## Focussing distance? When a focussing scale is intermediate between two marks, how do you calculate the proportional value? In the photo on the right, the distance mark is about 0.75 of the way from 3 metres to infinity: what distance is it focussed on? Click the Example button below to answer this question.

To use the calculator, enter the lower distance mark in the left box, and the further distance mark in the right box, then enter the relative position as a proportion of the way from the nearer mark to the further one. For infinity, click the button. (You can enter "Infinity", but it must have a capital 'I') This works for any units (metres, feet, parsecs, or ells), as long as both marks are on the same scale. I don't check anything, so if you enter a proportion outside the range 0 to 1, you can extrapolate as well as interpolating, but see below for how reliable this is. "NaN" is Javascript for "Not-a-number", and means someone has done something silly.

 Nearer mark Further mark - - - - - - - - - - - - Click intermediate button, or enter proportion value Proportion: Calculated focussing distance:
##### "Everything to do with lenses works in reciprocals. That's all you need to know, really."

Look at the three equally-spaced marks visible in the window of the lens above: 1.5 - 3 - Infinity (metres). Why these values? Because if you take reciprocals, they are equally spaced apart: 1/1.5=2/3 - 1/3 - 0. So to find the value for halfway between the 3 and Infinity marks, we take the mid-point of the reciprocals (1/3 and zero, making 1/6) for a focussing distance of 6 metres. Three-quarters of the way corresponds to the reciprocal between 1/6 and zero, that is 1/12.

##### And is this correct? Is it accurate?

It's a pretty good approximation. Do we need this calculation in practice? Well, not often, but someone asked this question on photo.net because they had no intuition as to any values intermediate between (say) 3 metres and infinity. In such cases it's rather useful. (Most responses, by the way, concentrated on telling the questioner that this was the wrong question to ask, and similar evasions, although one person attempted to find the "average" of 3 metres and infinity by claiming 100[?] times the focal length as the Official Value of infinity.)

But let's just work out how accurate the calculator is, or rather, the approximation it requires to be reasonable. First of all, notice that the focussing marks are around the barrel of the lens, but the focussing mechanism normally depends on a conventional screw thread, so the (rotational) distance between marks is indeed proportional to the (longitudinal) movement of the lens. So here's the justification for this "reciprocal averaging". Suppose we have two sets of subject distance (u, u') and image distance (v, v') for this lens: this gives us two 'copies' of the standard lens formula:

1/u + 1/v = 1/f
1/u' + 1/v' = 1/f

Since the left hand sides are equal, if we add them we get 2/f, and if we take the mean of the left hand sides, we get 1/f again:

(1/u + 1/u')/2 + (1/v + 1/v')/2 = 1/f

So the new values on the left also fit the formula, and give us a new focussing point. The value here is what we used for the calculation for the distance to subject, but unfortunately not quite what we used on the image distance (i.e. the lens position), because we took the simple average. So effectively, we're assuming that:

(1/v + 1/v')/2 approximates 1/((v+v')/2)

But of course, this is a reasonable approximation if we are talking about adjacent marks on a focussing scale. Suppose the two marks represent distances 50 and 52 mm from the film plane: then the left (LHS) and right (RHS) hand sides are given by:

LHS = (1/50 + 1/52)/2 = (0.0200000 + 0.0192307)/2 = 0.0198153
RHS = (1/((50+52)/2)) = 1/51 = 0.0198078

These are close! We need six significant figures to see any difference. But of course, if you try to extrapolate (entering a proportion value of 10, or the opposite ends of the focussing scale of a macro lens), this breaks down.

Just for the exercise, let's consider a 50mm lens, with focussing marks for 5 m and Infinity, and find the exact focussing distance for the midway point. We'll subtract the 50mm from the lens-to-subject distance to account for the fact that camera manufacturers measure the focussing distance not to the lens, but to the film plane (but we'll forget the extra fraction of a millimetre!) So u' = 4950. Now calculate v', the lens-to-film-plane distance for the 5 m focussing mark.

1/4950 + 1/v' = 1/50
So 1/v' = 1/50 - 1/4950 = 99/4950 - 1/4950 = 98/4950
So v' = 4950/98 = 50.510204

Obviously, the infinity mark means v=50, so the position midway between the two marks (we'll call v") is:

v" = (50.510205 + 50) / 2 = 50.2551

So now find the precise focussing distance u":

1/u" + 1/v" = 1/50
1/u" = 1/50 - 1/50.2551 = 0.02 - 0.0198984 = 0.0001016
So u" = 1/0.0001016 = 9842.6

And to get the "official version", the distance to the film plane, we add the 50mm back on, and get the Answer:

Midway focussing = 9842.6 + 50 = 9892.6mm, just under 11 cm short of the 10 metres the calculator gives. Conclusion: it works! 