Magnification factors for macro lenses

[William W]

A simple exercise to (approximately) validate the magnification of my EF 100 F/2.8 USM Macro Lens
and if you have the EF 100 F/2.8 L IS USM Macro Lens then you can follow the same procedure:

Tools:
Macro lens; suitable format camera; bit of scrap 1mm graph paper; pencil; straight edge; (tripod and head if you want to make the measurements, steps 7 & 8)

Method:
1. I grabbed my EF 100 F/2.8 USM and a 5D Series camera
2. I drew a 36mm x 24mm rectangle on a piece of graph paper
3. I stuck the diagram on the glass shower door
4. I took a sample photo NOT at the Minimum Focus Distnce (Image 1)
5. I set the lens to the MINIMUM Focus Distance
6. I moved the camera in and out until the graph lines were in focus and made a photo (Image #2, note: what you see here is the full frame, i.e. NOT cropped, of the JPEG SOOC, just resized and converted to B&W for easier viewing)

Extra stuff just for fun:
7. I observed the Focus Distance for Image #2 as being 312mm (Image #3)
8. I observed the Working Distance for Image #2 as being 154mm (Image #4; note front element is recessed about 3mm inside the hood mounting ring)

Images:

IMAGE #1


IMAGE #2


IMAGE #3


IMAGE #4


Conclusions:

1. Upon counting the 1mm grids in image #2, I conclude that the Macro Lens provides 1:1 or x1.0 magnification when it is at the Minimum Focus Distance

2. Upon observing the Minimum Working Distance and the Minimum Focus Distance of the Lens when making Image #2, I conclude that:

2a > the actual MWD is 156mm

2b > the actual MFD is 312mm

Note: These figures would allow for any measurement error by me (especially error measuring the MWD) and also these figures would provide the computation to reflect the Lens's ability to produce 1:1 (i.e. x 1.0 Magnification).

3. I conclude the Canon Data Sheet probably round the MFD from 312mm to 310mm (actually the data sheet I have states the MFD as "0.31m" and that figure that implies an accuracy to the nearest centimeter, if so, then the rounding is correct in its procedure.)

WW

All Images © AJ Group Pty Ltd Aust 1996~2016 WMW 1965~1996

[george013]

UPDATE AND (maybe) CORRECTION - depending upon what ens it is that Rob has . . .


I just re-read the OP. Excerpt here -



I originally thought that was a typo meaning "EF 100mm F/2.8 USM" but I think it now is a typo meaning "EF 100 F/2.8 L IS USM"

If Rob has the latter lens then, yes, the data sheets state that the MFD for that lens is 300mm (not 310mm as I stated assuming a different lens).

From my diary notes when I tested the EF 100 F/2.8 L IS USM, the MWD data stated was 144mm, (i.e. NOT 149mm as per my EF 100 F/2.8 USM).

That means the EFFECTIVE Focal Length at the Maximum Magnification for the EF 10 F/2.8 L IS USM is about 72mm.

Hence putting those figures into the CiC calculator we get a magnification = x 0. 67 (again!).

If I get the time I will still do a rough lab test on my own lens: as mentioned previously the Canon Data Sheets are probably in error by a couple of mm and that is enough to throw the calculator (or the user of the calculator) into an (unnecessary) meltdown.

***

Rob,

You can do a really easy and very simple test yourself to clear up any fears or reservations that you might have about your lens: simply get a bit of 1mm graph paper (or any paper) and draw a rectangle 24mm (h) x 36mm (w) and maybe colour it in, if you do not have graph paper.

Then get your 100mm Macro Lens (which ever one it is) and set it to the closest focus distance and then creep up on the rectangle and make a picture when it is in precise focus . . . (please use a 135 Format camera)

a) if the rectangle sits EXACTLY in your image file's frame - then you have 1:1
b) if the rectangle spills over the edge of your image file's frame - then you have greater than 1:1
c) if the rectangle has the outside showing in the your image file's frame - then you have less than 1:1

WW

POST SCRIPT: I did the exercise as an example for you and anyone else interested in checking their own macro lens - see my comment #21 below . . .

There might come more corrections.

First of all read my first post again. Specially the first sentence.

Doing macro, magnification, the distance to the film/sensor plane is used. There's a mark on the camera from which one can measure.

Then the note under the calculator.

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).

Then the specs of the lens.

Closest Focusing Distance


0.99 ft./0.3m (maximum close-up magnification: 1x)

There're now two different questions: 1) what is the theoretical magnification with a given focal length and focal distance and 2) what is the focal length given a magnification and focal distance.

