How to test sunglasses' uv blocking levels using UV meter?
September 10, 2010 3:50 AM Subscribe
I have access to a UV meter. Using this tool how can I test the sunglasses I have actually filters out the light rays that are dangerous. I don't have any technical knowledge to interpret the numbers displayed by the tool.
The sensor displays what it is measures (i think it is irradiance) in two ranges
1 μW/cm2 ~ 9999 μW/cm2
0.01 mW/cm2 ~ 40.00 mW/cm2
For example when I test it against a normal incandescent lamp from a reasonable distance it measures 261 μW/cm2 and behind the sunglasses it drops to 240 μW/cm2
Is this drop in the value indicative that sunglasses are providing the protection? How can I interpret these numbers in general?
The sensor displays what it is measures (i think it is irradiance) in two ranges
1 μW/cm2 ~ 9999 μW/cm2
0.01 mW/cm2 ~ 40.00 mW/cm2
For example when I test it against a normal incandescent lamp from a reasonable distance it measures 261 μW/cm2 and behind the sunglasses it drops to 240 μW/cm2
Is this drop in the value indicative that sunglasses are providing the protection? How can I interpret these numbers in general?
Response by poster: Thank you,
I will test it in the sun as soon as it comes out here.
Should the value drop to 0 if the uv filter (sunglasses) are functioning properly or is there a safe threshold?
posted by neworder7 at 5:01 AM on September 10, 2010
I will test it in the sun as soon as it comes out here.
Should the value drop to 0 if the uv filter (sunglasses) are functioning properly or is there a safe threshold?
posted by neworder7 at 5:01 AM on September 10, 2010
There is undoubtedly a safe threshold, though I don't know what it is.
The small drop you observe with a desk lamp is probably due to the desklamp already effectively filtering most of the UV (250 uW/cm^2 is a fairly small number).
posted by JMOZ at 5:06 AM on September 10, 2010 [1 favorite]
The small drop you observe with a desk lamp is probably due to the desklamp already effectively filtering most of the UV (250 uW/cm^2 is a fairly small number).
posted by JMOZ at 5:06 AM on September 10, 2010 [1 favorite]
As far as I can tell, the UV specification "100% UV protection" means it filters out 99-100% of all UV A and B light. So if you get the sunglass lens very close to the sensor (so there is no stray light coming in the sides), the sunglasses should filter out nearly all of the UV light. So yes, I think it should drop down to close to zero*. What does your detector read when you put it under some blankets in the dark or something, so there is definitely no UV light reaching it?
*that is, it should drop down close to whatever the display reads when you put something thick and opaque covering the sensor.
posted by Salvor Hardin at 5:19 AM on September 10, 2010
*that is, it should drop down close to whatever the display reads when you put something thick and opaque covering the sensor.
posted by Salvor Hardin at 5:19 AM on September 10, 2010
Looking at the manual, it appears that it does not measure ultraviolet band C (UVC), which is at 280 nm–100 nm.
Standard black lights you can buy easily peak at either 350 nm or 370 nm, with respective widths of frequency of 40 nm and 20 nm. You can buy a black light from Spencer's or a cat urine light just to check to make sure your detector is working. Then, pass your sunglasses between the black light (probably good to put this in a cardboard box) and the detector to see how much is cut down. The ratio between before and after is called transmittance. That will have tested your UVA protection, at least to some degree.
A UVB source is harder to grab. There are medical black lights that are technically UVB, but they're very, very close to being UVA anyway.
I have a UV source, somewhere, a terrifying instrument I have yet to plug in. I think it might be UVC.
Really crappy UV meters, the cheap ones, are more "is light falling on me?" detectors than anything else, but that doesn't look like what you have here.
posted by adipocere at 7:05 AM on September 10, 2010
Standard black lights you can buy easily peak at either 350 nm or 370 nm, with respective widths of frequency of 40 nm and 20 nm. You can buy a black light from Spencer's or a cat urine light just to check to make sure your detector is working. Then, pass your sunglasses between the black light (probably good to put this in a cardboard box) and the detector to see how much is cut down. The ratio between before and after is called transmittance. That will have tested your UVA protection, at least to some degree.
A UVB source is harder to grab. There are medical black lights that are technically UVB, but they're very, very close to being UVA anyway.
I have a UV source, somewhere, a terrifying instrument I have yet to plug in. I think it might be UVC.
Really crappy UV meters, the cheap ones, are more "is light falling on me?" detectors than anything else, but that doesn't look like what you have here.
posted by adipocere at 7:05 AM on September 10, 2010
Response by poster: Thank you all for your comments.
When I cover up the sensor with an opaque object it goes down to 0 indoors and to 30 μW/cm2 outdoors.
In the direct sunlight the display goes beyond max
Outdoors in the shadow the value it measures is 4200 μW/cm2
Outdoors in the shadow with Rayban sunglasses filter it drops to 3000 μW/cm2
Outdoors in the shadow with cheap sunglasses filter it drops to 2400 μW/cm2
I was wondering how these values correlate to the UV-A/B intensity (low/moderate/high/extreme)? It is hard to conclude that the sunglasses are giving the 100% UV protection which they claim am I right?
posted by neworder7 at 3:59 PM on September 10, 2010
When I cover up the sensor with an opaque object it goes down to 0 indoors and to 30 μW/cm2 outdoors.
In the direct sunlight the display goes beyond max
Outdoors in the shadow the value it measures is 4200 μW/cm2
Outdoors in the shadow with Rayban sunglasses filter it drops to 3000 μW/cm2
Outdoors in the shadow with cheap sunglasses filter it drops to 2400 μW/cm2
I was wondering how these values correlate to the UV-A/B intensity (low/moderate/high/extreme)? It is hard to conclude that the sunglasses are giving the 100% UV protection which they claim am I right?
posted by neworder7 at 3:59 PM on September 10, 2010
Well, they're supposed to block 100% of HARMFUL UV, which may be different wavelength ranges than your meter is measuring. Also, I'm a bit mistrustful of your meter. I would try using a light with little/no UV (for example, a non-blue LED) and see what you get.
posted by JMOZ at 5:40 AM on September 12, 2010
posted by JMOZ at 5:40 AM on September 12, 2010
This thread is closed to new comments.
Since the UV spectrum you're actually trying to protect yourself against is coming from the sun, I would measure the difference using the sun as a UV source. Just make sure to do it on a cloudless day, so there isn't any natural variation in the UV light that could skew your results.
posted by Salvor Hardin at 4:53 AM on September 10, 2010