What is the formula for calculating radiant intensity from an LED onto a body?

AI Thread Summary
To calculate the radiant intensity from an LED onto a body, one can use basic geometry and unit conversions. Given an LED with a radiant intensity of 80mW/Sr, a diameter of 5mm, and a beam divergence of +/- 20 degrees, the area of illumination can be approximated by treating the LED as a point source. The total radiant flux of 25mW can then be divided by the calculated area to determine the power per unit area. The discussion highlights the importance of understanding the geometry involved in the calculations. Ultimately, the correct radiant intensity per square centimeter is approximately 2.4mW/mm².
rayellam
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Hi,
I’m hoping some genius can help me out here. I’m trying to determine how many joules of energy per square centimeter (from an LED) are radiated onto a body given the following parameters.

LED radiant intensity = 80mW/Sr
LED diameter = 5mm
LED beam +/- 20 degrees

Wavelength = 880nM
Total radiant flux = 25mW
Distance from LED to body = 5mm

Thanks
Ray
 
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Simple geometry (and unit conversions) suffices, genius has nothing to do with it.
 
Hi Andy,
Well thanks for the help on this :-0...well I get the genius bit but unfortunately the information to my original question is a little on the ‘light’ side (pardon the pun). I was only looking for some pointers and a little help.

Ray
 
Well, since you know the angle of the emitted cone of light, you can calculate the size of the spot if the LED is 5 mm from your surface to be illuminated. Assume the LED is a point source, for simplicity; this will result in a good first-order approximation. Once you know that, basic geometry will give you the area of the spot. And since you're given the total radiant flux, you can divide one by the other to find the radiant power per unit area.

One Joule is equal to one Watt for one second.
 
OK, sorry may math is not good.
I calculated the beam radius of the point source based on a right angled triangle.

LED height = 5mm
Radiant flux = 25mW
Divergence = 20 degrees

So beam radius = (5 / tan70) = 1.82

Then
25mW/1.82mm = 13.73 mW/mm2

?

Ray
 
oops,
25/(PI R2)
=25mW/10.4 = 2.4mW/mm2
 
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