As part of my physics project, I investigated the relationship between a light bulb's illuminance and the distance from the measuring device to it.

Illuminance was measured in flux, and distance in metres.

I derived an inverse relationship. First of all, can I check if this is the relationship I was supposed to get? (I understand that there is an inverse square relationship between luminosity and distance, but since illuminance is measured in lumens/sq metre, then I think an inverse relationship is correct...)

Next, using this relationship, I need to extend my project so that I can find distances to stars using this relationship.

As long as 2 stars have a similar mass/size/density, then I can use this relationship to find the distance to the 2nd star as long as the illuminance of the stars are known, and the distance to the 1st star is known.

So I tried to do this with an example - the Sun and Tau Ceti, as they are reasonably similar.

I have values for the Sun's distance/illuminance - $1.5\!\times\!10^8 $km, $1.2\!\times\!10^5$lux. I now am trying to calculate the distance to Tau Ceti using this method.... But nowhere can I find a value for its illuminance.

Please can someone help me with this? Is there a way to convert from apparent magnitude or luminosity to illuminance?

Thank you

EDIT: The method I want to use from the inverse relationship is:


where $d_1$ and $l_1$ are the known distance and illuminance of the 1st star, whilst $d_2$ and $l_2$ is that of the second, and $d_2$ is unknown.

  • $\begingroup$ You're probably looking for Tau Ceti's en.wikipedia.org/wiki/Absolute_magnitude $\endgroup$
    – user21
    Commented Dec 2, 2015 at 12:49
  • $\begingroup$ I'm not looking for that, as I'm trying to convert to illuminance (measured in lux) $\endgroup$
    – Shuri2060
    Commented Dec 2, 2015 at 21:31
  • $\begingroup$ Using the absolute magnitude of Tau Ceti, you can determine how bright it is compared to the sun, which should let you determine the luminance, since you already know the Sun's luminance. $\endgroup$
    – user21
    Commented Dec 3, 2015 at 2:35
  • $\begingroup$ Thank you, but as stated above, I need to use the illuminance due to the nature of my project. $\endgroup$
    – Shuri2060
    Commented Dec 3, 2015 at 19:21

1 Answer 1


You can easily convert flux to luminosity and vice versa using the spherical nature of the propagation of light. The equation you need is:

$$ L=4\pi D^{2}f $$ Where $f$ will be your flux, and $D$ is the distance to the object, Tau Ceti in this case.

From Pijpers (2003) Tau Ceti's luminosity is $0.52\pm0.03L_{\odot}$. You should have everything you need there. $L_{\odot}$ is of course the luminosity of the Sun.

  • $\begingroup$ I don't quite understand (I'm not that good at Physics). I'm looking to convert to illuminance (measured in lux). Is this the same as flux? $\endgroup$
    – Shuri2060
    Commented Dec 2, 2015 at 21:30
  • $\begingroup$ Illuminance is the total luminous flux incident on a surface, per unit area. So $I=f/A$, so what is your area, $A$, of your surface? (then you can re-arrange my equation for to obtain the luminosity)... $\endgroup$ Commented Dec 2, 2015 at 21:46
  • $\begingroup$ I still don't quite understand and don't see how area is involved in this..... The original experiment involved me using a light meter and a light bulb with the distance increasing between them at each reading. The light meter read in lux, and I derived an inverse relationship..... But no area involved. I am told by Wikipedia that the illuminance of the Sun is around 120,000 lux at its brightest. Now, knowing also the distance to the Sun, I expected that if the illuminance of Tau Ceti was known, I could just use these 3 values to find the distance to Tau Ceti. $\endgroup$
    – Shuri2060
    Commented Dec 2, 2015 at 22:40
  • $\begingroup$ But I can't see how area is involved in this $\endgroup$
    – Shuri2060
    Commented Dec 2, 2015 at 22:41
  • $\begingroup$ So illuminance is defined as the flux per unit area. Your measuring device will have a light capturing device which has a well defined area, maybe a few square inches. So, you get your illuminance. If you convert to a flux by knowing the capture area of your device you can then use all the known variables and the equations given to start calculating distances. Maybe you can't use the method you have prescribed? $\endgroup$ Commented Dec 2, 2015 at 22:45

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .