| MadSci Network: Physics |
Hi Matt, The energy coming from the Sun can be partitioned into 3 parts, UV, visible, and infrared. Though UV is harmful, it is only a small part of solar radiation energy-wise. About half of Sun's enery in visible, which we can see and other half in infrared, which we cannot see but can feel. Add UV, visible, infrared together, the total solar enery is 1347 W/m^2 (energy per square meter). Passing through the atmosphere, half of energy losts. On the ground, total solar energy is about 680 W/m^2 on average. Note, this number may vary greatly depending on the time of year, location on Earth, and cloudiness of sky. Out of 680 W/m^2, about 300 W/m^2 is in the visible. If the "brightness" in your question literally refers to the brightness (help you to see), then you need light bulbs that can give you about 300 W/m^2. If you use incandescent bulbs, you need one 600 W light bulb. For a better effect, use 2 300W or 6 100W bulbs illuminating from different angles. You get the same brighness with a lower voltage fluorescent bulb, but at this moment I cannot give you more definite number. The experiment I did for myself suggested a fluorescent bulb of about 500W (but I had expected lower). Please let me know if this answers your question. Best Xiaodong
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