[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 N6Z{BLZ  
function sjr=nfdre(~) &GWkq>  
"zZIS6j  
%系统焦距及各镜间距输入，间距取负正负 G.U5)4_^  
l)+:4N?iVv  
f=input('f:'); Cef:tdk7  
d1=input('d1:'); eR!G[Cw-  
d2=input('d2:'); HT`1E0G8)  
d3=input('d3:'); o`r(`6@  
J7i+c];!<  
A=f^2/(d3*d2)-f/d1; #%S0PL"x U  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); '=cKU0 G#  
C=d3/d2-f/d1;
2RppP?M!  
Lu>H`B7Q"  
a1=(-B+sqrt(B^2-4*A*C))/(2*A);%α1 d$;/T('  
a2=d3/(a1*f);%α2 b7h+?!H]R  
b2=a1*(1-a2)*f/d2;%β2 _dU P7H (  
b1=(1-a1)*f/(d1*b2);%β1 WtX>Qu|  
AZZRa69=  
[N|xzMe  
%曲率半径 #s+X+fe  
F,v7ifo#f  
R1=2*f/(b1*b2) F=&;Y@t  
R2=2*a1*f/(b2*(1+b1)) m@oUvxcd  
R3=2*a1*a2*f/(1+b2) uu}-"/<~7  
Jgi Iq  
A1=b2^3*(a1-1)*(1+b1)^3; 9qH[o?]
 
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; Fh`-(,e?5  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; lfy7w|  
v*P[W_.  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); 2a[9h#  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); Bxv8RB  
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); jzrt7p*k}  
YsAF{  
CB=[C1 B1;C2 B2]; s)tpr
 
AB=[A1 B1;A2 B2]; Vj!WaN_  
AC=[A1 C1;A2 C2]; ahp1!=Z-=  
4 =Fg!Eu<  
%非球面系数 .;slrg(5F  
k2=-(det(CB)/det(AB)); $,P\)</VR  
k3=-(det(AC)/det(AB)); Mdky^;qq3;  
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 m0Z7N5v)  
k2=k2 s]bPV,"p  
k3=k3 yiO31uQt  
Ma_=-cD  
end (txt8q  
O0zi@2m?B  
%有中间像，焦距输入为正数 xSOoIsL[  
pe#*I/)b  
function sjr=yfdre(~) Mz}i[|U\  
)"-fHW+fy  
f=input('f:'); i;avwP<0  
d1=input('d1:'); Zx9.pFc"  
d2=input('d2:'); QqjTLuN  
d3=input('d3:'); &@"w-M  
Vs|sw  
A=f^2/(d3*d2)-f/d1; X{s/``n  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); 3z$HKG  
C=d3/d2-f/d1; s79q5  
Y 0]Kl^\A  
a1=(-B-sqrt(B^2-4*A*C))/(2*A); F*X%N_n  
a2=d3/(a1*f); V-:`+&S{^  
b2=a1*(1-a2)*f/d2; S8vx[<  
b1=(1-a1)*f/(d1*b2); !rz)bd
3$  
8`DO[Z  
%曲率半径 7_1W:-A7W  
OAY8,C=M  
R1=2*f/(b1*b2) C*$|#.l  
R2=2*a1*f/(b2*(1+b1)) u8xk]:%  
R3=2*a1*a2*f/(1+b2) 9ec0^T  
2h5tBEOX.s  
A1=b2^3*(a1-1)*(1+b1)^3; df!+T0  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; mQFa/7FX  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; Ko]QCLL  
xp~YIeSg  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); !W0P`i<  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); R9b
hC9NP  
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); EqVsxwa  
CGCQa0  
CB=[C1 B1;C2 B2]; B<(Pd  
AB=[A1 B1;A2 B2]; G]1(X38[si  
AC=[A1 C1;A2 C2]; id3)6}  
,I/2.Q})[  
%二次系数 te4F"SEf  
GJ.kkTMT  
k2=-(det(CB)/det(AB)); xq#U4E  
k3=-(det(AC)/det(AB)); G`9cd\^  
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 qPL^zM+  
k2=k2 #Tup]czO  
k3=k3 _{[k[]  
I93 ~8wQ  
end

谢谢分享，学习一下 l*F!~J3