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I've been studying for the pGRE for the past couple of weeks (what a load of... nevermind), and one of the questions requires the use of what is apparently deemed the illumination formula, used to calculate the something (I'm assuming something like percentage of photons detected). It is $\frac{\pi r^2}{4 \pi l^2}$, where $r$ is the radius of the detector and $l$ is the length between the point-source of radiation and the detector.

It looks like I'm dividing the area of two circles. If this is indeed a percentage-thing, the numerator is the area of the face of the detector (this could be something like the "amount" of photons detected). Thus, the denominator somehow represents the total number of photons emitted. (But this does not answer the $\frac{1}{4}$ comes from, or why it isn't some function of the arclength of a circle around the point-source.)

Where does this formula come from?

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  • $\begingroup$ Do you know the formula for the surface area of a sphere? $\endgroup$ – DJohnM Feb 7 '15 at 2:33
  • $\begingroup$ @User58220 Oh, damnit! I didn't even think of that. $\endgroup$ – AmagicalFishy Feb 8 '15 at 19:44
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It is an approximation that is ok so long as $l \gg r$.

The idea is that your detector will intercept some small fraction on the surface of a sphere of radius $l$, which has a surface area of $4\pi l^2$. If the luminosity $L$ is spread out isotropically over this sphere, then the flux at the edge of the sphere is $L/4\pi l^2$ per unit area.

Strictly speaking, the radiation travels radially outwards, so if you then impose a flat circular detector into the radiation field, it intercepts those rays travelling through the chord defined by the flat surface cutting the edge of the sphere. On the other hand, if $l \gg r$, then we can assume that that each point on the detector is at an identical distance $l$ from the source and so the received illumination is the area of the detector $\pi r^2$, multiplied by the flux at the edge of the sphere defined above.

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Presumably the emitted light is spread out uniformly. In that case at a distance $\ell$ the light is spread out over a sphere with a surface area of $4\pi\ell^2$. If your detector has an area $A$ then the fraction of the light that hits your detector is just $A/(4\pi\ell^2)$.

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