# How to find focal length from image magnification % ?March 13, 2011 4:50 PM   Subscribe

Math/Optics: How can I derive the focal length of a lens from the magnification percentage of an image?

Math/optics wizards, please help. I need a precise, mathematical answer, not a guesstimate or a measure-it-in-the-field solution.

I need to derive the focal length of a lens that would have the same view as a magnified image.

I have two photos with identical dimensions in pixels:

WidePhoto is a photo taken with a camera with its zoom lens all the way wide. I know the focal length used for this photo, as well as the angle of view.

ZoomedPhoto is an image created by selecting a small area of WidePhoto and magnifying it to the same dimensions as WidePhoto. I know the dimensions in pixels of the selection area taken form WidePhoto to create this image. I also know the magnification factor as a percentage.

I've uploaded an example image here.

What formula do I use to derive the focal length of the lens needed to produce the "zoomed in" image?

If you can tell me how to derive the Angle of View from the image, I can calculate focal length:

focalLength=SensorSize/(2*TAN((AngleOfView*PI/180)/2)).

(...that's right, isn't it?)

Thank you in advance for your help. My brain hurts from wrestling with this.
posted by ScarletPumpernickel to Technology (10 answers total)

I'll give you my simple view and let others throw stones. It was my understanding that doubling the focal length effectively doubles the optical zoom.

This would mean the if you take the zoom ratio of the two images (ZoomedPhoto/WidePhoto) then

Effective Focal Length = zoom ratio x Focal length of lens used for WidePhoto
posted by NoDef at 5:40 PM on March 13, 2011

Thank you for responding, NoDef.

I agree that 2*focalLength = 2*magnification.

However, does n*focalLength = n*magnification?

I'm not sure it's a linear progression. It may be a more complex relationship that just happens to cross through the sweet spot of 2*focalLength = 2*magnification.

Does anyone know the answer?

Thanks!
posted by ScarletPumpernickel at 6:03 PM on March 13, 2011

NoDef - thats not right, since a shorter focal length lens produces higher magnification... my recollection from high school is it's a plain inverse relationship: halve focal length to double magnification.
posted by russm at 6:06 PM on March 13, 2011

If you know what the first focal length is, and the magnification between the two images, then assuming all other things are the same, the focal length of the second image will be f1 * M.

f2/f1 = M where M is the "zoom factor" or magnification and f2 is the larger focal length.

Higher focal lengths on a camera lead to more zoomed in images. (russm, are you talking about the thin lens equation for a single lens? That has an inverse relationship, but cameras are not using simple (single) lenses.)
posted by achmorrison at 6:17 PM on March 13, 2011 [1 favorite]

ok, so

M = f / (f - S1)

where M is magnification, f is the lens focal length, and S1 is the distance from the lens to the object.

assuming S1 is much larger than f, M is negatively proportional to f. so yeah, if you're blowing something up by 200% then the focal length of the apparent lens is half the focal length of the real lens.
posted by russm at 6:19 PM on March 13, 2011 [1 favorite]

on non-preview. yeah, I was assuming that the thin lens equation was appropriate since a single "focal length" parameter was used.
posted by russm at 6:22 PM on March 13, 2011

At this level of accuracy, aside from some sign conventions, I think the equations are the same for an ideal thin lens or a composite lens (which is usually designed to mimic an ideal thin lens I believe).

Using russm's equation,

M= f/ (f-S1)

if S1 >> f then

M = -f / S1

thus, M is proportional to f. If I double f, I similarly double M. So I think the original suggestion that magnification should be proportional to focal length holds...

I think some of the confusion might be coming from thinking about a magnifying glass where magnification is inversely proportional to the focal length.

M ~ 25 cm / f

In this case, the distance from your eye to the lens (typically designed for 25 cm) is larger than the distance from the lens to the object you are looking at. If you increase the focal length in this case you effectively decrease the magnification of the magnifying glass.
posted by NoDef at 7:34 PM on March 13, 2011

Thanks, all.

Achmorrison, looking at the image here, if I measure the dimensions of WidePhoto and the RedCropBox, then

M=WidePhotoDimensions/RedCropBoxDimensions > 1

and

ZoomedPhotoFocalLength = WidePhotoFocalLength * M

Right?
posted by ScarletPumpernickel at 8:50 PM on March 13, 2011

If the final (zoomed) dimensions of the image made from your "RedCropBoxDimensions" is the same as the WidePhotoDimensions, which from the linked photo it seems like you are indicating that it is, then yes, both of your expressions are correct. That's part of assuming all other things are the same in both cases. If that is the case, then NoDef's original answer is also correct.

The thin lens equation really can't be used for zoom lenses. First of all, it's a multi-element lens. And generally, the lenses inside can't be considered thin. If you try to use the thin lens equation you might get close, but then you run into the issue that the image distance is changing at the same time the focal length is changing.
posted by achmorrison at 10:40 PM on March 13, 2011

Thank you!
posted by ScarletPumpernickel at 10:52 PM on March 13, 2011

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