Hello all..... This problem occurred to me the other day, so I decided to see if I could work it out, but am stuck. How much heat can be generated (in therory) using an ordinary magnifying glass to focus the sun's rays on a sunny day? Here's what I'm working with: A decent approximation for the sun's power on the surface of the earth is about 1200 W/m^2. I'm assuming a glass lens about 3" in diameter. Changing units and figuring the area gives 0.00456 m^ for the area of the magnifying glass. The magnifying glass will intercept roughly 5.5 W/m^2 available from that 100 W/m^2 the sun is providing. I think I'm satisfied with this so far, but what I'm having trouble reckoning is what to do next. I believe the kindling temp of paper is about 450 F, and I know paper easily burns by putting a mag. glass to it, and I still think it's cool! Anyway, seems to me that the smaller the spot the mag glasss gives the greater the heat (within reason) that can be generated. I'd like to get a little guidance please. This is getting under my skin and it is frustrating to have my mind bouncing off this without being able to proceed. Anyone have some input or ideas? Thanks!
You effectively know the power through the magnifying glass, all you need to know is the area of the spot, that will give you enough information to work out the irradiance. Claude.
You will need to consider the reflectivity of the surface you are trying to heat and the rate of dissipation of the heat generated by the absorbed light.
Well if you get the details worked out I'd be interested to look at them here. It IS a very interesting problem. I haven't studied heat for close to forty years, so this is especially challenging. Reflectivity and heat dissipation appear to be relevant, but so does specific heat. I think specific heat (say BTU/(lb)(degree F) gives you the means to convert power (watts to BTU/hr) to the 450 degree temp change you seek. Wood has a specific heat of 0.42 in the above units and that might be a good enough surrogate for paper. good luck!
Thanks for the suggestions so far, but I am clueless about this right now. I really don't know how to approach this.....and still trust the results. I realize a number of things about this.....that the focal point will be smeared out due to abberations in the lens, and I realize that it is the IR that is also getting refracted and doing the heating (??). I thought about trying to use Q = =mcT, but this does not help me get a definitive feel for the temp. You need to select a material...its 'c', and some amount of mass of the stuff. I dunno.....I've been fiddling with this for days and days now.....and I get frustrated when my thoughts keep traveling into the same ruts time after time. help lol
That's a little bit muddled. The magnifying glass will intercept 5.5 Watts, not 5.5 Watts/m^2. What do you mean by "how much heat.."? If you mean the power, then 5.5 Watts is the answer. If you mean temperature, then it is much harder to calculate. As you said, spot size will be determined in part by lens aberrations. You can get rid of chromatic aberrations by using a fancy compound lens, or more simply by using a parabolic mirror instead. If you get rid of all aberrations, then the spot size will be the focal length times the angular size of the sun in the sky (9.6 milliradians). So you can get the power density in the spot. To get temperature from this, you need a lot of other things, like the emissivity of the surface and the conductivity.
My take is that you can make some simplifying assumptions...pick an area for the size of the magnification spot, maybe 1/32" diam...make the paper black, then it won't reflect much. Used the specific heat of wood. Paper doesn't conduct well, so 99% of the heat from the focused light will stay put...it's not a conductor like metal. (I know because paper has never burned me unless it was flaming!!) I think most of the relevant factors have been suggested by KRAB...simplfy and see what you get...
There will also be a 4% loss due to Fresnel reflection off the surface of the magnifying glass. Of course, whether you wish to include this depends on how accurate you want to be. Claude.