1). The focal length is 100mm. The calculator wants a minimum focal distance of 4xfocal length so 400. Use those figures and the result is a magnification of 1.

2). The specs says a minimal focal distance of 300mm and a magnification of 1. In the lens formula the magnification is 1 if the optical subject distance equals the image distance. The the image distance is twice the focal length of the lens. See Ted's drawing. And yes, (300/2)/2=75, just as 300/4=75. That's what I meant with a difficult way to explain.

And no nodal points involved.
And a "effective focal length" is just the image distance.

George

[William W]

Hi George,

Yes, truly, I do understand all that you wrote.

However, I think that you are NOT understanding my point which is:

The OP's question was to determine what the magnification of his lens was by using the CiC Calculator.

In other words he did NOT want to accept as an opening statement that the Canon Data which stated x1.0: Magnification was the truth.

THEREFORE

If:

1. the data that the lens can achieve x1.0 Magnification is NOT accepted as necessarily true
AND
2. the OP wanted to use the CiC Calculator to find the Magnification of the lens
THEN
3. a simple way of determining the EFFECTIVE FOCAL LENGTH of the lens (to put into the CiC calculator) is to find out or measure the Minimum WORKING Distance of the Lens when it is at the Minimum Focus Distance.

WW

[dem]

After checking a couple of DoF/magnification calculators, I found that all of them imposed the 4xf limit as if magnifications beyond 1:1 were not theoretically possible. So how did Canon manage to make a 5:1 lens?

We'll need a bit of algebra to understand what is going on. If f is the focal length, f_o is the object distance from the lens and f_i is the image distance from the lens, then

1/f_i + 1/f_o = 1/f (thin lens formula)
m = f_i / f_o (definition of magnification)
f_o + f_i = L (object-sensor distance)

These three equations can be combined together and give:

2 + m + 1/m = L/f

The left hand side of this equation is greater than 4 for any value of m. That's it, L/f cannot physically be greater than 4.

This equation does not stop m from being greater than 1 though. Indeed, if m=0.5 is a root of this equation, then m=2 is also a root. This is essentially a quadratic equation and all calculator are assuming that f_i < f_o and are displaying only one value for magnification, only one of the two possible roots.

[george013]

......
Conclusions:

1. Upon counting the 1mm grids in image #2, I conclude that the Macro Lens provides 1:1 or x1.0 magnification when it is at the Minimum Focus Distance

2. Upon observing the Minimum Working Distance and the Minimum Focus Distance of the Lens when making Image #2, I conclude that:

2a > the actual MWD is 156mm

2b > the actual MFD is 312mm

Note: These figures would allow for any measurement error by me (especially error measuring the MWD) and also these figures would provide the computation to reflect the Lens's ability to produce 1:1 (i.e. x 1.0 Magnification).

3. I conclude the Canon Data Sheet probably round the MFD from 312mm to 310mm (actually the data sheet I have states the MFD as "0.31m" and that figure that implies an accuracy to the nearest centimeter, if so, then the rounding is correct in its procedure.)

WW

All Images © AJ Group Pty Ltd Aust 1996~2016 WMW 1965~1996

The working distance is the distance between the physical camera and object. The optical center of the lens is somewhere else in the lens. So you can't say the minimal focus distance is 2x the minimal working distance. All you can say is that if the minimal focus distance is 300 with a magnification of 1, that then the focal length is 75.
You made some wrong measurements. Big ones, relation working distance and focal distance in connection with a magnification of 1, and small ones, measuring with a lint like on the photo. If you want to measure with mm you'll have to calculate with a measure fault. Especial with the tool you used.

[george013]

Hi George,

Yes, truly, I do understand all that you wrote.

However, I think that you are NOT understanding my point which is:

The OP's question was to determine what the magnification of his lens was by using the CiC Calculator.

In other words he did NOT want to accept as an opening statement that the Canon Data which stated x1.0: Magnification was the truth.

THEREFORE

If:

1. the data that the lens can achieve x1.0 Magnification is NOT accepted as necessarily true
AND
2. the OP wanted to use the CiC Calculator to find the Magnification of the lens
THEN
3. a simple way of determining the EFFECTIVE FOCAL LENGTH of the lens (to put into the CiC calculator) is to find out or measure the Minimum WORKING Distance of the Lens when it is at the Minimum Focus Distance.

WW

William,

I didn't mention Canon in my first post. Only later to tell that a 100mm macro with a minimal focus distance of 300mm is not a 100mm macro.

