Question on magnification calculations.

[LordV]

One thing that has puzzled me with magnification calculations involving a macro lens is that when calculating magnification with ext tubes you seem to have to use the reduced focal length figure at min focus distance rather than the focal length as stated for the lens. (normally obtain this figure from min focus dist/4), but when you are doing a calculation for add-on dioptres or a reversed lens, you do not seem to need to use the reduced focal length figure.
Any explanation

calc for ext tubes = existing mag +( length of tubes/focal length of lens)

calc for diopter or reversed lens on another= existing mag + (focal length main lens/focal length diopter or reversed lens)

Brian V.

[rjlittlefield]

Your first calculation is easy to explain/verify. At any particular setting, every real lens behaves very much like a theoretical lens that is simple but "thick". With that theoretical lens, distance from sensor to rear principal plane is just d=(mag+1)*(focal length). Slog through the algebra to confirm that added magnification is equal to added extension divided by focal length.

Your second calculation is not so easy to explain, and it's only an approximation. When you stick a second lens on the front of a lens that is focused closer than infinity, then the result depends on the separation between the lenses. If the lenses were thin and the separation were zero, then it would be easy to calculate the combo, and that calculation would use the focal length of the rear lens at its current setting. When the lenses are thick and the separation is not zero, something more complicated happens. That more complicated something depends on the focal length of the lens at its current setting, not its rated focal length, but the calculation is ugly and depends on things that are hard to know, like where the principal planes are located.

I think what you've discovered is a simple formula that coincidentally happens to roughly fit what your lenses do. If you change the lens separation, or use different lenses, the fit will change.

As a contrived example, consider focusing a subject at close distance using your main lens on extensions, then mount an add-on diopter so far in front of your main lens that it is just barely in front of subject. The effect is...nothing! Well, not quite nothing, but small, certainly much smaller than your formula would predict.

--Rik

[LordV]

Thanks Rik - sounds like the reversed lens calc comes out roughly correct with other parameters accidently cancelling themselves out.

When people ask about this sort of thing, I always say the easiest most accurate way is to measure it taking a pic of the mm scale of a ruler but the above calcs do give some idea of likely mags with "normal" setups.

Brian v.

[rjlittlefield]

Brian, I just remembered that we've had a long-standing question about actual focal length of the MP-E 65 when set to its 5X position.

Any chance I could get you to measure field width at 5X and at 5X plus some tubes, so we can figure it out?

--Rik

[LordV]

Brian, I just remembered that we've had a long-standing question about actual focal length of the MP-E 65 when set to its 5X position.

Any chance I could get you to measure field width at 5X and at 5X plus some tubes, so we can figure it out?

--Rik

Perhaps if you can give me a clue how to do it
Brian v.

[ChrisR]

3 posts up, Brian. Perhaps your eyes do the same as mine - an athletic long-jump over formulae?!
(It's the only exercise I get - running things through my mind, skipping anything important, and jumping to conclusions).

SO I think if you measure field width with the lens on the camera to get magnification,
then add a tube of known length and measure again
you can work out the focal length from the difference magnification.

Er,
FL= (Mag2 -Mag1)/(Extension2 -Extension1)

In English, the focal length works out to be the length of tube you have to put in to take the magnification from 1 to 2, or 2 to 3, or 3 to 4... Which you would already know, that way round

(Remember your viewfinder probably isn't 100% so look at the file)

[LordV]

Ah in that case think I already know the answer.
With 65mm of ext tubes the mag goes from 5X to 6.7X.
So think by calc the focal length at 5X is 65/1.7 = 38mm ?
Brian v.

[rjlittlefield]

With 65mm of ext tubes the mag goes from 5X to 6.7X.
So think by calc the focal length at 5X is 65/1.7 = 38mm ?

That's the way I get it.

I would also be interested to know what happens to the effective f-number.

