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Question:
I am asking this question for my daughter (13 years old, in grade 9), as she is considering the idea for a science fair project. The project involves determining what sort of protection from UV is most effective (eg, cotton, wool, protective creams or sun-screens, etc.). The difficulty seems to be : how to measure the amount of UV that "passes through" the protection. Any suggestions?



Replies:
as far as I know, there are not any cheap or otherwise readily available sensors for UV that are not also sensitive to visible light. You would need a cut-off filter (not cheap) and a light meter as used by photographers. One possible idea: certain kinds of plastics degrade over time due to UV exposure-- you might pose this question and the plastic idea in the chemistry section. Sorry about the run-around...

John Hawley


A good absorber of radiation will heat up in accordance with the amount of radiation it absorbs. Glass, I understand, absorbs UV; quartz does not. So perhaps you can make use of the differential heating through glass and quartz to evaluate the effectiveness of your UV absorbers. I think that quartz windows are used in lamps that shine on plants. At least these might be interesting areas for your 13 yr old to investigate, and if they do not lead directly to the instrument she is looking for, maybe they will lead to something else that is interesting. That is the nature of research.


The previous responses make good points, but there is something else you need to consider: not all uv wavelengths cause sunburn or skin cancer. Wavelengths from 400 - 320 nanometers (nm) are relatively harmless. These are called UVA. It is the UVB (320 - 280 nm) in sunlight that is the culprit. Instruments that measure only the UVB are expensive, because they need special filters and detectors. I cannot think of an inexpensive way to do it. Maybe someone else can. Good luck.

John Hawley



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