[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  P\m7 -  
function sjr=nfdre(~) }ucIH@U{  
OWjJxORB  
%系统焦距及各镜间距输入，间距取负正负 }-p[V$:S  
&[R8Q|1j  
f=input('f:'); 1MsWnSvzf  
d1=input('d1:'); j`*N,*ha  
d2=input('d2:'); m4w')r~  
d3=input('d3:'); \YJy#2K  
eJ8]g49mD6  
A=f^2/(d3*d2)-f/d1; * A|-KKo\  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); 5ABhj*7  
C=d3/d2-f/d1; \t4tiCw  
zJlQ_U-!  
a1=(-B+sqrt(B^2-4*A*C))/(2*A);%α1 j=+"Qz/hr_  
a2=d3/(a1*f);%α2 \uOdALZ  
b2=a1*(1-a2)*f/d2;%β2 ;o
Ov/3  
b1=(1-a1)*f/(d1*b2);%β1 *ZSdl0e  
6R UrF  
.aOnGp  
%曲率半径 &N
ZfJs  
;$j7H&UNQj  
R1=2*f/(b1*b2) H5>hx{  
R2=2*a1*f/(b2*(1+b1)) h&@R| N  
R3=2*a1*a2*f/(1+b2) ybdd;t}&1  
&p UZDjo?  
A1=b2^3*(a1-1)*(1+b1)^3; X<OSN&d  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; Px@/Q  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; pI[ZBoR~  
D<{{ :7n  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); Z(0@1l`Z-`  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); &xZyM@  
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); X26   
=ziwxIo6  
CB=[C1 B1;C2 B2]; W1aa:hEf  
AB=[A1 B1;A2 B2]; d^ ZMS~\*  
AC=[A1 C1;A2 C2]; E .6HpIx  
8G%yB}pa  
%非球面系数 S*)1|~pRvQ  
k2=-(det(CB)/det(AB)); Tsb{25`+  
k3=-(det(AC)/det(AB)); @R~5-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 >cmE t  
k2=k2 F9q8SA#"  
k3=k3 h:\oly\  
w2(guL($  
end ^,Ydr~|T  
s Wjy6;  
%有中间像，焦距输入为正数 ~=r^3nZR/J  
c>bq%}  
function sjr=yfdre(~) CR<`ZNuWz  
he3SR@\T  
f=input('f:'); >n5:1.g  
d1=input('d1:'); ni&*E~a  
d2=input('d2:'); (kBP(2V  
d3=input('d3:'); 9<CG s3\  
_cDF{E+;  
A=f^2/(d3*d2)-f/d1; 96Wp!]*  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); C"T1MTB  
C=d3/d2-f/d1; ;+"f  
Pj#'}ru!  
a1=(-B-sqrt(B^2-4*A*C))/(2*A); %rKK[  
a2=d3/(a1*f); vW,snxK6y&  
b2=a1*(1-a2)*f/d2; 5m$2Ku  
b1=(1-a1)*f/(d1*b2); q!'
rz  
c/W=$
3  
%曲率半径 |Yi)"-  
[BbutGvj  
R1=2*f/(b1*b2) c2SC|s]  
R2=2*a1*f/(b2*(1+b1)) U4?(A@z9^  
R3=2*a1*a2*f/(1+b2) Doze8pn  
(AY9oei>  
A1=b2^3*(a1-1)*(1+b1)^3; fg%&N2/(.B  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; p 5u_1U0  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; kQdt}o])  
V)o,1  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); 6&v?)o  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); 0CvsvUN@  
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); vy@rQC %9  
v"u^M-_  
CB=[C1 B1;C2 B2]; UnWW/]E  
AB=[A1 B1;A2 B2]; r~[vaQQ6L  
AC=[A1 C1;A2 C2]; $e%2t^ i.g  
1-SVCk -  
%二次系数 8am`6;O:!  
@* 1U{`  
k2=-(det(CB)/det(AB)); &FdWFt=X  
k3=-(det(AC)/det(AB)); ye6H*K  
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 Ru!He,k7  
k2=k2 Rxpn~QQ  
k3=k3 "LhUxnll  
s3s4OAY  
end

谢谢分享，学习一下 wHjLd$ +o