They rate the lens at f/2.8, but that's presumably based on an entrance pupil diameter of 23 mm (65/2.8 ). If the entrance pupil stays at 23 mm while the focal length shortens to 38 mm, then the lens would essentially become f/1.65. At 5X, the calculated effective f-number would be f/16.8 using rated f/2.8, but it would be only f/9.9 using f/1.65, assuming pupil factor = 1 in both cases. Or it could be something entirely different if the lens has a pupil factor different from 1.

The effective f-number can be determined fairly accurately by comparing the exposure times needed to get equal brightness images of a uniformly lit gray card, as seen through the MP-E 65 at 5X and 1X, and as seen through an ordinary infinity-focus lens.

--Rik

[LordV]

With 65mm of ext tubes the mag goes from 5X to 6.7X.
So think by calc the focal length at 5X is 65/1.7 = 38mm ?

That's the way I get it.

I would also be interested to know what happens to the effective f-number.

They rate the lens at f/2.8, but that's presumably based on an entrance pupil diameter of 23 mm (65/2.8 ). If the entrance pupil stays at 23 mm while the focal length shortens to 38 mm, then the lens would essentially become f/1.65. At 5X, the calculated effective f-number would be f/16.8 using rated f/2.8, but it would be only f/9.9 using f/1.65, assuming pupil factor = 1 in both cases. Or it could be something entirely different if the lens has a pupil factor different from 1.

The effective f-number can be determined fairly accurately by comparing the exposure times needed to get equal brightness images of a uniformly lit gray card, as seen through the MP-E 65 at 5X and 1X, and as seen through an ordinary infinity-focus lens.

--Rik

Rik - probably not quite what you wanted. but this is the average exposure time of 4 runs of the MPE-65 at ISo800 F2.8 set focused on a backlit piece of china surfaced paper (fairly smooth illumination ) in a fairly dark room
Not any easy way I could compare to another lens with differences there would be in setup.

1:1 ,1/650th
2:1, 1/325th
3:1, 1/199th
4:1, 1/121th
5:1 1/90th
I would have guessed that the F2.8 value was nominal and the lens would be F5.6 at 1:1 ?

Brian V.

[rjlittlefield]

I would have guessed that the F2.8 value was nominal and the lens would be F5.6 at 1:1 ?

Lenses are normally rated as focal length divided by entrance pupil diameter. This corresponds to their effective f-number at infinity focus, even if the lens physically cannot focus at infinity.

I'm presuming that the MP-E 65 is truly 65 mm focal length at some setting, probably 1:1. Under that assumption, the entrance pupil diameter would be 65/2.8 = 23 mm diameter.

Whether the lens then acts like f/5.6 at 1:1 depends on its pupil ratio. If the pupil ratio is 1, then it does. If the pupil ratio is different from 1, then it doesn't.

The behavior at other settings depends on how the effective focal length, the entrance pupil diameter, and the pupil ratio change together. Potentially, it's an awful mess!

The numbers you provide indicate that while it's a bit complicated, it's not an awful mess.

Take the simplest possible assumption, that it acts like a thin lens that focuses entirely by extension (which we know it does not!). Under that assumption, the ratio of exposure times between 1:1 and 5:1 would be 1 to 9. Your measurements indicate 1 to 7.2. F-number goes as the square root of exposure times, so 1 to 3 for the simple thin lens versus 1 to 2.69 for the MP-E 65.

For practical purposes that's an insignificant difference, only about 1/6 of an f-stop.

I guess this is telling us that the pupils shrink as the lens gets shorter.

--Rik

[mgoodm3]

I played with a MPE-65 a couple times. Just using the detector-object distance I figured that it was running around 40 mm at 5:1 (not necessarily accurate depending upon the principal plane spacing). I think that I got about 62 mm at 1:1

[LordV]

I played with a MPE-65 a couple times. Just using the detector-object distance I figured that it was running around 40 mm at 5:1 (not necessarily accurate depending upon the principal plane spacing). I think that I got about 62 mm at 1:1

Pretty similar result for the focal length at 5X !
Brian V.