I was wanting to compare to a more common spec of 4x NA 2.0 with effective aperture at f/2.4, if I got that right, and the Canon 65mm f/2.8 MP-E gives an effective aperture of f/34 at 5x with it's sharpest setting at f/5.6.
I'm not at all sure I've got that right though.
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PaulFurman
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Hi Paul
Aparently the maual for the MP-E 65mm gives this formula:
Effective Aperture = (Aperture Setting) + (Aperture Setting x Magnification)
So wide open we get 2.8 +2.8X5 = f/16.8. Not so shabby.
I believe that formula is a good starting point for effective aparture of any modern macro.
I think you meant to say 4X NA 0.2. While I have been disagreed with, I would give that as roughly 4/(0.2x2) = f/10.
However you are probably asking a rather different question - which gives the better quality? Not so easily answered as NA or F number only gives the theretical maximum resolution. In general terms objectives get far closer to perfection than lenses set to there widest. I would put my money on the objective, especialy if it was the infamous (due to its ebay price) Nikon 4X 0.2NA 160/-. Still, The MP-E is designed to handle a full frame sensor while the Nikon is quoted as good for the diagonal of a cropped sensor. If you are fortunate enough to own a full frame canon camera then, at least in terms of theretical resolution, they are about equal. My guess is that somewhere less than 5X and more than 1X the MPE will pull ahead in pure quality of picture terms.
Aparently the maual for the MP-E 65mm gives this formula:
Effective Aperture = (Aperture Setting) + (Aperture Setting x Magnification)
So wide open we get 2.8 +2.8X5 = f/16.8. Not so shabby.
I believe that formula is a good starting point for effective aparture of any modern macro.
I think you meant to say 4X NA 0.2. While I have been disagreed with, I would give that as roughly 4/(0.2x2) = f/10.
However you are probably asking a rather different question - which gives the better quality? Not so easily answered as NA or F number only gives the theretical maximum resolution. In general terms objectives get far closer to perfection than lenses set to there widest. I would put my money on the objective, especialy if it was the infamous (due to its ebay price) Nikon 4X 0.2NA 160/-. Still, The MP-E is designed to handle a full frame sensor while the Nikon is quoted as good for the diagonal of a cropped sensor. If you are fortunate enough to own a full frame canon camera then, at least in terms of theretical resolution, they are about equal. My guess is that somewhere less than 5X and more than 1X the MPE will pull ahead in pure quality of picture terms.
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rjlittlefield
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I see that Blame has posted while I was writing, but I think the discussions are complementary, so I'll post what I had.
There are no NA 2.0 microscope objectives. Theoretically, the best you can do with a "dry" lens (imaging through air) is NA 1.0. Real lenses top out a little below that, around 0.9. Oil immersion objectives can go higher. The theoretical limit is the refractive index of oil and glass, around 1.6. In practice, you can buy NA 1.4 oil immersion objectives.
I suspect you mean 4X NA 0.2 (not 2.0). That value is available in high end objectives such as plan apochromats.
An objective specified as 4X NA 0.2 is equivalent to an ordinary symmetric lens rated at f/2.0. At 4X, both would be running at an "effective aperture" (working aperture on the camera side) of f/10.
In contrast, this new lens is specified as 4X NA 0.47. That is equivalent to an ordinary symmetric lens rated at f/0.85. At 4X, both would be running at an "effective aperture" (working aperture on the camera side) of f/4.26.
I don't know where your f/2.4 comes from. Perhaps from a discussion of pupil factor, which can easily make a lens rated at f/2.0 act as if it were really f/2.4.
I know, lots of numbers. Do they make sense? Let me know if not.
--Rik
Technical footnote:
The formula for relating NA to the rated f-number of an equivalent lens with pupil factor P=1 is
f_rated = (1/(2*NA)) * (m/(m+1))
The first part, 1/(2*NA) is the classic formula for relating numerical aperture to working f-number. The second part, m/(m+1), is an adjustment that accounts for the difference between working f-number and rated f-number, as a function of magnification.
Paul, your description of the Canon MP-E is correct. The rest is a bit scrambled. Let me see if I can sort it out.PaulFurman wrote:I was wanting to compare to a more common spec of 4x NA 2.0 with effective aperture at f/2.4, if I got that right, and the Canon 65mm f/2.8 MP-E gives an effective aperture of f/34 at 5x with it's sharpest setting at f/5.6.
I'm not at all sure I've got that right though.
There are no NA 2.0 microscope objectives. Theoretically, the best you can do with a "dry" lens (imaging through air) is NA 1.0. Real lenses top out a little below that, around 0.9. Oil immersion objectives can go higher. The theoretical limit is the refractive index of oil and glass, around 1.6. In practice, you can buy NA 1.4 oil immersion objectives.
I suspect you mean 4X NA 0.2 (not 2.0). That value is available in high end objectives such as plan apochromats.
An objective specified as 4X NA 0.2 is equivalent to an ordinary symmetric lens rated at f/2.0. At 4X, both would be running at an "effective aperture" (working aperture on the camera side) of f/10.
In contrast, this new lens is specified as 4X NA 0.47. That is equivalent to an ordinary symmetric lens rated at f/0.85. At 4X, both would be running at an "effective aperture" (working aperture on the camera side) of f/4.26.
