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Calculate scale using extension tubes with 50mm lens


jen_martin1

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<p>Hi all,<br>

<br />I am a grad student getting ready to do field work in just over a week. Part of my work includes using extension tubes to take close-up images of sediments as they are exposed in a geologic outcrop. </p>

<p>I have a Fuji S-2 Pro camera with a Nikon 50 mm lens (made for a film camera, it was purchased about 15 years ago) and I will be using the standard Kenko extension tubes (also purchased about 10-15 yrs ago, originally for a 35 mm film camera). Is there a way to calculate the scale of a full-frame image given the lens, camera body, and a particular tube or combination of tubes? <br>

<br />Thanks so much for your help,<br>

<br />Jen</p>

 

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<p>Why don't you photograph a millimeter scale with the various tube combinations, and determine the scale by the size of the field of view captured with each setup. It would only take a few minutes, and the data would be correct for sure. It's a better way than calculation, since the lens focal length may not be precisely what is printed on the lens.</p>
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<p>Here is a jpeg of a chart that gives scale (and exposure for before-TTL days). It should be printed at 150 ppi - that is 4.52" by 9.4" -</p>

<p>It's from <em>Modern Photography</em> 1967-04 p. 73</p><div>00bniM-541172484.thumb.jpg.df9270cf08f883fbfa8d3fa899c5d976.jpg</div>

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<p>Hi Jen. I'm not sure if this is the question you're asking, but... Assuming the 50mm lens acts like a "thin lens" (which may or may not be terribly valid here especially if the lens has floating focus elements, but it's a start), the "<a href="http://en.wikipedia.org/wiki/Thin_lens_formula#Imaging_properties">thin lens formula</a>" says that:<br />

<br />

1/(lens-to-subject distance) + 1/(lens-to-image distance) = 1/(lens focal length).<br />

<br />

That will tell you how far the (nodal point of the) lens needs to be from the sensor in order to focus at a given distance. Put a possibly-more-useful way around:<br />

<br />

lens-to-subject distance = 1/(1/(lens focal length) - 1/(lens-to-image-distance))<br />

<br />

So, for example, a 50mm lens that's 50mm from the sensor is focussed at infinity (1/(lens focal length) - 1/(lens-to-image-distance) = 0). A 50mm lens that's 100mm from the sensor is focussed 100mm from the lens (1/50 - 1/100 = 1/100) - so this is how far away you'll be with the lens focussed as if it were at infinity, but after you've added 50mm of extension tubes.<br />

<br />

As for how large the image is, take the sensor size (<a href="http://www.dpreview.com/products/fujifilm/slrs/fuji_s2/specifications">23x15.5mm</a> in the case of your S2Pro) and scale it by the ratio of the distance between the lens and subject and the distance between the lens and camera. In the above case with 50mm of extension, the lens-to-subject and lens-to-sensor distances are the same, so you're at 1:1. Add another 50mm of extension and you'll be focussing 75mm from the lens (1/50 - 1/150 = 1/75). With the subject 75mm from the lens and the sensor 150mm from the lens, you're at 2:1 magnification (150/75 = 2), and you'll be looking at an area 11.5 x 7.75mm (23/2 and 15.5/2).<br />

<br />

If, instead, you add 25mm of extension tubes to the lens, you'll be focussing 150mm away (1/50 - 1/75 = 1/150). At this distance, you'll capture an image <i>twice</i> the size of the sensor - 46mm x 31mm.<br />

<br />

Again, all this is with the lens set to "infinity focus". You can get closer if you focus with the lens as well.<br />

<br />

I hope that helps. Have fun with a calculator!<br />

<br />

(And JDM: Cool chart!)</p>

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<p>Different combinations of extension rings with scale 50mm camera lens.<br>

Subject distance 6 inches (152.4mm) to 38 inches (965.2mm).<br>

Magnification life-size 1X to 0.1X<br>

<a href="http://books.google.com/books?id=gClySvAhrzAC&pg=PT128&dq=bellow+allow+you+to+maximum&hl=en&sa=X&ei=F4LYUZXiM-fi0QHZ_IDoAg&ved=0CC0Q6AEwAA">http://books.google.com/books?id=gClySvAhrzAC&pg=PT128&dq=bellow+allow+you+to+maximum&hl=en&sa=X&ei=F4LYUZXiM-fi0QHZ_IDoAg&ved=0CC0Q6AEwAA</a></p>

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