# Converting lux to watts with known wavelength

I was wondering if there was a possible way to convert lux to watts accurately if the wavelength is known.

I know that for the sun there is an approximate conversion of 0.0079W/m$^2$ per Lux using standard 555nm wavelength but I can't find or work out how to do the conversion.

• Wavelength of what? A single-wavelength source like a laser? – garyp Aug 30 '17 at 18:32
• You need the value of the luminosity function for that wavelength: are you interested in low light or well light areas? Because the luminosity function aren't the same. – user154997 Aug 30 '17 at 18:46
• @garyp the wavelength of light that has passed through a coloured filter – Terminus Est Aug 31 '17 at 13:06

Let's give it a try. The flux per unit of surface is given in lux as

$$\Phi = 683 \int_0^{+\infty} V(\lambda)P(\lambda)d\lambda$$

where $P(\lambda)$ is the spectral flux per unit of surface and $V(\lambda)$ is the luminosity function. The latter takes into account the perception of light by the human eye whereas the former is a property of the light source, i.e. $P(\lambda)d\lambda$ is the power emitted in the wavelength range $[\lambda, \lambda+d\lambda]$ per square meter. Your assumption of monochromaticity would amount to use a delta function for $P$. Thus denoting by $\lambda_0$ the wavelength, and by $P_0$ the power in W/m$^2$,

$$\Phi = 683 V(\lambda_0) P_0.$$

So, we just need to know the luminosity function. The standard reference is the Judd-Vos function [JV78]. Wikipedia has the curves for low light (green curve) and well light (black curve) conditions.

[JV78] J. J. Vos. Colorimetric and photometric properties of a 2◦ fundamental observer. Color Research & Application, 3(3):125–128, 1978.

• That has really helped, thanks! However I am confused with the dimensions- the maximum luminous efficiency of 683 has units of lm/W, the power has units of W/$m^2$, wavelength would be in meters and the luminosity function is dimensionless meaning lux would have dimensions lm/m rather than lm/$m^2$ – Terminus Est Aug 31 '17 at 13:00
• I think you misinterpreted my last equation: $V(\lambda_0)$ means the value of the luminosity function at the wavelength $\lambda_0$, not the product of $V$ by $\lambda_0$, as your comment suggest you have interpreted my formula. Or am I missing something? – user154997 Aug 31 '17 at 13:08
• Ah yes that makes sense, I misinterpreted. Does this mean that finding the wattage per unit area is simply a case of rearranging to P0=Φ/683V(λ0)? – Terminus Est Aug 31 '17 at 13:59
• Yes, indeed. I forgot you asked the conversion in that direction. – user154997 Aug 31 '17 at 14:01
• Not quite a delta function. You forgot the band width. I think it should be $\Phi = 638\,V(\lambda_0)P(\lambda_0)\Delta\lambda$ where $\lambda_0$ is the peak transmission of the filter and $\Delta\lambda$ is the passband. Assuming, of course, a narrow band filter. Note that the units of your expression are W/m${}^2$/m ... flux density per wavelength interval. – garyp Aug 31 '17 at 14:36