# Help calculating watts/m2 from radiant flux



## thepaan (Feb 24, 2010)

I'm trying to calculate the watts/m2 of an LED. I found many a page telling how to calculate lux but the spec sheet lists radiant flux (Φ) instead of lumens. What I'm unclear on is calculating the loss in intensity based on distance. For example, one spec sheet for a blue LED lists 5W Φ. Does this mean, if I focus all the light to a one square meter area, at one meter there will be 5W Φ? What is the distance those 5W Φ were measured at and over what area?


----------



## Benson (Feb 24, 2010)

thepaan said:


> I'm trying to calculate the watts/m2 of an LED. I found many a page telling how to calculate lux but the spec sheet lists radiant flux (Φ) instead of lumens. What I'm unclear on is calculating the loss in intensity based on distance. For example, one spec sheet for a blue LED lists 5W Φ. Does this mean, if I focus all the light to a one square meter area, at one meter there will be 5W Φ? What is the distance those 5W Φ were measured at and over what area?



5W is the total; it doesn't matter what distance and area. If it all hits one square meter, it'll be 5W/m^2 average; if it hits 100m^2 it will be 0.05W/m^2 average.

If you're looking for a peak, rather than average, intensity, you'll need to look at the beam plot (relative intensity vs. off-axis angle), model it as cos(theta) or similar, and integrate that over a hemisphere to get the ratio of peak intensity to average over a hemisphere, and go back to the total output to get W/sr at the center. Then since 1 m at a distance of 1m is 1sr, you can get the beam-center intensity in W/m^2 at any distance by the inverse-square law...

Hope that made sense...


----------



## thepaan (Feb 24, 2010)

So, if I have two cylinders with 100% reflective walls and put a 5W LED inside one end. If one cylinder is 1 meter long and the other is 10 meters long then there will be 5W of light comming out the other end of both?

That doesn't seem right to me.


----------



## znomit (Feb 25, 2010)

thepaan said:


> So, if I have two cylinders with 100% reflective walls and put a 5W LED inside one end. If one cylinder is 1 meter long and the other is 10 meters long then there will be 5W of light comming out the other end of both?



Thats correct.


----------



## spencer (Feb 25, 2010)

Only assuming the air is perfectly clean. If anything absorbs any light at all (like dust or something) then 5W won't be coming out the other end.


----------



## gcbryan (Feb 25, 2010)

thepaan said:


> So, if I have two cylinders with 100% reflective walls and put a 5W LED inside one end. If one cylinder is 1 meter long and the other is 10 meters long then there will be 5W of light comming out the other end of both?
> 
> That doesn't seem right to me.



That's because you can't have more light coming out than the led produces.


----------



## bshanahan14rulz (Feb 25, 2010)

think of it this way: energy and matter are the same thing, and you can't delete it. energy either turns into matter, or a different kind of energy, but it has to go somewhere

In your perfectly reflective tube example, if your LED is putting out 5W of light, and all light is directed into the tube, then yes, 5W of light will come out the other end. Add air, and reflective losses, and less than 5W of light will come out the other end; the rest turns into heat as it is absorbed into the tube or air.


----------



## thepaan (Feb 25, 2010)

I guess it makes sense - I was just expecting it to be a bit more complicated than that....


----------



## znomit (Feb 25, 2010)

thepaan said:


> I guess it makes sense - I was just expecting it to be a bit more complicated than that....



Note that the "5W" label usually refers to the power in, not the light out.


----------



## thepaan (Feb 26, 2010)

Yup. I was just using that for an even number. The one I'm looking at is labeled 10W but puts out 1.8W red at 700mA.


----------

