Calculate the intensity of a solid object made up of point light sources?

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SUMMARY

To calculate the intensity of light from a solid object composed of point light sources, integration is necessary. Each differential area element (dA) within the object contributes to the total luminosity observed at a specific point (p,q) based on the inverse square law. The luminosity from each dA is calculated as (m/[(x-p)²+(y-q)²])dA, where m represents the luminosity per unit area. Integrating this expression over the entire area yields the total luminosity received by the observer, although the integral can be complex.

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  • Understanding of inverse square law in light intensity
  • Familiarity with differential calculus and integration
  • Knowledge of luminosity and its measurement
  • Basic concepts of geometric shapes, particularly circles
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ericwithakay
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So I understand that to calculate the intensity a distance d away from a light-bulb you would treat the light-bulb as a point source and do a gaussian-sphere-type-of-thing to figure out the intensity at the desired location.

My problem involves a solid object made up of "point" light sources and I'm wondering whether I need integration to be able to solve the problem.

Any advice would be greatly appreciated.

Thanks

-E
 
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Yeah, sounds like you'll need to integrate. I think you could get this with a little more work.
You want your differential element of your body, be it area, volume, what have you, to have some measure of luminosity. So say you have the unit circle that has luminosity m per unit area. Then you subdivide your circle into little differential elements dA. Then for each dA the luminosity you observe at point (p,q) is (m/[(x-p)^2+(y-q)^2])dA lumens from each dA or whatever measure of light intensity you are using (that's the Guassian sphere thing you mentioned). That is, the luminosity falls according to the inverse square of the distance. If you integrate over the area of the circle you get the total luminosity an observer at (p,q) gets. Not an easy integral though.
 

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