The basic idea is that, once you have figured out your image magnification (more about that later), the exposure factor E can be computed as follows:
M = magnification
E = (M + 1)(M + 1) = (M + 1)square
The table gives some examples. The first column gives the magnification M, and the second column gives the corresponding exposure factor. By multiplying a given shutter time with the exposure factor, you obtain the shutter time you have to use in order to compensate for the loss of light. The third column gives an example. If normally (in the given light situation) you would use 1/1000 as shutter time, with a M of 0.4 you have to double the exposure time to 1/500.
M (M+1)² 1/1000 : (M+1) f/8 : stops 0.1 1.21 826 1.10 8.8 0.3 0.2 1.44 694 1.20 9.6 0.5 0.3 1.69 592 1.30 10.4 0.8 0.4 1.96 510 1.40 11.2 1.0 0.5 2.25 444 1.50 12.0 1.2 0.6 2.56 391 1.60 12.8 1.4 0.7 2.89 346 1.70 13.6 1.5 0.8 3.24 309 1.80 14.4 1.7 0.9 3.61 277 1.90 15.2 1.9 1 4.00 250 2.00 16.0 2.0 1.1 4.41 227 2.10 16.8 2.1 1.2 4.84 207 2.20 17.6 2.3 1.3 5.29 189 2.30 18.4 2.4 1.4 5.76 174 2.40 19.2 2.5 1.5 6.25 160 2.50 20.0 2.6 1.6 6.76 148 2.60 20.8 2.8 1.7 7.29 137 2.70 21.6 2.9 1.8 7.84 128 2.80 22.4 3.0 1.9 8.41 119 2.90 23.2 3.1 2 9.00 111 3.00 24.0 3.2
This is all very useful when you are working with available light, but it becomes somewhat cumbersome when you want to use a flash, and you don't have a TTL-flash mode on your camera, or for one reason or another don't want to use the TTL-flash.
When using flash light as the primary source of light, shutter time has no role whatso-ever (provided you stay below or at the synchronisation time). With flash light, we are first and foremost interested in apertures, and we have to compensate the light loss by changing the aperture value and/or changing the power/distance of the flash.
Of course, to return to the example, if working with a M=0.4 implies that 1/1000 becomes 1/500, we can deduce that M=0.4 needs a compensation of one stop (either by halving the shutter time, or by opening the aperture one value).
However, this can be done in a more straightforward way. In the fourth column of the table, we calculate (M + 1) instead of (M + 1)(M + 1).
Now, that is convenient. By multiplying the working aperture value (as indicated on our objective), say f/8, with (M+1), we obtain the effective aperture for the given magnification. The fifth column gives some examples for aperture f/8. An aperture of f/8 with a magnification of M=0.4 gives an effective aperture of f/11, basically what we could expect from the previous example.
Yet, the effective aperture values will not always be a practical solution in applied work. Better objectives may allow you to choose the aperture in ½ or even 1/3 of a stop, but even then it would be far more convenient to have the relationship between magnification and exposure compensation directly in stops, instead of having to decide whether a value such as 18.4 is closer to f/16 + 1/3, or is closer to f/16 + 2/3.
Well, this can done. The next graph gives the exposure compensation in stops for magnifications between 0.1 and 10. This is definitely more convenient : at a M equal to 10, the exposure factor is (10+1)(10+1)=121, obliging you to do some mental calculations when you actually only wanted to make pictures. In my opinion, it is much more easy to know you have to compensate for 7 stops in that case. It is also a much more general approach, since it allows you to compensate either by reducing time or by opening the aperture, whatever comes in handy.
In order to use these graphs, it is not necessarily to know how they are obtained, so feel free to skip the next section. However, for those, like me, who don’t accept without (some) understanding, here is how I calculated them (the formulas are also convenient if you want to reproduce the graphs in your own spreadsheet).
When you look at the second column in the table, and concentrate on the bold figures, you will notice that when you loose a stop of light, the exposure factor E goes consecutively from 2 (well, 1.96) to 4, to 8. These are all powers of 2. What you get there is that the power to which you have to raise the number 2 in order to obtain the exposure factor E for a given M, is in fact the number which says how much stops you loose for that given magnification. If this number of stops is denoted as S, and using some (long forgotten ?) highschool maths, we get :
E = 2^S (2 to the power S).
log 2 (E) = S
By converting this to a more common base,
log 2 (E) = log 2 (e)* log e (E)
and some rearrangements, we get
S = (1/ln(2))*ln(E)
which can be readily introduced in most spreadsheet software.
How to determine the magnification M ?
In 35mm photography, the negative will be 24mm times 36mm. So if you are wondering what magnification you are working with, just hold a ruler in the position that is in focus, and count the number of mm on the ruler that are visible on your ground glass. If you only see 18 mm (with the camera and the ruler in the same horizontal position), the implication is that you have a M equal to 36/18, i.e. 2/1.
However, don’t forget that there are only a few cameras that really give you a 100% view on the ground glass. Consult your manual. If you e.g. only get a 82% view, you will have to reduce the 36mm accordingly.
Some notes about the limitations of the above formulas
The formula which states that exposure compensation is equal to (M+1)(M+1) is in fact only valid for symmetric objectives. Loosely speaking, an objective is symmetric if the aperture you see when looking to the front and looking to the back, is approximately equal. This is generally the case for normal focal distances (i.e. the neighbourhood of 50mm in 35mm-photography). This definitely is not the case with wide-angle objectives (at least if they have a retro-focus design, which is most of the time).
If the objective is not symmetric, one has to take into account the pupillary magnification, and the formula for exposure compensation is then as follows :
E = (1 + (M/P)) square
where P is the ratio of the (diameter of the) exit pupil over the (diameter of the) entrance pupil. What does this mean ? With a symmetric objective, the exit and the entrance pupil will by and large be the same, implying that P is approximately equal to one. M divided by one remains M, so the P-factor can be discarded.
If however I make a visual inspection of my 28mm wide angle objective, I can see that the entrance pupil (the side of the object) is markedly smaller than the exit pupil (the image side) (note that the P-ratio is independent of the aperture value chosen). I « guestimate » that the ratio is something like 2/1.
At first sight this seems promising, since dividing M by a P equal to 2 will substantially reduce the exposure factor, suggesting that the high magnifications you can obtain with a wide angle 28mm objective (e.g. M=5 with « only » 14 cm bellows extension) won’t become too dark, relatively speaking.
Until you realise that in such a set-up, you have to reverse your wide angle, thereby reversing exit and entrance pupil. For a M equal to 5, this will give an E equal to (1+10)(1+10)=121, which, as we saw a few minutes ago, corresponds with the exposure factor that in a symmetric objective only occurs at a M of 10.
I have to admit that I have never used a wide angle on bellows, so my conclusion is entirely based on what I « read » in the formula. If someone could corroborate the conclusion on the basis of personal experience, that would be very nice.
In the end, notwithstanding all these nice formulas, it will always remain necessary to do your own tests with your personal material. At least the formulas will give you a clue as to where to start, so you don’t have to start in complete darkness.
