# Candlepower Vs. Lumens

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One of the greatest joys in most any young person's life is looking out at a clear night sky, seeing all of those pinpoints of far-away light in the evening constellations, and having a sense for the first time of the sheer vastness of the universe. Without visible light, and the invisible electromagnetic radiation emitted by stars like the sun, life on Earth and everywhere else would be impossible.

Physicists need ways to precisely keep track of all of the visible radiation ("light") as well as invisible radiation bombarding the Earth from all directions at all times. They might want to know about its visible qualities, or they may be more concerned with its energy. To help with these tasks, scientists have come up with the candela and the lumen.

## Basic Physical Concepts of Irradiance

For purposes of these kinds of problems, which are concerned with the qualities of the radiation from a given spot reaching a particular area of space, the light source is treated as a single point, and the light or energy it emits is assumed to radiate equally in all directions. Thus all same-sized sections an invisible sphere with the light source at its center would experience the same flow, or flux, of energy through that selection.

The "patch" of space through which the radiation from the source passes is treated as perpendicular to the electromagnetic rays, unless other conditions are specified.

## Candle Power and the Candela

First, know that the term "candle power" has fallen into the dustbin of physics history. The candlepower has been replaced by the candela (cd) and can be regarded as essentially the same unit.

It is not important for you to commit this to memory, but the candela measures luminous intensity, denoted by I, with 1 cd being the luminous intensity of a source that emits a single frequency of radiation (540 x 1012 hertz, or cycles per second) and has a radiant intensity of 1/683 of a watt per steradian, or curved "patch" of the invisible sphere through which radiation passes that has been chosen for examination.

The irradiance E of a surface is given by the relationship E = I / r2 for radiation traveling perpendicularly through the steradian.

## The Lumen

When thinking in terms of lumen vs. candela, think in terms of the total energy emanating from a source vs. that portion of it that the human eye happens to be equipped to register.

The lumen (lm) is more diverse than the candela in that is takes into account radiation that they eye is unable to see. The lumen may can be defined as the luminous flux emitted onto a steradian by a point source having a luminous intensity, I of 1 candela. A lux is a unit equal to 1 lm/m2.

Thus while the lumen and the candlepower are not amenable to easy conversions, the fact that they change in the same direction is helpful. For reference, a typical 100-watt lightbulb serves up a luminous flux of 150 lm, while a standard automobiles high-intensity headlight checks in at around 150,000 lm.

## Converting Between Candelas and Lumens

The candlepower vs. lumens (or these days, candela to lumens) problem has vexed many a student. This is because you cannot convert one to the other directly, as they do not represent the same physical thing. You can, however, work with the two at the same time and draw comparisons.

Ignoring units:

\text{lm} = \text{cd} × 2π(1 − \text{cos}(θ/2))

Here, θ represents the cone apex angle, or the angle between the circle at the base of an invisible "cone" of any chosen proportions radiating outward from the light source and the rays themselves. This "circle" is the "surface" through which the light rays "flow" to contribute to the flux (lm) and also where they "shine" to contribute to lm. You will be give this angle when asked to solve problems like these.

In the case of a point light source radiating equally in all directions, which is what is being considered here, the problem is simpler. Since the maximum value of [1 − cos (θ/2)] is 2, which occurs when cos (θ/2) = −1,

\begin{aligned} \text{lm} &= 2π(1 −(−1)) \text{cd} \\ &= 4π \;\text{cd} \end{aligned}

Thus for an isotopic sphere, lumens is just candelas times 4π.

#### About the Author

Kevin Beck holds a bachelor's degree in physics with minors in math and chemistry from the University of Vermont. Formerly with ScienceBlogs.com and the editor of "Run Strong," he has written for Runner's World, Men's Fitness, Competitor, and a variety of other publications. More about Kevin and links to his professional work can be found at www.kemibe.com.