# Talk:Magnification

## comment moved from article

"magnification of light(i.e. sunlight)to produce energy.where does this fit?" - edit by 70.189.159.109 to the main article. I moved it here. SeventyThree(Talk) 16:56, 22 July 2005 (UTC)

Well i think you can not magnify the sunlight whatever we are getting from the sun, but you can concentrate or condense at a point to produce higher temperature at a point. it is just making the image of the sun at a focal point.

## Magnification and focal length

I want to ask in the equation M = f/f-d0 Where is the value of h0 —Preceding unsigned comment added by 202.141.36.66 (talk) 05:35, 21 May 2008 (UTC)

I've puzzled over the M = f/f-do formula a number of times, trying to grasp the relationship between empirical magnification and 'abstract' focal length. It seems like a remarkably tautological and inflexible formula. Unable to do differentiation, or find a proof that I could reverse engineer, the closest I've come up with is this :
Verifying the effective focal length (f) of a camera lens using similar triangles and three empirical measurements : the height of the object (ho), the height of the image of the object (hi) and the distance between the object and the image (df) (also known in photography as the "focus distance"[1]?).
By definition, the focal length is the distance between the principle plane and the image plane when focused at infinity (≈ hyperfocal distance). In a camera lens system, each individual element has a principle plane, but when we describe the optical power of the whole system as a single effective focal length, there will be only one corresponding effective principle plane.
So we need to calculate where the 'effective principle plane' bisects df, which will give the value for di = EFL. From similar triangles, we know that
hi / ho = di / do,
and di + do = df.
From di + do = d, we can derive the relationship between di and d as di / (do + di), and its (measurable) conjugate: hi / (ho + hi) [is this correct ?]
Therefore di / df = hi / (ho + hi) and so
f = df x hi / (ho + hi)
Eg. If image height = 24mm, object height = 1,800mm, and distance between image and object = 3,800mm. 3800x24/1824 = 50mm. It seems to work in practice (accurate measurements are fiddly but possible).
However, maths is not among my talents and I still can't figure out how this formula (if correct) relates to M = f/f-d0 or how f can be defined as a function of M and df ? I can see how you might say, when the object distance is several orders of magnitude larger than the focal length, M = -di / do ≈ -di / (di + do) which is like saying M ≈ -f / df ... ? --Redbobblehat (talk) 19:47, 3 March 2009 (UTC)
I've checked it and I think it is correct, although the English was garbled (now corrected). Note that the convention used is that real, inverted images have negative magnifications (which is not intuitively easy). Note also that if the distance to the object is relatively large (eg most photographs), the magnitude of the magnification (ie ignoring the minus sign) is nearer zero than one. Man with two legs (talk) 17:55, 9 March 2009 (UTC)
Thanks for checking the formula (for typos !), and the english does read much better now :). I think I found a couple of sources relevant to the M = f/f-do formula. Both versions derive from the Gaussian model, so the distance do is measured between the object and the front principal plane.
This source [2], apparently using the "cartesian" sign convention, (where s = do) gives
• do = (f (1-M)) / M ... which rearranges to ... M = f / (do + f)
These sources [3], [4] and [5] give essentially the same formula but seem to be using the "intuitive" sign convention, which I will denote as M' (where u = d'o):
• d'o = f (1 + 1/M') ... which rearranges to ... M' = f / (d'o - f)
I still can't figure out where M = f/f-do comes from ? -Redbobblehat (talk) 17:49, 10 March 2009 (UTC)
Er, the last two formulae differ only in that one is minus the other so they simply use different sign conventions (assuming we correct for the technical error of writing M = f/f-do which should have been M = f/(f-do), equivalent to the formula in the article. Taken literally, M = f/f-do implies M = 1-do which is obviously wrong). When I derived it, I got M = f / (do - f) but I was not using the same sign convention as the article. Man with two legs (talk) 10:22, 11 March 2009 (UTC)
Ah eureka! It wasn't the brackets thing - I just hadn't realised that (do - f) * -1 = (f - do) ... [weeps into beer] ... So does that mean M = f / (do + f) is wrong for either sign convention (see [6]) ? --Redbobblehat (talk) 00:26, 12 March 2009 (UTC)

## reference diagram

Basically, black dimensions are real, grey are virtual. The direction of the arrows can be used to describe cartesian +/- signage ... left or down = negative, right or up = positive.

