1
$\begingroup$

I recently started as designer at a making custom signage (small signs viewed up close, so there are lots of sizing limitations) and none of the other designers have any solid rules to rely on for building these signs when they are light through light diffusing acrylic. Often signs will look patchy, with individual LED's being distinguishable through the acrylic, due to the variation in lighting, but not the consistency of the acrylic.

Are there any established equations/relationships between the light diffusing properties of a material and the resulting scattering and intensity of the light? How can I measure or estimate the light diffusing properties? The variables I can currently estimate or measure are: light intensity of the LED's, their spacing, thickness of material its shining through and the resulting light intensity, as well as, grading the resulting light pattern.

One of the issues we face is that our supplier does not send a consistent opacity of light diffusing acrylic, so a thickness that would normally look fine, will now either not let enough light through, or it will look patchy (because of point source LED's.) To be clear, the variation is batch to batch, not on the same sheet.

Ideally, these relations/equations would help me understand the difference in thicknesses required. Any starting point would be helpful.

$\endgroup$
4
  • $\begingroup$ Know also that there is a difference between cast acrylic and extruded acrylic in terms of optical "distribution" of light. You may be using extruded acrylic in those instances where the light is uneven. Laser engraving results are different between the two types as well. $\endgroup$
    – fred_dot_u
    Mar 17 '19 at 10:55
  • $\begingroup$ The formulas, which are by the way quite complex, dont mean much without reliable measurements. $\endgroup$
    – joojaa
    Mar 17 '19 at 12:51
  • $\begingroup$ @joojaa any good starting points? If I have a point to start my research from I might be able to answer my own question. $\endgroup$
    – inund8
    Mar 17 '19 at 17:27
  • $\begingroup$ Well you might go over to computer graphics these guys study the making of light simulators. I dont have time to introduce you to the intricasies of developing a raytracer. $\endgroup$
    – joojaa
    Mar 17 '19 at 18:26
2
$\begingroup$

You seem to face a few issues.

First is to account for the potential to have patchiness in any one your sheets. Theoretical equations for light transmission through a system will be easiest when applied for isotropic materials, not for "patchy" systems. This problem has to be solved at the distributor end.

Second is to account for the transmission (opacity) for a given thickness in an isotropic sheet. For an isotropic material that is uniformly absorbing light at a constant amount per unit volume, the transmission falls off exponentially with thickness. For a given thickness, the total amount of light that is absorbed is a function of such factors as the concentration of absorber and its absorbing or scattering efficiency. The theoretical formulations are found under themes such as Beer's law for absorption of light. In essence, the light transmission equation is as below, with $I$ as the transmitted intensity, $I_o$ as the incoming intensity, $\alpha$ as an absorption factor, and $t$ as thickness.

$$ \frac{I}{I_o} = \exp(-\alpha t) $$

This equation neglects reflection at the interfaces.

Third is to account for scattering to cause opacity. In polymers, light scatters from crystalline region. A perfectly glassy polymer is transparent. Uniform scattering comes from uniformly sized, uniformly distributed regions of crystallinity. A rough "off the top of my head" equation is to model the opacity as a rule of mixtures approach using the volume fraction of uniformly-sized, uniformly distributed crystalline regions $f_c$.

$$ O \equiv \frac{I}{I_o} = 1 - f_c $$

This models a polymer with a "full volume distribution" of crystalline regions $f_c = 1$ will be fully opaque. You are free to set $f_c$ as some other metric for linearity between scattering and opacity. Patchiness in any given sheet of polymer indicates uneven distribution of sizes or number density of crystalline regions. This problem has to be solved at the distributor end.

Fourth is to account for inconsistent stocks from sheet to sheet. When for example the concentration of absorber, type of absorber, or size/number density of scattering regions varies from sheet to sheet, $\alpha$ or the scattering potential will vary from sheet to sheet. The light transmission will therefore vary from sheet to sheet even when the sheet has the same thickness $t$. This problem has to be solved at the distributor end.

Finally is account for the LEDs as point sources. Even in the best cases, LEDs are point sources not uniform sources of light when compared to tubes or bulbs. One suggestion to get around this is to invert your LEDs. Point them toward a diffuse scattering surface. Have that surface project the scattered light outward.

$\endgroup$
7
  • $\begingroup$ I updated the question to be more clear. The sheets ARE isotropic, the variations in opacity are from batch to batch. Bulbs and tubes are not possible, due to being restricted to low voltages and small sizes. We do often flip LED's on their sides. Thank you for your answer, its at least a starting point. $\endgroup$
    – inund8
    Mar 23 '19 at 19:25
  • $\begingroup$ This does not sound too useful as OP clearly wants to investigate light scatters rather than absorption. But yeah i agree biggest challenge is to get good measirements. $\endgroup$
    – joojaa
    Mar 25 '19 at 14:00
  • $\begingroup$ Yes, sortof, its just not possible to use same formula. $\endgroup$
    – joojaa
    Mar 25 '19 at 16:01
  • $\begingroup$ @joojaa OK. I've added a note about light scattering in polymers and an "off the top of my head" expression, hopefully to help the OP further appreciate the distinction between absorption and scattering of light. $\endgroup$ Mar 27 '19 at 21:39
  • $\begingroup$ thats better, +1 for being responsive. But this is still a local model on the direct light path for it to be any use in askers situation he needs some info on how the scattering behaves outside that line. $\endgroup$
    – joojaa
    Mar 29 '19 at 6:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.