# I can't understand luminance

1. Sep 14, 2011

### a b

Hello everybody,
i previously posted a topic about an exercise on luminance that was (appropriately, I admit) moved in the homework section. But nobody has still answered me.
But the reason for which I didn't understand that exercise was that I was not sure about what luminance actually is, so I would like to understand it better. Can you help me? I've searched the Internet and asked my schoolmates (in this period i can't find the teachers) but nobody convinced me.
I've read that luminance is the flux divided by the surface divided by the solid angle; but what does this exactly mean? In which way is luminance meant to be used?
If i have a spherical lamp with uniform luminance in each of its points, indipendent of direction, and therefore the sphere is emitting isotropically, can I obtain the total luminous flux radiated by the sphere if I multiply the luminance by the total surface of the lamp and then by 2 pi (not 4 pi, because i suppose that the sphere is emitting only outside of its surface) ?
Please don't move this topic, it is actually a theoretical question even if i don't know other way to express it than this exercise-like shape.

2. Sep 14, 2011

### Drakkith

Staff Emeritus
Well, do you know what luminous flux and the solid angle mean?

3. Sep 14, 2011

### JeffKoch

If this relates to a problem in a book, I suggest looking it up in the same book. Optics terminology is muddled, and the same word is sometimes used in different ways - so for example you trying to "define" it by reference to flux and solid angle is not precise, you really need a paragraph and a picture like what is probably presented in your book.

4. Sep 15, 2011

### Ken G

Here's the key thing you need to know about luminance, as you've defined it (I would just call it the frequency-integrated specific intensity, but what's in a word). It is the single number that will tell you what a detector of certain attributes and at a certain distance will read, expressed in a way that is independent of both the distance of the detector and its area. So it is the way to characterize the "brightness" of the source that is independent of both the distance of your detector, and the area of your detector, which would otherwise of course affect the rate you detect light. Taking the case of a spherical source of total luminosity L (energy/sec), you can see that a detector of area A at distance D will detect energy at the rate $$L*A/(4*\pi*D2)$$, if you just think about the fraction of L that will pass into A. Since the solid angle $$\Omega$$ of a distant source obeys $$\Omega/4\pi = \pi R^2 / 4 \pi D^2$$, for a source of radius R, we can eliminate D and get that the detector will absorb energy at the rate $$LA\Omega/4\pi^2 R^2$$. In that expression we see the solid angle of the source and the area of the detector explicitly, so if we divide them out, we get just $$L/4\pi^2 R^2$$, and that is the energy per second per area per solid angle (steradian). So that's just the luminance, and the point is, it is the number that characterizes the source brightness in a way that is independent of my detector, yet lets me calculate what I'll detect if I know the solid angle of the source and the area of the detector.