I have cobbled together a diagram (from PAR's jpgs at real image and virtual image) showing the "dimensions" of a virtual and real image. This is a first draft, so I present it hear for comment before adding it to the article (if we agree it's useful). So comments welcome, but remember this took time and effort, so please be nice :-) ... and no, I don't know how to make SVGs --Redbobblehat (talk) 23:28, 12 March 2009 (UTC)

Hi RBH, this looks like a great diagram. Sounds like you've already been nagged about SVGs, but really they are astonishingly more useful thanks to the ease of editing and resizing. Have you tried Inkscape? It's a highly intuitive and easy-to-use free SVG-editing program. I've made tons of diagrams for WP with it. There's some helpful information on using it with Wikipedia/Commons at Commons:Help:Inkscape. Moxfyre (ǝɹʎℲxoɯ | contrib) 20:37, 12 June 2009 (UTC)
As there have been no objections in nearly 2 years, I've moved this diagram into the article. --Redbobblehat (talk) 19:48, 4 February 2011 (UTC)

## sign convention

I have also tried to rewrite the "photography" paragraph to make it clearer. Not sure if I succeeded though; it's basically all about sign conventions ... perhaps a section entitled "sign conventions" would be better ?

• Photography: The image recorded by a photographic film or image sensor is always a real image and is usually inverted. When we measure the height of an inverted image using the cartesian sign convention (where the x-axis is the optical axis) the value for hi will be negative, and as a result M will also be negative. However, the traditional sign convention used in photography is "real is positive, virtual is negative"[1]. This means that in photography: Object height and distance are always real and positive. When the focal length is positive the image's height, distance and magnification are real and positive. Only if the focal length is negative, the image's height, distance and magnification are virtual and negative. Therefore the photographic magnification formulae are traditionally presented as:
$M = {d_i \over d_o} = {h_i \over h_o} = {f \over d_o-f} = {d_i-f \over f}$

--Redbobblehat (talk) 23:35, 12 March 2009 (UTC)

After nearly 2 years without objections, I have moved this version into the article. --Redbobblehat (talk) 20:01, 4 February 2011 (UTC)
I'm not sure if the sentence "Only if the focal length is negative, the image's height, distance and magnification are virtual and negative." is helpful here. I think it helps to clarify the point about sign convention, but AFAIK a camera lens with negative focal length would not produce a usable image ... ie no photographic lens system would have a negative focal length! Comments welcome --Redbobblehat (talk) 20:01, 4 February 2011 (UTC)
I also considered adding some guidelines for using the formulae in close focusing applications. Specifically : to measure d and d' from the front and rear nodal points of the lens - which are not easy to locate, and pointing out that when focused at infinity, (d'-f)/f = 1/f. I hesitate because I'm not sure that this article is the correct place to provide such detail. Comments welcome --Redbobblehat (talk) 20:01, 4 February 2011 (UTC)

## Telescope angular magnification

The article lists telescope angular magnification as f_e/f_o. But since f_o > f_e in telescopes, that results in a magnification of <1! That makes no sense to me. These facts about magnification need citations, anyway. PenguiN42 (talk) 00:45, 15 May 2010 (UTC)

## Comment transferred from the main article

This was moved here from the main article where it was posted by user:Robert M Dewey:

Consider who your readers are what what they may want to know. Some like me want to know the basic concepts not the mathematics. What are concave and convex lenses? Does one magnify and the other minify. Which lenses does one use to make a telescope -- concave or convex or one of each or what? can you do this by holding each lens with your fingers? These are not childish questions. They get at the underlying concepts of the physics of optics. The technical mathematics is good too. But don't forget that if you can't explain the basic facts underlying the mathmatics of optics simply, you probably only know the mathematics by rote. I came for a simple explanation of how magnification and minification works. I'd provide the information myself if I knew it. But that's what I came to your article to find out. It isn't there.

