Friday

April 18, 2014

April 18, 2014

Posted by **Fabiola** on Friday, March 30, 2007 at 12:01am.

Let x be the distance from the stronger source, in feet. The distance from the weaker source is therefore (10 - x). Let the total radiant power of the brighter souce be 3P and the other equal to P.

The received illumination is

Y = 3P/x^2 + P/(10-x)^2

Differentiate that to find where dY/dx = 0. Mimimum illumination will be there. You won't need to know P; it will cancel out.

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