= Spectral radiant flux
{title2=$\Phi_{e,\lambda},\ \Phi_{e,\nu}$}
<Spectral radiant flux> resolves <radiant flux> by <wavelength> or <frequency>:
$$
\Phi_e=\int_0^\infty\Phi_{e,\lambda}(\lambda)\,d\lambda
=\int_0^\infty\Phi_{e,\nu}(\nu)\,d\nu.
$$
Its units are $\mathrm{W\,m^{-1}}$ per metre of <wavelength> or $\mathrm{W\,Hz^{-1}}$ per hertz of <frequency>. A density quoted per nanometre must be integrated with <wavelength> intervals in nanometres; its numerical value is $10^{-9}$ times the per-metre density. The <frequency and wavelength spectral radiance conversion> applies equally to <spectral radiant flux>, through the <change of variables formula>:
$$
\Phi_{e,\nu}(\nu)=\Phi_{e,\lambda}(c/\nu)\frac{c}{\nu^2},
$$
with $c$ the <speed of light> in <vacuum> and $\Phi_{e,\lambda}$ expressed per metre. The <Jacobian determinant> preserves integrated <power> rather than the height or location of an <optical spectrum>'s peak.
For two explicitly synthetic <optical spectra>, let $x=\lambda/(1\,\mathrm{nm})$ and quote the following densities in $\mathrm{mW\,nm^{-1}}$:
$$
f_A(x)=\begin{cases}0.020(1-|x-500|/100),&400\leq x\leq600,\\0,&\text{otherwise},\end{cases}
\qquad
f_B(x)=\begin{cases}0.010,&600\leq x\leq800,\\0,&\text{otherwise}.\end{cases}
$$
The <triangle> under $f_A$ and the <rectangle> under $f_B$ both have <area> $2\,\mathrm{mW}$: $200\times0.020/2=200\times0.010=2$. Their band <radiant fluxes> differ: $450$–$550\,\mathrm{nm}$ contains $1.5\,\mathrm{mW}$ of $A$, while $650$–$750\,\mathrm{nm}$ contains $1.0\,\mathrm{mW}$ of $B$. These are specified functions, not measured source spectra.
\Image[/radiant-flux-synthetic-spectral-integrals.png]
{title=Equal total <radiant flux> can hide different band <radiant fluxes> and detector signals}
{id=radiant-flux-synthetic-spectral-integrals}
{description=Original synthetic <optical spectra> with exact total <radiant flux> $2\,\mathrm{mW}$ each. Darker shading marks the stated 100 nm bands. The lower panel uses a stipulated linear <spectral responsivity> $R(\lambda)=\bigl(0.2+0.001(x-400)\bigr)\,\mathrm{A\,W^{-1}}$; integrating it against the two <spectral radiant fluxes> gives dark-subtracted <electric currents> $0.60$ and $1.00\,\mathrm{mA}$. The response is illustrative and is not a calibration curve for a real <photodiode>.}
The dark-subtracted <electric current> of a linear <photodiode> is
$$
i=\int R(\lambda)\Phi_{e,\lambda}(\lambda)\,d\lambda,
\qquad [R]=\mathrm{A\,W^{-1}},
$$
using its <spectral responsivity>. The figure's stipulated $R$ gives $i_A=0.60\,\mathrm{mA}$ and $i_B=1.00\,\mathrm{mA}$ despite equal total <radiant flux>. Symmetry places $A$'s mean <wavelength> at $500\,\mathrm{nm}$ and $B$'s at $700\,\mathrm{nm}$; a linear $R$ averages to its value at that mean. A single <electric current> therefore cannot determine an unknown broadband <radiant flux> without additional spectral information or a sufficiently flat calibrated response.
Back to article page
