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The luminous efficacy of radiation is luminous flux divided by total radiant flux, in $\mathrm{lm}{\mathrm{W}}^{-1}$, for the specified visual conditions. For nonzero radiant flux, it is a spectral weighted mean of spectral luminous efficacy. Specify the denominator: a lamp's luminous efficacy relative to electrical input power also includes conversion losses and is a different ratio. ^[ref]

To isolate conversion losses, suppose a converter receives 100 blue photons at ${\lambda }_b=450\mathrm{nm}$, absorbs a fraction $a=0.60$, and has fluorescence quantum yield $\Phi =0.90$: 0.90 converted photons per absorbed blue photon. Stipulate that every converted photon has ${\lambda }_c=560\mathrm{nm}$, that unabsorbed blue is transmitted, and that both outputs escape without further losses. The output is then 40 blue and 54 converted photons. The total photon yield is $0.94$, while the retained optical energy fraction isThe lower energy of each converted photon matters even if $\Phi =1$. Combining this stipulated converter with the preceding chosen light-emitting diode gives $P_{\mathrm{after}}=16.5430\mathrm{mW}$ and an overall optical/electrical power ratio $0.258485$. These chosen narrow lines and photon counts illustrate conversion accounting; they do not establish a white appearance, color quality or luminous efficacy. Those require the full spectrum and a model of human visual sensitivity.

The emitted optical power divided by electrical input power. For the bare-device calculation in light-emitting diode efficiency, $P_{\mathrm{elec}}=I{V}_f$; a complete lamp also consumes power in its driver. Luminous efficacy, measured in lumens per watt, additionally weights the spectrum by human visual sensitivity and is a different quantity.