Lens Magnification Question

[qdsf]

Hi

I am looking at calculators for magnification factors when using extension tubes and diopters. Cambridge in Colour have both. My problem is I cannot reconcile the magnification factors I am getting from the calculators to a real test.

The lens I am testing is the Canon EF 24-105mm f/4 1:4 L IS USM. With no extension tubes I get, in a real test, exactly 1:4; i.e. a 15mm coin at the MFD of 450mm is 3.75mm at Focal Length 105mm. My sensor is 22.5 x 15. Thus I expect lens magnification calculators to return a value of 0.25 or 1:4. They don't. Both Cambridge and another calculator return (for a Focus distance of 450mm and Focal Length of 105mm a magnification of 0.59x.

I clearly must be missing something here, and would appreciate any help.

[DanK]

Welcome to CiC. Can you please go to the top right and enter your real name and location into your profile? We use real names here, and locations often help people target their advice.

Mod: should this be moved to the digital cameras and equipment or general discussion forum? I don't think it will get much traffic here.

qdsf: sensor size is irrelevant. Magnification won't be changed by altering the size of the surface onto which you are projecting the image.

I did a quick informal test with my 24-105 (1st generation) and got similar results to yours: the MFD is about 45 mm, and the magnification was about 1:4, which matches the specs I found online. The only reason I can think of for this is a fairly extreme amount of focus breathing, that is, an effective focal length of about 70mm at closest focus. That surprises me a bit, although it is standard on internally focusing macro lenses. E.g., my 100mm macro has an effective FL of somewhere around 70mm at closest focus (I don't recall precisely). This is why the standard extension tube formulas don't work with macro lenses if you use the nominal FL. For example, 68mm of extension with my 100mm gives me roughly 2:1, which is more than the formula with 100mm would indicate.

[Manfred M]

Agreed Dan - the question should attract more viewers here in the Digital Cameras & Equipment Forum.

[qdsf]

Thanks for response. I am still confused though.

If Magnification of a lens is a function of Focal Length / Focus Distance then for then 24-105mm lens with a MFD of 450mm (manufacturer specs and tested):

At the following Focal Lengths, magnification should be:

at 105mm = 105/450 = 0.23x (agrees with Canon specs for this lens)
at 70mm = 70/450 = 0.16x
at 24mm = 24/450 = 0.05x

Yet the calculator returns:

at 105mm = 105/450 = 0.59x (way different)
at 70mm = 70/450 = 0.24x (not so close)
at 24mm = 24/450 = 0.06x (close)

I get the same results from https://www.kielia.de/photography/calculator/

So, before I even think about adding extension tubes into any calculation, I think I must be missing something and/or misunderstanding the results from the calculators.

[george013]

Thanks for response. I am still confused though.

If Magnification of a lens is a function of Focal Length / Focus Distance then for then 24-105mm lens with a MFD of 450mm (manufacturer specs and tested):

At the following Focal Lengths, magnification should be:

at 105mm = 105/450 = 0.23x (agrees with Canon specs for this lens)
at 70mm = 70/450 = 0.16x
at 24mm = 24/450 = 0.05x

Yet the calculator returns:

at 105mm = 105/450 = 0.59x (way different)
at 70mm = 70/450 = 0.24x (not so close)
at 24mm = 24/450 = 0.06x (close)

I get the same results from https://www.kielia.de/photography/calculator/

So, before I even think about adding extension tubes into any calculation, I think I must be missing something and/or misunderstanding the results from the calculators.

Magnification is image distance/subject distance.
Be aware that the minimal focus distance is measured from the sensor. In most formulas it's measured from the optical centre.

George

[qdsf]

Magnification is image distance/subject distance.
Be aware that the minimal focus distance is measured from the sensor. In most formulas it's measured from the optical centre.

George

The Cambridgeincolour calculator defines focus distance as sensor to subject:

https://www.cambridgeincolour.com/tu...-magnification

Notes: the "focusing distance" is measured as the distance between camera sensor and subject, and the "lens focal length" is the actual lens focal length (without multipliers).

It is this calculator that gives me the 0.59x magnification for a 105mm focal length with a focus distance of 450mm.

So I am still confused

[DanK]

where did you get this?

If Magnification of a lens is a function of Focal Length / Focus Distance

As George wrote, magnification is image distance/subject distance.