I don't know where your f/2.4 comes from. Perhaps from a discussion of pupil factor, which can easily make a lens rated at f/2.0 act as if it were really f/2.4.
I know, lots of numbers. Do they make sense? Let me know if not.
--Rik
Technical footnote:
The formula for relating NA to the rated f-number of an equivalent lens with pupil factor P=1 is
f_rated = (1/(2*NA)) * (m/(m+1))
The first part, 1/(2*NA) is the classic formula for relating numerical aperture to working f-number. The second part, m/(m+1), is an adjustment that accounts for the difference between working f-number and rated f-number, as a function of magnification.
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PaulFurman
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OK thanks, I was initially asking about effective working aperture at the sensor side, so that's f/4.5, compared to f/10 for an f/2 macro lens or a 0.20 NA objective. Yes I mis-typed the NA.
So the f/0.85 compares to f/2 in a more ordinary lens, calculated for infinity use (which none of these can do but it makes sense compared to normal photographic lenses). Perhaps I found the f/2.4 number referring to this way of stating it and yeah, it might have included pupil factor - I had just searched the group to snip examples so may have lost some context.
If I use (1/(2*NA)) to convert to infinity f-rating; .20 gives f/2.5.
At 4x; (inf. f-stop(mag+1)) gives f/12.5 like this: 2.5(4+1)
Two discrepancies:
You say (m/(m+1)) but I know that first m as f, meaning f-stop as rated for infinity focus as marked on the aperture ring of most camera lenses.
You say .20 NA is f/2 for infinity rating but when I use your formula, I get f/2.5. Pupillary magnification?
My way of calculating shows the monster lens as f/1 for infinity and f/5 on the sensor at 4x, for purposes of diffraction and pixels.
I calculate the MP-E at 4x as f/14 on the sensor, wide open and f/28 with f/5.6 on the ring, so there must be a pupillary factor involved or I'm completely messed up, which sounds likely.
So the f/0.85 compares to f/2 in a more ordinary lens, calculated for infinity use (which none of these can do but it makes sense compared to normal photographic lenses). Perhaps I found the f/2.4 number referring to this way of stating it and yeah, it might have included pupil factor - I had just searched the group to snip examples so may have lost some context.
If I use (1/(2*NA)) to convert to infinity f-rating; .20 gives f/2.5.
At 4x; (inf. f-stop(mag+1)) gives f/12.5 like this: 2.5(4+1)
Two discrepancies:
You say (m/(m+1)) but I know that first m as f, meaning f-stop as rated for infinity focus as marked on the aperture ring of most camera lenses.
You say .20 NA is f/2 for infinity rating but when I use your formula, I get f/2.5. Pupillary magnification?
My way of calculating shows the monster lens as f/1 for infinity and f/5 on the sensor at 4x, for purposes of diffraction and pixels.
I calculate the MP-E at 4x as f/14 on the sensor, wide open and f/28 with f/5.6 on the ring, so there must be a pupillary factor involved or I'm completely messed up, which sounds likely.
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rjlittlefield
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Paul, the "m" is magnification.
Let's walk through a specific case with simple numbers.
Suppose that the 4X objective has focal length 40 mm.
Then at 4X it will be 50 mm away from the subject and 200 mm away from the sensor.
Because the objective is rated at NA 0.2 when it is 50 mm away from the subject, its aperture diameter will be 0.2*50*2 = 20 mm.
If you now specify that lens in standard photography terms, its f-number is 20 mm aperture / 40 mm focal length = f/2.0
Plugging numbers into the formula I gave, the calculation goes as
f_rated = (1/(2*NA)) * (m/(m+1)) = (1/(2*0.2)) * (4/(4+1)) = 2.0
giving the same result.
Pupil factor is not involved.
Your "(1/(2*NA)) to convert to infinity f-rating" is a decent rule of thumb, but it's not an accurate calculation. 1/(2*NA) gives the working (effective) f-number from the standpoint of the subject, at the rated magnification. It's not the infinity f-rating. 1/(2*NA) is bigger than the infinity f-rating of the lens, because the lens had to be moved away from the subject in order to focus at finite distances. NA is calculated from the aperture diameter and the distance as focused; rated f-number is calculated from the aperture diameter and the lens's focal length.
--Rik
Let's walk through a specific case with simple numbers.
Suppose that the 4X objective has focal length 40 mm.
Then at 4X it will be 50 mm away from the subject and 200 mm away from the sensor.
Because the objective is rated at NA 0.2 when it is 50 mm away from the subject, its aperture diameter will be 0.2*50*2 = 20 mm.
If you now specify that lens in standard photography terms, its f-number is 20 mm aperture / 40 mm focal length = f/2.0
Plugging numbers into the formula I gave, the calculation goes as
f_rated = (1/(2*NA)) * (m/(m+1)) = (1/(2*0.2)) * (4/(4+1)) = 2.0
giving the same result.
Pupil factor is not involved.
Your "(1/(2*NA)) to convert to infinity f-rating" is a decent rule of thumb, but it's not an accurate calculation. 1/(2*NA) gives the working (effective) f-number from the standpoint of the subject, at the rated magnification. It's not the infinity f-rating. 1/(2*NA) is bigger than the infinity f-rating of the lens, because the lens had to be moved away from the subject in order to focus at finite distances. NA is calculated from the aperture diameter and the distance as focused; rated f-number is calculated from the aperture diameter and the lens's focal length.
--Rik