Units of Wavelength

For many applications, particularly when dealing in the infrared region of the spectrum, the preferred spectral unit is wavelength in μm, \(\lambda\) = 10⁶ c/\(ν\). We can deduce the spectral radiance per μm, \(L_\lambda\), from (1) by noting that

\[L_{\lambda}|d \lambda|=L_{v}|d v| \hspace{5cm} L_{\lambda}=\left|\frac{d v}{d \lambda}\right| L_{v}=\frac{10^{6} c}{\lambda^{2}} L_{v}\]

With this, and substituting \(ν = 10^6 c/\lambda\) into (1), the spectral radiance per μm is:

\[L_{\lambda}=\frac{2 \times 10^{24} h c^{2}}{\lambda^{5}} \frac{1}{e^{10^{6} h c / 2 k T}-1} \mathrm{W} \mathrm{m}^{-2} \mathrm{sr}^{-1} \mu \mathrm{m}^{-1} \label{g}\tag{7}\]

To find the wavelength of the peak, we set the derivative to zero:

\[0=\frac{d L_{\lambda}}{d \lambda}=\frac{-10^{25} h c^{2}}{\lambda^{6}} \frac{1}{e^{10^{6} h c / \lambda k T}-1}+\frac{2 \times 10^{24} h c^{2}}{\lambda^{5}} \frac{\left(10^{6} h c / \lambda^{\lambda} k T\right) e^{10^{6} h c / \lambda k T}}{\left(e^{10^{6} h c / \lambda k T}-1\right)^{2}}\]

\[0=5-\frac{10^{6} h c}{\lambda k T} \frac{e^{10^{6} h c / \lambda k T}}{e^{10^{6} h c / \lambda k T}-1}\]

Letting \(x = 10^6hc/ \lambda k T\), we arrive at the transcendental equation \(5(1 - e^{-x}) = x\), whose numerical solution, \(x = a_5 ≈ 4.96511423174\) provides

\[\lambda_{\text {peak}}=\frac{10^{6} h c}{a_{5} k T} \quad \mu \mathrm{m} \label{h}\tag{8}\]

The peak value, found by substituting (\ref{h}) into (\ref{g}), is

\[L_{\lambda, \text { peak }}=\frac{2 \times 10^{24} h c^{2}}{\left(\frac{10^{6} h c}{a_{5} k T}\right)^{5}} \frac{1}{e^{10^{6} h c /\left(\frac{10^{6} h}{a_{5} k T}\right) k T}-1}\]

\[L_{\lambda, \text { peak }}=\frac{2 a_{5}^{5} k^{5}}{10^{6} h^{4} c^{3}} \frac{1}{e^{a_{5}}-1} T^{5} \quad \mathrm{W} \mathrm{m}^{-2} \mathrm{sr}^{-1} \mu \mathrm{m}^{-1} \label{i}\tag{9}\]

As we did above with spectral units of Hz, we can derive these radiometric quantities in terms of photons per second. Dividing (\ref{g}) by the energy of a photon, \(10^6 hc/ \lambda\), gives the spectral photon radiance,

\[L_{\lambda}^{p}=\frac{2 \times 10^{18} c}{\lambda^{4}} \frac{1}{e^{10^{6} h c / \lambda k T}-1} \text { photon } \mathrm{s}^{-1} \mathrm{m}^{-2} \mathrm{sr}^{-1} \mu \mathrm{m}^{-1} \label{j}\tag{10}\]

The wavelength where this peaks is found by differentiating:

\[0=\frac{d L_{\lambda}^{P}}{d \lambda}=\frac{-8 \times 10^{18} c}{\lambda^{5}} \frac{1}{e^{10^{6} h c / \lambda k T}-1}+\frac{2 \times 10^{18} c}{\lambda^{4}} \frac{\left(10^{6} h c / \lambda^{2} k T\right) e^{10^{6} h c / \lambda k T}}{\left(e^{10^{6} h c / \lambda k T}-1\right)^{2}}\]
\[\begin{array}{l} 0=1-\frac{10^{6} h c}{4 \lambda k T} \frac{e^{10^{6} h c / \lambda k T}}{e^{10^{6} h c / \lambda k T}-1} \quad \text { let } \quad x=10^{6} h c / \lambda k T \newline 4\left(1-e^{-x}\right)=x \newline x=a_{4} \approx 3.92069039487 \end{array}\]
\[\lambda_{\text {peak }}^{P}=\frac{10^{6} h c}{a_{4} k T} \quad \mu \mathrm{m} \label{k}\tag{11}\]

The peak spectral photon radiance is

\[L_{\lambda, \text { peak }}^{P}=\frac{2 \times 10^{18} c}{\left(10^{6} h c / a_{4} k T\right)^{4}} \frac{1}{e^{\frac{10^{6} h c}{\left(10^{6} h c / a_{4} k T\right) k T}}-1} \label{l}\tag{12}\]

\(L_{\lambda}\) and \(L_{\lambda}^{P}\) are shown in Fig. 2. Note that as with units of Hz, the spectral radiance and spectral photon radiance have different behaviors, and distinctly different temperature dependences.

Fig. 2-- Spectral radiance, \(L_{\lambda}\), (top) and the spectral photon radiance, \(L_{\lambda}^{P}\), (bottom) as a function of wavelength, \(\lambda\), for various temperatures. The small black dots indicate the wavelength and value of the peak, at 10 K temperature intervals. Note that \(L_{\lambda}\) and \(L_{\lambda}^{P}\) have different wavelength dependences. Although the peak wavelength is inversely proportional to \(T\) for both quantities, \(L_{\lambda}^{P}\) peaks at a longer wavelength than \(L_{\lambda}\). Furthermore, the peak value of \(L_{\lambda}\) increases as \(T^5\), whereas the peak value of \(L_{\lambda}^{P}\) increases as \(T^4\).