject enlargement - or the answer is just simply NO(?)

  • 4
    \$\begingroup\$ I... uh.... wow, that's a lot of words and numbers. What's the actual question here? \$\endgroup\$
    – mattdm
    May 22, 2014 at 18:19
  • 3
    \$\begingroup\$ I think you've got at least 3 questions here. Best to split them out if you want answers to them - those that aren't already answered anyway. \$\endgroup\$
    – MikeW
    May 22, 2014 at 20:22
  • 2
    \$\begingroup\$ In the real world the focal length of a lens is measured by the distance colimated light (light at infinity) is focused behind the entrance pupil. Most real lenses change focal length by a higher percentage than the the deviation you have cited in your question when focused on objects as close as the examples in your question. \$\endgroup\$
    – Michael C
    May 23, 2014 at 4:23
  • 1
    \$\begingroup\$ You can't have the (1/5000) as a constant. When you switch from 50mm focal length to 500mm focal length, you must reduce subject distance from 5000mm to 4550mm. Re-calculate all your values now. \$\endgroup\$ May 26, 2014 at 12:44
  • 3
    \$\begingroup\$ The answer is indeed no - for short distances the magnification is not proportional to the focal length (it is at infinite distance). For the relatively short distances you are mentioning the FOV is changing when you change focal length as the effective thin lens moves forward as focal length increases. This would account for the approx 10% difference going from 50mm to 500mm lenses over a 5m distance. \$\endgroup\$
    – John
    May 27, 2014 at 12:41

5 Answers 5


The mathematics and geometry your working with apply to SIMPLE LENSES. A single lens element with a simple convex curvature of negligible thickness. They also technically only apply at hyperfocal distance in air (vs. water, or vacuum, etc.) The Wikpedia states this:

As mentioned above, a positive or converging lens in air will focus a collimated beam travelling along the lens axis to a spot (known as the focal point) at a distance f from the lens.


If the distances from the object to the lens and from the lens to the image are S1 and S2 respectively, for a lens of negligible thickness, in air,

That's a long list of limitations. In those SPECIFIC circumstances, a simple, thin lens in air will perform according to the formula specified.

As your focal plane moves closer to the lens, even a simple lens is going to behave differently than the ideal mathematics for a simple, thin lens in air would indicate. There are a multitude of problems involved in the use of simple lenses, in addition to the magnification factors your asking about there are also numerous optical aberrations that occur making simple lenses less than optimal.

A real camera lens is much more complex, involving multiple lens elements that bend light in such a way as to maintain not only a flat field and the sharpest possible focus from corner to corner, but they also maintain the proper magnification ratio, entrance pupil size, etc.

I've used both 400mm and 600mm lenses. The expected change in subject magnification between those two is 2.25x. At 600mm, the subject should be 2.25x larger than at 400mm. In practice, that is generally the case. There are small variances...for example the Canon 100-400mm lens actually tops out at around 390mm instead of actually being 400mm, and depending on your exact focal plane, there can be other shifts. In general, though, my experience is that with the EF 600mm f/4 L II lens, subjects are indeed about 2.25x larger than with the EF 100-400mm lens.

Complex multi-element lens designs don't have the same geometric effect on light as simple lenses do. There are a multitude of factors that go into choosing what elements to include in a camera lens, not the least of which is choosing a front element design that is large enough and curved enough to gather all the necessary light from the right angles to support the specified FoV. Additionally, the front element and all the elements before the diaphragm must produce the necessary entrance pupil magnification to actually achieve the specified F-Ratio. All the elements in concert, including floating elements that move with focus or with changes in zoom, must work together to produce the required subject magnification.

Sometimes compromises must be made in order to achieve good IQ, or to achieve the desired zoom range, etc. which in turn often result in less than ideal behavior (even at hyperfocal distance). High quality lenses will usually be near to ideal, however lower end lenses may not be ideal. Changing focus may change focal length, changing zoom may affect focus (i.e. the lens is not parfocal), etc. Regardless, the geometry and math involved in complex lenses is likewise more complex than with a simple lens. The results can be similar, however they are not necessarily the same. For demonstration purposes, we often assume that a complex lens behaves like a simple lens...at hyperfocal distance. Even at hyperfocal distance, while they will perform largely like a simple lens, they will rarely perform identically. At closer focus distances, they can perform quite differently than a simple lens, depending on the actual design.

  • \$\begingroup\$ Thanks for answer, confirming the answer is "no", and giving me detailed information about the real lens structures. Do you mean that, image at 600mm is 2.25x linearly of image at 400mm? Or linearly 1.5x resulting 2.25x in the 2D image? \$\endgroup\$
    – midnite
    May 28, 2014 at 1:12
  • \$\begingroup\$ The subject area increases by 2.25x (600mm shrinks the FoV in both width AND height), linear scaling is 1.5x. Think about it, though...the increase in area is more relevant, as it means you have that many more pixels on your subject. Pixels on subject is really where IQ comes from...the more pixels you can get on your subject, the sharper, less noisy, etc. for any given output magnification. \$\endgroup\$
    – jrista
    May 28, 2014 at 3:56
  • \$\begingroup\$ Thanks again for the observation. This means that, in your experience, 1.5x of focal length does match 1.5x of (linear) image magnification. Yes i also agree the Thin Lens Formula is quite far from real life cases. May be i will try dig into the "Thick Lens Formula" later. \$\endgroup\$
    – midnite
    May 28, 2014 at 6:54
  • 1
    \$\begingroup\$ Aye, linear magnification is the same as the ratio of focal length. \$\endgroup\$
    – jrista
    May 28, 2014 at 23:09

