I Lux & Lumens -- how do they work?

[Xenon02]

Yes. At an introductory teaching level, the lux values are usually average ones; and this is not usually made clear.

Good to know :D So I was also right with adding every surface together ? Floor + walls + ... like in the example I gave ? To use this funny equation ?

But in the real world, the average lux value for a large area is often not useful. However, for fun, here a short problem...

The square floor of a room measures 4m x 4m.
A 300 lumen bulb hangs 2m above the centre of the floor.
The bulb emits light equally in all directions.
What is the average illuminance (lux) of the floor?

From those 3 dots I can say that I am quite tough to teach me :D

The light bulbt emits the light in all directions which is not only the floor but also walls ect ?
If the walls to then it's hard for me to tell because I don't know how much lumen receives the floor.
But I'll tell that the floor receives all 300 lumens. Then surface is 16 m².

Lux = 300/16 = 18,75 lux per m².

And, if you are interested, the lux-level at the centre of the floor will be more than three times bigger than the lux level near a corner of the floor. So knowing the average value is of limited use.

I can see it now that Those lux are only average value.

I think it also works with the flash lights.
If I look for example here :

The one with 200 Lux. It would be also average because the center of this area is more brighter and the edges. Soo 200 Lux is also average.

 

[Steve4Physics]

Good to know :D So I was also right with adding every surface together ? Floor + walls + ... like in the example I gave ? To use this funny equation ?

Yes, you were right - providing you really want the average lux value for the different surfaces. There is usually no point in calculating this average. Try answering this question:

The average lux value for a room is 100 lux. What is the lux value for the floor?
a) More than 100lux
b) 100lux
c) Less than 100 lux
d) Can't tell

Click the 'Spoiler' to check your answer:

Spoiler

The correct answer is d). You can't tell the floor's lux level. Knowing the average value for the whole room doesn't help. It is useless in this situation.

The light bulbt emits the light in all directions which is not only the floor but also walls ect ?

Yes. In my problem, the light hits the ceiling, walls and floor. For example, if you used a simple old-fashioned filament bulb.

If the walls to then it's hard for me to tell because I don't know how much lumen receives the floor.

There is a way to work out how much light reaches the floor. For a hint, click the spoiler:

Spoiler

The 300 lumen bulb emit light equally in all directions. Imagine the bulb is at the centre of a cube. How much light passes through each face of the cube?

But I'll tell that the floor receives all 300 lumens. Then surface is 16 m².

Lux = 300/16 = 18,75 lux per m².

So that's not the answer we want. Can you do it using the hint given above?

I can see it now that Those lux are only average value.

I think it also works with the flash lights.

Yes. For a simple flashlight beam, falling squarely on a wall, the average is useful.

 

[Xenon02]

Yes, you were right - providing you really want the average lux value for the different surfaces. There is usually no point in calculating this average. Try answering this question:

The average lux value for a room is 100 lux. What is the lux value for the floor?
a) More than 100lux
b) 100lux
c) Less than 100 lux
d) Can't tell

Click the 'Spoiler' to check your answer:

I'm fully aware that average lux value has no point here. It was just curiosity.
Also it's logical that I can't say what is the lux value on the floor. Because on the floor it could be 300-500 lux. but average of every lux value on every surface is 100 lux. Some can have 50 lux some even 2 lux (maybe).

There is a way to work out how much light reaches the floor. For a hint, click the spoiler:

Spoiler

The 300 lumen bulb emit light equally in all directions. Imagine the bulb is at the centre of a cube. How much light passes through each face of the cube?

thinking of the cube then 300/6 (4 walls, 1 floor, 1 ceiling).
So 50 lumen on the floor. So looking at it this way then. 50lumens/16m² = 3,125 lux average.

Yes. For a simple flashlight beam, falling squarely on a wall, the average is useful.

it's easier more practical because all lumens are focused in this one area, instead of part of lumens in one area.

 

[Steve4Physics]

I'm fully aware that average lux value has no point here. It was just curiosity.

Sorry. I wasn't sure if you had realized.

thinking of the cube then 300/6 (4 walls, 1 floor, 1 ceiling).
So 50 lumen on the floor. So looking at it this way then. 50lumens/16m² = 3,125 lux average.

Yes. Well done!

I'd think about it like this. A cube has 6 equal-area faces. If the 300 lumen (evenly spread) bulb is at the centre of any cube, then, by symmetry, exactly the same amount of light must fall on each face of the cube. So each face recives 300/6 = 50 lumen.

For the problem in Post #29, the floor measures 4m and 4m and the bulb is 2m above the centre of the floor. The floor must receive 1/6 of the total light, i.e. 50 lumens.

(Note, if the bulb were at a different height, the problem gets *much* more complicated!)

 

[Xenon02]

(Note, if the bulb were at a different height, the problem gets *much* more complicated!)

Because the light then isn't evenly spread as I can assume. And it will be more complicated to calculate the lumens that the floor receives.