[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 n6f  
function sjr=nfdre(~) HI}9"(t}  
h*JzJ0X  
%系统焦距及各镜间距输入，间距取负正负 Q-:IE T  
ub* j&L=  
f=input('f:'); s/"?P/R  
d1=input('d1:'); |d B`URP  
d2=input('d2:'); Pfv| K;3i  
d3=input('d3:'); 4tb y N  
+9[/> JM  
A=f^2/(d3*d2)-f/d1; jbU=D:|  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); dmkd.aP4  
C=d3/d2-f/d1; ojA i2uz  
ief~*:5  
a1=(-B+sqrt(B^2-4*A*C))/(2*A);%α1 V?*\ISB`}  
a2=d3/(a1*f);%α2 RTgR>qI&)  
b2=a1*(1-a2)*f/d2;%β2 a9=pZ1QAG  
b1=(1-a1)*f/(d1*b2);%β1 V#Px  
v_$'!i$  
=(^-s Jk  
%曲率半径 A"`^Abrm  
8a;I,DK=j  
R1=2*f/(b1*b2) #`>46T  
R2=2*a1*f/(b2*(1+b1)) W]t!I}yPR  
R3=2*a1*a2*f/(1+b2) "&,Gn#'FG  
 d Xiv8B1  
A1=b2^3*(a1-1)*(1+b1)^3; %bp8VR sY  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; lOc!KZHUp  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; \ M_}V[1+  
79?%g=#=  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); DD4fV`:kG  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); aNLkkkJg<;  
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); 2KUm(B.I  
*XCid_{(  
CB=[C1 B1;C2 B2]; /_`f b)f  
AB=[A1 B1;A2 B2]; Y}/e"mp  
AC=[A1 C1;A2 C2]; d0;$k,  
Y)-)owx7  
%非球面系数 6^oQ8unmS  
k2=-(det(CB)/det(AB)); g@<E0 q&`$  
k3=-(det(AC)/det(AB)); .5;Xd?  
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 _7O;ED+  
k2=k2 1ud+~y$K  
k3=k3 aa'u5<<W  
VGVZ`|  
end  ;m;a"j5  
qJQ!e
  
%有中间像，焦距输入为正数 .!kO2/:6  
Jf/X3\0N7  
function sjr=yfdre(~) ~is$Onf99#  
h|MTE~   
f=input('f:'); /%#LA  
d1=input('d1:'); F%8W*Y699  
d2=input('d2:');  !IZbMn6  
d3=input('d3:'); Q UQ"2oC  
(\Iz(N["G  
A=f^2/(d3*d2)-f/d1; :< )"G&  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); lYS+EVcR  
C=d3/d2-f/d1; :DxCjv  
x}`]9XQ  
a1=(-B-sqrt(B^2-4*A*C))/(2*A); .)7r /1o  
a2=d3/(a1*f); Xy74D/ocui  
b2=a1*(1-a2)*f/d2; ~4YLPMGKl  
b1=(1-a1)*f/(d1*b2); ,<^7~d{{3m  
n>_EEw2/  
%曲率半径 HOn,c@.9Y  
:%!}%fkxH  
R1=2*f/(b1*b2) "71,vUW  
R2=2*a1*f/(b2*(1+b1)) #SHmAB  
R3=2*a1*a2*f/(1+b2) rcC}4mNe  
O:=%{/6&D  
A1=b2^3*(a1-1)*(1+b1)^3; tA?cHDp4E  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; Y4\BHFq  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; $Vi[195]2  
|NbF3 fD  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); Lv`*+;1K  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2);
-`iXAyr)m  
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);
JYY:~2  
"5KJ /7q!  
CB=[C1 B1;C2 B2]; ];-DqK'  
AB=[A1 B1;A2 B2]; Y:~A-_  
AC=[A1 C1;A2 C2]; o)X(;o  
D^?jLfW8  
%二次系数 +0w~Skd,  
!besMZ  
k2=-(det(CB)/det(AB)); I^M%+\  
k3=-(det(AC)/det(AB)); }@6/sg  
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 QFP3S(  
k2=k2 5/v,|  
k3=k3 -xJ_5  
B?Vr9H7n  
end

谢谢分享，学习一下 Q*S|SH-cZ0