Could someone explian to me what a Candella is?

[Starwatcher16]

could someone explain to me what a Candella is? I am not really getting it.

Okay, so it is a Watt/Steradian. Okay.

Lets say that when we get a flux of 100 photons passing through a surface per second, that that is equal to one watt.

So if we have, let's say 50 Candella, we have a flux of 5000 photons passing through a surface whose area equals r^2 per second.

What I do not get, is, if the Candella is defined this way, then doesn't that mean that the candella is dependent on r? If I were to double r, then i am quadrupling the surface area, and therefore my Candella will go down to a quarter its previous value.

So what does it mean, when you say a lamp has a candella rating of X, without giving the r that was used to determine that?

I mean, if I choose a big enough R, I could get the sun to have a candella of 1E-1000

 

[negitron]

No, the power per steradian is a constant. If you move father from your light source (assuming an isotropic radiation pattern), the power per steradian doesn't change, but the area occupied by that steradian does. See the following figure:

http://imagine.gsfc.nasa.gov/docs/science/try_l2/inverse_square.gif

Note as your distance from the source, R, increases, the larger your area of illumination becomes, but the angle remains constant. A steradian is simply a two-dimendional angular measurement.

 

[cepheid]

First of all, a candela is not exactly a watt per steradian. It would be so nice if the physical quantity measured by candelas was just "power emitted per unit solid angle." However, this is not the case. You need to distinguish between radiometric units and photometric units. Radiometric units are units that measure physical quantities that are used as objective ways of quantifying electromagnetic radiation (e.g. power emitted, power emitted per unit solid angle, etc.). They pertain to all forms of electromagnetic radiation.

In contrast, photometric units units measure the intensity of visible light as perceived by the human eye. The human eye is not equally sensitive to all wavelengths, therefore, this unit that is being measured in candelas, which is called luminous intensity, is the power per unit solid angle WEIGHTED by the responsivity of the human eye as function of wavelength. If I recall correctly, this visual sensitivity curve has its peak (defined to be = 1) at 550 nm. So a source emitting 1 W/sr of yellow light would be 1 cd, but a source emitting 1 W/sr of red light would be less than 1 cd (by a factor determined by that weighting curve).

A good article to read would be:

http://en.wikipedia.org/wiki/Photometry_(optics )

paying special attention to the two tables of units, SI photometric units versus SI radiometric units. The section on "Watts vs. Lumens" in that article is pretty good as well. What is being said there explains how you can buy a 10 W compact fluorescent light bulb that that is stated to appear just as bright (in lumens) as a typical 60 W incandescent bulb (it says so right on the package). This is possible because an incandescent bulb shines because it is hot. Therefore, a great deal of the power is being wasted as heat, rather than producing useful (visible) electromagnetic radiation. Indeed, the visual sensitivity curve is obviously 0 at infrared wavelengths, so much of the energy due to EM radiation is not visible and therefore not contributing to the illumination.

I'll post again in a second regarding your query about solid angle.

 

[cepheid]

As for the power per unit solid angle (known as radiant intensity, I, if you look in the link I posted), it is defined as:

I = P/Ω​

where P is the power and Ω is the solid angle. Solid angle is a 2D analog of a regular angle: if an area A is a distance R from the source, then the solid angle Ω subtended by the area at the source is given by:

Ω = A/R²​

Let's say we're interested in a solid angle corresponding to a fraction f, of total surface area of a sphere at any given distance. Then:

Ω = (f4πR²)/(R²)​

yielding:

I = P(R²) / (f4πR²)

I = P/4πf​

We can see that the R²'s cancel, leaving a constant. If the radiant intensity is a constant, then the power emitted into a given solid angle will be a constant (indepenent of distance).

As a final note, this result is not at all at odds with the idea that the power per unit area decreases with distance, because a given solid angle corresponds to a larger and larger physical area with increasing distance (EDIT: see diagram in negitron's post). That is just a statement in words of my equation above (the distance dependencies cancel each other out).