Converting Radiometric to Photometric Units

When first approaching photographic science a photographer is often confused by the unfamiliar units used.  In highs school we were taught energy and power in radiometric units like watts (W) – while in photography the same concepts are dealt with in photometric units like lumens (lm).

Once one realizes that both sets of units refer to the exact same physical process – energy transfer – but they are fine tuned for two slightly different purposes it becomes a lot easier to interpret the science behind photography through the theory one already knows.

It all boils down to one simple notion: lumens are watts as perceived by the Human Visual System.

They both indicate a certain amount of energy transferred per second, except that the lumen counts only the portion of energy that can be perceived by humans.  Once this concept is absorbed the rest is easy because every single photometric SI unit is just a lumen over a period of time or space – therefore if one knows how to convert watts to lumens one can convert virtually all radiometric units relevant to photography to photometric units in exactly the same way (space and time are assumed to stay the same:-)

For instance the units of exposure are lx-s which are none other than lm/m^2 \cdot s; units of Luminance are cd/m^2 which are just lm/m^2 \cdot sr); and so on.

Converting  a radiometric function \Phi_e(\lambda) in watts (per whatever) to a photometric function \Phi_v in  lumens (per whatever) is simple: it is the energy flux in watts (per whatever) weighted by the photopic eye response function V(\lambda).  In mathematical terms for daylight, photopic vision:

(1)   \begin{equation*} \Phi_v = K_m \int\limits_{380}^{780} \Phi_e(\lambda) V(\lambda}) d\lambda \: lumens \end{equation*}

where \Phi_v in lumens (per whatever) is the photometric version of \Phi_e(\lambda) in watts (per whatever) and K_m a conversion constant equal to 683.002 lm/W.

The key to conversion is the adapted eye response function V(\lambda) which looks as follows for daylight as measured by Sharpe at all (2005)

Photopic Sharpe

Because it is multiplied together with the radiometric  function that we are trying to convert, when V(\lambda) is zero the output in lumens is also zero no matter what the input in watts.  It is virtually zero below 395nm and above 718nm, where its value is 10 stops below the peak at 556nm.  So for instance a 1mW laser that emits all its power at 556 nm produces 0.683 lumens, while an infrared 1mW laser that emits all its power at 750nm produces virtually none.  You get the idea: same power, different lumens based on eye response.

That’s for lasers which are not often found in nature by a photographer:-)  In the typical case of broad spectrum sources one needs to know their Spectral Power Distribution in watts per whatever as a function of wavelength \lambda and plug it into formula (1) as \Phi_e(\lambda) to convert it to photometric units.  Next a practical example.