If I understand correctly the term "spectral flux density" describes radiant flux for a given wavelength, right?
Like the given wavelength is the $\nu$ in:
$$F(\mathbf{x},t;\nu)=\oint_\Omega I(\mathbf{x},t;\hat{\mathbf{n}},\nu)\hat{\mathbf{n}}\,d\omega(\hat{\mathbf{n}})$$
For example, the EM wavelength of green light is about $500\,\mathrm{nm}$.
So can we talk about the spectral flux density $D$ of green light at a given point on a surface? Or have I got it wrong?
If I've got it right – what I don't understand is why is it per unit meter wavelength? Do we need to divide through by $500 \times 10^9$ to get the value of D in SI units?
Let's take a concrete example: say there are $1000\,\mathrm{lx}$ (ie $1000\,\mathrm{lm/m^2}$) of uniformly white light hitting a surface. What is the spectral flux density of green light in the SI unit of $\mathrm{Wm^{-3}}$?
Best Answer
Spectral flux density is required because the number of photons (or amount of energy) at some precise wavelength is zero. You have to integrate over a finite range of wavelengths to get a finite flux. The thing that is integrated over a range of wavelengths must therefore be expressed per metre (or more conveniently per whatever unit of wavelength you are using). i.e. To work out the power received in total you have to integrate the flux density over a wavelength bandwidth.
It is difficult to answer your second question which mixes up units of perceived brightness with physical units of power. It would depend on the conversion factor from lumens to Watts and what the actual spectrum of a perceived white light source was. But let's assume your white light is a scaled solar spectrum. Direct sunlight has about 100 lumens/W and a power per unit area of 1000 W/m$^2$, so your light source is about 1% as bright as direct sunlight.
The spectral flux density of direct sunlight peaks in the green at about 1.3 W/m$^2$ per nm, so your source would be 100 times fainter/smaller than that. If you then really want to express that as W/m$^3$, multiply by $10^9$ to get $1.3\times 10^7$ W/m$^2$ per m (or W/m$^3$).