calculate the illuminance of a cone with solid angle 0.2 deg

[chrisewolf]

Hey guys, I am (again) a bit confused about a topic in photometry and I hope someone can help me out!

 

I am taking measurements under certain angles to the surface normal of an LED and measure the Luminance L(cd/m^2)=(lm/m^2/sr). To obtain the illuminance I have to integrate over the solid angle. I know from the detector that the measurement angle is 0.2 degrees. The luminance does not change over azimuth so I integrated that out right away and I centered the azimuth at x.

 

lets say I took a measurement under an angle to the surface normal of 50 degrees and I measured L=150 cd/m^2

 

can someone check the attached calculation and comment if that is right?

 

Thank you very much in advance

 

Cheers,

Chris

[Alper Ülkü]

To me it does not seem correct. I think you got the solid angle wrong. 0.2 steradian solid angle defines a detector with apex angle 14.5 degrees, quite uncommon.

 

Illuminance = Omega. Luminance . cos (theta)

 

If Omega as you say is 0.2 then Ev = 0.2 * Lv * cos (theta) = 19.28 lm / m2 (Ev/Lv factor is not 2/25:0.08 but is 0.1285)

 

If 0.2 degree is the angle subtended by the detector which is generally the case; then Omega becomes = 2pi (1 - cos 0.2/2) = 9.5e-6 sr, Ev = 150 x 9.5e-6 x cos (50) = 9.16e-4 lm/m2