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	The inverse square law		2018-06-05
		Amy pose la question : Question about inverse square law ; Hi, I'm trying to understand some nuances about this law and have been reading about it a lot online and trying out some homework for personal interest (hobby ) - not school related. I understand that the simple formula is ; 1/d^2 I was wondering about what appears to be an oddity to me, that I came across when I was working with a test example I found at this link ; http://www.softschools.com/formulas/physics/inverse_square_law_formula/82/ The problem that caught my interest was this one on that page ; "1) If a bright flashlight has a light intensity of 15.0 candela at a distance 1.00 m from the lens, what is the intensity of the flashlight 100.0 m from the lens?" So I have a few different questions about this, but the most important one is, what does it mean if we replace the "1" in ; 1/d^2 with a different value such as a ratio in the form of a decimal value, so instead we have something like this ; 0.75/d^2 ??? Does this ratio represent a curve then? I am thinking that the "1" in the normal formula represents a straight line Some of the links I was researching ; http://wisptools.net/book.php?c=3&s=2 http://www.softschools.com/formulas/physics/inverse_square_law_formula/82/ https://www.nde-ed.org/GeneralResources/Formula/RTFormula/InverseSquare/InverseSquareLaw.htm Penny Nom lui répond.
	What is the intensity 5m below the surface?		2007-03-31
		david pose la question : I have this question which I am supposed to set it up and solve as a differential equation. I know how to solve the diffrential equation but I am having hard time understanding this question. Here is the question: The intensity of light in the ocean decreases the deeper you dive. In fact, the rate at which the intensity decreases is proportional to the current intensity. Setup the corresponding differential equation and solve for I(Y), the intensity I as a function of current intensity Y. If the light intensity 2m below the surface is 25% of the intensity at the surface, what is the intensity 5m below the surface. Can you please explain to me what does it mean by current intensity and how do I set this equation up. Thanks for the help. Penny Nom lui répond.

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