For a second time the note under the calculator

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).

George

[dem]

The next question is whether the above formula (post #24) can be applied to an actual Macro Lens that is far from being infinitely thin?

Yes, it can if the object-image distance L is adjusted to discard the "thickness" of the lens. Say the distance between the front and the back nodal planes is d, then

2 + m + 1/m = (L-d)/f

is the exact formula that describes the magnification of a "thick" lens. The problem is that we can only measure L and m (as Bill did), but have no way of directly measuring d and f. Plus both d and f vary with focus distance.

What are we going to do if we want to design a calculator? First, let's ignore d as it is normally about 10 times smaller than L. Second, let's use nominal value of the focal length (100 mm) rather than the effective focal length (74 mm, 85 mm or whatever it is at a given focus distance). And that's what all the calculators do. They only give a rough estimation of the magnification the real camera is going to achieve. This estimation gets worse at closer focus distances.

Bottom line: do not trust the calculator, do your own tests to know what your gear can achieve.

[george013]

The next question is whether the above formula can be applied to an actual Macro Lens that is far from being infinitely thin?

Yes, it can if the object-image distance L is adjusted to discard the "thickness" of the lens. Say the distance between he front and the back nodal planes is d, then

2 + m + 1/m = (L-d)/f

is the exact formula that describes the magnification of a "thick" lens. The problem is that we can only measure L, but have no way of directly measuring d and f. Plus both d and f vary with focus distance.

What are we going to do? First, let's ignore d as it is normally about 10 times smaller than L. Second, let's use nominal value of the focal length (100 mm) rather than the effective focal length (74 mm, 85 mm or whatever it is at a given focus distance). And that's what all the calculators do. They only give a rough estimation of the magnification the real camera is going to achieve. This estimation gets worse at closer focus distances.

Bottom line: do not trust the calculator, do your own tests to know what your gear can achieve.

The magnification calculator is a calculator designed for primes. Though a macro lens is called a prime, it's not. It acts like a zoom lens. Some groups are not fixed, but floating. But working with a camera and by example a 50mm lens with an extension ring, it works well.

Can you tell me how you get to 2 + m + 1/m = L/f? I've seen this formula more but was not able to know how to get there.

George

[dem]

The magnification calculator is a calculator designed for primes.

The calculators are designed for an ideal thin lens. Neither primes nor zooms fall into this category. I agree that primes normally are less susceptible to "focus breathing" than zooms though.

Can you tell me how you get to 2 + m + 1/m = L/f? I've seen this formula more but was not able to know how to get there.

Sure. You start with these three equations:

1/f_i + 1/f_o = 1/f (1. thin lens formula)
m = f_i / f_o (2. definition of magnification)
f_o + f_i = L (3. object-sensor distance)

Then both f_o and f_i can be expressed in terms of m and L:

From (2), we get f_i = m * f_o, then substitute this into (3) to get f_o + m * f_o = L, from which f_o = L / (1+m).

Next, f_i = m * f_o = m * L / (1+m).

Substitute f_o and f_i to (1) to get: (1+m) / (m*L) + (1+m) / L = 1/f. Now times both sides by L to get

1/m + 1 + 1 + m = L/f

or

2 + m + 1/m = L/f

[ajohnw]

The next question is whether the above formula (post #24) can be applied to an actual Macro Lens that is far from being infinitely thin?

Yes, it can if the object-image distance L is adjusted to discard the "thickness" of the lens. Say the distance between the front and the back nodal planes is d, then

2 + m + 1/m = (L-d)/f

is the exact formula that describes the magnification of a "thick" lens. The problem is that we can only measure L and m (as Bill did), but have no way of directly measuring d and f. Plus both d and f vary with focus distance.

What are we going to do if we want to design a calculator? First, let's ignore d as it is normally about 10 times smaller than L. Second, let's use nominal value of the focal length (100 mm) rather than the effective focal length (74 mm, 85 mm or whatever it is at a given focus distance). And that's what all the calculators do. They only give a rough estimation of the magnification the real camera is going to achieve. This estimation gets worse at closer focus distances.

Bottom line: do not trust the calculator, do your own tests to know what your gear can achieve.

It isn't possible to do an accurate calculation what ever sums are used other than having what is usually referred to as prescription of the lens. That will contain all of the physical data on the lens. Spacings, glass types, curves and etc. Then it can be ray traced. Once up on a time the ray tracing was done using log tables and early calculating machines. These days it's done with software such as this

https://www.zemax.com/

There are a number of similar packages and most include iterative optimisation facilities.

