Is 1 lux unrealistically low for a cell phone camera flash? I have a light source which is similar to a typical cell phone camera flash. Based on a measurement and some calculations, I estimated that its illuminance is about 1 lux, or close to twilight values in the table here. This seems way too low since the flash hurts to look at but twilight doesn't. Now, the measurement was made about 1 foot from the source, the illuminated area is about the size of a postage stamp, and the flash duration is short. Still, when the flash fires, the illuminated area seems as bright as if it were daylight, which is about 10,000 lux here, so my estimate seems wrong by several orders of magnitude. Is that correct? 
 A: Yes, I think $1\,\mathrm{lx}$ is indeed absurdly low for a flashgun of any reasonable capacity.
A very brief intuitive argument as to why this is is that the flashes from flashguns last for under $1/1000\,\mathrm{s}$ and yet, distances of the order of $1\,\mathrm{m}$, they can light scenes brightly enough in that time to be equivalent to an exposure of $1/100\,\mathrm{s}$ of the same scene in full sunlight.

Below I have tried to put a numerical value on how bright a flash must be for a particular (non-exceptional) model.  I suspect that this contains errors, but it bears out the intuitive argument above.
Although I don't know how bright cellphone flashes are, it's possible to work out how bright camera flashes are since they publish enough information.  As an example, I took the Pentax F-360 FGZ II flashgun (manual: this was just the first I found in a page of flashes and it looks like a non-stupidly-powerful one: I presume it is brighter than cellphone flashes).  This has a 'guide number' of $36$ (in metres, at ISO $100$: see below), and at full brightness it lights for $1/1200\,\mathrm{s}$.
The first thing to sort out is how large an area the flash illuminates at a given distance, $d$.  I will assume that it covers enough area for a $50\,\mathrm{mm}$ lens on $35\,\mathrm{mm}$ film with a fudge factor of $2$ (so, twice the area).  Such a lens has an angle of view, $\alpha \approx \pi/8$ (45 degrees in old money).  So assuming a standard $3/2$ film frame, I get that the area covered, $A$ is
$$
A = 2 \times \frac{24}{13}\left(d\tan\left(\frac{\alpha}{2}\right)\right)^2
$$
Where the  $24/13$ comes from fitting a film frame into the image circle of the lens, and the $2$ is the fudge factor (the flash covers some rectangle which is going to be significantly bigger than the film frame or there will be vignetting), which is probably really quite a lot larger than $2$.  And I may have made mistakes in this but it's good enough.
So now, rather than think about lumens or lux which will require knowing the albedo of whatever the camera is looking at, I will work out how bright the flash must be compared to full sunlight.  This is helped by knowing the famous sunny 16 rule, known to all photographers of a certain age:

in full sunlight, you get reasonable exposures with a film of ISO speed $S$ by exposing for $1/S\,\mathrm{s}$ at $f/16$.

And now we come to the nightmare of flash guide numbers, which are not numbers but quantities with dimensions.  The basic definition of a guide number, for a given film (or sensor) speed $S$ is that
$$
G_S = \frac{d}{a}
$$
where $a$ is the aperture of the lens.  $a$ is dimensionless (it's the ratio between the focal length and the effective diameter of the lens), so the units of $G_S$ are length.  This definition works because the light that gets into the lens goes as the square of the aperture $a$, and the amount of illumination falls of as the square of distance $d$.  And really I should factor in $S$ here but I've just defined $G$ at a given speed, $S$.
So for the flash referenced at the top of this answer, we know that $G_{100} = 36\,\mathrm{m}$, and since we're interested in $f/16$ because we know the sunny 16 rule, we get:
$$
\begin{aligned}
d &= \frac{G_{100}}{16}\,\mathrm{m}\\
  &= \frac{36}{16}\,\mathrm{m}\\
  &= \frac{9}{4}\,\mathrm{m}
\end{aligned}
$$
In other words, as $f/16$ & using ISO $100$ film, this flash will light something a bit over $2\,\mathrm{m}$ away.
But it does this in $1/1200\,\mathrm{s}$, and in fact I will add another fudge factor of $2$ here, because the flash isn't a perfect top-hat function, it will ramp up and down.  So I'll assume that it does it in $1/2400\,\mathrm{s}$.  In other words, at this distance, it lighting the scene $24$ times more brightly than the Sun, because the Sun takes $1/100\,\mathrm{s}$ to do the same thing.
And finally we can work out the area it must illuminate from above: we now know that $d = 9/4\,\mathrm{m}$ and $\alpha = \pi/8$:
$$
\begin{aligned}
A &= 2 \times \frac{24}{13}\left(\frac{9}{4}\tan\left(\frac{\pi}{16}\right)\right)^2\\
  &\approx 0.74\,\mathrm{m^2}
\end{aligned}
$$
So based on the lux table here, full sunlight might be $50\times 10^3\,\mathrm{lx}$, so this flashgun must be putting about $1200\times 10^2\,\mathrm{lx}$ into an area of about $0.74\,\mathrm{m^2}$.
I'm kind of uncomfortable with this working, because I'm unwilling to believe flashguns are that bright (note there are two fudge factors here which could reduce things by a factor of $4$ though).  But I think it is very clear that, yes, $1\,\mathrm{lx}$ is absurdly low as the brightness of a flashgun.
