[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 vq1u!SY  
function sjr=nfdre(~) z`!f'I--!  
7*@BCu6  
%系统焦距及各镜间距输入，间距取负正负 nxs'qX(D  
i)@H  
f=input('f:'); Dj{=Y`Tw  
d1=input('d1:'); ^`fqK4<  
d2=input('d2:'); EO"G(v  
d3=input('d3:'); r[3 2'E  
CzI/Z+\  
A=f^2/(d3*d2)-f/d1; ]f*.C9Y  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); ):nC&M\W~  
C=d3/d2-f/d1; ~X,ZZ 9H  
zZiga q"  
a1=(-B+sqrt(B^2-4*A*C))/(2*A);%α1 s[}cj+0  
a2=d3/(a1*f);%α2 ~y1k2n  
b2=a1*(1-a2)*f/d2;%β2 LuWY}ste  
b1=(1-a1)*f/(d1*b2);%β1 l:j>d^V*&x  
'19kP.  
!gj_9"<  
%曲率半径 ]>,Lw=_[_  
+z+u=)I  
R1=2*f/(b1*b2) +,J!xy+~,  
R2=2*a1*f/(b2*(1+b1)) d8VFa'|  
R3=2*a1*a2*f/(1+b2) u=ZZ;%Rvd  
Duq.`XO  
A1=b2^3*(a1-1)*(1+b1)^3; YRU#/TP  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; S~V?Qe@&Z  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; 8)="Ee  
J:0`*7  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); _nec6=S6(  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); [~k!wipK  
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); 2 5Ia  
>(a[b@[K  
CB=[C1 B1;C2 B2]; tTPjCl  
AB=[A1 B1;A2 B2]; g]U!]  
AC=[A1 C1;A2 C2]; goc"+K  
+g%
Ah  
%非球面系数 /}=cv>S5V  
k2=-(det(CB)/det(AB)); S{zl<>+  
k3=-(det(AC)/det(AB)); $#FA/+<&$  
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 @"0n8y  
k2=k2 Y|aaZ|+  
k3=k3 VX e7b  
=8EGB\P  
end #w;%{C[D  
?]0bR]}y  
%有中间像，焦距输入为正数 td|O#R  
@<B$LJ|jdG  
function sjr=yfdre(~) W+#Zmvo  
d*Kg_He-  
f=input('f:'); I1!m;5-c9k  
d1=input('d1:'); g>-u9%aa  
d2=input('d2:'); R^Y_i  
d3=input('d3:'); [+b&)jN*2  
K b z|h,<  
A=f^2/(d3*d2)-f/d1; m=%yZ2F;  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); 8C<%Y7)/  
C=d3/d2-f/d1; MzI
n~[\  
]gmkajCzD  
a1=(-B-sqrt(B^2-4*A*C))/(2*A); C^ Oy.s  
a2=d3/(a1*f); =7-@&S=?s  
b2=a1*(1-a2)*f/d2; YT)@&HaF  
b1=(1-a1)*f/(d1*b2); lNB<_SO  
AmBLZ<f;  
%曲率半径 W4:#=.m  
<[tU.nh  
R1=2*f/(b1*b2) -qJO6OM  
R2=2*a1*f/(b2*(1+b1)) hmJa1fw=  
R3=2*a1*a2*f/(1+b2) AHn Yfxv_  
N6!$V7oT  
A1=b2^3*(a1-1)*(1+b1)^3; !k8j8v&  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; &U yQ<O>  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; VHx:3G  
Og(|bs!6  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); "M=1Eb$6=  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); ]gDX~]f[  
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); P#qQde/y  
 @+!u{  
CB=[C1 B1;C2 B2]; p-B |Gr|  
AB=[A1 B1;A2 B2]; /t_AiM,(  
AC=[A1 C1;A2 C2]; "i;"  
$ 1v'CT  
%二次系数 q 1+{MPJ  
7SjWofv  
k2=-(det(CB)/det(AB)); zl@hg<n  
k3=-(det(AC)/det(AB)); `_>44!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 N3?h
u}  
k2=k2 oPR?Ar  
k3=k3 YUQKy2  
N6%M+R/Q  
end

谢谢分享，学习一下 D@o8Gerq~