Spectral mismatch correction ($F={\Phi }_v/{\Phi }_{v,\mathrm{indicated}}$)

For collected spectral radiant flux $f(\lambda )$ and a linear photodiode with spectral responsivity $r(\lambda )$, the dark-subtracted electric current is $i=\int rfd\lambda$. A reference optical spectrum $f_A$ sets the conversion $C_A$ from electric current to collected luminous flux. Applying it to another optical spectrum $f_T$ givesAssume nonzero calibration and test signals. Then $F{\Phi }_{v,\mathrm{indicated}}$ is the corrected luminous flux. Multiplying either optical spectrum by an overall positive scale leaves $F$ unchanged. If $r$ is proportional to $V$, $F=1$ for every optical spectrum. Uniformly overfilled aperture measurements use wavelength-resolved irradiance instead, multiplying the electric current expression by aperture area; the same ratio cancellation gives the illuminance correction. This is the algebra behind source-dependent calibration, rather than a new measured correction curve. ^[ref]

Use the triangular $w$ in equal radiant power under visual weighting and stipulate the detector's spectral responsivityThis model has nonnegative response: $w$ is zero outside $405$–$705\mathrm{nm}$. For a $1\mathrm{mW}$ ideal calibration line at $555\mathrm{nm}$, $i_A=0.5\mathrm{mA}$ and ${\tilde{\Phi }}_{v,A}=0.683\mathrm{lm}$, so $C_A=1366\mathrm{lm}{\mathrm{A}}^{-1}$. An equal-radiant flux test line at $630\mathrm{nm}$ gives $w=1/2$, $r=0.375\mathrm{A}{\mathrm{W}}^{-1}$ and $i_T=0.375\mathrm{mA}$. The unchanged conversion indicates $0.51225\mathrm{lm}$, whereas the model's intended weighted result is $0.3415\mathrm{lm}$. The required spectral mismatch correction is $F=2/3$: the uncorrected result is 50% too large. All these numbers are declared model outputs, not instrument measurements or standard photometry data.