meself 发表于 2004-10-12 03:04:00

!!!向各位大哥求救(一定要帮我)!!!

<P>请帮我解答此题:</P>
<P>在1.52的bk7上镀一层光学厚度为:入0/10 (入0=500 nm)的1.38的MgF2,求垂直入射时在入=600 nm处的膜层的反射率(不考虑内表面反射)?(紧急求解)</P>

ic5 发表于 2004-10-12 03:39:00

<P>我用软件模拟大约有3.5%左右</P>

zxp163a 发表于 2004-10-12 05:05:00

我也算了一下，和楼上结果一样

meself 发表于 2004-10-13 20:44:00

谢谢2位大哥:2楼的能贴出解答过程吗?

handsome 发表于 2004-10-20 03:20:00

<P 0in 0in 0pt"><FONT face="Times New Roman"><FONT size=2>解：<p></p></FONT></FONT></P><P 0in 0in 0pt"><FONT size=2><FONT face="Times New Roman">在</FONT>l<FONT face="Times New Roman">=600 nm 处的位相为：</FONT>j<FONT face="Times New Roman">=knd=(2</FONT>p<FONT face="Times New Roman">/</FONT>l<FONT face="Times New Roman">)</FONT>l<FONT face="Times New Roman">o/10=(2</FONT>p<FONT face="Times New Roman">/600)500/10=</FONT>p<FONT face="Times New Roman">/6<p></p></FONT></FONT></P><P 0in 0in 0pt"><FONT face="Times New Roman"><FONT size=2>y(1)=Yo/Y1=1.52/1.38=1.1,<p></p></FONT></FONT></P><P 0in 0in 0pt"><FONT size=2><FONT face="Times New Roman">y(1)’=[(y(1)+itan(</FONT>p<FONT face="Times New Roman">/6))/(1+iy(1)tan(</FONT>p<FONT face="Times New Roman">/6)]=1.0452-0.08767i<p></p></FONT></FONT></P><P 0in 0in 0pt"><FONT size=2><FONT face="Times New Roman">y(2)= y(1)’Y1/Y2=1.38(1.0452-0.08767i)=1.445-0.121i<p></p></FONT></FONT></P><P 0in 0in 0pt"><FONT size=2><FONT face="Times New Roman">r=(1-y(2))/(1+y(2))=-0.18432+0.04i<p></p></FONT></FONT></P><P 0in 0in 0pt"><FONT size=2><FONT face="Times New Roman">R=0.0356<p></p></FONT></FONT></P>Over
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