Going beyond 90mm for Macrophotography

[Stagecoach]

John, (ajohnw)

I have just realised I made a boob last night in that having taken that shot and others and for records taking a 'rule' shot (6mm equating to 4:1 magnification) with the 105mm at 1:1 I had totally overlooked throughout the exercise the focus setting of the reversed 50mm on the front

Perhaps a repeat tonight to to see the affect of the 50mm at various focus distance.

Grahame

[george013]

Some of that relates to a simple single lens George not a compound system of lenses. One aspect of that is when ever the object is moved the image plane moves too and the image circle size will also change as well.. Does that happen on a camera lens? I some how suspect you are also relating magnification with light loss on a surface area basis and trying to pull that into diffraction effects. Light captured is purely a function of the "equivalent" area of the glass that is capturing the light. Magnification based light loss doesn't have any significant effect on diffraction otherwise microscope objectives would have a terrible time. On those F ratio still sets resolution but they have the advantage that they are designed for a very specific fixed object distance. Very high resolution ones need focusing to with 1/2 a micron.

It's pointless keeping this up. You need to spend some time studying this subject from some reliable source - not odd ball comments that you may have picked up on the web some where.

John
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John,
Don't forget that you and me are both also part of that internet at this very moment.
So a last time. The formules used below are the same as in your book references.

http://en.wikipedia.org/wiki/Airy_disk

Cameras

If two objects imaged by a camera are separated by an angle small enough that their Airy disks on the camera detector start overlapping, the objects can not be clearly separated any more in the image, and they start blurring together. Two objects are said to be just resolved when the maximum of the first Airy pattern falls on top of the first minimum of the second Airy pattern (the Rayleigh criterion).

Therefore the smallest angular separation two objects can have before they significantly blur together is given as stated above by

\sin \theta = 1.22\ \frac{\lambda}{d}

Thus, the ability of the system to resolve detail is limited by the ratio of λ/d. The larger the aperture for a given wavelength, the finer the detail which can be distinguished in the image.

Since θ is small we can approximate this by

\frac{x}{f} = 1.22\ \frac{\lambda}{d}

where x is the separation of the images of the two objects on the film and f is the distance from the lens to the film. If we take the distance from the lens to the film to be approximately equal to the focal length of the lens, we find

x = 1.22\ \frac{\lambda f}{d}

but \frac{f}{d} is the f-number of a lens. A typical setting for use on an overcast day would be f/8 (see Sunny 16 rule). For blue visible light, the wavelength λ is about 420 nanometers (see cone cells for sensitivity of S cone cells). This gives a value for x of about 4 µm. In a digital camera, making the pixels of the image sensor smaller than this would not actually increase optical image resolution. However, it may improve the final image by over-sampling, allowing noise reduction.

As you can see above, marked by me, the airy disk is calculating using the image-distance but for easyness it's said to be equal the focal-distance


http://en.wikipedia.org/wiki/Numerical_aperture

Working (effective) f-number

The f-number describes the light-gathering ability of the lens in the case where the marginal rays on the object side are parallel to the axis of the lens. This case is commonly encountered in photography, where objects being photographed are often far from the camera. When the object is not distant from the lens, however, the image is no longer formed in the lens's focal plane, and the f-number no longer accurately describes the light-gathering ability of the lens or the image-side numerical aperture. In this case, the numerical aperture is related to what is sometimes called the "working f-number" or "effective f-number." A practical example of this is, that when focusing closer, with e.g. a macro lens, the lens' effective aperture becomes smaller, from e.g. f/22 to f/45, thus affecting the exposure.

The working f-number is defined by modifying the relation above, taking into account the magnification from object to image:

\frac{1}{2 \mathrm{NA_i}} = N_\mathrm{w} = (1-m)\, N,

where N_\mathrm{w} is the working f-number, m is the lens's magnification for an object a particular distance away, and the NA is defined in terms of the angle of the marginal ray as before.[3][5] The magnification here is typically negative; in photography, the factor is sometimes written as 1 + m, where m represents the absolute value of the magnification; in either case, the correction factor is 1 or greater.

The two equalities in the equation above are each taken by various authors as the definition of working f-number, as the cited sources illustrate. They are not necessarily both exact, but are often treated as if they are. The actual situation is more complicated — as Allen R. Greenleaf explains, "Illuminance varies inversely as the square of the distance between the exit pupil of the lens and the position of the plate or film. Because the position of the exit pupil usually is unknown to the user of a lens, the rear conjugate focal distance is used instead; the resultant theoretical error so introduced is insignificant with most types of photographic lenses."[6]

Conversely, the object-side numerical aperture is related to the f-number by way of the magnification (tending to zero for a distant object):

\frac{1}{2 \mathrm{NA_o}} = \frac{m-1}{m}\, N.

And in the above, by me marked, the f-number is corrected for the image-distance.

All formules in photography like here are based on the thin lens formule. They are good enough for the photographer. It would be something else if you are constructing an objective.

And the f5.6 photo is still better than the f16 photo.

George

[ajohnw]

I give up George. I suggest you go take some macro photographs and also get into optics at a much higher level that 1st order thin lens theory - which actually is not good enough for photographers in extremes. They are nothing more than approximations. Poor ones really especially relating to diffraction.

I sometimes get the feeling that you are trolling really.

If you want to know more about diffraction I suggest you look round here but bear in mind that as far as extended objects are concerned even a 1/4 wave error has a noticeable effect,. 1/2 wave is appalling. Sad that resolution is modelled by assuming a number of point sources making up an image but it's as good as we can get and is in real terms only a model.

http://www.telescope-optics.net/diff...berrations.htm

http://www.telescope-optics.net/diffraction_image.htm

You will need to think about how some of that relates to photography which is aimed at giving a visually acceptable final result.

John
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