Eventually at large distances, any finite object will have the apparent radius decay as 1/r and solid angle as 1/r^2.
I think in this case it could be atmospheric absorption or just getting near a peak point of emission.
The coherent case is similar I believe, but because for an antenna (emitter) the field is usually tangent to the surface, you get some additional trigonometric factors (not purely solid angle). But by my calculations it's still 1/r^2.
For gravity, it's clear the orientation of a small surface doesn't influence its gravitational force. In fact the source can be replaced by a point of equivalent mass.
For radiation, it seems the angle of a surface matters? For example, lambertian surfaces (everyday diffuse surfaces) do depend on their angle for total incoming radiation? Does only reflection exhibit lambertian behavior? There's something funny at work here.
In the lambertian case, there's a cosine term making your terms go to zero and sum converge.
When observing a small oblique (lambertian) surface, you get less radiation by a factor of cos(angle). I believe this is true for black body radiators as well (think a small glowing plate emitting red light). I'm not completely sure it would be true for x-ray and high energy radiation, and intuitively I suspect it indeed isn't. It seems like for high energy emissions, or generally isolated emissions (discounting opaqueness), only the total power is relevant like in gravity. I think self-absorption might be the culprit here.
If we say self-absorption (material absorption) in the high energy case is roughly negligible, then it becomes more strongly non-lambertian and the received power is simply a 1/r^2 integral (which would also explain why radiation gets stronger as it approaches the surface).
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Here's a simple proof of the solid angle formula I've found I believe is correct for lambertian surfaces: chop your object into infinitesimal pieces (limit goes to 0). For each piece, it tends to behave like a point, so we expect the radiation scaling with distance to be 1/r^2. But solid angle also scales as 1/r^2 for small enough pieces. (it's probably missing many technicalities to make the proof work, but I think it works). Note that for oblique surfaces their solid angle is a function of cos(angle), which follows Lambert's cosine law exactly as well.
I made a mistake and assumed it was Beta or Gamma radiation that travel easier in air.
It will still decrease with distance. If you represent the wall as a collection of point sources distributed over a plane, as the distance to the plane diminishes, the distance to individual points do not diminish from and to the same distance as each other, and do not diminish at the same rate as each other. The distances to all sources still diminish but not in the same way, so the appropriate function describing the sum of the contributions from each point is no longer 1/d^2. Also notice that although far away points hardly change in distance at all, their contribution is also very small, i.e the rate of change and intensity are inversely correlated, which is why it's not constant as you were suggesting.
There will be some continuous equation to describe this... someone more familiar with radiance feel free to chime in!
I'm not sure in practice how much of a difference this makes compared to air absorption though.
I feel like I want to plot this now to build an intuition.
The situation is similar for an opaque light source (apart from the lack of atmosphere in the way, I don't think the sun would be much brighter if you were much closer to it and looking through a pinhole), but I'm not sure it applies to gamma radiation which is emitted uniformly in all directions from each point and treats most stuff as pretty transparent - unless there's some weird interference you don't get to cancel out the parts coming from different directions.
What I'm suspecting is that under a certain distance for a given finite plane it's almost constant similar to the infinite version, but outside of that distance there must be some non uniform falloff function.
A rectangle is a mess. It's easier with a circle. The approximations at short and long distance are the same, but a circle has an "easy" formula in between. http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elelin.h... 1-x/sqrt(x^2+R^2) where R is the radio of the circle. Let's use R=10 to keep it simple.
For a short distance 0<x<R/10, then it's almost a constant https://www.wolframalpha.com/input?i=1-x%2Fsqr%28x%5E2%2B10%...
For a long distance, x>4R it's almost like R^2/(2*x^2) https://www.wolframalpha.com/input?i=1-x%2Fsqr%28x%5E2%2B10%...
There are some trick to add more term to the approximations to reduce the middle part where both approximations are bad, like using A+Bx or A+Bx+Cx^3 for short distance and D/x^2+E/x^4 or D/x^2+E/x^4+F/x^4 for long distance. It depends on how much you care about the precision and how many calculations you want to do.
In some particular cases like a circle or the shell of a sphere there are closed formulas for the intermediate distances. In other cases there is no nice formula.
If you have a 1/r^2 decreasement from a point source, the if you have an infinite wall it becomes a constant value that does not decrease with the distance.
Clearly the wall is not infinite, but when the distance from the wall is much smaller (1/5?) than the distance to the border of the wall, the approximation is good enough. The wall is like 8 tiles tall , so I expect the radioactivity at 1/2 tile (15cm, 5in) to be almost equal to the radioactivity when the counter is touching the wall.