Early information on using an optical bench may allow a lens to be "measured" in several ways that would allow a more accurate approximation but it would still be an approximation.

John
-

[george013]

The calculators are designed for an ideal thin lens. Neither primes nor zooms fall into this category. I agree that primes normally are less susceptible to "focus breathing" than zooms though.



Sure. You start with these three equations:

1/f_i + 1/f_o = 1/f (1. thin lens formula)
m = f_i / f_o (2. definition of magnification)
f_o + f_i = L (3. object-sensor distance)

Then both f_o and f_i can be expressed in terms of m and L:

From (2), we get f_i = m * f_o, then substitute this into (3) to get f_o + m * f_o = L, from which f_o = L / (1+m).

Next, f_i = m * f_o = m * L / (1+m).

Substitute f_o and f_i to (1) to get: (1+m) / (m*L) + (1+m) / L = 1/f. Now times both sides by L to get

1/m + 1 + 1 + m = L/f

or

2 + m + 1/m = L/f

Thanks for the explanation how to get to the last formula.

Given this lens having a magnification of 1 with a focal distance of 300 it will be
2 + 1 + 1 = 300/f
f = 300/4 = 75.

George

[William W]

The working distance is the distance between the physical camera and object.

That statement is wrong.

Working Distance in Macro Photography is the distance from the front of the LENS (specifically the front nodal point of that first bit of glass of the Lens) to the Plane of Sharp Focus, just as I measured it.

***

The optical center of the lens is somewhere else in the lens. So you can't say the minimal focus distance is 2x the minimal working distance. All you can say is that if the minimal focus distance is 300 with a magnification of 1, that then the focal length is 75.
You made some wrong measurements. Big ones, relation working distance and focal distance in connection with a magnification of 1, and small ones, measuring with a lint like on the photo. If you want to measure with mm you'll have to calculate with a measure fault. Especial with the tool you used.

George I have never mentioned "optical center of the lens" . . . if you think I made wrong measurements and flawed conclusions that's fine, you keep that opinion. I shall not attempt to explain any further.

WW

[William W]

Thank you for explaining how the DoF/magnification calculators perform.

*

. . . So how did Canon manage to make a 5:1 lens?

To address this question, it might also assist to consider the design of this particular lens: MP-E65mm-f/2.8 1~5x-Macro-Lens.

In simple terms it is a very good macro optic at the end of extendable bellows.

Similarly, the EF 50/2.5 (native x0.5 Magnification) is EXTENDED when the "Life Size Converter" is added to it, providing x1.0 Magnification

Similarly (all three) of the EF 100/2.9 Macro Lenses will achieve GRETAER THAN x1.0 Magnification, if Bellows or Extension Tubes are added.

WW

[george013]

William,

It starts to becomes a discussion of definitions.
When you say in post 13:

There is (complicated) Mathematics to arrive at the appropriate formula to use, however: if the lens actually produces x1 Magnification (i.e. 1:1), then the simplest, correct formula to find the EFFECTIVE Focal Length is 1/2 of the WORKING Distance (Subject to Front Nodal Point).

The "effective focal distance" is the image distance: distance optical center to sensor plane.With a magnification of 1 it's 2*focal length.
The working distance to me is the space between the front of the camera, whatever that might be, and the subject. A working distance of 10 cm means that I can move a box of 10cm between them. It's not an optical value, just practical. There is no relation to whatever nodal point, optical center or whatever. I know you disagree and I know you've much more knowledge of photography than me, but still I disagree.

ByThoms "quick & dirty guide to macro" http://www.bythom.com/qadmacro.htm

Be aware of "working distance". Working distance is defined as the distance from the front of the lens to the subject. A broad generalization is that the more focal length in your macro lens, the larger the working distance. Thus, a 35mm macro lens will get 1:1 at very, very close distances to the subject, while a 200mm macro will be a much greater distance from the subject. Nikon has gotten coy lately and is no longer putting working distance on their primary lens product pages. Here they are for 1:1:

60mm: <2"
105mm: ~6"
200mm: ~10"

From Sigma https://blog.sigmaphoto.com/2013/mac...robert-otoole/

The working distance of a macro lens, not to be confused with minimum focus distance, is the distance between the front of your lens and the subject. This is different from the minimum focus distance which instead means the distance to the subject as measured from the focal plane mark on the camera body, not from the front of the lens. Working distance is a more important figure since it tells how much space you have between the front of your lens and your subject. Working distance generally increases with longer focal length lenses, shorter lenses usually have shorter working distances.

