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[求助] targets cone angle 是怎么使用的?

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发表于 2011-7-8 12:43:40 | 显示全部楼层 |阅读模式
哪位大哥 能和我说下, D  B% {3 m  z1 P) T( S
TFCalc targets cone angle是怎么使用的$ x6 H5 H6 d% T) x- j
谢谢; K# f# b( \0 ^$ X
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发表于 2011-7-8 13:16:51 | 显示全部楼层
本帖最后由 gds 于 2011-7-8 21:19 编辑
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给你个官方的例子:
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Broadband AR for a Cone of Light
& B, L( W8 T' G% W# KHere we consider a coating on glass that reflects only 0.5% of a cone of light for a range of wavelengths from 420 to 680 nm. The cone axis is normal to the glass. The cone has an F-number of 0.778, which means the half-angle is 40 degrees. An uncoated glass surface reflects 4.4% of this cone of light. We use a feature new to version 3.5: cone-angle targets. As optimization targets we use 1 g* ]& x1 a' u3 ?
R < 0.5% for wavelengths 420, 425, 430,..., 680 and for a 40-degree cone
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In TFCalc, these targets are easy to enter by using the "Generate Cone Targets" command on the Options menu of the "Targets - Cone Angle" window. We use TFCalc's needle/tunneling optimization to design the coating from "scratch". That is, we begin the design process with a single thin layer of TiO2 and allow TFCalc to increase the thickness and number of layers in the design. TFCalc finds a 6-layer design in a few minutes. The performance of this design is shown below. The little circles indicate the optimization targets.
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, j. q! {- M; c* Y3 j! f- Q Here is the design, starting with the layer closest to glass, and with thicknesses given in nm:
1 T$ Q/ K2 r* T% e; hTIO2          11.04. \$ P% l, k1 |" E3 l" x1 b
SIO2          44.43
$ x. b3 O$ X0 D: l1 t7 U  c4 ?TIO2          34.99
, Y* d( V9 ^' G# S& O2 i' V' `5 z* fSIO2          28.25: U, L4 C0 T) j% T3 f& `9 v5 H
TIO2          30.46
0 C1 r) @9 x! N) USIO2         104.92- m4 g: |' _0 Z

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发表于 2012-9-7 02:35:36 | 显示全部楼层
多角度的?还是圆锥角啊?
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