# Illuminance at a given distance from a diffuse emitting light box

Therapeutic light boxes purport to deliver 10,000 lux to a user, typically sitting ~24" from a box. I was wondering how many lumens are necessary to achieve this illuminance, e.g. what lamps would be necessary to build such a box.

Suppose the light box has a flat, square diffuser, where the square has a total area $$A$$. The box emits a total of $$l$$ lumens. Given the distance $$d$$ to the light source, can we have a formula for the illuminance received by a viewer centered in front of the light box, facing it directly? Here's my thinking so far, by referring to Photometric quantities and this handbook. It doesn't quite seem to work.

We need to first determine the luminance $$L$$ of the box diffuser's surface (lm/m^2/sr). It has a luminous exitance of $$l / A$$ (lm/m^2). Apparently for a Lambertian (diffuse) surface, we should divide by $$\pi$$ to get luminance: $$L = l / A / \pi$$ (lm/m^2/sr).

Multiplying this luminance by the area $$A$$ of the box, it has a luminous intensity of $$L \cdot A = l / \pi$$ (lm/sr). The box is viewed from distance $$d$$ away, so its apparent size will be approximately(1) $$A / (4 \pi d^2)$$ steradians. Multiplying luminous intensity by solid angle, the viewer receives $$l/\pi \cdot A / (4 \pi d^2)$$ (lm).

But wait -- the units seem wrong now. I should have gotten lm/m^2 (lux), not lumens. A single point only receives an infinitesimal amount of lumens. Where did this go wrong?

(1) In reality, the corners of the boxes are farther, so it appears a smaller area. I wonder how much effect this has.

Imagine that the light box is instead an isotropic light source. Then at distance $d$ the $l$ lumens would be spread out over an area of $4 \pi d^2$, and the illuminance would therefore be $l/4 \pi d^2$. But it only shines into half the hemisphere, and the total light emitted in any direction goes like $cos(\theta)$ so as you say, you need to multiply by 4 if the observer's position is more or less directly in front of the box. So you get $l/\pi d^2$ lux. The area of the box doesn't matter--a bigger box will cover more of the observer's field of view, but won't look as bright.

There is an inconsistency in your last step. You have calculated the luminous intensity for light leaving the entire box (lm/sr). You then multiplied by the solid angle subtended by the box relative to the viewer. This tells you the total lumens incident on an area the size of the box at the location of the viewer (if the viewer had the same area as the box, it's the total amount of light to hit the viewer). If you want the illuminance, you need to divide by this area.

• So for a point light source emitting $l$ lumens in one hemisphere, the illuminance would be $l/2\pi d^2$. Could you elaborate on why you multiplied by 4 instead of 2? And I'm not quite clear - can you show mathematically how the box can be reduced to a point source?
– PBJ
Jan 13, 2018 at 0:07
• Thanks for pointing out the mistake and conceptual misunderstanding. Dividing by area produces $l / (4 \pi^2 d^2)$ (lm/m^2). But that is also making the assumption that the lumens are spread out uniformly across that area, which is not true - the edges of the area will be less lit than the center.
– PBJ
Jan 13, 2018 at 0:17
• I think fundamentally it doesn't make sense to have a luminous intensity of the entire box, right? Because it's a large area and not a point?
– PBJ
Jan 13, 2018 at 0:20
• You don't get exactly the same amount of light from the extended source as you would from a point source, it's just an approximation. You could integrate over the whole area of the box, but for the purpose of deciding what lamps to use that doesn't seem necessary. The reason it's 4 instead of 2 is that a perfect diffuse looks just as bright from any angle. Since the projected area is smaller when viewing from the side, the total amount of light emitted in a given direction goes like $cos(\theta)$ where $\theta$ is the direction normal to the surface. Jan 13, 2018 at 3:57
• I think it's ok to talk about luminous intensity as long as you understand that it's a property of the source as a whole, integrating the light coming from every part of the surface. Jan 13, 2018 at 3:58