I have a table of data containing irradiance of light at different wavelengths. This is how it looks like for 300.5 nm:

• Wavelength, nm: 300.5
• Wavelength, $\mu$m: 0.3005
• W/m$^2$/$\mu$m: 403
• W/m$^2$/nm: 0.403

My question is, how can I convert W/m$^2$/nm or W/m$^2$/$\mu$m to W/m$^2$? And what exactly do those other units mean (W/m$^2$/nm)?

• The close vote is harsh. I recall being puzzled by this myself in my days of fledgling nerddom. – John Rennie Nov 4 '14 at 17:20
• Maybe the simplest short explanation is: $W/m^2$ is the power per unit area. The other denominator, $/nm$ , refers to the bandwidth, or range of wavelengths in question. While the choice of units looks odd, just remember that they represent different things (area vs. wavelengths). To get total power, integrate over wavelengths. – Carl Witthoft Nov 4 '14 at 18:18

All sources of light have a spread of wavelengths. There is no such thing as a light source that produces light of exactly one wavelength. Let's assume that the power emitted by your light source looks like this:

I just made up this curve, but the shape of the curve doesn't matter for this discussion. The $y$ axis shows the spectral irradiance and as you say this has units of $\text{W}/\text{m}^2/\text{nm}$. The power emitted at exactly $300.5$ nm is zero, but the power emitted over all wavelengths between $\lambda = 300.3$ nm and $\lambda = (300.5 + \delta\lambda)$ nm is the area under the curve between the dashed lines, that is:

$$W(300.5\text{ to }300.5+\delta\lambda) \approx I(300.5)\delta\lambda$$

And the units of $I(\lambda)\delta\lambda$ are indeed $\text{W}/\text{m}^2$ as we'd expect for power.

More generally, to get the power emitted over a range of wavelengths from $\lambda_1$ to $\lambda_2$ you have to integrate the spectral radiance:

$$W = \int_{\lambda_1}^{\lambda_2} I(\lambda) d\lambda$$

• I dont get it, why would the power emitted at a particular wavelength be zero? If there was a light source, that emitted only light at 300.5 nm. Its power output on a surface would be... zero? – user1806687 Nov 4 '14 at 17:48
• From a practical perspective, any use of irradiance occurs over some range of wavelengths. Something, somewhere, limits the range, and the analysis above must be carried out. The data presented in the OP gives the spectral irradiance at only one wavelength. This might imply that in the context of wherever the data came from, the spectral range is expected to be small. (Examples of such a things would be light through narrow band filters, or spectrometer slits.) Of course, one would then use the first of the two expressions. – garyp Nov 4 '14 at 17:52
• @user1806687: All light sources have a finite width of wavelengths. If nothing else, the uncertainty principle ensures we can never get perfectly monochromatic light. If you take only light between wavelengths $\lambda$ and $\lambda + \delta\lambda$ then you are taking only part of the emitted light, so the power is less than the total power emitted. As you let $d\lambda \rightarrow 0$ the percentage of the total power you are taking goes to zero. – John Rennie Nov 4 '14 at 17:55
• @user1806687: this puzzled the hell out of me when I first encountered it. And there wasn't even a Physics SE back in the late seventies to help me with it :-) – John Rennie Nov 4 '14 at 18:06
• @JohnRennie, yeah, but there was Hecht and Zajac :-) – Carl Witthoft Nov 4 '14 at 18:20