Scattered Intensity of Light vs Angles of Incidence and Emission So I just derived an expression for the scattered intensity of light from an atmosphere with ideal isotopic scattering: $I/F=\frac{cos(i)}{4(cos(e)+cos(i))}$, where $I$ is intensity, $F$ is flux, $i$ is the angle off incidence, and $e$ is the angle of emission ($i,\ e=0$ are normal to atmosphere).

When I plot the scattered intensity as a function of $i$ with $e$ constant, it decreases while the angle increases. Why is this? My theory is that at a large incidence angle, the light encounters a larger surface area, so reflected light will have a higher variation of emission angle, meaning that scattered light at a given emission angle will be lower. Does this make sense? Or is there another reason for this?

Next, I plotted scattered intensity as a function of $e$ with $i$ constant. This time, the intensity increases with increases emission angle. I don't have a theory as to why this is the case, nor an intuition to back it up. Why would this be the case?
It is easiest to understand if you fix the incident angle and explain why the emergent intensity is largest at highly oblique angles.  Since the light comes from the outside, it only penetrates so far, and this in turn causes the atmosphere to act like a source of scattered light.  But the source of scattered light is brightest near the top of the atmosphere, since that's where more of the external light penetrates.  Whenever the sources are brighter near the top, it produces what is called "limb brightening," where if you look from highly oblique angles, you mostly probe those higher, brighter regions.  Looking down the normal is where you see the deepest into the atmosphere, where the external light does not penetrate as well. 

You can also think about what the individual photons are doing, and ask what is their distribution over emergent angle.  If cos(i)=0, all the photons scatter right at the surface, so it is tantamount to introducing an isotropic radiation field right at the surface.  The photons that go outward will of course have an isotropic distribution, and an isotropic incident radiation field must scatter isotropically (that is a consequence of the principle of reciprocity).  So the emergent distribution is isotropic, but intensity is also per solid angle, so accounts for the foreshortening, and that's where the 1/cos(e) in the intensity comes from.  If cos(i)=1, on the other hand, the incident photons tend to penetrate more, and must diffuse their way out, which gives less of an advantage to low cos(e) after the foreshortening is included.

As for fixing the angle at which you are looking and altering i, here your result says that the intensity always peaks as cos(i) rises.  Your expression claims that is true at every e, so to me this suggests a normalization error.  You want to keep the incident F fixed, but that requires you must get the same outgoing F if you integrate over all e.  Your result says the I is higher at all e if cos(i) is higher, but that contradicts the idea that you are keeping the incident F the same.  Maybe your result is actually comparing the incident and emergent intensities, not the emergent intensity to incident flux.  Then at higher cos(i), for the same incident I, the incident F is falling, explaining your rise in I/F.