[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 ||3%REliC  
function sjr=nfdre(~) S~L$sqt  
[:<CgU9C  
%系统焦距及各镜间距输入，间距取负正负 7YT%.ID  
@NlE2s6a  
f=input('f:'); /.r|ron:e  
d1=input('d1:'); rZ`ob x\S  
d2=input('d2:'); 9qS~-'&q#  
d3=input('d3:'); vI3L <[W  
0o~? ]C  
A=f^2/(d3*d2)-f/d1; Z18T<e  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); 0nUcUdIf+  
C=d3/d2-f/d1; l&l&
eOE  
rOd<nP^`\  
a1=(-B+sqrt(B^2-4*A*C))/(2*A);%α1 ?145^ w  
a2=d3/(a1*f);%α2 4ti\;55{W  
b2=a1*(1-a2)*f/d2;%β2 (os}s8cIh  
b1=(1-a1)*f/(d1*b2);%β1 Bfe#,  
~t7?5b?*\  
Hx[YHu KL^  
%曲率半径 t}L kl(  
>d-By  
R1=2*f/(b1*b2) wSoIU,I  
R2=2*a1*f/(b2*(1+b1)) Vg'vL[Y  
R3=2*a1*a2*f/(1+b2) x`n$4a'7b  
B1)gudP`  
A1=b2^3*(a1-1)*(1+b1)^3; -|;{/ s5  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; r >bMx~a]  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; {FRUB(68b  
Kn-
cwz5  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); -7E)u  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); gDbj!(tm  
C2=b2*(a1-1)^2*(1+b1)*(1-b1)^2/(4*a1*b1^2)-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)*(1-b2)^2/(4*a1*a2*b1^2*b2^2)-b2*(a1-1)*(1-b1)*(1+b1)/(a1*b1)-(a2*(a1-1)+b1*(1-a2))*(1-b2)*(1+b2)/(a1*a2*b1*b2)-b1*b2+b2*(1+b1)/a1-(1+b2)/(a1*a2); T-e'r  
nmIos]B  
CB=[C1 B1;C2 B2]; _8x:%$   
AB=[A1 B1;A2 B2]; Tbf't^O
t$  
AC=[A1 C1;A2 C2]; lId}sf  
!9A6DWAE$  
%非球面系数 _9yb5_  
k2=-(det(CB)/det(AB)); .*\TG/x  
k3=-(det(AC)/det(AB)); ipzv]c&  
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 |f5WN&c  
k2=k2 JSz;>  
k3=k3 8@;|x2=y  
,:6gp3  
end #dyz  
r7B.@+QK  
%有中间像，焦距输入为正数 wEd+Ds]$  
%siBCjvo=  
function sjr=yfdre(~) ay~c@RXW  
wW2b?b{*Z  
f=input('f:'); XC44]o4jx  
d1=input('d1:'); |2RoDW  
d2=input('d2:'); ]q!,onJ  
d3=input('d3:'); dCq-&3?t  
{jz?LM  
A=f^2/(d3*d2)-f/d1; FuqMT`  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); z[Sq7bbYO  
C=d3/d2-f/d1; iCd$gwA>F  
sVXIR  
a1=(-B-sqrt(B^2-4*A*C))/(2*A); gbvM2  
a2=d3/(a1*f); {Hc [H-  
b2=a1*(1-a2)*f/d2; 25EuVj`zL  
b1=(1-a1)*f/(d1*b2); m~7[fgN2  
NsbC0xLd  
%曲率半径 nV`n=x  
_(A9k{  
R1=2*f/(b1*b2) tE|W8=be/  
R2=2*a1*f/(b2*(1+b1)) :+?eF^5  
R3=2*a1*a2*f/(1+b2) (]cM;  
wWq(|"  
A1=b2^3*(a1-1)*(1+b1)^3; iakqCjV  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; 2=R}u-@6p  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; p7SX,kpt>  
^7b[spqE  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); 5&Y%N(  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); SNLZU%jan  
C2=b2*(a1-1)^2*(1+b1)*(1-b1)^2/(4*a1*b1^2)-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)*(1-b2)^2/(4*a1*a2*b1^2*b2^2)-b2*(a1-1)*(1-b1)*(1+b1)/(a1*b1)-(a2*(a1-1)+b1*(1-a2))*(1-b2)*(1+b2)/(a1*a2*b1*b2)-b1*b2+b2*(1+b1)/a1-(1+b2)/(a1*a2); qH1k  
Evjvaa^  
CB=[C1 B1;C2 B2]; Tt^PiaS!  
AB=[A1 B1;A2 B2]; IZ9L ;"}  
AC=[A1 C1;A2 C2]; Q4~/Tl;  
W^(:\IvV  
%二次系数 :eHD{=  
8]G  
k2=-(det(CB)/det(AB)); 
VIIBw  
k3=-(det(AC)/det(AB)); *yx5G-#?  
k1=(k2*a1*b2^3*(1+b1)^3-k3*a1*a2*(1+b2)^3+a1*b2^3*(1+b1)*(1-b1)^2-a1*a2*(1+b2)*(1-b2)^2)/(b1^3*b2^3)-1 JXT%@w>I  
k2=k2 RC[mpR;2  
k3=k3 :A,g:B  
yM_ta '^$  
end

谢谢分享，学习一下 Q4H(JD1f)