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I just bought some UV LEDs
I bought them for testing banknotes and identity papers. I read the documentation and saw some warnings about UV light in which the vendor advises not to look directly at the light and not to expose skin to the light.
Can I use these LEDs safely ?
The limits of "safe" emitted light are very complicated. You can read about the basics here.
A rule of thumb I have heard is if the emitted power is over 5mW, protection should be used. Since the LEDs you linked are capable of 10mW, yes they can be harmful. Do not use them until you understand how they can be harmful and how to prevent it.
UV is particularly dangerous because we can't see it so our blink reflex won't help. To safely work with these LEDs you should get a pair of glasses that block the possible wavelengths the LED can emit, 390 to 405nm ±2.5nm. Examples here.
This paragraph answers an edited out question in the OP wondering why he had a UV pen with no warnings. As for your UV pen light question, it is likely the power was low enough that it was not harmful. Less than 0.39mW (roughly) is considered eye safe so under that no warning would've been required.
When in doubt or dealing with health and safety matters, I strongly advice to avoid at all means opinions, rules of thumbs or any other common sense wisdom. Do not accept any advice which is not backup by an authoritative reference. Good practice in science and engineering mandates going in first place to the source of contrasted and accepted knowledge.
For this case, I refer you to the excellent ICNIRP (International Commission on Non-Ionizing Radiation Protection) paper on the matter:
Guidelines on limits of exposure to incoherent UV ratiation between 180 and 400nm
The paper may be a little bit daunting for those lacking a little bit of background on optics and power calculation, however, it is very clear and well written.
Here follows a summary of my interpretation of these guidelines, for your specific case:
The ICNIRP guidelines indicate a maximum recommended exposure of 10KJ/m2.
Finally:
$$ I_p = \frac{P_{tot}}{\pi} \frac{\sin^2\theta_1}{\sin^2\theta_2} = \frac{10\text{mW}}{\pi} \frac{\sin^2 (60^{\circ})}{\sin^2 (7.5^{\circ})} \approx 140 \frac{\text{mW}}{sr} $$
Lets's calculate two cases:
The operator stares at the LED at 0 degrees (frontal view) at 1m of distance:
$$ T_{\text{max}} = \frac{10\text{KJ}/\text{m}^2}{140\text{mW}/\text{m}^2} \approx 20\;\text{hours} $$
The operator stares at the LED at 0 degrees (frontal view) at 10cm of distance:
$$ T_{\text{max}} = \frac{10\text{KJ}/\text{m}^2}{14\text{W}/\text{m}^2} \approx 12\;\text{minutes} $$
The emitted light power of those LEDs is UVA typically 10mW with a 15° 50% power angle. There's no maximum specified, but probably we could say more than 15-20mW would be very unlikely.
That's within the range where caution is called for. Here is a very detailed safety report from a large University. I'll reproduce a chart below, but note that it is based on exposure with 'breaks' to allow cellular repair to take place. Also, I would expect it's conservative.
If you get a bit of distance from then end of the LED, the power density drops with the square of the distance. If you allow an irradiance of 0.1mW/cm^2 (30 second exposure limit according to the chart), then you need to be at least distance \$d\$ from the LED:
$$d = \left(\sqrt\frac {P_{mW}}{Ir_{MAX}}\right) \frac{1}{\tan(7.5°)}$$
So if you allow 20mW at the LED you'd need a distance of more than about 1m to be safe for a ~30s exposure (and assuming the power is fairly well distributed over the center 'spot').
