[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 6]sP"  
function sjr=nfdre(~) #M)+sK$H%f  
.%EYof  
%系统焦距及各镜间距输入，间距取负正负 :2.<JUDM  
+'$5Jtz  
f=input('f:'); )*XD"-9  
d1=input('d1:'); ka_(8  
d2=input('d2:'); 4nrn Npf`b  
d3=input('d3:'); FyQr$;r  
bE;c&g  
A=f^2/(d3*d2)-f/d1; q5G`q&O5  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); Ae[fW9
7  
C=d3/d2-f/d1; P-/"sD  
*M^<oG  
a1=(-B+sqrt(B^2-4*A*C))/(2*A);%α1 Xp.$FJ1)  
a2=d3/(a1*f);%α2 PX*}.L *x  
b2=a1*(1-a2)*f/d2;%β2 bC/Ql  
b1=(1-a1)*f/(d1*b2);%β1 Q2!5  
TwsI8X  
P1R5}i  
%曲率半径 M@z_tR'3\  
3 "iBcsLn  
R1=2*f/(b1*b2) ?<` ;lu/eL  
R2=2*a1*f/(b2*(1+b1)) Q5b9q$L$  
R3=2*a1*a2*f/(1+b2) q BIekQT  
ed2r<H$  
A1=b2^3*(a1-1)*(1+b1)^3; >6R3KJe  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; uBl&{$<  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; & 'CUc/,  
guG&3{&\s  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); )8!*
,e=4  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); JW!.+ Q  
C2=b2*(a1-1)^2*(1+b1)*(1-b1)^2/(4*a1*b1^2)-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)*(1-b2)^2/(4*a1*a2*b1^2*b2^2)-b2*(a1-1)*(1-b1)*(1+b1)/(a1*b1)-(a2*(a1-1)+b1*(1-a2))*(1-b2)*(1+b2)/(a1*a2*b1*b2)-b1*b2+b2*(1+b1)/a1-(1+b2)/(a1*a2); iu?gZVyka  
a^8PB|G  
CB=[C1 B1;C2 B2]; U:z5`z!  
AB=[A1 B1;A2 B2]; e#)NYcr6  
AC=[A1 C1;A2 C2]; Bf.iRh0Q5  
Qvty;2$o@  
%非球面系数 W4,'?o  
k2=-(det(CB)/det(AB)); S:XsO9
:{  
k3=-(det(AC)/det(AB)); W*Si"s2  
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 Ze[,0Y!u&  
k2=k2 !1=*"H%t  
k3=k3 nEUUD3a  
!,dp/5 V  
end 9GRQ^E  
,kf.'N  
%有中间像，焦距输入为正数 zE<Iv\Q  
 Q6RTH  
function sjr=yfdre(~) L9<\vJ
 
i)'tt9f$  
f=input('f:'); |dz"uIrT  
d1=input('d1:'); 4BSqL!i(  
d2=input('d2:'); )o<^6Ic%7  
d3=input('d3:'); E#\Oe_eq~N  
b8_F2  
A=f^2/(d3*d2)-f/d1; yVgC1-8i*  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); fZj,Q#}D  
C=d3/d2-f/d1; yIcTc  
=7JSJ98  
a1=(-B-sqrt(B^2-4*A*C))/(2*A); q
-+:1E  
a2=d3/(a1*f); F}7sb#G  
b2=a1*(1-a2)*f/d2; NYr)=&)Ke.  
b1=(1-a1)*f/(d1*b2); 
KzP{bK5/  
i!RfUod  
%曲率半径 7FJ4;HLQ  
+ c+i u6+"  
R1=2*f/(b1*b2) ``P9f
d  
R2=2*a1*f/(b2*(1+b1)) 33EF/k3vW  
R3=2*a1*a2*f/(1+b2) h=
0a9vIXF  
!:m.-TE  
A1=b2^3*(a1-1)*(1+b1)^3; !\<a2>4$T  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; XPU>} 4{  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; pY@QR?F\  

v7SYWO#  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); p0WUF\"  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); &92/qRh7  
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); M2-`p  
4qz+cB_  
CB=[C1 B1;C2 B2]; \yu7,v  
AB=[A1 B1;A2 B2]; D*ZjoU  
AC=[A1 C1;A2 C2]; l'/`2Y1  
vUVFW'-  
%二次系数 _FYA? d}  
`!/[9Y#Hp  
k2=-(det(CB)/det(AB)); ~1%*w*  
k3=-(det(AC)/det(AB)); ]c~yMA+]FZ  
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 L FkDb}  
k2=k2 K^U=
"  
k3=k3 B=r DU$z  
aTTkj\4  
end

谢谢分享，学习一下 rP3tFvOH