Optical output power of LED

[daPoseidonGuy]

Hi, i was doing a science project on LEDs, and i needed to calculate optical output power versus input electrical power to find relative wall plug efficiency.
This is the only good formula i found
Optical output power of LED (watts) =Nlinearfactor × Voltage drop across resistor (volts)
Pout = N × Vres
What is N exactly? I do not know how to calculate optical output power as I dont know what help is. Id appreciate help a lot, thanks.
All the values I have are experimental. Im trying to calculate the relative wall plug efficiency of an LED. heres my data right now.
I do not have a data sheet and the steps Im following at this point are somewhat modeled of those shown here:
http://www.sciencebuddies.org/science-fair-projects/project_ideas/Energy_p003.shtml#procedure
in the testing and data collection section.

Voltage across resistor (V) ± .01 = 2.49

Distance from photocell to light (cm) ± .05 = 4.00

Voltage across light (V) ± .01 = 5.75

Current intensity (mA) ± .01 = 360
If N is just something I have to leave as a variable, then what is the point of measuring the distance between the photocell and the light bulb? In the experiment I moved the breadboard closer or further to get 2.5v across the resistor, as thats what I understood from the procedure. Was I supposed to do that? What do I do with the value for the distance between the photocell and light bulb?
This project comes from here btw:
http://www.sciencebuddies.org/science-fair-projects/project_ideas/Energy_p003.shtml#procedure

Edited December 27, 2015 by daPoseidonGuy

[studiot]

OK so do you understand what the first equation you posted means and where it come from?

 

 

Optical output power of LED (watts) =Nlinearfactor × Voltage drop across resistor (volts)

 

I ask because in your project instructions they tell you where it comes from immediately before they quote it.

 

Why are do they have you measuring the voltage across a resistor?

What does that tell you?

Edited December 27, 2015 by studiot