# Illuminance of a non-punctual source

When working with a punctual light source, I get the formula (for the normal incidence)

$$E=\frac{I}{r^2}$$

This is how I got it:

First I identify that the expression for the solid angle is

$$d\Omega=dS/r^2$$

Then knowing that $$dF=Id\Omega$$ and $$dE=dF/dS$$, I just substitute to get the formula.

However, I can't tell what step is not valid for the case of the source being non punctual.

Also, in that case, what would the formula be? Or would it be dependent on the exact geometry of the source?

• ... added some details as requested. Mar 19, 2023 at 0:54

You would divide your source into small "infinitesimal" punctual sources. If one such small source, of infinitesimal intensity $$dI(\vec r_s)$$, is located that $$\vec r_s$$, then radiance of that source at some point $$\vec r_p$$ in space is given by $$dE= \frac{dI(\vec r_s)}{\vert \vec r_p-\vec r_s\vert^2}$$ The total integrated radiance is then $$E=\int_VdE = \int_V \frac{dI(\vec r_s)}{\vert \vec r_p-\vec r_s\vert^2} \tag{1}$$ where the integration is over the volume containing your source. The denominator factor can be expanded, resulting in a multipole expansion similar to the expansion of a potential created by a macroscopic charge distribution.

In the case of a single punctual source, there is no integration since everything is concentrated at one point. Assuming your source is at the origin, you get the usual result with $$\vec r=\vec r_p$$.

The radiance does depend on the distribution through the factor $$dI(\vec r_s)$$ since this factor is not necessarily constant, i.e. the value $$dI(\vec r_{s1})$$ at $$\vec r_{s1}$$ is not necessarily the value of $$dI(\vec r_{s2})$$ at $$\vec r_{s2}$$. This factor essentially tells you that the light coming from point $$\vec r_s$$ can be considered as a point source of strength $$dI(\vec r_s)$$ located at that point. The integration in Eq.(1) basically adds all the contributions.

Thus, with the assumption that $$\vert \vec r_p\vert>\vert \vec r_s\vert$$ (the point $$r_p$$ is outside the source), we have

\begin{align} \vert \vec r_p-\vec r_s\vert^2&= (\vec r_p-\vec r_s)\cdot (\vec r_p-\vec r_s)\\ &=r_p^2+r_s^2-2r_pr_s\cos\theta_s\\ &=r_p^2\left(1-2\frac{r_s}{r_p}\cos\theta_s+\frac{r_s^2}{r_p^2}\right)\, ,\qquad \frac{r_s}{r_p}<1\, , \end{align} so that \begin{align} \frac{1}{\vert \vec r_p-\vec r_s\vert^2}&=\frac{1}{r_p^2} (1-2\frac{r_s}{r_p}\cos\theta_s+\frac{r_s^2}{r_p^2})^{-1}\\ &\approx \frac{1}{r_p^2}\left(1+2\frac{r_s}{r_p}\cos\theta_s\right) \end{align} so the leading term is in $$1/r_p^2$$ and yields $$E\approx \int_V\frac{dI(\vec r_s)}{r_p^2}=\frac{I}{r_p^2}\, ,$$ recovering the point source result.

Note that here I have assumed incoherent sources else there could be interference effects in the same way that constructive or destructive interference effects can occur when combining sound waves at a particular point.

• One question, what is the volume you are integrating on? Is it the volume of the source? For example, if the source were to be an LED light, would $V$ be the volume of the LED or would it be the space it illuminates? Mar 19, 2023 at 1:15
• the volume of the source. This is why there is a $\vec r_s$ in there: $I(\vec r_s)$ is the strength of the course at point $\vec r_s$ in the source. Mar 19, 2023 at 2:07
• @MikelSolaguren of course if your source is not a volume but a surface you would integrate over the surface of the source, and if your source was a line you would integrate over the length of the source. Mar 19, 2023 at 2:10