= Phosphor-converted white light-emitting diode
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 is
$$
\frac{P_{\rm after}}{P_{\rm before}}
=(1-a)+a\Phi\frac{\lambda_b}{\lambda_c}
=0.40+0.60\times0.90\times\frac{450}{560}
=0.833929.
$$
The lower <energy> of each converted <photon> matters even if $\Phi=1$. Combining this stipulated converter with the preceding chosen <light-emitting diode> gives $P_{\rm after}=16.5430\,\mathrm{mW}$ 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.
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