# Homework Help: Relation between lux and W/m^2

1. Feb 25, 2016

### says

1. The problem statement, all variables and given/known data
I posted this in anther forum but I think it might be better to post it here.

I was wondering if anyone could help me understand the conversion between lux and W/m2 with relation to luminous efficacy (lm/W).

Thanks :)

2. Relevant equations

3. The attempt at a solution

2. Feb 25, 2016

### haruspex

It's not clear where your difficulty lies. Do you understand the relationship between lux and lumens?
Watts measure power, while lumens measure the response of the eye to that power. This response will depend on the wavelengths involved.

3. Feb 25, 2016

### Merlin3189

I'm no expert on this, but as I understand it
1 Lux is 1 Lumen per m2
1 Lumen is 1 Candela . Steradian
1 Candela is equivalent to $\frac{1}{683}$ watt per steradian of green light (555nm)

So 1 lux is $\frac{1}{683}$ watt/m2 for that green light, but this gets a bit complicated, because the definition of the candela in watts is at a specific frequency. The lumen weights different frequencies according to their visual effect (sensitivity of the eye), so other colours need greater powers.
I think this article (Energy Efficiency of LEDs) might give you the sort of info you need. It explains the intricacies of luminous efficacy (both sorts) and gives examples of the sort of data you're interested in.

4. Feb 26, 2016

### says

I got asked how to convert lux to W/m2, and how it relates to luminous efficacy (lm/W). I didn't think you could convert lux to W/m2, hence my confusion.

It's related to a solar energy class. I think @Merlin3189's reply was what I was looking for though.

5. Feb 26, 2016

### haruspex

Ok. If you had written that in the first place I would have replied lm/W=(lm/m2)/(W/m2).

6. Feb 26, 2016

### Merlin3189

Haruspex is right of course. That's what makes light such a messy business.
Lux, lumen and candela all refer to the effect of light on an eye, which does not have a simple relation to its power.

"I didn't think you could convert lux to W/m2, ..." and it is indeed a dubious matter, because lights of equal lux might (probably would) have different W/m2 values.
But you can calculate corresponding values for a given light source. If the power in each frequency step is multiplied by "the" eye's sensitivity at that frequency, the sum of these divided by 683W will tell you the lux value of that light.
So two sources emitting the same power in the visible spectrum can have different luminance because their spectral distributions are different. For example a daylight and warm light source might emit the same power in visible light, but have different luminance.

What the validity is of any of these measurements, I wonder. I think a 5W, 395 lumen daylight LED looks brighter than a 5W, 370 lumen warm white LED, but they're different colours as well. So if I tried to compare them with a Bunsen or Joly photometer, I would never get a point where they were indistinguishable.
The ubiquitous electronic photometers used to measure luminance now depend on filters to correct the spectral response of silicon sensors or use CdS sensors, which can be similar to "the" eye's sensitivity (and can of course be further corrected by filters.) Their measurements should be impartial, though comments about filter optimisation for different light sources, makes me wonder just how well matched they are.