# 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?

• How did you arrive at that result? It has some surprising elements-- first of all, it is simple, and few things are with scattering, and second of all, it doesn't have anything in it about the thickness of the atmosphere, yet scattering should depend on that. – Ken G Apr 22 '18 at 14:26
• The full expression I derived was $I/F=\frac{\varpi_0\,\mu_0}{4(\mu+\mu_0)}P(\alpha)(1-exp[-\tau(1/\mu+1/\mu_0)])$, where $\varpi_0$ is the single-scattering albedo, $\mu_0 = cos(i)$, $\mu = cos(e)$, $P(\alpha)$ is the normalized phase function, and $tau$ is the optical depth. I got this expression by integrating over layers of an atmosphere. In my question, I'm using a bunch of simplifying assumptions to look at how scattered intensity depends on incidence and emission angles. – Spuds Apr 22 '18 at 15:04
• OK, so it's a semi-infinite atmosphere with some simplifying assumptions attached to get a nice result. Now I think I can answer. – Ken G Apr 22 '18 at 15:11

• Your answer for the first part, with emission angle really helped! However, I found the inverse about the incident angle. $I/F$ decreased with increased $cos(i)$. – Spuds Apr 24 '18 at 1:08
• Woops I meant to say that it decreases with increased $i$, not $cos(i)$. – Spuds Apr 24 '18 at 1:25