calculating minimum focus distance? (300mm lens + 430mm bellows = ??)

<p>Hi All,</p>

<p>I have bellows that extend 43cm or so, and I'm wondering if there's a way to calculate the minimum focus distance (and, hence, magnification factor) for a 300mm lens (it has been many years since I last studied optics in high school physics :)</p>

<p>Is there a simple-ish equation? Or does someone know from personal experience?</p>

<p>best,<br /> MK</p>

<p> </p>

<p>Core formula is:<br>

1 / f = 1 / u + 1 / v<br>

Where f is the focal length of the lens, u is lens to subject distance, and v is lens to film distance.<br>

More properly, v and u are measured from the focal point of the lens. For a symmetric lens, that's usually smack-dab in the middle of the lens. For a telephoto lens, it's way in front of the lens -- that's the definition of telephoto. For a retrofocus lens, it's way behind the lens.<br>

Magnification is v / u.</p>

 

<p>thanks so much---this is very helpful!</p>

<p>John gave the basic focusing equation for a thin lens. That equation and equations for magnification are given on the Lens Tutorial at http://www.photo.net/learn/optics/lensTutorial. For a thick / complex lens, the same equations work, but the distances need to be measured to the principal points. For most lenses, other than true telephotos (which exist for LF photography) and true retrofocus (used with SLRs), one can obtain reasonable accuracy for most purposes by measuring to the center of the lens. If you need high accuracy, you can consult the manufacturer's datasheet for the location of the principal points. A useful measurement on a datasheet is the Flange Focal Length or Distance, which is the distance from the image to the back of the shutter (front of lens board) when the lens is focused on infinity.</p>

<p>It is also going to depend on the flange focal length of the lens. A 300mm tele requires much less bellows draw to reach infinity then a Sironar or Symmar or process lens type 300mm.</p>

<p>Bob is right, if your lens is a true telephoto - such as a Tele-Artar, Tele-Xenar or Nikkor-T - then it'll have a much shorter back-focus than a normal 300mm lens. Typically, tele lenses have a back-focus of about half their focal length, something like 160mm for a 300mm lens. So you need to measure your bellows extension with the lens focused at infinity, to find out what extension you have in hand. Once you've done that, you can then use the conjugate focii formula given above to work out the minimum focusing distance.</p>

<p>OK. Say your lens does have an infinity back-focus of 160mm; that means you have (430-160) 270mm of bellows to spare. Now we can either add 300mm to that figure to use in the formula 1/f=1/v+1/u, or we can divide the extension by the lens focal length to get the magnification directly. The magnification is simply 270/300 or 0.9, which is pretty close to life-size. Of course if we're talking about a 300mm long focus lens, then the magnification will be closer to 130/300 = 0.4333 or just under half size.</p>

<p>Hi !<br>

Just to add to the confusion :<br>

I have a Tele Xenar of 360 mm focal length. When focused at infinity, I can substitute my 210 Symmar lens and this lens is quite perfectly focused. So they share quite the same flange distance !<br>

But when I unscrew the front element of the 210 symmar which is a convertible this gives me a 370 mm focal length. I've to set the flange distance to 420 mm !!! to get this 370 lens in focus at infinity !<br>

So you better measure things if you've odd lenses ;-)</p>

<p>Simply and roughly, it takes the f/l of the lens in bellows draw to get to infinity focus. It further takes an additional f/l equivalent of bellows draw to get to 1:1 on film. With a non telephoto design (and a symmetrical design lens), if you have 130mm of bellows draw remaining beyond infinity focus for a 300mm lens, you'll get a smidgen bigger than 1:3 (1:3.1).</p>

<p>Ivan, how are you arriving at that magnification figure of 1:3.1? The magnification for any lens, telephoto or otherwise, is given by its extension from infinity focus divided by its focal length, which as I worked out above is 130/300 = 0.4333 for a 300mm lens with 130mm of extension. As a ratio this is 1:2.3, not 1:3.1.</p>

<p>If we work it out using the conjugate focii formula we get a figure of 992.3mm for the minum focusing distance, and since we know that the bellows extension is 430mm then the magnification must be 430/992.3 = 0.43333. Again this works out to a ratio of 1:2.3.</p>

<p>What I suspect has happened is that you've divided 992.3 by 300 = 3.3, and this is not the correct way to calculate the magnification. Magnification is given by subject-to-lens distance divided by total lens extension (for the thin lens model).</p>

<p>Oops-- my fingers got ahead of my mental calculator. Yup, I went the wrong way on factoring 30mm extension beyond 1:3 on film. As Rodeo Joe correctly points out, 130:300 is 1.3:3 is .43X, which indeed is a smidgeon under half life-size on film.</p>

<p> </p>

<p>Ok I am not sure I am following the math here. So if anyone is still looking at this please help me out. I have a Calumet CC400 and from what I can tell it has a 16" rail which translates to about 400mm I think. It is actually 406mm but for ease we'll call it 400. According to the formula Rodeo gave that gives a 1/3 magnification on film. On a 4x5 that means to fill the frame you are looking at an object about 12 by 15 inches. All this using a standard 300mm lens. Ok so the question based on the previous equation is, Given what I would have I would plug in the numbers 1/300 = 1/u + 1/400? So essentially 1/300-1/400 = 1/u and u would be 1200 so my minimum focus is 1200mm or just under 4 feet?</p>