Quote:
Originally Posted by jambutty
Very true but how much heat does a torch with a blue filter over the lens produce that can have an affect on my calculator? Undoubtedly some infrared will filter through as well as will other spectral colours because the blue filter is unlikely to be a perfect blue.
The bottom line is that my calculator will be activated by light from any source and it is the brightness (or amount measured in lumins) of light falling on the solar panel that will determine at what point the calculator will become operative. Where the light comes from or what colour it is, is irrelevant. The blue end of the spectrum has more energy than the red end and therefore I conclude that more energy means a brighter light. With a 10% differential between the outer limits of the spectrum as shown by http://www.iee.cz/iwtpv04/6-Pociask.ppt slide number 32 it is hardly a major factor.
In any case the point being made applies to, presumably, a modern solar cell. My calculator is more than 20 years old. |
The only way we could completely say that it's not just the IR that is responsible for power production in your calculator we'd need to block the infrared light from hitting the solar panel without blocking the visible As you can see at -
http://members.misty.com/don/irfilter.html - it's not the cheapest nor easiest thing to do.
I hesitate to suggest it, but a sheet of glass may be a reasonable makeshift blocker of infrared light. Placing say your magnifying glass over the solar panel might wield some interesting results but it's nowhere near a perfect filter. What we could do instead is find a source which produces the same luminosity but doesn't emit as much infrared and see what effect that has. Which would be... a LE bulb. Luminosity and radiance are two different things. The incandescant bulb is more radiant but the two are both equally (as near as will make a difference to your calculator) luminous. I have to go though now cause I have a lecture soon.
Your argument about blue light = more energy = more power produced by the calculator is not correct as I pointed out. The blue photons are not able to produce an electron hole as they have *too much* energy and hence cannot be absorbed. You can research all this yourself I'm sure and show me a webpage or quote a textbook. Big library here I'm sure I'll find a copy. Nice debate, it's causing me to remember my quantum phys from a-level. As well as teaching me various things about solar panels and LE bulbs.