# Calculating Luminous Flux from Intensity

A few years ago a colleague asked me how to calculate total luminous flux emitted from an LED, when the only information you have is maximum intensity. Often those cagey LED manufacturers omit this critical information which you may need in order to design for a certain target light output.

First we’ll review the short answer and later we will go into more detail.

It turns out that for a lambertian source the relationship between max intensity and total lumens is very simple.

Where

• Φ is total luminous flux emitted from the lED in lumens (lm)
• Imax is the maximum intensity of the LED in candelas (cd), which is usually specified on the datasheet if intensity is specified at all.
• the units of the constant pi is steradians (sr)1

See below for a derivation of Equation 1.

We will use as our starting point the fact that intensity, I is flux per solid angle, or, in differential form:
$I = \frac{\delta \Phi}{\delta \omega }$                                                         Eq. 21

let’s talk for a moment about δω.

Fig. 1 differential solid angle. The source is located at the origin, and the luminous flux emitted into the entire hemisphere is calculated.2

δω is the solid angle of the squarish-surface feature of the hemisphere enclosed by the lines δv and δh.  It is defined as follows:
$\delta \omega = \frac{\delta v\delta h}{r^{2}}$                                           Eq. 3
First lets look at δv:

Fig. 2

from figure 2 we can see that

$\delta v = r\delta \theta$                                             Eq. 4
similarly, for  δh, looking at a side view of the gray area shown in figure 1:

Fig. 3

since the radius of the circle that δh lies on is equal to rsin(θ),

thus, combining equations 3, 4, and 5:

Now lets go back to equation 2 and integrate:
$\Phi = \iint_{0}^{2\pi }I(\theta ) sin(\theta )\delta \theta \delta \gamma$              Eq. 7

We integrate δγ from 0 to 2π because we are interested in the light emitted around the entire circumference of the source, and our integration results as follows:

now, if the source is lambertian, then we know from a previous discussion that

Ilambertian(θ) = Imaxcos(θ), so our integral reduces to:
$\Phi = 2\pi I_{max}\int cos(\theta ) sin(\theta )\delta \theta$      Eq. 9

by the double-angle formula, this can be rewritten as
$\Phi = \pi I_{max}\int sin(2\theta )\delta \theta$                  Eq. 10

evaluating θ from 0 to π/2, you get:
$\Phi = \pi I_{max}\left [ -\frac{1}{2}cos(\pi)+\frac{1}{2}cos(0) \right ]$            Eq. 11
This reduces to equation 1 above,
$\Phi = \pi I_{max}$                                                     Eq. 1

Next time, we will talk about what happens when you want to calculate flux for an intensity distribution other than lambertian. Questions?  Contact me at Rachel@HowYouLightIt.com or comment below.

Works Cited:

1. Murdoch, Joseph B. Illuminating Engineering: From Edison’s Lamp to the LED. 2nd ed. New York: Visions Communications, 2003.

2. Sahiner, A. V. Computer Graphics, Boğaziçi University, Spring 2010. Web. 15 January 2011. (Radiometric Terms and Radiance at: http://www.cmpe.boun.edu.tr/courses/cmpe535/spring2010/)

# Lambertian Sources Explained

Often light sources or reflecting objects are referred to as “lambertian”, describing the three-dimensional distribution of light in space as it leaves the emitting source or reflecting object.  Matte wall paint is roughly lambertian, as are skin, white printer paper and most high-power LEDs1,2.

Is there a quick test to tell if something is lambertian?

You may be able to approximate a source  as lambertian by looking for three things:

1. no glare
2. no reflection
3. no change in brightness as viewing angle changes

Let’s verify that white printer paper is lambertian: first make sure that it is not illuminated from behind.  Hold the paper so that its entire 8.5″ by 11″ area is facing you.  As you rotate the paper so that its apparent area is reduced to a 8.5″ or 11″ line, observe whether you see any glare, any reflections, or whether the brightness of the paper appears to change.  The answer for all three criteria is no.

Try it again with your computer screen.  Since there is a layer of glass or plastic on the outermost surface of the screen, there is bound to be a bit of glare or reflection as you change the viewing angle, therefore, your computer screen is not lambertian, but is described by some other BRDF.

Let’s see at what a lambertian source or reflector looks like graphically:

(a) luminance, L , (b) projected area, Ap, and (c) intensity, I, as a function of angle - lambertian source1

Thus, from the graphs it is evident that for a perfectly diffusing source (i.e., a lambertian source), luminance is constant over angle (L(θ) = K, where K is a constant), and projected area and intensity are a cosine function of angle (that is, Ap(θ) = Amaxcos(θ), and I(θ) = Imaxcos(θ).

Let’s review graph (b) to make sure we understand how to interpret these polar plots.  When you held the 8.5″ x 11″ page so that its entire surface area was visible, you had your eyes where the circle intersects the vertical line.  As you rotated the page, the projected area of the page was reduced as a cosine function of angle until it reached zero.

Questions about lambertian sources? contact me at Rachel@HowYouLightIt.com or comment below.

Works Cited:

1. Murdoch, Joseph B. Illuminating Engineering: From Edison’s Lamp to the LED. 2nd ed. New York: Visions Communications, 2003.

2.Ryer, Alexander D. Light Meaurement Handbook. Newburyport: International Light Inc., Technical Publications Department, 1997.