[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 $b8>SSz  
function sjr=nfdre(~) MKH7d/
x  
dxU[>m;  
%系统焦距及各镜间距输入，间距取负正负 _I-0[w  
Z>{8FzP.F  
f=input('f:'); JLeV@NO  
d1=input('d1:'); )GR^V=o7,Y  
d2=input('d2:'); sF|lhLi  
d3=input('d3:'); ;j(xrPNb  
S=my;M-  
A=f^2/(d3*d2)-f/d1; zxj!ihs<  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); YnNei 7R  
C=d3/d2-f/d1; g$:2c7uL  
c8yD-U/-  
a1=(-B+sqrt(B^2-4*A*C))/(2*A);%α1  (0k0gq;  
a2=d3/(a1*f);%α2 bEy%S"\<  
b2=a1*(1-a2)*f/d2;%β2 kg-%:;y.  
b1=(1-a1)*f/(d1*b2);%β1 O ,DX%wk,  
@!F9}n AP  
6qw_|A&g  
%曲率半径 Gis'IX(  
@Xh4ZMyEx  
R1=2*f/(b1*b2) k"n
#4o:  
R2=2*a1*f/(b2*(1+b1)) e;gf??8}  
R3=2*a1*a2*f/(1+b2) 5Ut0I]h|z  
2/bck)p=  
A1=b2^3*(a1-1)*(1+b1)^3; bLUn0)c  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; ;DGWUK.U[H  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; Y>z(F\  
cJty4m-  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); u!X2ju<  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); [s2%t"H-y  
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); gNO$WY^  
hqE#BnQxP,  
CB=[C1 B1;C2 B2]; k%LE"Q  
AB=[A1 B1;A2 B2]; &hF>}O  
AC=[A1 C1;A2 C2]; }KOu  
LyQO
_mT2  
%非球面系数 -Z
w"o>  
k2=-(det(CB)/det(AB)); q6,xsO,+  
k3=-(det(AC)/det(AB)); 6Z'zB&hM}  
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 @hv9=v+  
k2=k2 |, :(3Ml  
k3=k3 z,NHH):~  
UjfB+=7I{L  
end qH(2 0Z!  
q&wMp{  
%有中间像，焦距输入为正数 k'\RS6M`L  
WAQv4&xGM  
function sjr=yfdre(~) "5e]-u'  
]>:>":<:  
f=input('f:'); .Kv@p jOr  
d1=input('d1:'); 4@VX%5uy
 
d2=input('d2:'); `q/y|/v<  
d3=input('d3:'); NR3IeTd  
L Y
M`  
A=f^2/(d3*d2)-f/d1; 6jnRC*!?  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); >0m-S :lk  
C=d3/d2-f/d1; R52!pB0[  
c.XLEjV|  
a1=(-B-sqrt(B^2-4*A*C))/(2*A); <^&'r5H  
a2=d3/(a1*f); 1"e=Zqn$)  
b2=a1*(1-a2)*f/d2; g>Z1ZK0;M  
b1=(1-a1)*f/(d1*b2); y}3V3uqK  
EXzY4D ^  
%曲率半径 EYQ!ELuF  
?^7
~|?v  
R1=2*f/(b1*b2) QoW3*1o  
R2=2*a1*f/(b2*(1+b1)) :DZiDJ@  
R3=2*a1*a2*f/(1+b2) w8on3f;6n#  
l%)XPb2$J  
A1=b2^3*(a1-1)*(1+b1)^3; {wm  `  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; m760K*:i\  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; $X{& KLM[  
FqiK}K.~/  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); 1HZexV  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); Z86[sQBg  
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); ?oF@q :W  
%1lLUgf3G/  
CB=[C1 B1;C2 B2]; o 1b#q/  
AB=[A1 B1;A2 B2]; `?>OY&(  
AC=[A1 C1;A2 C2]; 1n,JynJ  
F%Xq}LMd  
%二次系数 )Qo6bei!  
Zw3|HV(so  
k2=-(det(CB)/det(AB)); :4A^~+J  
k3=-(det(AC)/det(AB)); Xak~He  
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 yWs/~5[F  
k2=k2 x^ Wgo`v)  
k3=k3 J|>P,x#G  
+yTL  
end

谢谢分享，学习一下 d9^h YS{