Metabones commercialize it for adapting SLR lenses to mirrorless, but didn't invent the focal reducer. Astronomers have been using them for quite some time, and Enterprising DIY users were toying with them and SLR lenses/cameras as well.
A lens is circular, and will produce a circular image (usually - funny baffles and built in hoods not withstanding) - hence it is "creatively" called the "image circle". The focal reducer will make the size of this circle smaller (while keeping minimum and maximum focus distance constant).
In making the circle smaller, it packs rays of light more tightly together. The effect of this is it will make the image brighter, and can actually increase the MTF (think sharpness of details) of the lens on that camera.
The other effect is since the image circle is smaller, parts of the image circle (if it was large enough) would come into view, making the lens appear wider.
There's 2 ways to think about it. You can think of it as a wider, faster lens. Or you can think of it as making your camera sensor larger. The camera will see the lens and focal reducer as ONE UNIT. Both of them together is just a lens to the camera. If we look at a focal reducer on the camera from the lens' perspective, light rays just see a bigger sensor. This is also evident if you just look down at the camera. Here's a picture to explain that:
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| The lines are to show dimension. you can try counting pixels if you really want. you'll find x/x' and y/y' both roughly equal 0.72 |
For example - my 55mm F1.2
I'm using a focal reducer on my Micro Four Thirds camera, of 0.72x
The crop factor is 2 (1/4 the area)
The equivalent angle of view compared to full frame is 110mm.
The equivalent depth of field compared to full frame is 110mm @ F2.4
Light gathering stays F1.2 - but smaller sensors are getting less light, so raising ISO (sensitivity) generally introduces noise much faster.
You'll see tons of arguments one way or the other. These are just numbers. They're neither good nor bad. Crop sensors (sensors smaller than fullframe) exist for a reason.
Anyway, so my 55mm F1.2, effective looks like a 110mm F2.4 on a MFT camera (but light gathering stays F1.2).
Putting on a focal reducer (0.72x) my 55mm F1.2, becomes a (55 x 0.72)mm F(1.2 x 0.72) lens - i.e. 39.6mm F0.864!
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| With and without the focal reducer - think of it as changing the lens to something wider and faster! This was a cheap generic chinese adapter selling for $100 on ebay. |
The other way to think about a focal reducer is it makes the sensor effectively larger from the perspective of the lens. Lets do the math.
My crop factor 2 becomes a crop of 2 x 0.72 or 1.44. That's even bigger than APS-C sensors.
Assuming the crop of 1.44, the calculations are applied to each lens used - (55 x 1.44)mm F(1.2 x 1.44) = 79.2mm F1.728.
Light concentration gain? Still have to consider sensor original size or the focal reducer strength here F(1.2 x 0.72) or use the effective aperture and divide by original crop factor = F1.728/2 = F0.864.
I've expressed light gain in terms of aperture. Here's the math to understand what that means in terms of actual light.
change in light = (old F-Stop/new F-Stop)² - I'll explain later
So the F0.864 light gathering compared to F1.2 is (F1.2/F0.864)² = 1.93 times brighter. In other words, we're a stop brighter - effective ISO is doubled.
An alternate way to calculate would be direct from the focal reducer parameter of 0.72x. (1/0.72)² = 1.93.
Finally, here's the math to understand "change in light = (old F-Stop/new F-Stop)²"
lets say we're using a lens of focal length 100
the aperture is a particular size say, 50mm, so our F-stop is F2.
Light coming in is proportional to the area of the aperture.
Area = πr²
since we're working with diameter and diameter is 2x radius:
Area = (1/4)πd²
light concentration is proportional to aperture area, therefore:
Concentration = k(1/4)πd²
Since k is arbitrary, let's wrap up pi and 1/4 into a general constant:
Concentration = jd²
F-Stop = Focal length[L]/aperture[d]
so...
d=L/F
Now lets relate light concentration with F-Stop
Concentration = j(L/F)²
Since L is a constant for a given prime, let's wrap that into the other constant.
Light concentration = k/(F²)
i.e the amount of light is inversely proportional to the square of the F-Stop.
Comparing new and old F-stops:
to show concentration gain = (old F-Stop/new F-Stop)²
Light concentration new = k/(F'²)
Light concentration old = k/(F²)
The amount of light I'm getting compared to before is
Light concentration new / Light concentration old
i.e.
[k/(F'²)] / [k/(F²)] = F²/F'²
Where F is old F-Stop and F' is new F-Stop.
Basically, one f-stop is twice as bright as another, if it's 0.707 the value of the other (1/√2).
FUN RIGHT!!!!!
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| The 55mm F1.2 on my EP3 using a cheap chinese focal reducer |
Generally F1.0 and faster lenses cost an arm and a leg for the first installment. This adapter gives me an effective ultrafast lens adding only $100 (adapter cost). The Canon SSC FD 1.2 is about $350 (usually less) on ebay. A brass mount replacement to EF is about $130, so about $580 for a ~40mm F0.85. Very useful for getting less depth of field at a comfortable angle of view, and of course doubling light gain. Cheaper to go Nikon here normally.
Another use is taking good APS-C lenses, and making them incredibly wide on Micro Four Thirds - e.g. the 11-16mm Tokina, will be reduced to ~8-11.5mm, giving a 16mm equivalent wide end. Sigma's 8-16mm would become an astonishing 11.5mm. Of course this would only work well if the lenses are capable of manually being focused. Aperture works with manual lenses since they're designed with the aperture ring.
I'll be playing around with this in the future, but so far, I'm surprised with the optical quality. I don't think details are diminished at all, and I've not yet seen bad flare. Sharpness of this combo is good, but doesn't compare with the 25mm F1.4 Leica that was designed for MFT from the start. It does however make it look like a dark lens by comparison, which is an amazing accomplishment.
