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How can the irradiance of a point source behave proportionally to 1/R^2, where R is the range of the source yet the irradiance of an extended source be independent of range? It doesn't make any sense to me.

Consider two formulas I found online looking for irradiance of a point source vs extended source:

point source: Irradiance = dpsi/dA = I / R^2, where I is the intensity (W/sr) dpsi is the power and dA is an area element.

extended source: Irradiance = dpsi/dA = pi/4 L (d/f)^2 where d is the diameter of a lens, f its focal length, and L the radiance of the object in W/m^2 sr.

Take an object of uniform radiance (spatially) and approximate it with a point source. As the object gets closer the estimate based on the point source approximation increases quickly where the irradiance calculated from the extended source doesn't change at all. Now, I know the point source is a simplification, but looking at these formulas it sure looks like a bad one!!! What am I missing? Because I know it is used a lot and looking at it now, I'm not sure why.

I have a heavy mathematical background but very little physics. So, go ahead and lay it on me with the math in your answer but give context to the physics.

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  • $\begingroup$ where did you find that equation for the extended source? I am a bit confused as to why there should always be a lens involved ... $\endgroup$
    – Sanya
    Sep 15 '17 at 23:35
  • $\begingroup$ I don't think there has to be a lens, but the formula in the following resource uses one. I think without the lens you just get something like irradiance is proportional to L (not dependent on R). I'm using the first PPT from LSU that comes up as a reference: google.com/… $\endgroup$ Sep 15 '17 at 23:37
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If a source cannot fit within the angular view of a telescope or other lens, then it doesn't matter how far away you are from it as long as it has uniform radiance.

If you move further from that object, you can see more of it, and this value is proportional to $R^2$. This balanced with the inverse relationship of intensity with distance.

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  • $\begingroup$ So the extended source formula is only true when you are so close that the entire object cannot fit in the FOV? $\endgroup$ Sep 16 '17 at 0:48
  • $\begingroup$ I am mostly familiar with this effect with electric fields, but it should be true of any source that has a $1/R^2$ intensity and is sufficiently large. For example, this is why the electric field between two charged plates in a capacitor is constant, except near the fringes. $\endgroup$ Sep 16 '17 at 1:41
  • $\begingroup$ Suppose we are far enough away to see all of it, but each pixel in your camera may see only a part of it. Then, does the irradiance incident on the dA that corresponds to each pixel behave as an extended source but the sum of the irradiance behaves like a point source? Is it possible to show this R^2 dependence mathematically? $\endgroup$ Sep 16 '17 at 19:50
  • $\begingroup$ Think of the net flux of light contained in an object you are viewing. That is, how many photons are being emitted. For a sphere of a given radius, $R$ from which you observe, the flux will be evenly spread across the surface, which is $4\pi R^2$. The area you view from will be constant, regardless of how far away you are. Each pixel might not see all of it, but when you combine the pixels together, your average irradiance would be reduced according to your distance from the source. $\endgroup$ Sep 16 '17 at 20:51

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