The Cambridgeincolour calculator defines focus distance as sensor to subject:

That is correct. Typically, in discussions of magnification, "focus distance" is subject to sensor, while "working distance" is subject to front of lens.

[george013]

I think as soon we're talking of macro with the focus distance the distance to the sensor is meant. There's a sign on your camera indicating the place. It had practical benefits when working with elbows and primes.
In other cases the distance to the optical centre is used as in dof-calculators.
As far as I figured out.
In the M calculator here distance to the sensor is used. You can check it by using a focus distance which is 4x focal length. M must be 1.

When using the dof calculator you can't use a focus length less as the focal length. You would get a negative image. So here the distance to the optical center is used. Just try it.

George

[William W]

(My bold and underlined in the sections I have quoted is now for emphasis)

*

qdsf wrote:

“If Magnification of a lens is a function of Focal Length / Focus Distance then for then 24-105mm lens with a MFD of 450mm (manufacturer specs and tested)”

then -

Dan K wrote:

where did you [qdsf] get this?

If Magnification of a lens is a function of Focal Length / Focus Distance

As George wrote, magnification is image distance/subject distance.

*

It occurs to me that this paragraph from CiC Tutorial might have been misinterpreted or misunderstood:

“Magnification is controlled by just two lens properties: the focal length and the focusing distance. The closer one can focus, the more magnification a given lens will be able to achieve — which makes sense because closer objects appear to become larger. Similarly, a longer focal length (more zoom) achieves greater magnification, even if the minimum focusing distance remains the same.”

REF: CiC Tutorial, [LINK]

***

From first principles, for a Simple Lens:

1. When
f = focal length of lens
u = distance of Object from Lens
v = distance of Image formed from Lens
Then:
1/f = 1/v – 1/u

2. When Magnification of Lens = M
Then:
M = v/u

*

In simple terms, ignoring the fact that when we consider modern camera lenses we are discussing a COMPLEX Lens System, we can still apply the above Simple Lens Formula to understand the principles which are involved and perhaps make a more precise interpretation and meaning of the words that are written in the CiC Tutorial.

In a closed camera system, the distance of Image formed from Lens (v) is (more or less) constant. That is to say the Sensor Plane or Film Plane is at a constant distance from the “Lens”.

Therefore, if v is a constant, then the Magnification (M) is indeed “controlled by just two lens properties: the focal length (f) and the focusing distance (u)”

BUT "M" is NOT "Focal Length / Focus Distance"

I think this is source of the confusion.

WW

[mikesan]

1. When
f = focal length of lens
u = distance of Object from (the optical center of) Lens
v = distance of Image formed from (the optical center of) Lens


In a closed camera system, the distance of Image formed from Lens (v) is (more or less) constant. That is to say the Sensor Plane or Film Plane is at a constant distance from the “Lens”.

WW

You may want to rethink this. All of my cameras achieve focus by varying (v), the distance between the optical center of the lens and the Sensor Plane.
from your formula:
1/f = 1/v + 1/u

at any given focal length, if (u) changes then (v) must, of necessity, change.

[William W]

You may want to rethink this. All of my cameras achieve focus by varying (v), the distance between the optical center of the lens and the Sensor Plane.
from your formula:
1/f = 1/v + 1/u
at any given focal length, if (u) changes then (v) must, of necessity, change.

Indeed.

In a COMPLEX LENS SYSTEM the OPTICAL CENTRE of the lens requires consideration, however, I don’t think that I need to rethink what I wrote.

The purpose of what I wrote was to throw light on why the OP might be confused and for that explanation to be as simple as possible.

Please re-read my words and note that I emphasized the formula that I quoted was applicable to a SIMPLE LENS.

Further, reading on, please note that I emphasized that I was applying that formula to a COMPLEX LENS SYSTEM, solely for the purposes of explaining why the OP might be confused and thought that the CiC tutorial meant M = Focal Length/Focus Distance.

The key sentence is here, re-quoted below and if was not explicit in the first, then the meaning should be clear now:

"In simple terms, ignoring the fact that when we consider modern camera lenses we are discussing a COMPLEX Lens System, we can still apply the above Simple Lens Formula to understand the principles which are involved and perhaps make a more precise interpretation and meaning of the words that are written in the CiC Tutorial."

