physics question

[VJ]

Hello,

A friend od min just had an odd question: suppose you have 4 spotlights in a room without mirrors. Now suppose that you put a mirror on a wall close to those spotlights.
Is there more lux than without the mirror?

I think there is more, as less light will be absorbed by the walls. But I'm not convinced.

What do you think?
And can you prove it mathematically?

Thanks!

Jörg

[Umfriend]

Is there more lux, where? Exactly?

[Dr Mordrid]

Light is energy. Without additional sources in the room the total number of source photons cannot increase, but the light distribution and absorption by the walls due to the addition of the mirrors will change the apparent light level in the room.

[VJ]

I suspect in the room in general. Let's suppose the lights are close to a wall (a bit like in a bathroom), and that you measure in the centre of the room.

Jörg

[VJ]

Light is energy. Without additional sources in the room the total number of photons in the room cannot increase, but the distribution and absobtion by the walls with new mirrors will change. This could change the apparent light level in the room.

I agree, but will the altered absorbtion and distribution change the readout of a lightmeter in the room?


Jörg

[Dr Mordrid]

Yes, and the more reflective the walls are made the more it will change until they're 100% mirrored. Of course there will also be differences according to the placement & orientation of the meter.

[rylan]

Also depends on the type of light meter... is it a spot meter or ambient lux meter.

[VJ]

Also depends on the type of light meter... is it a spot meter or ambient lux meter.

Would a spotmeter aimed at e.g. the opposite wall keep the same readout?


Jörg

[Dr Mordrid]

Depends on where it's aimed. I would doubt a corner would read the same as pointing straight at a wall. It sure doesn't in my studio without the use of fill lamps.

[Jon P. Inghram]

Here's a simply way to look at it: Imagine looking at the room from the viewpoint of one of the spot lights and/or from the meter. If the mirror is reflecting a dark surface (like dark carpet or whatever,) then it's going to reduce the visible light. If the floor is light then it's not going to make much difference, except that lighting reflected from the floor isn't as comfortable or useful as from walls or the ceiling.

[VJ]

Depends on where it's aimed. I would doubt a corner would read the same as pointing straight at a wall. It sure doesn't in my studio without the use of fill lamps.

No, I meant aimed at the same point, but with and without the mirror.

Jörg

[Jon P. Inghram]

You know, there are these machines called "computers" that can "compute" things like how light bounces around in a room.

Describe the room, preferably with a MSPaint sketch, and I'll make a quick render with Art of Illusion.

[az]

Hello,

A friend od min just had an odd question: suppose you have 4 spotlights in a room without mirrors. Now suppose that you put a mirror on a wall close to those spotlights.
Is there more lux than without the mirror?

I think there is more, as less light will be absorbed by the walls. But I'm not convinced.

Yes. lx is the amount of light that falls onto a certain surface. You usually measure the light that falls onto a table, horizontal illuminance. More light will fall onto it when you use mirrors than when you use black walls.

"Proof":

Lux is lumens (lm) per square meter. The lumens is the flux of light that gets produced by the lighting, the square meters is the desk. Angle is factored in as sin(angle) for all light not hitting the surface at 90°, which means that light "hitting" the surface from 0° (meaning a spot with no height directly from the side) would not illuminate a surface at all. So we have lux = lm / m^2 * sin(angle)

Let's assume the Mirror is 50% reflective and the black wall 5%. The light sources in this room are not spots aimed at a non-reflective table with no light scattering, but are rather omnidirectional.

The light from them hits the table at close to 90°, which means that lux(direct) = lm(source) / m^2.

Added to that is lux(reflected) = lm(reflected) / m^2* sin(angle). You can see that when lm(reflected) is greater, lm(total) and thus lux(total) must be greater as well (compensated by the angle at which it hits the table).

[az]

Would a spotmeter aimed at e.g. the opposite wall keep the same readout?

Not if it's accurate enough: There are no truly non-reflective surfaces, so the light would bounce around until it would reach that wall and from there bounce onto the light meter, and it will make a difference if part of the light was bounced back by a mirror, or if part of the light was bounced back by a wall that absorbed more of it.

[Dr Mordrid]

[resisting a long post about resonant coupling ]

It's a whole mish-mash of effects involving wavelength absorption, reflectivity, texture of the surfaces etc. etc. Sometimes makes setting up a shot a royal pain even with a good meter, which is why you bracket your shots....even with digital cams.