Why is a wall brighter when light hits it at a right angle than an oblique angle?

Question

When light from a point source is shone on a wall, the wall is brightest at points where the light is hitting it at a right angle, and seems to get darker as the angle becomes shallower. See for example this image:

Under a simplistic ray model of light where we imagine that all rays are equally bright, and that every point hit by a ray of light re-emits rays in all directions, we can't predict this behavior.

What is the simplest mathematical model of light that allows us to explain this phenomenon?

Answer 1

inverse square law states that a wave dissipates in energy proportional to the inverse square of distance. That is to say, as you get further away, light gets less intense.

if you consider a point source that's one meter away from a wall, then the center of where that light shines will be 1 meter away, but as you get further away from the center, the point on the wall is further away from the source (just some trigonometry with pythagorean theorem). So it will be less intense. If you put the source at the center of a spherical room, it would be lit equally at all points.

Answer 2

Imagine a bundle of parallel rays striking a surface. If the bundle meets the surface at a 90 degree angle, then it's going to illuminate a certain area. But if the bundle meets the same surface at a shallow angle, it's going to illuminate a larger area. Same number of lumens divided by a larger area means fewer lumens per unit area.

Answer 3

I would imagine the flaw in that argumentation here is that the light is not absorbed and re-radiated in all directions, but it is rather reflected mainly in one direction (simple geometric reflection at a surface). The reflected ray does not meet your eye, hence at points where the rays hit the wall at an angle, the reflection should appear darker.