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Thread: Relative magnification math - need help

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    Relative magnification math - need help

    Hi everyone... I have been a lurker for some time, glad I finally joined!


    I have a magnification question regarding magnifiers / reading glasses, etc. I think what I present below is accurate, but I have been fooled many times by "relative magnification" issues, so any input or confirmation would be greatly appreciated.


    I am trying to reconcile these two scenarios.

    1) I am using a 50mm fl film loupe to view photographic slide film on a light-box. The film is 55mm square, the loupe sits right atop the film. Pretty simple...

    2) This film is then digitized, and eventually must go back to film, which is not an easy process. So as an intermediate close-inspection step, the original 55mm sq. image is printed on back-lit film on an inkjet printer. The proposed print size is 10" square. It will be viewed on a light box through a 2" wide round pipe. They eye to print distance is 5.7" and the person inspecting the image will obviously need reading glasses. This exercise is to allow the user to inspect the image at a greater magnification than the first scenario above.

    So, I am trying to reconcile - How does the magnification of these two scenarios compare to each other? Here is how I calculated such...

    Scenario 1 - Using the 25cm reference magnification distance, I will conclude the film loupe is magnifying the 55mm square film 5x (250/50mm fl). The film loupe (magnifier) sees the entire film area without moving the loupe.


    Scenario 2 -

    10 inches sq. inkjet print / 55mm film image (2.1") = 4.7x magnification of the original film.

    The user views a 2" diameter circle of this 10" square image, through a 2" wide pipe at 5.7" eye to image distance.

    So 10"/5.7" = 1.75x added magnification vs. 10" view distance.

    These two magnifications combined, 1.75 x 4.7 = 8.3x total magnification.


    So my conclusion is, when the user is looking through the film loupe at a given area for close inspection, they will see a given area at 5x magnification, but when viewed through the 2" wide pipe at 5.7" eye to image distance, they will see the same area at 8.3x. So the inspection process has added 1.7x added magnification (8.3 / 5).

    The person viewing would require the following diopter reading glasses...

    10" / 5.7" = 1.75x

    4 diopter reading glasses will allow a person to focus at 10", so I am assuming, 7 Diopter reading glasses will allow the person to focus at 5.7" (4 x 1.75). Is this right?

    I am also assuming that these 7 diopter reading glasses will provide NO additional magnification, as they do NOT act as a true magnifier, but rather magnify through proximity, i.e., I have accounted for their added proximity magnification by the 10" / 5.7" component above. Do I have this right?

    If I wanted to place an achromat doublet atop the 2" pipe, so people do not have to wear reading glasses, I am assuming it will require a 143mm fl, 250/143 = 1.75 magnification. Is this accurate?

    Some may wonder why we are looking through a round pipe - this is another long story, which is not very Germain to the magnification comparison issue, so I figured it was best to avoid confusion. TYIA

    Any other "gotchas" or "bewares" please feel free to comment.

  2. #2
    Master OptiBoarder Darryl Meister's Avatar
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    Here is a short article I wrote on the magnification of simple vision aids, which might be of assistance. I should note that the article deals primarily with the magnification of single plus lenses of negligible thickness that are positioned at or within the primary focal length of the lens, as is typically the case for low vision aids and loupes, for instance.
    Attached Thumbnails Attached Thumbnails Magnification of Simple Low Vision Aids.pdf  
    Last edited by Darryl Meister; 01-06-2008 at 08:29 PM.
    Darryl J. Meister, ABOM

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    Darryl, keep trying to download the file, but no luck?

    can you try it?

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    Master OptiBoarder Darryl Meister's Avatar
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    I was able to download it, myself, but I'm happy to e-mail you a copy if you want to send along your e-mail address...
    Darryl J. Meister, ABOM

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    Thank you Darryl, I assume it was a broadband problem on my end, cause it downloaded perfect today....

    Now, I have to digest it. At first glance, it looks like a very interesting paper..... and your approach is obviously more sophisticated vs. the simplistic approach I presented.

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    Master OptiBoarder Darryl Meister's Avatar
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    At first glance, it looks like a very interesting paper..... and your approach is obviously more sophisticated vs. the simplistic approach I presented
    It just discusses the basic principles and calculations involved.

    In your particular situation, you just need to calculate the magnification of the loupe at the intended viewing distance in the first case (5.0), the magnification of the spectacle lens at the intended viewing distance in the second case (1.7 for a +6.91 D lens), and the magnification achieved by scaling the original object size by printing it larger (4.6), which should then be multiplied by the magnification of the spectacle lens. Your final relative magnification then becomes Spectacle Mag x Object Mag / Loupe Mag, or roughly 1.7 x 4.6 / 5.0 = 1.6 times greater.
    Darryl J. Meister, ABOM

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    Thank you Darryl, I was buried in your white paper when I got this response! OK, so it seems we are on the same page here, I appreciate the confirmation.

    Could you advise if this is correct also....

    > If I wanted to place an achromat doublet atop the 2" pipe, so people do not have to wear reading glasses, I am assuming it will require a 155mm fl, 250/155 = 1.6x magnification. Is this accurate?

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    Master OptiBoarder Darryl Meister's Avatar
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    Yes, a viewing distance of 5.7" (0.14478 m) would require a focal power of 1 / 0.14478 = +6.91 D, if the object is positioned at the focal length of the lens. Compared to a reference distance of 25 cm, this represents a magnification of 6.91 / 4 = 1.7. You will also need to ensure that a 2" aperture with a +6.91 D lens will still allow you to see enough of the magnified object at 5.7".
    Darryl J. Meister, ABOM

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    thanks again for the confirmation Darryl...

    I will begin experimenting...

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