Trying to understand Magnification, please help

[Vernian]

First off, sorry if this is in the wrong section – it seemed a bit too technical for general chat, but I can’t quite find the right place for asking this kind of question.

I’m thinking of getting my first DSLR. But knowing that doing this I will be giving up my “macro” functionality until such time as I buy a macro lens, I wanted to try and see what effect this will have. Thus I’ve used the following calculations to see how small a similar image would be with a standard kit lens. However the calculations is giving me an image size on my current camera that is quite a bit larger than I expected, so could someone please look at my calculations and tell me where I’m going wrong

Using 1/f= 1/v + 1/u
M = v/u
f = Current camera focal length (using real not effective): 5.1 mm = 0.0051 m
u = Distance to subject: 5 cm = 0.05 m
then v = 1/ ( (1/f) – (1/u) ) = 1/ ( 1/0.0051 – 1/0.05) = 0.0056 m
Magnification M = 0.0056/0.05 = 0.112

So far so good, now for the bit I’m not quite sure about. My subject/object is 2.5 cm high (OH) , my sensor is 6.2 mm high (SH) and my photos have a total pixel height of 3648 (p) in portrait orientation. So my thinking is that the following formula should give me how many pixels high (OP) the object should be (with all length converted to meters again):
OP = ((OH * M)/SH)*p
= 1647 pixels

But that means my image should fill half the image (height wise at least) and that’s not what I’m seeing on my actual photos. Am I going off on entirely the wrong tangent here?

Alternatively it may be easier if I could just see examples of objects taken at minimum focal distance for a 18-55 and 70-300 kit lenses (currently thinking of cannon 500d, so it’s a 1.6 crop factor), if anyone has links/ photos that show this I would really appreciate it.

Oh and a final question, though I’m not sure if this is the right part of the forum – if a kit lens is marked as 18-55 is that actual focal length or, focal length as adjusted for crop factor?

Thanks!

[RichardTaylor]

(1) The focal length marked on Canon lenses is the real ones.


RE close focussing distances.
Think maximum magnification.

For the Canon EF-S-18-55mm-f-3.5-5.6-IS it is 0.34.
Canon EF-S 18-55mm f/3.5-5.6 IS Lens Review
That means the image will captured at a maximum of approx 1/3 life size at minimum focussing distance. (see the link above)

I am not aware of 1 70-300 kit lens however the Canon-EF-70-300mm-f-4-5.6-IS-USM-Lens maximum magnification is 0.25 which is 1/4 life size.

See here. (there are also some comparisons with other lenses.
Canon EF 70-300mm f/4-5.6 IS USM Lens Review

[wulf]

Originally Posted by Vernian

First off, sorry if this is in the wrong section – it seemed a bit too technical for general chat, but I can’t quite find the right place for asking this kind of question.

Macro technique... I've moved it there.

Wulf

[Vernian]

Originally Posted by wulf

Macro technique... I've moved it there.

Wulf

Thanks!

Originally Posted by RichardTaylor

(1) The focal length marked on Canon lenses is the real ones.

RE close focussing distances.
Think maximum magnification.

Thanks for the information - maximum magnification did it for me. My current camera/lens has a maximum magnification of 0.34 too, so if I understand it correctly I could be getting the same image, just with having my camera a a different distance from the object.

I've also been given the good advice to buy a memory card in store and use their equipment to take some photos, take the memory card home and look at the photos to be 100% sure.

[AnEye]

I have a Sony cyber-shot DSC-H2, 12x zoom, 6.0 mega pixels. I am new to this.... and toying around with what I have trying to per-fect the pix I take, I also have a 58mm macro lens which I'm learning to use, but also have a 58mm 2+ close-up lens I do not know how to use... what is the distance I can shoot from with the close-up lens and the macro lens? I get some really awesome pix with the macro, but then other times they are a total blurr.... I have found that my ISO setting takes the best pix, haven't experimented with the other settings to much. need some advice on those too, but another time. advise plz?

[ravncat]

Originally Posted by Vernian

Using 1/f= 1/v + 1/u
M = v/u
f = Current camera focal length (using real not effective): 5.1 mm = 0.0051 m
u = Distance to subject: 5 cm = 0.05 m
then v = 1/ ( (1/f) – (1/u) ) = 1/ ( 1/0.0051 – 1/0.05) = 0.0056 m
Magnification M = 0.0056/0.05 = 0.112

What lens is giving you a focal length of 5.1mm ?
What distance are you measuring to get 5cm?

I think, another thing to remember is that a photographic lens is not a "thin lens" or a "single lens" so things get a bit more complicated. (Here's what it looks like for a two lens system Two lens system - Image distance and magnification - Most kit lenses probably have around seven elements - that move in relation to each other.

using 55mm focal length and a minimum focus distance of 11 inches (.28 m) you end up with M of .244 which is alot closer to the maximum reproduction ratio of a kit nikon 18-55 (1:3.2)

There are a few other things that come in to hamper us further - the first is that many lenses have focal length breathing - especially when they focus at a close distance - or in other words, the focal length changes as you focus more closely - this makes calculating the magnification even harder.

The second is what you tried to solve above - the sensor size vs number of pixels - as those are like magnification as well. You're solution there was fine. As a general rule, if the subjects are at normal (not macro photography distance) , Take the focal length of the lens used divided by the camera to subject distance. That should give you an approximate reproduction ratio. so a 200mm lens focused at 2 meters should give you a 1:10 reproduction ratio.

For macro distances - it's easier to take a photograph of a ruler - and see how many millimeters you captured, and compare that to the sensor size - so if you capture 25.1mm across on a 25.1mm x 16.7 sensor - you have a reproduction ratio of 1:1. for capture. then you can have the fun of figuring out how much magnification is happening when you change the pixels per inch size for the screen)

Most times you can find the reproduction ratio in the specs for a lens. I don't mean to discourage trying to calculate what it will be for yourself - but i've not seen a nice concise equation for it based on a physical dslr lens - which are often too complex to use the thin lens equation for (especially if it is a retrofocus wide angle or a telephoto lens).

The equation relating D, distance from sensor, F, focal length, and
M, magnification is: D=F(1+M)(1+M)/M
Solving that for M requires the quadratic - and leads to
M = (D - 2 F - sqrt[D^2 - 4 D F])/(2 F) or (D - 2 F + sqrt[D^2 - 4 D F])/(2 F) - use whichever is real.


There are equations and guides out there for figuring out how extension tubes, reversing lenses and diopter additions will affect your magnification factor. A nice site Lens Magnification and Depth of Field Calculator has a calculator with the basic lens equation programmed in - it's easy to use though - it sometimes has trouble with exotic lenses