[原创]在框架结构确定的情况下，基于matlab的消四种像差的三反系统初始结构的求解 [复制链接]

%无中间像，焦距输入为负数 &>/nYvuq-  
function sjr=nfdre(~) "~ `-Jkm   
07#!b~N  
%系统焦距及各镜间距输入，间距取负正负 p[wjHfIq  
5|b/G  
f=input('f:'); i6-K!  
d1=input('d1:'); *vb)
d0}P  
d2=input('d2:'); #i,O "`4  
d3=input('d3:'); `Ctj]t  
UvI!e4_  
A=f^2/(d3*d2)-f/d1; ^>"?!lv  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); |WX4L7yrhK  
C=d3/d2-f/d1; v!uLd.(  
,..b)H
5n  
a1=(-B+sqrt(B^2-4*A*C))/(2*A);%α1 5YCbFk^  
a2=d3/(a1*f);%α2 V O
X>Sl  
b2=a1*(1-a2)*f/d2;%β2 D%A-& =  
b1=(1-a1)*f/(d1*b2);%β1 Hke\W'&  
?/wloLS47  
f[RnL#*xJU  
%曲率半径 Xx^c?6YM  
1K`A.J:Uy  
R1=2*f/(b1*b2) /[s$A?  
R2=2*a1*f/(b2*(1+b1)) `#ztp)&  
R3=2*a1*a2*f/(1+b2) jt3SA [cy  
x#^kv)  
A1=b2^3*(a1-1)*(1+b1)^3; P#xn!fMi  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; .EdV36$n  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; H,}?YW  
I?"5i8E  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); 8n)Q^z+ K  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); $enh45Wy  
C2=b2*(a1-1)^2*(1+b1)*(1-b1)^2/(4*a1*b1^2)-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)*(1-b2)^2/(4*a1*a2*b1^2*b2^2)-b2*(a1-1)*(1-b1)*(1+b1)/(a1*b1)-(a2*(a1-1)+b1*(1-a2))*(1-b2)*(1+b2)/(a1*a2*b1*b2)-b1*b2+b2*(1+b1)/a1-(1+b2)/(a1*a2); ,&-[$,  
UQFuEI<1-  
CB=[C1 B1;C2 B2]; PG!vn@b6  
AB=[A1 B1;A2 B2]; J\V(MN,  
AC=[A1 C1;A2 C2]; ,b4&$W].  
X&IY(CX  
%非球面系数 2}b bdXx  
k2=-(det(CB)/det(AB)); H #BgE29  
k3=-(det(AC)/det(AB)); H4,yuV  
k1=(k2*a1*b2^3*(1+b1)^3-k3*a1*a2*(1+b2)^3+a1*b2^3*(1+b1)*(1-b1)^2-a1*a2*(1+b2)*(1-b2)^2)/(b1^3*b2^3)-1 =m:W  
k2=k2 K="+2]{I  
k3=k3 5glGlD6R  
MF`'r#@:wa  
end .p`'^$X^  
X\%3uPQ  
%有中间像，焦距输入为正数 5*$Zfuf  
9x!y.gx  
function sjr=yfdre(~) :X f3wP=  
OxZ:5ps  
f=input('f:'); LnGSYrx1  
d1=input('d1:'); sXxO{aeev  
d2=input('d2:'); gwJu&HA/  
d3=input('d3:'); w C"%b#(}  
CCOg1X_  
A=f^2/(d3*d2)-f/d1; k 9rnT)YU  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); {9 O`/|  
C=d3/d2-f/d1; qS al~  
S)yV51^B  
a1=(-B-sqrt(B^2-4*A*C))/(2*A); 'QF>e  
a2=d3/(a1*f); ?C35   
b2=a1*(1-a2)*f/d2; "Ycd$`{Vgt  
b1=(1-a1)*f/(d1*b2); @XC97kGWp  
R:N-y."La.  
%曲率半径 R:?vY!  
6^;!9$G|D*  
R1=2*f/(b1*b2) (N4(r<o;  
R2=2*a1*f/(b2*(1+b1)) %<yM=1~>  
R3=2*a1*a2*f/(1+b2) $:!T/*p*  
^<0NIu}  
A1=b2^3*(a1-1)*(1+b1)^3; IviQ)hp  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; RCXSz  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; PqLqF5`S  
<t% A)L%  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); EG4bFmcs  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); ]bb}[#AY  
C2=b2*(a1-1)^2*(1+b1)*(1-b1)^2/(4*a1*b1^2)-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)*(1-b2)^2/(4*a1*a2*b1^2*b2^2)-b2*(a1-1)*(1-b1)*(1+b1)/(a1*b1)-(a2*(a1-1)+b1*(1-a2))*(1-b2)*(1+b2)/(a1*a2*b1*b2)-b1*b2+b2*(1+b1)/a1-(1+b2)/(a1*a2); <z2mNq  
+Q31K7Gr  
CB=[C1 B1;C2 B2]; +z0}{,HX  
AB=[A1 B1;A2 B2]; -oBas4J  
AC=[A1 C1;A2 C2];  **w~  
8p{  
%二次系数 2EwWV0BS  
Cf[F`pFM  
k2=-(det(CB)/det(AB)); ?3,tG z)  
k3=-(det(AC)/det(AB)); h./vTNMc  
k1=(k2*a1*b2^3*(1+b1)^3-k3*a1*a2*(1+b2)^3+a1*b2^3*(1+b1)*(1-b1)^2-a1*a2*(1+b2)*(1-b2)^2)/(b1^3*b2^3)-1 E:=KH\2f  
k2=k2 zN/~a
)  
k3=k3 % 3-\3qx*  
&iy(oM  
end

谢谢分享，学习一下 WfT)CIKs