[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 \LB =_W$  
function sjr=nfdre(~) x;NCW  
[V`j@dV  
%系统焦距及各镜间距输入，间距取负正负 A3%s5`vNvH  
Ij>x3L\-  
f=input('f:'); *JXiOs  
d1=input('d1:'); y4`<$gL   
d2=input('d2:'); SJ1 1LF3)  
d3=input('d3:'); [T', ZLR|  
pfW0)V1t  
A=f^2/(d3*d2)-f/d1; )f4D2c&VE  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); 't=\YFQ*v  
C=d3/d2-f/d1; M-KjRl  
rkA0v-N6v  
a1=(-B+sqrt(B^2-4*A*C))/(2*A);%α1 nf!RB-orF  
a2=d3/(a1*f);%α2 4cK6B)X  
b2=a1*(1-a2)*f/d2;%β2 P[PBoRd2  
b1=(1-a1)*f/(d1*b2);%β1 ,)A^3Q*  
ywlN4=  
b7>^w<ki  
%曲率半径 R}4o{l6  
hd1H  
R1=2*f/(b1*b2) +M%i3A  
R2=2*a1*f/(b2*(1+b1)) XJnDx 09h  
R3=2*a1*a2*f/(1+b2) K^AX=B  
dwks"5l  
A1=b2^3*(a1-1)*(1+b1)^3; O4FW/)gq  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; ann!"s_  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; )F 6#n&2  
vTYI ez`g  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); 8Dpf{9Y-E  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); MJ[#Gq\0R  
C2=b2*(a1-1)^2*(1+b1)*(1-b1)^2/(4*a1*b1^2)-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)*(1-b2)^2/(4*a1*a2*b1^2*b2^2)-b2*(a1-1)*(1-b1)*(1+b1)/(a1*b1)-(a2*(a1-1)+b1*(1-a2))*(1-b2)*(1+b2)/(a1*a2*b1*b2)-b1*b2+b2*(1+b1)/a1-(1+b2)/(a1*a2); FQ?,&s$Bmd  
:qy`!QPUm  
CB=[C1 B1;C2 B2]; V#!ihL/>  
AB=[A1 B1;A2 B2]; B+|E|8"  
AC=[A1 C1;A2 C2]; RsU=fe,  
"/hM&  
%非球面系数 )NZ6!3[@  
k2=-(det(CB)/det(AB)); CtVY;eG  
k3=-(det(AC)/det(AB)); cH6ie?KvAo  
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 k%#`{#ni  
k2=k2 _GK^7}u  
k3=k3 -i|qk`Y  
hNUAwTH6  
end N]: "3?%  
xEaRuH
c  
%有中间像，焦距输入为正数 +4ax~fuU  
)'\Jp 7*3  
function sjr=yfdre(~) w.J[3m/  
qVC_K/w 7  
f=input('f:'); `(1em%}  
d1=input('d1:'); EDvK9J  
d2=input('d2:'); tA$,4B?  
d3=input('d3:'); NAhV8  
La?q>  
A=f^2/(d3*d2)-f/d1; { yU1db^  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); @lAOi1m,,  
C=d3/d2-f/d1; =If% m9  
nL@ "FZ`(  
a1=(-B-sqrt(B^2-4*A*C))/(2*A); i0,{*LD%^  
a2=d3/(a1*f); ?UQVmE&  
b2=a1*(1-a2)*f/d2; 8YraW|H  
b1=(1-a1)*f/(d1*b2); oM-{)rvQd  
0.O pgv2K  
%曲率半径 )gV+BHK  
wNDLN`,^H  
R1=2*f/(b1*b2) `|wH=  
R2=2*a1*f/(b2*(1+b1)) ,T"pUeVJ  
R3=2*a1*a2*f/(1+b2) -2|D( sO  
;_K+b,  
A1=b2^3*(a1-1)*(1+b1)^3; sl|s#+Z  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; q#v.-013r  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; zc]
F  
v83uGEq(  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); hxx,E>k  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); (}O)pqZ>  
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); [e3|yE6  
|K'{R'A  
CB=[C1 B1;C2 B2]; UA{sUj+?  
AB=[A1 B1;A2 B2]; [6 wI22  
AC=[A1 C1;A2 C2];  ?1r@r  
<qZXpQ#  
%二次系数 B P
"PUl:  
]l+Bg;F#V  
k2=-(det(CB)/det(AB)); %9[GP7?  
k3=-(det(AC)/det(AB)); wc)[r~On(5  
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 y4
,2Xs9,  
k2=k2 51.F,uY  
k3=k3 Y+iC/pd  
<?52Svi}}  
end

谢谢分享，学习一下 10}oaL S