This colorSpec vignette is a response to the question:
Is there any program that given the lumen and the colour spectrum (relative intensity at any wavelength) of a light source and it outputs the total amount of photons per wavelength?
that was asked on a stackexchange forum, see @174314.
Featured functions in this vignette are: actinometric()
and photometric()
.
library( colorSpec )
Read the SPD for a domestic LED bulb from the
Lamp Spectral Power Distribution Database, see @LSPDD.
The unique ID is
LED_A19_dim_domestic-use_softwhite-color_CREE-brand_6W_120V_75Lum_2700K_bernard
.
wave = 275:899 path = system.file( "extdata/sources/Cree-LED.txt", package="colorSpec" ) bulb = readSpectra( path, wavelength=wave ) par( omi=c(0,0,0,0), mai=c(0.6,0.8,0.1,0.1) ) plot( bulb, main='' )
The energy unit is unknown. But it does not matter, because we are now going to scale so that the luminous flux of the SPD is 75 lumens, as specified by the manufacturer.
luminous.flux = photometric( bulb ) ; luminous.flux
The unit is lumen, but the first one - photopic1924
- is the
appropriate official standard.
Scale spectrum to have 75 lumens.
bulb = multiply( bulb, 75/luminous.flux[1] ) ; bulb # check that the luminous flux is now 75 lumens photometric( bulb )
From the man page for photometric()
we know that the unit for bulb
is
$\textrm{watt} * \textrm{nm}^{-1}$.
par( omi=c(0,0,0,0), mai=c(0.6,0.95,0.1,0.1) ) ylab = expression( 'Radiant Power' ~~~ '[watt * ' ~ nm^-1 ~ ']' ) plot( bulb, main='', ylab=ylab )
The unit is currently energy-based (energy of photons, aka radiometric),
but we need photon-based (number of photons, aka actinometric).
From the man page for actinometric()
we know that the output unit is
$\mu \textrm{mole} * \textrm{sec}^{-1} * \textrm{nm}^{-1}$.
bulb = actinometric( bulb ) par( omi=c(0,0,0,0), mai=c(0.6,0.9,0.1,0.1) ) ylab = expression( 'Photon Flux' ~~~ '[' ~ mu ~ 'mole * ' ~ sec^-1 ~ nm^-1 ~ ']' ) plot( bulb, main='', ylab=ylab )
This plot looks similar but note that $\lambda_{max}$ has moved slightly higher. For the total number of photons/sec from this bulb, compute the integral over $\lambda$.
bulb
So the total photon flux of the bulb is $1.102559 ~ \mu \textrm{mole/sec}$. To convert this to exaphotons, multiply by $0.602214 ~ \textrm{exaphotons/} (\mu \textrm{mole of photons} )$ to get $0.663976 ~ \textrm{exaphotons/sec}$.
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