[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 kd9rvy0oK  
function sjr=nfdre(~) ai% fj*  
{uEu ^6a5  
%系统焦距及各镜间距输入，间距取负正负 1_uq46  
ye1kI~LO(  
f=input('f:'); +D|y))fE  
d1=input('d1:'); p(UUH3%W  
d2=input('d2:'); Hw8`/'M=%5  
d3=input('d3:'); 7-81,ADv(  
kKnz F  
A=f^2/(d3*d2)-f/d1; ckRWVw   
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); jj8AV lN  
C=d3/d2-f/d1; `As
.1@  
VJ P]Jy_  
a1=(-B+sqrt(B^2-4*A*C))/(2*A);%α1 {,p<!Jq~G  
a2=d3/(a1*f);%α2 RO,  
b2=a1*(1-a2)*f/d2;%β2 u= ( kii=/  
b1=(1-a1)*f/(d1*b2);%β1 HgY"nrogt$  
hRZ9[
F[[  
+hg|!SS@5  
%曲率半径 0%.l|~CE&  
Kl~jcq&z  
R1=2*f/(b1*b2) s<x2*yVUA  
R2=2*a1*f/(b2*(1+b1)) n\aG@X%oq  
R3=2*a1*a2*f/(1+b2) Px:
PoOw\  
\:C@L&3[  
A1=b2^3*(a1-1)*(1+b1)^3; ZiodJ"r  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; FQSepUl  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; #kT3Sx  
G@b|{!  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); /m%Y.:g  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); 'l2'%@E>  
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); s)W^P4<  
*/ZrZ^?o  
CB=[C1 B1;C2 B2]; n?vrsqmZ  
AB=[A1 B1;A2 B2]; RajzH2j+>  
AC=[A1 C1;A2 C2]; 0}jB/Z_T  
5=/H2T!F  
%非球面系数 \"i2E!  
k2=-(det(CB)/det(AB)); {Cnz7TVB  
k3=-(det(AC)/det(AB)); G (Fi  
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 *KSQ^.sYh  
k2=k2 #
(B <n  
k3=k3 k8w }2Vw  
h{I)^8,M  
end J
SQNx2VqQ  
IBr?6_\%"4  
%有中间像，焦距输入为正数 #WlIH7J8Tc  
JX@/rXFY}  
function sjr=yfdre(~)  6!])\Ay  
gy/z;fB  
f=input('f:'); ?"[b408-  
d1=input('d1:'); dM8`!~#&PI  
d2=input('d2:'); vFXih'=_  
d3=input('d3:'); UOy9N  
_!kL7qJ"  
A=f^2/(d3*d2)-f/d1; m[,! orq  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2);  U=MFNp+  
C=d3/d2-f/d1; .<j\"X(  
v)>R)bzqe  
a1=(-B-sqrt(B^2-4*A*C))/(2*A); B$"CoLC7+  
a2=d3/(a1*f); j-@3jFu  
b2=a1*(1-a2)*f/d2; |13UJ v
R
 
b1=(1-a1)*f/(d1*b2); &m_4#  
i/`N~r  
%曲率半径 <rV3(qb#]J  
{Sm^F  
R1=2*f/(b1*b2) gH5E+J_$  
R2=2*a1*f/(b2*(1+b1)) +gh6eY8  
R3=2*a1*a2*f/(1+b2) -Zd0[& ']  

deO/`  
A1=b2^3*(a1-1)*(1+b1)^3; ~]P_Yd-|  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; 5%j !SVW  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; )J#7:s]eo  
;HRIB)wF  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); fF37P8Ir  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); @DG$  
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); $-'p6^5  
jFfuT9oId  
CB=[C1 B1;C2 B2]; `b 6j7  
AB=[A1 B1;A2 B2]; Y S )Q#fP  
AC=[A1 C1;A2 C2]; B*
#lkMr  
uc4#giCD  
%二次系数 WVlyR\.  
zX{K\yp  
k2=-(det(CB)/det(AB)); dq[X:3i  
k3=-(det(AC)/det(AB)); Kk|uN#m  
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 7xidBVx  
k2=k2 ]Z%9l(  
k3=k3 U
{Xg#UN  
JG-\~'9  
end

谢谢分享，学习一下 5{?J5