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Thread: Close-up filters, extension tubes, distance and magnification

  1. #1
    darekk's Avatar
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    Dariusz Kowalczyk

    Close-up filters, extension tubes, distance and magnification

    What is magnification of such triple combination:
    100 mm lens with distance on the ring set to 50 cm
    + 30 mm extension tube
    + close-up filter 4dpt ?
    I know one method of calculations, but not sure if it is correct. It gives 0.55 without the filter and 1.17 with the close-up filter.

  2. #2
    FrankMi's Avatar
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    Frank Miller

    Re: Close-up filters, extension tubes, distance and magnification

    Hi darekk, as this sounds like an odd request, I am curious as to why you ask?

    If you want to see how well it will perform when you take a picture, I'd say just go ahead and take a test shot and see. If you need the answer for a test, most likely it is in the training materials you were provided.

  3. #3

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    Hero

    Re: Close-up filters, extension tubes, distance and magnification

    Two things to keep in mind. 1) the shorter the lens the less useful a close-up lens becomes 2) same goes for longer lenses and extention-tubes.
    That said, this link might help you out figuring out what works and how well: http://eosdoc.com/jlcalc/

  4. #4

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    robert

    Re: Close-up filters, extension tubes, distance and magnification

    I wish i had had Hero's helpful magnification formula when i bought the extension tubes for my 70-200mm F4L. I like to discover the wonders of the macro world & attempt to capture its images. I have a tack sharp lens & wanted to see if i could expand its utility into macro using extension tubes since i had exhausted the use of my canon 250D 4X diopter added to the front of the lens. The limited exten tube instructions had mag values for a 50mm lens, but that didn't assist me, so I methodically shot a test pattern i conceived at 70mm,100mm,135mm lens focal settings to determine both the lens to subject distance & the image magnification that resulted using varied tube combinations for the above lens settings, i even added the 250D diopter to the combos to see if it was viable for greater than 1:1 image reproduction. I recorded the results on a card & laminated it, this way i have lens specific data as a starting point for what i'm trying to accomplish.

  5. #5
    darekk's Avatar
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    Dariusz Kowalczyk

    Re: Close-up filters, extension tubes, distance and magnification

    This is not needed for anything in practice ...

    β - magnification of the camera lens alone (+ eventually extension tubes). For the object distance set to infinity equals 0.
    βtotal - magnification of the combined system camera lens - close-up filter
    s' - image distance of the camera lens or combo (from the lens to the film or sensor)
    Δs' - extension of the image distance (for example extension tubes + extension resulting from setting object distance in the camera lens other than infinity)
    f - camera lens focal length
    fclose-up - close-up filter focal length
    ftotal - focal length of the combo lens - close-up filter
    Dclose-up - optical power of the close-up lens(es) [dpt]
    The optical power is additive.

    fclose-up = 1000 / Dclose-up [mm]
    ftotal = ffclose-up / (f+fclose-up)
    ftotal = f1000 / (fDclose-up + 1000) [mm]


    According to such book I have, the image distance (s') of the combined system lens + filter is the same as for the lens alone (what is a little suspicious). If this is true

    βtotal = s'/ftotal - 1
    s' = f + Δs'

    where s' - image distane of the camera lens alone (you can calculate it i.e. from the thin lens equation for the camera lens without the close-filter)

    Based on that assumption (regarding the image distance), you can derive stream of equations, for example:

    βtotal = Dclose-up/1000f(β+1) + β

    (this is my equation, but probably correct if the image distances are the same)

    β ≈ f / (object distance set on the lens - f)
    β = Δs' / f

    Example 1:
    f = 100 mm
    β = 1 (1:1)
    Dclose-up = 1 dpt

    βtotal = 1/1000100(1+1) + 1 = 1.2

    Example 2:
    f = 50 mm
    β = 0.1
    Dclose-up = 5 dpt

    βtotal = 5/100050(0.1+1) + 0.1 = 0.375
    Last edited by darekk; 15th November 2011 at 03:28 PM.

  6. #6
    Mark Lawrence's Avatar
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    Re: Close-up filters, extension tubes, distance and magnification

    I think you are getting way to technical.Over the top.
    Just use your set of tubes and lenses in various configurations and have fun,you might be impressed by what you find.

  7. #7
    herbert's Avatar
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    Re: Close-up filters, extension tubes, distance and magnification

    Lens optics are pretty complex. The focusing design has a big effect. For instance an interanl focusiing lens reduces its focal length as you move from infinity to close objects. As such the formulas vary with each lens. They are a good approximation but for the actual reproduction ratio you will have to measure it yourself (unless you can get a chart from the manufacturer).

    The reproduction ratio (or magnification) is a measure of how big the object is in real life compared to how big it is on your sensor. Therefore you can work it out practically using an object of known size. Thus measuring the reproduction ratio requires a ruler and some patience.

    Take photos of the ruler at different focussing distances. Then see what size the ruler is in your image and compare it to the sensor size.

    Magnification = size of sensor (mm) / ruler coverage (mm)

    E.g. If I can see 33mm on my ruler and the sensor size is 22mm then:

    Magnification = 22 / 33 = 2/3 = 0.667x

    You can find out some common sensor sizes here:

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

    If you do this for all your combinations and make a chart you will have a nice reference for labelling your images. Now you just need to take some images.

    Alex

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