WW

PS you have a typo in your quote it doesn't matter much for the purpose of this conversation as it is basically a matter of mathematical semantics, but the simple lens formula that I used was:

1/f = 1/v – 1/u (not 1/f = 1/v + 1/u)

[george013]

Indeed.

In a COMPLEX LENS SYSTEM the OPTICAL CENTRE of the lens requires consideration, however, I don’t think that I need to rethink what I wrote.

The purpose of what I wrote was to throw light on why the OP might be confused and for that explanation to be as simple as possible.

Please re-read my words and note that I emphasized the formula that I quoted was applicable to a SIMPLE LENS.

Further, reading on, please note that I emphasized that I was applying that formula to a COMPLEX LENS SYSTEM, solely for the purposes of explaining why the OP might be confused and thought that the CiC tutorial meant M = Focal Length/Focus Distance.

The key sentence is here, re-quoted below and if was not explicit in the first, then the meaning should be clear now:



WW

PS you have a typo in your quote it doesn't matter much for the purpose of this conversation as it is basically a matter of mathematical semantics, but the simple lens formula that I used was:

1/f = 1/v – 1/u (not 1/f = 1/v + 1/u)

It is +.

George

[george013]

**********

From first principles, for a Simple Lens:

1. When
f = focal length of lens
u = distance of Object from Lens
v = distance of Image formed from Lens
Then:
1/f = 1/v – 1/u

2. When Magnification of Lens = M
Then:
M = v/u

*********

Beside the wrong formula there're 2 unknown variables in Quentin's example: object distance and image distance or v and u.

George

[William W]

To George:

Hi, I do not want to bog the conversation down with a to and fro between us - but I shall address the two responses that you just made because they were both directed at my commentary;

Firstly regarding the formulae: 1/f = 1/v – 1/u - OR - 1/f = 1/v + 1/u. Either can be 'correct' dependent upon the discipline applied when one makes the measurements: specifically whether the distance "u" is expressed as negative integer, or a positive integer. As I mentioned in Post #11, that is merely a matter of Mathematical Semantics and it doesn't change the meaning of my comments..

Regarding the second point that you raised and please note that Mike, ("mikesan") already raised that issue in Post #10 and I answered in Post #11

But again and explained differently -

Yes, Quentin is indeed using a COMPLEX LENS.

Yes, for a COMPLEX LENS the values 'v' and also 'u' do indeed change.

NOTE that 'v' changes by a relatively SMALL amount, but yes, it does change.


But that's all irrelevant to what my comments were addressing - I was explaining what I think is the cause of Quentin's confusion - and I think the confusion stems from a misunderstanding of the meaning of the CiC Tutorial.

I used the formula for a SIMPLE LENS to show the only the basic PRINCIPLES

This SIMPLE LENS FORMULA certainly is NOT the formula to arrive at the worked solution for the calculation of MAGNIFICATION in a COMPLEX LENS SYSTEM - and (importantly) I think that fact has been very clear all along.

I trust that explains my input yet better again than Post #11, if not, then I think it best for me leave this conversation be, because I feel that I don't have the words to explain in any other or better fashion.

However I do hope it has been useful to Quentin.

WW

[TonyW]

The calculator is assuming that the calculation is for a simple lens.

The reason that Quentin has different values to the calculator is that the ratios should have been of different quantities.

Remember that the focus distance is measured from the sensor to the object. Let's call that d. Then the distance u from the object to the centre of the lens plus the distance v from the centre of the lens to the image (i.e. to the sensor) is the focal distance, given that we have a simple thin lens, i.e.

u + v = d.

Then we have the general lens equation: 1/u + 1/v = 1/f.

Given f = 105 and d = 450, we can solve these two equations for u an v and hence get the magnification u/v. I got 0.5896, which agrees with the calculator.

For f = 70, I got 0.2387 and, for f = 24, I got 0.0599, both of which agree with the calculator.

In practice, of course, we are not dealing with a simple lens. To do the calculation more accurately we would need to know the distance between the front and rear nodes of the lens.

[george013]

The calculator is assuming that the calculation is for a simple lens.

The reason that Quentin has different values to the calculator is that the ratios should have been of different quantities.

Remember that the focus distance is measured from the sensor to the object. Let's call that d. Then the distance u from the object to the centre of the lens plus the distance v from the centre of the lens to the image (i.e. to the sensor) is the focal distance, given that we have a simple thin lens, i.e.

u + v = d.

