I know there is already a question about EV, but it doens't quite answer my question. I understand EV as a relative measurement, 1 EV being one stop (doubling or halving the amount of light), but what is 0 EV? What does it mean when Nikon claims that a camera can autofocus in -2 EV?
The easy part is the absolute answer as usually defined.
LV is related to the the light level falling on the subject.
EV is the light level inside the camera.
EV0 is the light level at the camera to achieve correct exposure with
EV values are logarithmic so light level at EV8 = 2^8 x EV0 level = 256 x EV0 level.
So eg correct exposure at EV8 would be F16 at 1 second with ISO 100
So a given EV at the camera will have various LV levels depending on mean scene reflectance. [That trips off the tongue nicely :-) ].
Below is an EV chart, from this useful page
BUT a level of EV-2 would be correctly exposed at f1.4, 8s, ISO 100 - which is pleasantly surprising if the D800 can do it. (I think I saw that spec in the D800 announcement material few days ago).
Wikipedia is useful on LV & EV but takes some wading through.
Table 2 shows full moonlight as EV-2 to EV-3 which is consistent with the above table.
I attempted to equate the above to illuminance levels in lux (lumen/m^2) - which requires assumptions on reflectance and opens a large pandora's box. The Wikipedia discussion of calibration constants here on the "light meters" page rapidly shows that this is a well trodden battle ground with no one answer.
It seems about right to say that you get about EV0 at the camera with 1 lux of illumination and a typical subject reflectance mix. Very bright Moonlight is typically around 0.5 lux giving EV -1 by the above assessment which is only one EV off the statement cited previously - which is probably close enough to exact in this sort of area.
Other useful musings:
Ken Rockwell - EV & LV - notes that Ken's definion for LV0 is the same as what I said above for Ev0. I think my reference is the more correct one but, I may be wrong :-). He directly equates LV and EV at ISO 100 which seemeth not right.
Sekonic lux to EV table - they SHOULD know what they are talking about :-). They ay Ev0 is 2.5 Lux or about 1 EV brighter than I said above. They say:
To understand what EV 0 and EV -2 mean, perhaps it is useful to understand what it is not. It is not complete darkness or the absence of light. EV 0 is defined as f/1 @ 1 second. When manufacturers use the term to define an illuminance level, it is usually assumed that the amount of light is what is needed to produce a nominally correct exposure for that EV with a film/sensor sensitivity of ISO 100. Since the EV scale is logarithmic and each step is double or one half depending on which direction you are stepping, it means that each step is an additional power of 2.
Negative values in a exponential expression simply mean the number is the inverse of the same number raised to the same positive power. If 2^3=8, then 2^-3=1/8. Since an EV number is based on EV 0 raised to a power of 2, then negative EV numbers indicate a fraction of the base value, rather than a multiple of the base value as positive numbers indicate. If EV 3 requires 8 times as much light as EV 0 (2³=8) to produce the same exposure, then EV -3 requires 1/8 the light of EV 0 (2^-3=1/2^3=1/8) and EV -2 requires 1/4 the light of EV 0 (2^-2=1/2²=1/4) to produce the same exposure.
According to this EV 0 is defined as f/1.0 at 1 second (@100 ISO - see comment)
EV -2 equates to an exposure with an f/1.4 lens at 8 seconds, or f/5.6 at 2 minutes! Very dim indeed.