[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 X?SLYm@v  
function sjr=nfdre(~) cJ#%OU3p  
}-PV%MNud  
%系统焦距及各镜间距输入，间距取负正负 :Oc&{z?q  
&Ui&2EW  
f=input('f:'); \l?.VE
D  
d1=input('d1:'); 98!H$6k  
d2=input('d2:'); 3&
Fqd  
d3=input('d3:'); M7gM#bv>L  
rm!.J0 X  
A=f^2/(d3*d2)-f/d1; s/OXZ<C|  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); 8_uh2`+Bvb  
C=d3/d2-f/d1; ixJwv\6Y  
A5YS "i  
a1=(-B+sqrt(B^2-4*A*C))/(2*A);%α1 hJGWa%`  
a2=d3/(a1*f);%α2 % ^&D,  
b2=a1*(1-a2)*f/d2;%β2 =
ve, !  
b1=(1-a1)*f/(d1*b2);%β1 y:dwx*Q9I  
Ts3(,
Y  
0@2pw2{Ru  
%曲率半径 !gG\jC~n  
b*o,re)Dj  
R1=2*f/(b1*b2) nXxSv
~r  
R2=2*a1*f/(b2*(1+b1)) }9>X M  
R3=2*a1*a2*f/(1+b2) {-,^3PI\  
3bMUsyJ2  
A1=b2^3*(a1-1)*(1+b1)^3; ?r}!d2:dX  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; BXX1G  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; +( V+XT  

ATf{;S}  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); Q)ZkUmW  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2);
Z.mnD+{  
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); W'M\DKJ?  
+ywd(Tuzm  
CB=[C1 B1;C2 B2]; ^<nN~@
j  
AB=[A1 B1;A2 B2]; &Qq4xn+J  
AC=[A1 C1;A2 C2]; RYKV?f#[H  
bHH=MLZR:  
%非球面系数 U-GV^j  
k2=-(det(CB)/det(AB)); 6882:,q  
k3=-(det(AC)/det(AB)); vh6#Bc)i%w  
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 T#w *5Qf  
k2=k2 ?u8vK<2h  
k3=k3 /pDI \]  
IXmO1*o@  
end Is !DiB  
od~`q4p1(-  
%有中间像，焦距输入为正数
&-6D'@  
.j0]hn]  
function sjr=yfdre(~) QprzlxB  
3a#
j&]  
f=input('f:'); n)rF!a  
d1=input('d1:'); O5p]E7/e  
d2=input('d2:'); P1mPC  
d3=input('d3:'); r&-Ir3[  
vH^^QI:em  
A=f^2/(d3*d2)-f/d1; 7<VfE`Q3  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); Q.[^5 8  
C=d3/d2-f/d1; mLn =SU{#  
2_HPsEx  
a1=(-B-sqrt(B^2-4*A*C))/(2*A); ps6c>AN`A&  
a2=d3/(a1*f); B.J4
}Ua  
b2=a1*(1-a2)*f/d2; CfnRcnms  
b1=(1-a1)*f/(d1*b2); SAGLLk07G  
[{B1~D-  
%曲率半径 tr\Vr;zd  
^R4eW|H  
R1=2*f/(b1*b2) u!DSyHR '  
R2=2*a1*f/(b2*(1+b1)) =7`0hS<@F  
R3=2*a1*a2*f/(1+b2) K*Tvo`  
kM\O2ay  
A1=b2^3*(a1-1)*(1+b1)^3; ;zO(bj>  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; =pS\gLQu  
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}W06k>)%  
3f:I<S7  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); ~!!>`x  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); UI:{*N**Z  
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); `$\Y,9E}x  
_
r|$H_#  
CB=[C1 B1;C2 B2]; $V)LGu2(m  
AB=[A1 B1;A2 B2]; Q(KLx)  
AC=[A1 C1;A2 C2]; z,oqYU\:  
>9g`9hB  
%二次系数 9c=_p'G3Fw  
G K @]61b  
k2=-(det(CB)/det(AB)); >cD+&h34  
k3=-(det(AC)/det(AB)); Cs %-f"  
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 ^q,KRut  
k2=k2 Ga%]$4u  
k3=k3 hqY9\,.C  
soZw""|v  
end

谢谢分享，学习一下 7`tnoTUv