All, as I know so far, used formulas in photography are based on the lens formula. With the optical center in the middle. But for it's nearly practical impossible to know that point other formulas are deducted with more easy to know variables. By example the formula Den gave for the macro. Three variables: magnification, focal length of the used lens, focus distance meaning distance subject sensor. And when you measure the distance sensor to the front of the lens and subtract that from the focus distance you get the working distance.

About the mentioned calculator you want to use the "effective focal length", the image distance. And you give a wrong example how to calculate that, in my opinion. But the calculator doesn't ask for that. It asks for the "focusing distance".

Then the note under the calculator.
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).

I don't understand why you don't correct that.

George

[ajohnw]

Working distance is usually given as distance from sensor/film plane to subject George. This goes way back on macro lenses from a time when manufacturers actually produced tables particularly for bellows lenses.

The term nodal point is often used rather glibly. In terms of a single lens the wiki has an illustration showing that a real or sometimes called thick lens has 2 nodal points. It also states that photographers use the term incorrectly. In some ways that is true but a nodal point can have another meaning that may well apply to a single thick lens but more often applies to a lens consisting of several elements. It's a pivot point where a lens can be swung from side to side without the image at the centre moving. Not a thing that is easy to achieve precisely on complex lenses.

Thick lens optical calculations can be more precise than thin lens approximations but actual dimensional information and the type of glass needs to be known. This mostly applies to paraxial rays as it does to thin lens approximations. There are rays from point source forming a point source on the image plane or a source at infinity. Once the "rays" depart form that which means that there is a field width on the image plane things are even more of an approximation. These rays are usually called skew rays. They enter the lens at an angle when objects are at infinity.

While all sorts of sums for magnification, hyperfocal distance and etc may be discussed ad nausium the only accurate method available is ray tracing which as I mentioned earlier we can not unfortunately do as we don't have the information that is needed. We are stuck with rather rough approximations and shouldn't be surprised if they don't work out that accurately.

John
-

[george013]

Working distance is usually given as distance from sensor/film plane to subject George. This goes way back on macro lenses from a time when manufacturers actually produced tables particularly for bellows lenses.
....

-

John,
Read the links in my former post. That are not links to a simple club of hobbyist.
One more http://www.school-of-digital-photogr...-distance.html

Beside that, that's also not what William states.

George

[ajohnw]

John,
Read the links in my former post. That are not links to a simple club of hobbyist.
One more http://www.school-of-digital-photogr...-distance.html

Beside that, that's also not what William states.

George

Possibly George but all I have ever seen quoted on modern macro lenses is the distance from the focal plain to the subject. Old ones too but also in the form of tables for different levels of magnification.

John
-

[george013]

Possibly George but all I have ever seen quoted on modern macro lenses is the distance from the focal plain to the subject. Old ones too but also in the form of tables for different levels of magnification.

John
-

Again.
http://www.school-of-digital-photogr...-distance.html
And the other links. But they have no figures.

That's the minimal focus distance. The working distance is shown too.

George

[William W]

Working distance is usually given as distance from sensor/film plane to subject George.

That statement is incorrect.

Working Distance is the distance from the front of the lens (actually the front nodal point of the front element of the lens - NOT the Nodal Point of the total LENS CONSTRUCTION) to the Plane of Sharp Focus: This is as I have repeatedly stated in this conversation and as corroborated by George's cites in post #34 from Thom Hogan and the Sigma Website.

Yes, the term "Working Distance" does have its origins when lenses used bellows to make macro work, but it goes back previous to that when cameras had two 'standards' - the Front Standard and the Rear Standard and there were bellows in between.

The distance between the Film Plane (often signified by the symbol phi on the Camera) to the Subject (more accurately the Plane of Sharp Focus) is referred to as the Focus Distance

WW

[William W]

Working distance is usually given as distance from sensor/film plane to subject George. This goes way back on macro lenses from a time when manufacturers actually produced tables particularly for bellows lenses.

AND

Possibly George but all I have ever seen quoted on modern macro lenses is the distance from the focal plain to the subject. Old ones too but also in the form of tables for different levels of magnification.
John

Hi John

Considering your persistence that yours is the correct definition or even a definition which you state is ‘usually given’ and also considering George's passion for correct definitions: could you please cite some instances where it stated that the Distance from the Film Plane to the Subject is the "Working Distance".

Thanks.

WW