# What does spectral flux density mean "per wavelength"?

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 $$\mathrm{Wm^{-3}}$$?

• What has the equation (written in terms of frequency) got to do with the question? Jan 22 at 9:05
• I'm thinking spectral flux density would usually be watts per square meter (of incident area) per Hertz (of optical bandwidth). Jan 22 at 9:08
• "White light" is a very ambiguous term for a physicist. A lot of different spectra look white due to metamerism. And, to add to the confusion, white color is also not unique due to different color temperatures of the white point one may choose. What exactly do you mean by "uniformly white light"? Jan 22 at 9:12
• @ProfRob: See here: en.wikipedia.org/wiki/…' Jan 22 at 9:17
• @RogerWood: I believe the SI unit is formally Wm^-3. I assumed the extra per meter is for wavelength. See above link. Jan 22 at 9:19

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$$^3$$).