A 2X2 pano of a seed pod at 10X?

[elf]

Here's the first (successful) image using my BD Plan 10 objective. It was stitched by shifting the camera between frames. The camera was centered for the upper right frame, then shifted down, right, and up for the next frames. The left side was cropped quite a bit in the final image. Bellows draw step change was slightly less than .4mm.

Bellows extension varied from 170mm to 130mm.



p.s. I've been searching for the thread that says how to calculate the focal length and effective fstop for these objectives, but haven't found it yet.

Here's the objective mounted:

[Charles Krebs]

See this pdf for some info:

http://krebsmicro.com/mplan.pdf

Focal length is 21.19mm
And you would triple your working distance (up to 9mm from 3mm) if you removed the outer "BD" cylinder.

Are you shifting camera and lens together as a unit or keeping lens in same place and just moving camera body around? (The "quality" image circle on these is not really large enough to move just the camera body).

Your "f-number" would be about 2. Typically you see the relationship as:

f-number = 1/(2*NA)

Somewhere in a thread Rik gave a relationship with an additional term:

f-number = 1/(2*NA) * (m/(m+1))

--- which would work out to f1.8 if used at 10X

[rjlittlefield]

Somewhere in a thread Rik gave a relationship with an additional term.

Both relationships are correct but the two f-number's mean slightly different things.

1/(2*NA) is the effective f-number (working f-number) with respect to the subject. This is a concept that does not often occur in photography. The corresponding effective f-number with respect to the camera is simply m times larger, or m*(1/(2*NA)) = m/(2*NA).

1/(2*NA) * (m/(m+1)) is the nominal f-number that would have to be set on the aperture ring of an ordinary lens used at the same magnification.* Of course the effective f-number of an extended lens is just m+1 times its nominal f-number, so again the effective f-number with respect to the camera is m*(1/(2*NA)) = m/(2*NA).

If you want to think in terms of effective f-number with respect to the camera, the easiest way is to just use these formulas:

f_effective_at_camera = m/(2*NA) for a microscope objective, and

f_effective_at_camera = f_aperture_ring*(m+1) for an ordinary lens.

*These formulas for ordinary lenses ignore pupillary magnification factor, if any.

--Rik

[elf]

See this pdf for some info:

http://krebsmicro.com/mplan.pdf

Focal length is 21.19mm
And you would triple your working distance (up to 9mm from 3mm) if you removed the outer "BD" cylinder.

That's in the plan, but I'm using the outer cylinder to hold an annulus blocking the light leaks

Are you shifting camera and lens together as a unit or keeping lens in same place and just moving camera body around? (The "quality" image circle on these is not really large enough to move just the camera body).

So far, I've just been moving the camera. I haven't attached the lens to the lead screw yet.

Your "f-number" would be about 2. Typically you see the relationship as:

f-number = 1/(2*NA)

Somewhere in a thread Rik gave a relationship with an additional term:

f-number = 1/(2*NA) * (m/(m+1))

--- which would work out to f1.8 if used at 10X

Thanks, those were the numbers I needed. Now my spreadsheet, http://www.photomacrography.net/forum/v ... php?t=8054 , doesn't give me quite the expected numbers:

• Focal Length=21.19
  F stop=1.8
  Extension=210.020668
  Magnification=8.911310
  ObjectDistance=23.567877
  DOF=0.003966

I expected the magnification to be 10.0 at 210mm extension.

1/(2*NA) is the effective f-number (working f-number) with respect to the subject. This is a concept that does not often occur in photography. The corresponding effective f-number with respect to the camera is simply m times larger, or m*(1/(2*NA)) = m/(2*NA).

1/(2*NA) * (m/(m+1)) is the nominal f-number that would have to be set on the aperture ring of an ordinary lens used at the same magnification.* Of course the effective f-number of an extended lens is just m+1 times its nominal f-number, so again the effective f-number with respect to the camera is m*(1/(2*NA)) = m/(2*NA).

If you want to think in terms of effective f-number with respect to the camera, the easiest way is to just use these formulas:

f_effective_at_camera = m/(2*NA) for a microscope objective, and

f_effective_at_camera = f_aperture_ring*(m+1) for an ordinary lens.

Won't this give me a circular reference since I'm calculating the magnification from the object distance and DOF which use the fstop?

[rjlittlefield]

Now my spreadsheet, http://www.photomacrography.net/forum/v ... php?t=8054 , doesn't give me quite the expected numbers:

• Focal Length=21.19
  F stop=1.8
  Extension=210.020668
  Magnification=8.911310
  ObjectDistance=23.567877
  DOF=0.003966

I expected the magnification to be 10.0 at 210mm extension.

In the formula that 1/f = 1/o + 1/i, the distances o and i are measured from focal plane to a "principal plane" of the lens. But the principal plane of a lens is seldom located at the shoulder of the mounting threads, so you need to deal with the offset. For your purposes and this lens, the offset is 23.09 and the appropriate Extension is 233.09, giving an object distance of 23.309 and magnification = 10.

1/(2*NA) is the effective f-number (working f-number) with respect to the subject. This is a concept that does not often occur in photography. The corresponding effective f-number with respect to the camera is simply m times larger, or m*(1/(2*NA)) = m/(2*NA).

1/(2*NA) * (m/(m+1)) is the nominal f-number that would have to be set on the aperture ring of an ordinary lens used at the same magnification.* Of course the effective f-number of an extended lens is just m+1 times its nominal f-number, so again the effective f-number with respect to the camera is m*(1/(2*NA)) = m/(2*NA).

If you want to think in terms of effective f-number with respect to the camera, the easiest way is to just use these formulas:

f_effective_at_camera = m/(2*NA) for a microscope objective, and

f_effective_at_camera = f_aperture_ring*(m+1) for an ordinary lens.

Won't this give me a circular reference since I'm calculating the magnification from the object distance and DOF which use the fstop?

I guess your spreadsheet is set up to use f_aperture_ring as a basic input. In that case, ignore my comment, or better yet, remember it for possible use in the future.

--Rik

[elf]

In the formula that 1/f = 1/o + 1/i, the distances o and i are measured from focal plane to a "principal plane" of the lens. But the principal plane of a lens is seldom located at the shoulder of the mounting threads, so you need to deal with the offset. For your purposes and this lens, the offset is 23.09 and the appropriate Extension is 233.09, giving an object distance of 23.309 and magnification = 10.

That indeed is the source of the error. My spreadsheet now agrees with the observed behaviour.

Thanks.

[AndrewC]

Can you remind me again why you use "draw" to adjust the plane of focus rather than moving the camera or subject ?

Andrew

[elf]

Can you remind me again why you use "draw" to adjust the plane of focus rather than moving the camera or subject ?

Andrew

The main reason is the lens remains in a fixed position and this allows doing parallax free panoramas. If you rotate the camera and lens around the entrance pupil, there is no limit to the FOV vertically or horizontally. If you shift the camera left/right/up/down, you'll be limited by the image circle but even for larger magnifications, this should give you 3 or 4 times the FOV.

A minor advantage is also the length of movement is quite a bit larger, so it is more forgiving of sloppy step increments.

So far the real limiting factor is the sheer number of images required and the time it consumes just taking them. Most plants just aren't that stable and move around enough to ruin the panorama. I have a lot more failures than successes

[AndrewC]

Thanks.

Andrew

[ChrisR]

That's in the plan, but I'm using the outer cylinder to hold an annulus blocking the light leaks

A black rubber O ring works well