[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 yMyE s8  
function sjr=nfdre(~) n.wF&f'D]  
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iT]S9  
%系统焦距及各镜间距输入，间距取负正负 8d?g]DEN)6  
kHXL8k#T  
f=input('f:'); cU ?0(z7  
d1=input('d1:'); n9@ of  
d2=input('d2:'); [PQ?#:r  
d3=input('d3:'); hy}8Aji&  
~2<7ZtV=  
A=f^2/(d3*d2)-f/d1; uIcn{RZ_z  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); R>,:A%?^b5  
C=d3/d2-f/d1; u`y><w4i  
Of{/t1o?  
a1=(-B+sqrt(B^2-4*A*C))/(2*A);%α1 K)qF+Vb^j  
a2=d3/(a1*f);%α2 I"Ms-zs  
b2=a1*(1-a2)*f/d2;%β2 8CnRi  
b1=(1-a1)*f/(d1*b2);%β1 '
:gUOra|I  
b'J'F;zh>  
D@.tkzU@E  
%曲率半径 HFwN  
)N=NR2xB
Z  
R1=2*f/(b1*b2) {T4  
R2=2*a1*f/(b2*(1+b1)) e_s&L,ze  
R3=2*a1*a2*f/(1+b2) #[zI5)Meh  
\]P!.}nX#  
A1=b2^3*(a1-1)*(1+b1)^3; &8%e\W\K:/  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; V6t,BJjS  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; Vl_:c75"  
@?=|Y  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); ]4h92\\965  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); a;QMAd!  
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); m>4ahue$  
{.Z}5K  
CB=[C1 B1;C2 B2]; T%6&PrQ7  
AB=[A1 B1;A2 B2]; t]$P1*I  
AC=[A1 C1;A2 C2]; IB# @yH  
p!sWYui  
%非球面系数 pX&pLaF  
k2=-(det(CB)/det(AB)); !PrwH
;  
k3=-(det(AC)/det(AB)); o4*+T8[|5  
k1=(k2*a1*b2^3*(1+b1)^3-k3*a1*a2*(1+b2)^3+a1*b2^3*(1+b1)*(1-b1)^2-a1*a2*(1+b2)*(1-b2)^2)/(b1^3*b2^3)-1 0G7K8`a  
k2=k2 XK|R8rhg8`  
k3=k3  1WY/6[  
tjGd )  
end dDm):Z*`b  
yHHt(GM|o  
%有中间像，焦距输入为正数 ]l'Y'z,}  
vhsk0$f  
function sjr=yfdre(~) kqce
[hgs<  
qHNE8\9  
f=input('f:'); w"OP8KA:^T  
d1=input('d1:'); pY_s*0_  
d2=input('d2:'); *Ho/ZYj3  
d3=input('d3:'); |tv"B@`  
~><^'j[  
A=f^2/(d3*d2)-f/d1; h~(G$':^  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); "I=\[l8t  
C=d3/d2-f/d1; AtF3%Zv2  
,z;ky5Ct  
a1=(-B-sqrt(B^2-4*A*C))/(2*A); aYuD>rD  
a2=d3/(a1*f); OiE;B  
b2=a1*(1-a2)*f/d2; OE4+GI.r-  
b1=(1-a1)*f/(d1*b2); &VV~%jl;k  
87:!C5e}  
%曲率半径 GN!qyT  
*xON W  
R1=2*f/(b1*b2) v^zu:Z*  
R2=2*a1*f/(b2*(1+b1)) &^}6 9  
R3=2*a1*a2*f/(1+b2) vG;zJ#c  
]8T |f  
A1=b2^3*(a1-1)*(1+b1)^3; Zs}5Smjl;%  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; +hE(Ra#  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; 1Wm)rXW[x  
AJSx%?h:6  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); cs1l~bl  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); +T=(6dr  
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); 7\f\!e <  
d
'/TdVM  
CB=[C1 B1;C2 B2]; 5v?;PX  
AB=[A1 B1;A2 B2]; c>/
. ;p  
AC=[A1 C1;A2 C2]; Pb8^ b  
q^A+<d  
%二次系数 "TUe%o  
GrTulN?  
k2=-(det(CB)/det(AB)); $KLD2BAL  
k3=-(det(AC)/det(AB)); >m#e:[N  
k1=(k2*a1*b2^3*(1+b1)^3-k3*a1*a2*(1+b2)^3+a1*b2^3*(1+b1)*(1-b1)^2-a1*a2*(1+b2)*(1-b2)^2)/(b1^3*b2^3)-1 KN?6;G{  
k2=k2 ,RgB$TcE  
k3=k3 8E4mA5@  
DK=cVpN%s  
end

谢谢分享，学习一下 Xe#K{gA