[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 rBye%rQRq  
function sjr=nfdre(~) |lm  
U 3aY =8B  
%系统焦距及各镜间距输入，间距取负正负 Qv(}*iq]  
m]R< :_  
f=input('f:'); /zMiy?  
d1=input('d1:'); 8G`fSac`  
d2=input('d2:'); zGHP{a1O7  
d3=input('d3:'); EpB2?XGA  
dX-{75o5P  
A=f^2/(d3*d2)-f/d1; ):; &~  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); |
G&<@8O  
C=d3/d2-f/d1; ;| ##~Y.9  
1`)ie%=  
a1=(-B+sqrt(B^2-4*A*C))/(2*A);%α1 &Ey5 H?U!  
a2=d3/(a1*f);%α2 r&@#,g  
b2=a1*(1-a2)*f/d2;%β2 C=2  
b1=(1-a1)*f/(d1*b2);%β1 Uh'3c"  
AWsO?|YT  
Vcz ExP  
%曲率半径 mjO4GpG3  
/5 B{szf  
R1=2*f/(b1*b2) 8ME_O~,N  
R2=2*a1*f/(b2*(1+b1)) *&9_+F8ly  
R3=2*a1*a2*f/(1+b2) vQ>x5\r5O_  
89*CoQ  
A1=b2^3*(a1-1)*(1+b1)^3; 3?iRf6;n  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; lyNa(3  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; D^yZ!}Kl  
GGo)k1T|)  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); Ox7v*[x'  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); s%z\szd*  
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); }u:@:}8K  

_p<W  
CB=[C1 B1;C2 B2]; j|gQe .,1  
AB=[A1 B1;A2 B2]; wvBx]$SC  
AC=[A1 C1;A2 C2]; )8VrGg?  
}R4%%)j(Vj  
%非球面系数 !#j y=A  
k2=-(det(CB)/det(AB)); %K8Ei/p\t]  
k3=-(det(AC)/det(AB)); B{$4s8XU  
k1=(k2*a1*b2^3*(1+b1)^3-k3*a1*a2*(1+b2)^3+a1*b2^3*(1+b1)*(1-b1)^2-a1*a2*(1+b2)*(1-b2)^2)/(b1^3*b2^3)-1 4+e9:r]  
k2=k2 k FE2Vv4.  
k3=k3 z )s{>^D  
F$<>JEdX  
end smvIU0:K  
k,wr6>'Vt  
%有中间像，焦距输入为正数 E/2kX3}  
S+Z_Qf  
function sjr=yfdre(~) s kC*  
/tR@J8pV  
f=input('f:'); f1w&D ]|S+  
d1=input('d1:'); &6<>hqR^  
d2=input('d2:'); ]Q1
?Ox:'  
d3=input('d3:'); <wWZ]P2]  
aNLRUdc.  
A=f^2/(d3*d2)-f/d1; ^0vK >  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); Xu\FcQ{  
C=d3/d2-f/d1; .RFijr  
>]K:lJ]l  
a1=(-B-sqrt(B^2-4*A*C))/(2*A); {:BAh5e|  
a2=d3/(a1*f); Xg
L-t~_  
b2=a1*(1-a2)*f/d2; Z BjyQ4h  
b1=(1-a1)*f/(d1*b2); e/hA>  
6-#<*Pg  
%曲率半径 2L"$p?  
C#{s[l\]  
R1=2*f/(b1*b2) g$bbm}6S  
R2=2*a1*f/(b2*(1+b1)) $I*ye+a*{q  
R3=2*a1*a2*f/(1+b2) Jm8{@D%  
g(Q
)fw  
A1=b2^3*(a1-1)*(1+b1)^3; j#L"fW^GM  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; uqe{F+;8&  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; r+%:rFeX  
K8`Jl=}z%&  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); c6/+Ye =h  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); c1 aCN  
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); xPMTmx?2  
7|Z=#3INw  
CB=[C1 B1;C2 B2]; 7^1yZ1
(  
AB=[A1 B1;A2 B2]; 4@ EY+p  
AC=[A1 C1;A2 C2]; s zBlyT  
6r  
%二次系数 ~nYp*t C'  
|dNJx<-  
k2=-(det(CB)/det(AB)); ,@I_b  
k3=-(det(AC)/det(AB)); 3sr>?/>:  
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 rXl ~D!  
k2=k2 $Ro]]NUz|  
k3=k3 MI8f(
ZJK5  
+9mE1$C  
end

谢谢分享，学习一下 kpF")0qr