Your second diagram is exactly on point here. Or at least it is for traditional lenses (that is, lenses that are not internal focus). All you need to do is scale the diagram's width according to the focal length of the lens. Longer focal lengths mean that you are focusing closer (at distances less than infinity); that is, the distance between the lens and subject is smaller in proportion to the focal length at the same absolute distance, so the image will form proportionally further from the focal point.

When you are imaging things that are "at infinity" (that is, things that are sufficiently far away that they can be resolved properly when the lens is focused to infinity on all of the lenses you use), then doubling or halving the focal length of the lens will double or halve the size of the image (pus or minus the difference between the marked focal length of the lens and the actual focal length, which may be several millimetres). But that's only at infinity. Any closer, and you have to deal with bellows draw — moving the lens further away from the sensor/film/image plane — and how far you need to move the lens depends on the focal length. At a subject distance of 5 metres, the subject is 100 focal lengths away using a 50mm lens, but only 10 focal lengths away using a 500mm lens. The 500mm lens would need to be moved further away from the sensor (in proportion to its focal length) than the 50mm lens.

And yes, that does assume simple lenses of "normal" construction (even if that "simple" lens might have a relatively large number of elements, such as a modern apochromatic view camera lens). This behaviour becomes very apparent when you use large format cameras. A "normal" lens for an 8x10 is a 210mm lens — for landscapes. A headshot taken with an 8x10 is, for all intents and purposes, a macro shot: it will be a little more than half life size on the film. That means that the bellows will have to be extended to a little more than one and a half times the focal length of the lens in order to achieve focus, so the field of view of the focused lens will be equivalent to that of a lens 1.5 times the focal length. It's still a 210mm lens, but it has the field of view of a 315+mm lens when it's focused that closely. To take "the same picture" using an APS-C/DX camera, which has a sensor of around 16x24mm rather than 8x10 inches, you are so far away from the subject relative to the focal length of the lens that you would need to switch from your 35mm-ish "normal" lens to a 50mm to get approximately the same framing at the same subject distance. The difference between different focal length lenses on the same camera at the same subject distance is the same phenomenon, it's just a little more subtle at non-macro distances.

Things get a little different with internal focus lenses. They keep the same physical length, so it's obvious that the lens, as a whole, is not being moved further away from the image plane as you focus closer. They actually reduce their focal length in order to focus closer. Think of them as zoom lenses with the far end anchored in space. Ideally, they will maintain exactly the same subject size on the sensor as you focus closer (reducing the focal length will make the image smaller, but moving the back of the lens further away from the sensor will make it larger by the same amount, so it all comes out in the wash). The real world isn't quite ideal, though, and the relationship between lens elements required to correct various aberrations will change as the focal length and focus distance changes, so you may see a little bit of enlargement or shrinkage as you focus.

Things are also complicated a bit by telephoto and retrofocus designs, both of which have different focal lengths on the subject side of the lens than they do on the image side.

  • \$\begingroup\$ Internal focus lenses with the focusing element in front of the back focal point (the place where all rays cross) behave the way you describe with respect to focal length as you focus closer. This explains why the Nikon 70-200mm f/2.8 VR when set to infinity but focused very close is around 150mm in terms of actual focal length. Lenses with rear focusing elements located behind the back focal point can maintain the same focal length as you focus closer. This explains why the Canon EF 70-200mm f/2.8 series of lenses are still at around 196mm when focused at MFD. \$\endgroup\$
    – Michael C
    Dec 20, 2015 at 9:07

1) I'm not a math guy, but I do know that the focal lengths as marked on the lens are approximate and can vary from lens to lens and (as I understand) even the same lens depending on temperature and humidity. This isn't an answer, I understand. I guess I'm suggesting that worrying about the math isn't always helpful when all you are trying to do is take a picture.

2 --

A) I disagree with the statement that crop factor = subject enlargement. Crop factor indicates a difference in angle of view between different sensor sizes. You could take a 50mm lens on a DX camera and position a dime so it takes up a certain amount of space on the frame. Then if you took an FX camera placed it in the same space the dime would be the same size on the sensor, the FX camera would just show more of the space around the dime. Wikipedia -- http://en.wikipedia.org/wiki/Crop_factor

B) At the same time, I think most people understand that the crop factor isn't precise but a guideline. I'm not even positive that the various "half-frame" sensor sizes all have the same aspect ratio that a full-frame sensor has (or that all full-frame sensors are of the same size and aspect ratio). What's more important is what I see when I look though my lens with my camera. Right?

