Ir reflectivityformula

Hello, I am interested in Electro-polishing the inside of a 3003 aluminum tube. What kind of reflection percentages would I get in the 265nm range?

Ir reflectivityvs emissivity

% Reflectance (at 0.475 µm) and 45 Deg AOI RS = 94.32(S-polarized) RP = 88.96 (P-polarized) R = 91.64 (non-polarized, (RS+RP)/2 )

Aluminium has two uses in optics 1) It is often specified as a mirror substrate, as it is lightweight, easily processed and low cost, 2) It can be used as a thin film coating material. When vacuum deposited onto glass, metals, ceramics as a thin film coating of Aluminium, it means a metal or glass mirror can have the mechanical and thermal properties of the substrate, but the optical properties of Aluminium.

Emissivity vsreflectivity

This blog post is about the “specular” reflectivity of aluminium, i.e when polished to a smooth, bright, uniform mirror surface. If the Aluminium object you are reflecting your laser off is dull, or has a non uniform surface, then the specular reflectivity will be lower, possibly much lower, than the values I have calculated.

Ir reflectivitytest

% Reflectance (at 0.65 µm) and 45 Deg AOI RS = 93.25 (S-polarized) RP = 86.96 (P-polarized) R = 90.11 (non-polarized, (RS+RP)/2 )

Emissivity andreflectivityformula

Hi Skylyn, Here are the theoretical values for the %R of Aluminium at RGB wavelengths. I guessed at 45 Degrees angle of incidence. Hope this helps

Aluminium is rarely used as an uncoated reflective surface, the mechanical and environmental properties are poor. A protective layer of MgF2 or SiO2 is needed. This can slightly reduce the above values, or restrict the coating to UV / Visible use. Aluminium does have a dip in its reflectivity in the 700nm – 900nm spectrum.

Wavelength            % Reflectivity 248nm                      92.6 400nm                      92.0 532nm                      91.6 633nm                      90.7 800nm                      86.8 900nm                      89.0 1um                          94.0 3um                          98.0 10.6um                     98.7 20um                        99.0 100um                      99.4

emissivity +reflectivity= 1

Hi, Thank you for your interest. I can calculate for you the theoretical reflectance values based on experimental data for the real (n) and imaginary (k) refractive index of Aluminium. There are many other factors that can reduce the actual reflectivity, the most obvious being scatter and diffraction, and possibly chemical interaction with the surface or surface contaminants. We primarily work with Infrared applications so are not experienced with VUV applications. The values I have calculated are simply for personal interest only – we do not warrant them, claim any accuracy for them, or accept any responsibility for their use

hello, you say that you can calculate the reflactance of specific wavelenghts, do you have a formula for or are you checking results and making a conclusion of close wavelenghts to the one asked?

This assumes the only loss of energy on reflection is by absorption. At visible and shorter wavelength scattering and diffraction can cause losses, particularly so with diamond machined (SPDT) metal mirrors. Here at LBP Optics we polish our mirrors to extreme smoothness so scatter and diffraction are at the low levels seen with high quality fused silica optics for the UV

Ir reflectivitypdf

I’m sorry we have no experience at all of “millimetric wavelengths” and metal mirrors. We have customers using our mirrors at THz frequencies (say 100um wavelength) but GHz frequencies (100mm wavelength) are a “closed book” to me I’m afraid. A question for an RF engineer !

It is possible to calculate this from measured values of the complex refractive index of Aluminium, and we have done this in the table below. LBP can calculate the %R, phase shift and Polarisation changes on reflection for any wavelength and angle of incidence, just call us.

The values I have given are theoretical values calculated using the “refractive index” of the metal as reported in the literature. The calculated values and experimental values for %R agree quite well. The difficulty is for visible and IR wavelengths the refractive index is a “complex” number, having a real and imaginary part ( n + ik ). TRy a search for “Drude Theory” to see why this is the case. (I’m not sure I understand well enough to offer an explanation)

Amazing how your chart skips 3 important values fur use in accurate CGI and filmography. 475nm(blue), 510nm(green), 650nm(red). I’d like to know these values as this will help me tremendously in my personal projects.

% Reflectance (at 0.51 µm) and 45 Deg AOI RS = 94.17 (S-polarized) RP = 88.67 (P-polarized) R = 91.42 (non-polarized, (RS+RP)/2 )