Lens Formulae

Hi Guys,

I've found something else to worry my little grey cells.

My K10D 18-55 lens focuses down to 250mm and the specs says this is a magnification of 0.34 with the lens at 55mm. So far so good.

I've measured the 250mm to be from the target to just in front of the eyepiece adjustment lever (probably the sensor plane). My film camera did much the same.

Now come the query :-
Lens formulae gives
1/v + 1/u = 1/f (where u = target distance from centre of lens; v = image distance from centre of lens; and f = focal length)
M = u/v (where m = magnification)

So by inference u + v is equal to the target to image distance (ignoring the internodal space which would increase the distance anyway)

So is the focusing distance expressed as 250mm equal to u+v.
If so the above formulae give a target distance of 186mm and image distance of 63mm and to compound the query further the focal length comes out as 47mm

Whats happening?

regards
Bernard

The answer is probably stop fretting over nothing (that you can control anyway) and concentrate on pressing the shutter more, but my curiosity is awoken.

You have too much time on your hands.

Get out there and take photographs and leave such cogitations to the lens designers.

I did anticipate that, hence

regards
Bernard

Without resorting to even trying to think round your mathematics, focal length can change with focusing distance, and mostly does with imperfect lenses.

Your magnification might be easier to think of as about 1:3, which is actually usefully close and better than the 1:4 that is common to standard zoom lenses.

I'd spend more time taking images aas well. If you really want to go into the maths of lenses, then there are books on optics that will more than whet your appetite.

To be honest John it was book I picked up SH for 20p (Ilford Manual of Photography 196 that sparked the thought as I was reading it whilst sitting in front of a TV programme my wife was watching (tactful way of saying I was bored rigid).

I was also going to add other queries such as :-
"Has the circle of confusion been replaced by a single pixel"

but thought better of it

regards
Bernard

Phew!

Brilliant. Glad to know I'm not alone in having to sit through dreadful TV programmes in this way!

Your sums piqued my curiousity too; I had a play and tried solving your equations assuming a focal length of 55mm and made u=216mm. I'm not entirely sure what this means either, other than 216mm would be somewhere inside the lens/camera in your set up and so, unfortunately, it's probably not as simple as the book says!

I did a bit of optics during my degree, but as it was a while ago and concentrated on optical fibres not camera lenses, I can't claim any sort of authority on the subject.

However, If my memory serves me correctly, it was possible to model an optical fibre with a set of comparatively simple propagation equations but not very accurately. The more accurate you wish to make the model and the more physical factors you take in to account, the more complex the equations become.

For this reason, I would assume (without having read it) that the equations in the book may use a simplified model of the camera lens, making a set of assumptions about it (i.e. a simple, fixed focal length optical scheme) that mean it will never be perfectly accurate.

Considering that the kit lens is a zoom and contains a complex optical design with an aspherical element, I think that getting a result of 47mm is actually pretty good!

Like John says the focal length can change when focusing so it's quite normal.
I once read a review about the Canon MP-E 65mm, that one really change his focal length to get the 5x magnification but the focal length isn't important for this lens but the magnification is!

The Canon's reversed anyway.

Nope, circle of confusion will be at least a couple of pixels wide. Using the "standard" calculation that the circle of confusion is about 0.2mm on an 8x10 inch print, the size on the sensor is about 20 microns. The pixel size of the k7 is about 5 microns, so a single circle of confusion covers about 16 pixels in total (4x4).

I'm circularly confused

Heh, don't worry - I just saw this thread last night and couldn't resist contributing - I'm doing a lot of aberration and resolution calculations at work at the moment.

Hi Bernard

You describe u and v as distances from the "centre of the lens". This is more or less OK for a simple "thin" lens but your 18-55 is not simple or thin.

For a compound or thick lens the object and image distances have to be measured from the "principal planes". These are usually different for the object and image sides. The principal planes don't have to be within the lens, in fact at 18mm the image principal plane must be at least 25mm behind the rear element, and for a telephoto it can (confusingly!) be in front of the front element.

To get the object to image distance you need to take u plus v plus the distance between the principal planes.

This isn't completely academic for macro work as it means you may need a bit more space than you might think, though I don't lose a lot of sleep over it!

For magnification of .34 with a 55mm lens I get u + v = 217mm. Assuming the quoted magnification and focal length are accurate but rounded, this could vary by 5mm either way. The 250mm probably isn't exact, so I guess the distance between the principal planes would have to be about 37mm plus or minus 10mm which doesn't seem unreasonable given that your 18-55 is probably about 60mm thick.

Peter

I mean on a pentax slr, where the lens mount is 46mm in front of the sensor/film plane, and the rear element of the lens can't be more than a mm or two closer than this.

Bernard

Sorry, calculations in my previous post were wrong. 217mm was "u" only. I think u+v would then be 290mm assuming 55mm. Also tried my 18-55 and filled the frame with about 65mm of ruler which is about 0.363 x magnification. Theoretically u+v would then be 281mm but I measured the object to image distance at about 245mm.

Probably most likely John is right that focal length varies with focussing (especially as zooms often do this by moving elements around rather than the whole lens).

I think I should stop now and go take some photographs

Just one final apology as I see you had thought about the internodal space

Peter