3) I don't think zoom ranges are understood as the power of magnification, from a camera supply website:

"The focal length range from the near (or wide) end of the zoom to the far (or telephoto) end of the zoom. Zoom range is also measured with X, representing a multiplier. For example, a 35-105mm zoom is also listed as a 3X zoom."

  • \$\begingroup\$ Thanks for your answer. i am a photography guy too. i also agree that bits of differences wont matter our photos. But i also wish to get down into how my lovely lens works. For (2), if they have the same pixel counts, or adjusted to the same output/print size, it does mean magnification. \$\endgroup\$
    – midnite
    May 22, 2014 at 18:28
  • \$\begingroup\$ For (1), yes you are right. I have read a bit about thick lens webpage, not totally understand though. Even for a simple standard 50mm lens, while focusing on near subjects, its focal length changes! Not 50mm anymore! (to keep the same FOV) \$\endgroup\$
    – midnite
    May 22, 2014 at 18:34
  • \$\begingroup\$ If you want to get a better picture of the design of modern multi-element lens ... I'm really not sure where to go. A starting point might be advanced amateur astronomy, where there are still folks who shape their own lenses. \$\endgroup\$ May 22, 2014 at 20:18
  • 1
    \$\begingroup\$ On the subject of 2A (crop factor = subject enlargement), I understand what you are saying, but from the standpoint of the image projected onto the sensor in principal there is no magnification. Here is another website that covers the subject: luminous-landscape.com/tutorials/understanding-series/… A quote: 'From Ansel Adams in his classic book “The Camera” (New York Graphic Society 1980) -- "All lenses of the same focal length give images of the same size at any given subject distance."' It's a technical point, but you seem like a smart, technical person. \$\endgroup\$ May 22, 2014 at 20:23
  • \$\begingroup\$ Thanks for reply David. I do agree the images are (optically) the same size. That's why i wish to use them as a reference. Forget about DX/FX and crop factor stuffs. Images from 600mm isn't 1.5x larger than 400mm images. It is a bit (or significantly, in some cases) larger. I guess what i can do might be dig into the Thick Lens Formula, or get some lens for experiments, or ask in other topics. Thanks for your helps so far. \$\endgroup\$
    – midnite
    May 22, 2014 at 23:08

I believe the perspective mapping function is helpful here. It states that:

h' = -f*tan(theta)


  • h' is the image height measured w.r.t. the optical axis,
  • f is the effective focal length (EFL) of the lens, and
  • theta is the angular half field of view of the lens.

This is derived and valid only at infinity, but it holds to a fair approximation at finite distances that are still large w.r.t. f.

We must immediately take note of the fact that we have a tan dependency. For small angles, theta=sin(theta)=tan(theta) and so changes to efl will linearly alter the angle of view that fits on a fixed detector size. If you make the efl large enough the small angle approximation will be accurate to e.g. 7 digits and there will be "no" error.

However, for wide angles of view, e.g. a 10mm rectilinear lens for a full-frame camera, the crop or "punch in" will not be a linear effect. At 10mm and a 21.64mm image height we have a half-angle of view of 65.2 degrees. If we cut down to 14.42mm, APS-C, the half-angle of view is still 55.3 degrees. If we do this again at 1000mm, we get the full-frame angle of view to be 1.24 and the APS-C HFoV to be 0.82 degrees. 0.82*1.5 = 1.23 ~= 1.24 (they are basically the same if you keep enough sig figs).

The set of thick lens equations are not used to locate things like images. The thick lens equations are used to find parameters of a thick lens, notably its focal length and cardinal point locations. With this information it can be reduced to an equivalent thin lens and then lensmaker's equation used.

However, your lensmaker's equation is not in a nice form. This is generally to do with its introduction in many physics courses and a lack of respect for sign convention. The nicest form is:

1/s' = 1/s + 1/f

where, obeying sign convention,

  • s' is the image location
  • s is the object location
  • f is the EFL of the lens

We will also want to make use of:

m = h'/h = s'/s


  • m is the magnification
  • h' is the image height
  • h is the object height
  • s' and s are the image and object distances, respectively

So let's run through your examples and respect sign convention.

We have a 50mm lens and 5m object distance:

1/s' = -1/5000 + 1/50
  s' = 50.505mm.
   m = 50.505/-5000 
     = -0.0101x

The 100mm lens:

1/s' = -1/5000 + 1/100
  s' = 102.041mm
   m = 102.041/-5000 
     = -0.0204x

We are seeing nonlinear effects, just as you were. So sign convention has happened to work out for you by chance.
However the solve for s' (and by corollary, m) is a function of 1/x and a plot will reflect that.

enter image description here

By increasing f you move away from the right hand side of the plot towards the pivot point, s=-2f where s'=2f. At this point your slope is going to take off in the opposite direction.


I have to admit that I have not read all of your question, in fact I am not quite sure that I got what your question is. But you seem to have observed a simple fact: A 400m lens magnifies twice as much as a 200mm lens, provided that the focal point (i.e. your imagined “thin” lens) is in the same position. Of course that means that for the 400mm lens the sensor will have to be 20cm further from the subject. You also noticed that the difference that it makes when you instead keep the sensor in place is greater for nearer subjects. For far away subjects or short focal lengths you can usually disregard this difference.


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