Then we have the general lens equation: 1/u + 1/v = 1/f.

Given f = 105 and d = 450, we can solve these two equations for u an v and hence get the magnification u/v. I got 0.5896, which agrees with the calculator.

For f = 70, I got 0.2387 and, for f = 24, I got 0.0599, both of which agree with the calculator.

In practice, of course, we are not dealing with a simple lens. To do the calculation more accurately we would need to know the distance between the front and rear nodes of the lens.

How did you do that?

As I wrote before all calculators do use the simple lens formula. Stating that the're only valuable for simple lenses means I can't use any formula. A prime is also a complex lens, using different glasses.

Can you tell me that if using the nodal points in this example the correction of d with the distance between the two nodal points will be sufficient?

George

[JohnRostron]

Thanks to TonyWatts for getting to the nub of all this. I was screaming for someone to say it. The approximation m=f/u assumes that v is close to f. With close-ups, this becomes less and less true.
John

[TonyW]

How did you do that?

As I wrote before all calculators do use the simple lens formula. Stating that the're only valuable for simple lenses means I can't use any formula. A prime is also a complex lens, using different glasses.

Can you tell me that if using the nodal points in this example the correction of d with the distance between the two nodal points will be sufficient?

George

On reflection, I think that would not be sufficient. The reason is, I think, that the lens equation does not hold for a complex lens. People talk about "focus breathing" where the focal length of a lens changes with focus. That statement is not immediately clear, since it is hard to define the focal length of a lens that is not focused at infinity. What I think it implies is that the f used in the lens equation needs to be changed to make the equation true when u and v are both finite.

[george013]

On reflection, I think that would not be sufficient. The reason is, I think, that the lens equation does not hold for a complex lens. People talk about "focus breathing" where the focal length of a lens changes with focus. That statement is not immediately clear, since it is hard to define the focal length of a lens that is not focused at infinity. What I think it implies is that the f used in the lens equation needs to be changed to make the equation true when u and v are both finite.

When I use the formulas of you and eliminating one of the unknown I get something as 1/105=(1/(450-v))+1/v.
I don't know how to solve that. Either my start was wrong or I just don't see it.

I just tried it another way. M=(di-f)/f=(di-105)/105 --> 0.25*105=di-105 --> di=131.25
Lens has been focused on the shortest distance and no focus breathing has been assumed. Extension of the lens is 26.25mm.
Adding a ring of 30 mm would enlarge di to 161.25mm. Back to the formula M=(di-f)/f=(161.25-105)/f
105M=161.25-105 --> M=(161.25-105)/105=0.54
I hope I didn't make mistakes and I didn't check it with the calculators.
https://en.wikipedia.org/wiki/Magnification

I don't know what exactly is meant with complex lenses. A prime exists out off more elements but they are fixed relative to each other. In a zoom they move.

I remember a database from long ago and finally found it back. It's old but gives some idea using some figures. It looks to be pre-Canon. Nodal points can be outside the lens too. And I don't know from where they are measured.
http://netfrog.org/lens_table_1.htm

George

[xpatUSA]

Hi

I am looking at calculators for magnification factors when using extension tubes and diopters. Cambridge in Colour have both. My problem is I cannot reconcile the magnification factors I am getting from the calculators to a real test.

The lens I am testing is the Canon EF 24-105mm f/4 1:4 L IS USM. With no extension tubes I get, in a real test, exactly 1:4; i.e. a 15mm coin at the MFD of 450mm is 3.75mm at Focal Length 105mm. My sensor is 22.5 x 15. Thus I expect lens magnification calculators to return a value of 0.25 or 1:4. They don't. Both Cambridge and another calculator return (for a Focus distance of 450mm and Focal Length of 105mm a magnification of 0.59x.

I clearly must be missing something here, and would appreciate any help.

Quentin,

This may not help but I have a spreadsheet which uses real object size versus image plane size as inputs for macro work. Thin lens basis, nothing fancy.

Putting in F=105mm and your coin numbers, I get m=0.25, of course.

More interestingly, I get an extended focal length Fe of 131.3mm which comprises the marked focal length of 105mm plus the 26.3mm extension needed to focus the lens.

For your information the distance from the object to the image plane is calculated by the spreadsheet as 656.3mm, not 450mm, so there might be a clue there.

The extension necessary for your example may or may not be the cause of the inconsistent values you have observed.