Beer´s law does not apply if there is any scattering. In a pure absorber, if 1 layer absorbs 50 % of incident light and scatters 0 % then 2 consecutive layers transmit 25 % of light and absorb 75 %. And so on exponentially. 9 layers of pure absorbers would transmit 0,195% and absorb 99,8 %. But say that there is a pure scatterer layer which reflects 50 % of incident light back to the direction it came from. Then, assuming that 2 such layers consecutively are followed by darkness where transmitted light escapes or is absorbed, that double layer would transmit 33,3 % of incident light. 25% would be transmitted directly - but 6,25 % would be transmitted after double scattering (50 % reaches second layer, 25 % is reflected back to first layer, 12,5 % is reflected back forward to second layer, so 6,25% passes through second layer on second try), 1,5625% would be transmitted after quadruple scattering, etc. totalling 33,33%. And 66,666% would be reflected. Likewise, from 9 layers of pure scatterer, 0,195% would pass straight through - but 9,8% would be transmitted after various numbers of scatterings. At the limit of infinite optical depth, there would be no transmission - but unlike the case with absorber, in a scatterer the transmission at large optical depth undergoes harmonic decrease, not exponential decrease. But my mathematic above was simplified. I counted discrete scatterers, and only transmission or reflection straight back. Now imagine real scattering, in absence of any absorption. Like Rayleigh scattering, which can happen to any direction - light would be free to be scattered sideways till it is again Rayleigh scattered down again or back up. Does the qualitative conclusion hold - that in absence of absorption, scattering causes only harmonic decrease of transmission and, in the limit of infinite depth, complete reflection?