[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 k#:@fH4{PA  
function sjr=nfdre(~) ]7R&m)1
6  
yZ~eLWz  
%系统焦距及各镜间距输入，间距取负正负 I%Po/
+|+  
v6DxxE2n  
f=input('f:'); {{[jC"4AY  
d1=input('d1:'); k1Mxsd  
d2=input('d2:'); GKsL~;8"  
d3=input('d3:'); B/9<b{6  
0jJ28.kOp  
A=f^2/(d3*d2)-f/d1; 0@e}hv;  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); XG\a-dq[  
C=d3/d2-f/d1; b^l -*4  
yc%E$g  
a1=(-B+sqrt(B^2-4*A*C))/(2*A);%α1 ;aK !eD$  
a2=d3/(a1*f);%α2 V.QzMF"o  
b2=a1*(1-a2)*f/d2;%β2 QOh w  
b1=(1-a1)*f/(d1*b2);%β1 zEPx  
rb}wv16?  
N4`
9TN7  
%曲率半径 sg6w7fp>  
At
[n<8_|  
R1=2*f/(b1*b2) Va/@#=,q]  
R2=2*a1*f/(b2*(1+b1)) qvfAG
0p  
R3=2*a1*a2*f/(1+b2) #SihedWi  
Q!2iOvK  
A1=b2^3*(a1-1)*(1+b1)^3; qJFgbq4-  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; 2) /k`Na  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; 5nxS+`Pn.)  
F3Ak'h{Ay  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2);  IB.'4B7  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); XC/]u%n8](  
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); )*TW\v`B  
#-gGs
j;F  
CB=[C1 B1;C2 B2];
%M;_(jda  
AB=[A1 B1;A2 B2]; i^s`6:rNu  
AC=[A1 C1;A2 C2]; l`M5'r]l  
=FD`A#\C~  
%非球面系数 v-7Rb)EP  
k2=-(det(CB)/det(AB)); UU;-q_H6  
k3=-(det(AC)/det(AB)); iQm.]A  
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 5fj  
k2=k2 JJ N(M*;  
k3=k3 EYJi6#  
WDoKbTv  
end |?fW!y  
V$Xl^#tN  
%有中间像，焦距输入为正数 &` 00/p  
{sna)v$;  
function sjr=yfdre(~) /50g3?X,  
B::4Qme  
f=input('f:'); 
=mi:<q  
d1=input('d1:'); G1?0Q_RN  
d2=input('d2:'); 
z3vsz  
d3=input('d3:'); /mMAwx  
[ n0##/  
A=f^2/(d3*d2)-f/d1; Q2[prrk%j  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); 1o;g1Z/  
C=d3/d2-f/d1; A'~%
_}  
Hn#GS9d_?  
a1=(-B-sqrt(B^2-4*A*C))/(2*A); :Er^"9'A2  
a2=d3/(a1*f); 'd2qa`H'}B  
b2=a1*(1-a2)*f/d2; (/"K+$8'  
b1=(1-a1)*f/(d1*b2); *
793H\  
#LN5&i;s  
%曲率半径 v ]/OAH6D  
HvqF@/xh  
R1=2*f/(b1*b2) jsNH`"  
R2=2*a1*f/(b2*(1+b1)) V>FT~k_"  
R3=2*a1*a2*f/(1+b2) '#,e @v  

v
.l7Q  
A1=b2^3*(a1-1)*(1+b1)^3; Uw2,o|=O  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; /i-xX*  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; J0ZxhxX35  
Z{,GZT  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); gnQo1q{ 4  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); eq@am(#&kY  
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); ;t;Y.*&=S  
H[BD)  
CB=[C1 B1;C2 B2]; yyY~ *Le  
AB=[A1 B1;A2 B2]; jb|mip@` <  
AC=[A1 C1;A2 C2]; V-KL%  
AfbB~LlBq  
%二次系数 7Sg
weZ}"  
74@lo-/LY  
k2=-(det(CB)/det(AB)); ]^J+
-c  
k3=-(det(AC)/det(AB)); GI2eJK  
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 ^CZCZ,v  
k2=k2 w(76H^e  
k3=k3 Q0zW ]a  
Jv_.itc  
end

谢谢分享，学习一下 6g5PM4\