# 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 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.

• Nice drawing! To be picky & accurate, should include the Refraction of light beams when they cross the air-water interface; the radius change underwater – patta Aug 17 '19 at 11:35
• Yes, that’s true. What’s shown assumes no interface. – boyfarrell Aug 17 '19 at 11:40
• Thank you, @boyfarrell. However, wouldn't the dilution of the beam also have a retarding effect on the extent of attenuation (because, while the fractional loss for each infinitesimal increase depth stays constant, the absolute loss for each increase will be lower, so the total attenuation will be lower than it would be because of dilution)? Or is this accounted for by adjusting the value for the attenuation coefficient? – akbab Aug 18 '19 at 13:05
• Yes i think you could be right. Let me try another idea; will update my post. – boyfarrell Aug 18 '19 at 14:04
• Ok I re-thought the answer. Returning to the original differential equation of the Beer-Lambert law and include the surface area element as a function of distance travelled – boyfarrell Aug 18 '19 at 15:42