Question:
inverse square law help?
fanzomo
2012-02-06 11:06:07 UTC
im a bit confused about the "r" in the formula thst i think you use to work out the inverse square law 1/r^2. i think r is the distance from the light source (correct me if im wrong) but how do you know what r is, is there a set value that r will always be for example 1m? i probably worded this horribly but can someone try to explain this to me. thanks :)
Four answers:
ExTex
2012-02-06 12:37:21 UTC
The "r" is a "variable", like x the unknown in an algebraic equation, and in the inverse-square law, it just stands for a specific distance. That distance is how far it is from the center of the source of light, gravity, or whatever field gets weaker as r increases. As an example, consider an apparently small star and the eye that sees light from that star. The star is like a small point of light, and r is the distance from the center of the star to the retina of the eye that is seeing the star.

If the star is "x" in brightness at some distance r = y, then the energy of the light is proportional to the brightness x multiplied by the square of the distance r. That is Energy = A*x*r^2, where A is a constant that does not change for different distance and brighnesss amounts, due to energy conservation.

So for two different distances r1 and r2 and their brightnesses x1 and x2, the energy is the same (conserved) and

E = A*x1*(r1)^2 = A*x2*(r2)^2

Divide by A, which does not change either, to get

x1*(r1)^2 = x2*(r2)^2

Divide both sides by x2 to get

x1*(r1)^2/(x2) = (r2)^2

Then divide both sides by (r1)^2 to get

x1/(x2) = [(r2)^2]/[(r1)^2] = [(r1)/(r2)]^2

Notice how the 1 and 2 subscripts are upside-down, or "inverted" from each other on the two sides

of the last equation. And notice how the brightness ratio is not squared, but the distance ratio is squared. This is where the "inverse square law" name comes from.

If you know or are given any 3 of the 4 quantities involved in the equation, you can plug them into that equation and calculate what the 4th quantity must be.

Or if you even know or are given what the value of one of the ratios is, and one of the quantities of the other ratio, you can also calculate what the 4th value is by using the equation.



This is a rather long answer, but since you seemed to have so much trouble understanding what the law means, perhaps some part or all of it will help you understand better what the law means and what it is used for. For one thing, astronomers use it a lot to figure out how far away stars and galaxies are from the Earth, and that is just one of at least thousands of uses for the law.
anonymous
2016-05-16 07:16:46 UTC
The inverse squares law is vital to correct exposures. "View Cameras" usually have a bellows unit and racking this in and out to get a certain magnification of the subject changes the exposure drastically. If you wish to research the law further try visiting Wikipedia under the science banner type in "Inverse Square Law" where all will be explained
♘ Vader
2012-02-06 11:12:27 UTC
r is technically the radius .... since light radiates in all direction ....and relates to the surface area of a sphere .... imagine an expanding balloon



The physical quantity or strength varies inversely to the square of the distance qty ~ (1/d^2)
ry0534
2012-02-06 11:15:30 UTC
r is the distance from the source of the field to the point in question. And since fields are radiated outward in all directions (spherically) and the surface area of a sphere is linearly proportional to r^2, and intensity is inversely proportional to the area of the sphere.


This content was originally posted on Y! Answers, a Q&A website that shut down in 2021.
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