Spectral irradiance units conversion 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)?
 A: 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:
$$ W = \int_{\lambda_1}^{\lambda_2} I(\lambda) d\lambda $$
