If a source radiates $\newcommand{\e}{\mathrm e} \Phi_\e$ total energetic flux (in watts $\rm W$), then it is said to radiate luminous flux $\newcommand{\v}{\mathrm v} \Phi_\v$ defined by

$$\Phi_\v = K_{\rm cd} \int_{0\ \rm m}^\infty V(\lambda) \, {\partial\Phi_\e \over \partial\lambda} \, d\lambda$$


  • $K_{\rm cd} = 683\ \rm lm/W$ is the defining constant associated with photometric units;
  • $V(\lambda)$ is the luminosity function, a dimensionless weighting function scaled to unity at maximum that represents relatively how sensitive the average human eye is to any particular wavelength of electromagnetic radiation; and
  • $\lambda$ is the wavelength of the radiation.

What I understand

Luminous flux $\Phi_\v$ can be thought of as a measure of the ‘amount’ of a light a source emits. In addition, there are two important photometric quantities in terms of luminous flux:

  • Luminous intensity $I_{\v,\Omega}$, measured in $\rm lm/sr=cd$, represents the amount of luminous flux radiated through space per unit solid angle (where of course the solid angle is subtended with respect to the location of the source of radiation).
  • Illuminance $E_\v$, measured in $\rm lm/m^2 = lx$, represents the amount of luminous flux incident on a surface.


Why are both of these quantities of interest in photometry? It seems to me that they both measure ‘brightness,’ which decreases as the distance from a light source increases. If you could provide an answer with an example of how both might be necessary to understand a situation or with an explanation of how they are distinct with reference to human perception, I would appreciate that.


1 Answer 1


You have almost answered your question by stating that luminous intensity is to do with luminous flux emitted by a source whereas Illuminance is to do with the luminous flux from any number of sources hitting a surface.

Be careful about using the term brightness as that is a subjective quantity related to eye. For example think of what you would perceive about the brightness of a mobile telephone display when the display is in the dark as opposed to when the display is in sunlight.

  • $\begingroup$ [+1] Direct and concise. Speaking of perceived brightness, here’s this.... $\endgroup$ Apr 9, 2018 at 5:33
  • $\begingroup$ Followup question: Why do we not just look at the ‘lumens per steradian’ back off of a surface if we look at the ‘lumens per steradian’ form the source? $\endgroup$ Apr 9, 2018 at 14:27
  • $\begingroup$ @ChaseRyanTaylor Because the surface is not a "point" source. $\endgroup$
    – Farcher
    Apr 9, 2018 at 14:34
  • $\begingroup$ I drew a two-dimensional diagram and looked at “photons per steradian” versus “photons per metre of arc length” and everything—why luminous intensity is the same at all distances, why we look at illuminance—clicked for me! $\endgroup$ Apr 9, 2018 at 21:34

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