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I read that a blackbody at a temperature T, would emit thermal energy in the form of Electromagnetic waves. This thermal energy emitted per unit the area per unit time is called blackbody emissive power and is due to all possible wavelengths.

I then started with this term "spectral blackbody emissive power" which is the amount of thermal energy emitted in the form of EM waves of a particular wavelength, per unit area per unit time and per unit differential change in wavelength. I am not getting what's that differential change in wavelength.

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You can't sensibly talk about the power emitted per unit area at a single wavelength, $\lambda$, because at that one particular wavelength, no power will be emitted. Even a laser, commonly said to emit monochromatic light, emits over a narrow band of wavelengths, so since that band can be infinitely sub-divided, at any one discrete wavelength, zero power is emitted.

So we consider a very narrow band, $\Delta \lambda$, of wavelengths that is centred on $\lambda$. A finite amount of power, $\Delta P$, will be emitted in this band. But, the power will be proportional to the width of the band, so to get a figure that depends on the source itself and not on our choice of band-width, we divide the power within the band by $\Delta \lambda$. This, or strictly $\lim \limits_{\Delta \lambda\to 0} \frac{\Delta P}{\Delta \lambda}$, gives the spectral power per unit area in units of $\text W\ \text m^{-2}\ \text m^{-1}$.

Be aware that you could also give spectral powers per unit frequency band.

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  • $\begingroup$ Found this really interesting. This wasn't mentioned in any of my graduation level books. Could you provide me any source in which this is discussed in detail? $\endgroup$
    – Harshit Rajput
    Commented Mar 29, 2021 at 12:21
  • $\begingroup$ I don't think that any more detail is possible! I've looked through a few statistical mechanics textbooks, and they do seem to gloss over it. But it's implicit in the treatments of black body radiation given by Reif, Hill, Mandl, Huang... $\endgroup$
    – Philip Wood
    Commented Mar 29, 2021 at 12:43
  • $\begingroup$ Actually I was looking for an answer to why we can't define power for a single wavelength. $\endgroup$
    – Harshit Rajput
    Commented Mar 29, 2021 at 12:45
  • $\begingroup$ I've expanded the first bit of my answer. $\endgroup$
    – Philip Wood
    Commented Mar 29, 2021 at 12:53
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    $\begingroup$ Because if the emission is uniform across a band of wavelengths ($\lambda_2-\lambda_1$, say), the proportion of that power that will be emitted over a narrow band of width $\Delta \lambda$ somewhere between $\lambda_1$ and $\lambda_2$, will be $\frac {\Delta \lambda}{\lambda_2-\lambda_1}$, and if $\Delta \lambda$ is zero, that is for one discrete wavelength, the proportion of that power will be zero. $\endgroup$
    – Philip Wood
    Commented Oct 11 at 16:51

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