How can the loss of light irradiance due to both attenuation and the inverse square law be modelled? When considering the effect of depth on just light attenuation (is within  an aqueous solution), the Beer-Lambert law is used. From what I've found, the law is derived by determining the fractional loss of radiant flux, $\phi$ for an infinitesimal depth $dr$ and integrating over the depth $r$: 
$\int_0^r c\times dr = \int_0^r \frac{d\phi}{\phi}$      where $c$ is the attenuation coefficient 
Therefore I assume that to model the effect of the inverse square law on top of this the left half of the above equation has to be changed in some way? Given a light source of distance $d$ above the surface of the water, I have calculated an expression for the fractional loss due to the inverse square law: $k\frac{(r+d)^2}{r+d-dr)^2}$ (essentially the ratio of surface areas for the spheres with a radius difference of $dr$). 
This similar question, for sound attenuation, confirms that the phenomenon are independant
Am I on the right track? Or, more importantly, is there an existing model already? A simplification would also be useful as this is only for a high-school level 'extended essay'
Thank you 
 A: A pencil of radiation with intensity $I$ is travelling along direction $\hat{n}$ confined to solid angle $d\Omega$. 

We can describe the path taken simply by changing the radius $r$ from the coordinate system origin, the intensity of radiation $I$ will reduce due to attenuation by the material coefficient $\alpha$ and due to the expansion of the surface area element $dA=r^2 d\Omega$ associated with the solid angle.
We want to solve,
$$
\frac{dI}{dr} = - \alpha I dA = - r^2 \alpha I d\Omega
$$
Performing the angular integration gives a constant factor,
$$
\int_0 ^{\Omega_c} d\Omega = \Omega_c
$$
Giving,
$$
\frac{dI}{dr} = - r^2 \alpha I \Omega_c
$$
This is a first-order linear ODE which you can solve yourself, but the solution is proportional to 
$$
I(r) \propto \exp \left( -\Omega_c \alpha r^3 / 3 \right)
$$
If this was simply the Beer-Lambert law we would expect the exponential term to simply have an $r$ dependence. The solution has the $r^3$. Implying an additional $r^2$ due to the expanding surface area reducing the intensity faster. 
