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I found this article on hacking macro lenses. The author used a lens extension tube to increase the magnification of a standard lens into a macro lens. To my surprise the lens's Minimum Object Distance (MOD) was also significantly reduced. The photos he took appear to be slightly distorted and have shallow depth of field. Will this happen with all lenses and extensions? Will resolving power be reduced?

Here's how I would like to apply this concept. I am trying to hack a close-focus machine vision industrial camera using cheap parts. The camera needs to focus on a 4" x 5" region with a working distance of 4"-5" inches between subject and lens. I used an online calculator to determine that I need a 5-6mm focal length camera lens to maximize the subject in the photos. The problem I'm running into is that most lenses have a MOD greater than 8". So I would like to instead use a 3mm lens with a 2.5mm extension. Could this work?

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closed as off-topic by Gert, Sebastian Riese, Kyle Kanos, Qmechanic Dec 4 '15 at 0:40

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Generally, an extension ring does not affect the "magnification" of a lens, although it does increase the magnification of a focused image of a given object. Nor the focal length, either. What it does is to allow closer focusing, which allows the lens FOV to enclose a smaller object area. If you need a 6 mm lens, adding an extension ring to a 3 mm lens will not necessarily help. The relevant equation is the Lens Equation $$\frac{1}{Object Distance} + \frac{1}{Image Distance} = \frac{1}{Focal Length}$$By increasing the image distance you can decrease the object distance. Since the magnification M follows $$M =- \frac{ImageDistance}{ObjectDistance}$$ increasing the image distance with an extension ring provides greater magnification of a given object when it is in focus. In order to determine what will work, you need to know a) the focal length of the lens, b) the desired working distance, c) the final image distance, including the extension, d) the image sensor size, and e) the desired object size. The first 3 must be related to the last two to determine whether the system will do what you want.

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  • $\begingroup$ Since an extension ring is placed between the lens and sensor, I thought focal length would become the sum of the lens's native focal length and the extension ring length. But I now understand that focal length is determined by the curvature of the lens. I'm going to crunch some numbers and see if I can solve this. hyperphysics.phy-astr.gsu.edu/hbase/class/phscilab/imagei.html $\endgroup$ – skibulk Dec 4 '15 at 0:44
  • $\begingroup$ If I plugin the following figures to the equation, it appears that I actually need to shorten the image distance of the lens (It's back focal length is too deep. I suppose I would have to cut some off.) 1/115mm + 1/X = 1/6mm which solves to x= 6.33mm. The lens comes with a default focal length of ~10mm. $\endgroup$ – skibulk Dec 4 '15 at 1:06
  • $\begingroup$ @skibulk - Sorry, but 10mm cannot be a focal length, default or otherwise. That is 6 mm. I assume you mean the effective lens-to-sensor range (which is not exactly the physical distance) is 10 mm. $\endgroup$ – WhatRoughBeast Dec 4 '15 at 1:43
  • $\begingroup$ You are right. I mean to say the back focal length is ~10mm, by default $\endgroup$ – skibulk Dec 4 '15 at 1:48

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