# Does a parabolic diffuse surface reflect more light than a flat, rectangular surface in 2D (see figure)? Which surface reflects more light into the black collecting surface, considering both the light collecting surfaces have the same length? Is there an equation that could help me determine this? I'm not an optics expert, I work in radiation detection (I'm working on a scintillator, and I'm trying to optimize geometry, but I'm struggling with some stuff).

• What's the source of light? If the incoming light beams are all roughly parallel this has some implications regarding the angle of incidence on the reflective surface. What materials are involved in constructing the surfaces? What's the wave length of the radiation? – R. Romero Nov 21 '19 at 20:04
• @R.Romero Since the material is a scintillator, the light source is assumed to be randomly distributed all over the geometry (inside of it) – Betsy Nov 21 '19 at 20:23
• @CuriousOne Is there an equation that gives me "amount of light collected"? I'm trying to write a code that models the parabolic surface coordinates (also the box), and then i would reflect light off the diffuse surfaces and calculate how much reaches the collecting surface. I have found equations providing reflectance, but not sure how to calcualte "amount" or intensity of light collected. – Betsy Nov 21 '19 at 20:38

## 2 Answers

Using the cosine law aka as Lambertian reflection , yes. https://en.m.wikipedia.org/wiki/Lambertian_reflectance A Lambertian reflector looks the same intensity from any direction.

• The question is about the amount of light reaching the "light collecting surface" not about the light reflected away from the collecting surfaces... – CuriousOne Nov 21 '19 at 20:21
• @CuriousOne I have no clue how any light reaches the collector, unless the source is inside the object. – my2cts Nov 21 '19 at 20:28
• I clearly have no idea what a scintillator is haha, sorry about that – CuriousOne Nov 21 '19 at 20:35
• @CuriousOne a scintillator is a crystal that produces light when struck by charged particles like proton, electrons. But this is more optics related, where I'm totally blind (excuse the pun :D). – Betsy Nov 21 '19 at 20:40
• The image shows a "reflecting diffuse surface". I don't see a scintillator although the text mentions one. – my2cts Nov 21 '19 at 20:43

If light is coming from below, and assuming both surfaces are 100% reflective, light will always eventually escape through light collecting surface, regardless of its geometry. Even if the reflector is just flat.

I think you might want to clarify where the light is coming from and how exactly you quantify amount of light going through a surface.