Timeline for Anodized aluminium reflectivity at 10.6 um
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 7, 2014 at 16:06 | vote | accept | texnic | ||
| May 2, 2014 at 20:18 | comment | added | texnic | I realize now that I did not take surface reflection into account, which is probably responsible for 15 % loss at short wavelengths in your plot: 2 * ((1.75-1)/(1.75+1))^2 = 0.15. But the result (huge absorption at >8 um) stays the same. | |
| May 2, 2014 at 20:12 | comment | added | texnic | From your plot. Transmission is 3 %. $I'=I_0\exp(-\alpha t)$, $0.03 = \exp(-\alpha \times 0.5)$, $\alpha = -2\ln(0.03) = 7.01$. | |
| May 2, 2014 at 15:21 | comment | added | Chris Mueller | @texnic Seems pretty reasonable. I don't know where you got the absorption coefficient at 7.5 um though. | |
| May 2, 2014 at 15:18 | comment | added | texnic | Interesting. Thanks for explaining about the name, I was confused by that. I've done some math based on your curves. At 7.5 um, for 0.5 mm thickness, transmission is ~3 %. Then the absorption coefficient is ~7 1/mm, leading to a loss of 25 % for double-pass in a 20 um layer. For λ = 10 um it will be even higher, unless there is a local transmission peak there. Do you agree with this understanding? | |
| May 2, 2014 at 13:45 | history | answered | Chris Mueller | CC BY-SA 3.0 |