He's got a point. Man with two legs (talk) 09:42, 28 May 2010 (UTC)

I've now added a section Examples of magnification. Man with two legs (talk) 10:10, 28 May 2010 (UTC)

## Curved mirror

There are no informations about the Magnification of Curved mirrors. Though, they sell with Labels from ~"2x" to "7x" in stores. Can anyone give information about how this is calculated? --Itu (talk) 02:51, 26 November 2010 (UTC)

Agreed, this article needs to have some discussion of curved mirrors. 75.37.70.77 (talk) 16:58, 3 May 2011 (UTC)

## Unit magnification

In optics, does "unit magnification" simply mean M=1, or is there more to it than that ? Redbobblehat (talk) 18:32, 4 February 2011 (UTC)

## angular magnification

Currently the article states :

• Angular magnification — For optical instruments with an eyepiece, the linear dimension of the image seen in the eyepiece (virtual image in infinite distance) cannot be given, thus size means the angle subtended by the object at the focal point (angular size). Strictly speaking, one should take the tangent of that angle (in practice, this makes a difference only if the angle is larger than a few degrees). Thus, angular magnification is given by:
MA = tanE' / tanE
where E is the angle subtended by the object at the front focal point of the objective and E' is the angle subtended by the image at the rear focal point of the eyepiece.

My first question is : why measure angular object size (E) about the front focal point and not the entrance pupil ?

My guess is : it's just plain wrong - the "view" of anything rarely has anything to do (directly) with the focal or cardinal points ... it should say "Where E is the angular size of the object (the angle subtended by the object at the centre of the entrance pupil) and E' is the angular size of the image (the angle subtended by the image at the centre of the exit pupil)". -- Redbobblehat (talk) 21:09, 11 February 2011 (UTC)

My second question is : given that M == tanE' / tanE == transverse magnification, why have a separate definition for angular magnification ?

My guess is that under paraxial approximation, E' / E is "close enough to" tanE' / tanE, so in first-order optics angular magnification is usually given as = E / E' (Hecht 2003 Optics p.210). If so, perhaps it should state that "Whilst transverse magnification, M = tanE' / tanE, for calculations using paraxial approximation (where neither E or E' are more than +/- 10° from the axis) we can get away with M = E' / E". --Redbobblehat (talk) 21:09, 11 February 2011 (UTC)

## Virtual-object magnification?

The article discusses linear magnification and angular magnification, but not the magnification of the virtual object that occurs when the virtual object isn't at infinity. In terms of being able to see detail with a magnifying glass, angular magnification is what matters since the angular resolution of your eye is limited. However, this doesn't answer the question "How much bigger does the thing behind the magnifying glass appear to be?" If you hold a magnifying glass one focal length from your subject and look through it, the object appears to be at infinity in terms of focus accomadation and stereo vision, so the virtual object size is infinite. If you hold that lens to your eyes, the angular magnification is unity, but the object still appears to be infinitely big even though the images on your retinas look the same -- its' just that your eyes are focused and converged to infinity so it looks infinitely far away. If the object is less than one focal length from the lens, then in terms of accomidation and vergence, the object appears to be closer than infinity, so the virtual object height is finite. In the limit that the lens is on the object, it has no effect.

So: What's the formula for virtual-object magnification as a function of eye-lens distance, lens-object distance, and lens focal length? I don't see it here and there are no equations on virtual image. —Ben FrantzDale (talk) 13:19, 2 December 2011 (UTC)

1. ^ Sidney F. Ray Applied Photographic Optics: Lenses and Optical Systems for Photography, Film, Video, Electronic and Digital Imaging, Focal Press, 2002, ISBN 0240515404, p.40[7]