[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 O{:_-eI&d  
function sjr=nfdre(~) 4
Y2l]86  
Lx6C fR  
%系统焦距及各镜间距输入，间距取负正负 [}-CXB  
P4@<`Eb  
f=input('f:'); .hd<,\nW  
d1=input('d1:'); yyCx;  
d2=input('d2:'); t_%6,?S6  
d3=input('d3:'); QbA+\  
;[9WB<t  
A=f^2/(d3*d2)-f/d1; eJD!dGa  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); Q%:#xG5AmE  
C=d3/d2-f/d1; 46^LPC"x  
_o'_ z
]  
a1=(-B+sqrt(B^2-4*A*C))/(2*A);%α1 l;_zXN  
a2=d3/(a1*f);%α2 7[aSP5e>T  
b2=a1*(1-a2)*f/d2;%β2 yf5X=f.%@  
b1=(1-a1)*f/(d1*b2);%β1 N&ZIsaK,j  
rBG8.E36J  
H]>b<Cs  
%曲率半径 |_7nvck  
&aD]_+b  
R1=2*f/(b1*b2) U6SgV 8  
R2=2*a1*f/(b2*(1+b1)) ETQ.A< v  
R3=2*a1*a2*f/(1+b2) l'h[wwEXm{  
:"BZK5{8  
A1=b2^3*(a1-1)*(1+b1)^3; (5AgI7I,  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; U)mg]o-VE  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; cEzWIS?pp\  
=pHW
qGOD  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); _c
|aRRW  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); P5{|U"Y_  
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); u`GzYG-L  
haj\Dm  
CB=[C1 B1;C2 B2]; @k.j6LKbc  
AB=[A1 B1;A2 B2]; 1%W|>M`  
AC=[A1 C1;A2 C2]; oB$7m4xO\  
K5(:UIWx  
%非球面系数 (W.euQy  
k2=-(det(CB)/det(AB)); E*rnk4Y  
k3=-(det(AC)/det(AB)); %*4
Gx +b  
k1=(k2*a1*b2^3*(1+b1)^3-k3*a1*a2*(1+b2)^3+a1*b2^3*(1+b1)*(1-b1)^2-a1*a2*(1+b2)*(1-b2)^2)/(b1^3*b2^3)-1 7|=*z  
k2=k2 L_$M9G|5n  
k3=k3 G}.t!"  
Ya$JX(aUe  
end 9D 2B8t"a  
8GC(?#Kb  
%有中间像，焦距输入为正数 9n][#I)a3  
M+Rxt.~6  
function sjr=yfdre(~) 5$SO  
%DJxUuh  
f=input('f:'); N"d*pi#h  
d1=input('d1:'); `a.1Af;L  
d2=input('d2:'); XsE] Z4  
d3=input('d3:'); 0:<dj:%M  
\A-w,]9^V  
A=f^2/(d3*d2)-f/d1; )2c[]d/a4  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); 3W*O%9t7  
C=d3/d2-f/d1; M[9]
t("  
Yjo$^q  
a1=(-B-sqrt(B^2-4*A*C))/(2*A); 0Me*X  
a2=d3/(a1*f); Q<]~>cd^  
b2=a1*(1-a2)*f/d2; gB\ a  
b1=(1-a1)*f/(d1*b2); q#W7.8 Z@  
G[V?#7.
 
%曲率半径 FBfyW- 7  
4%5H<:V7  
R1=2*f/(b1*b2) enu",wC3  
R2=2*a1*f/(b2*(1+b1)) UnjUA!v  
R3=2*a1*a2*f/(1+b2) i2<dn)K[~-  
J?Kgev%  
A1=b2^3*(a1-1)*(1+b1)^3; nLZT3`@~,  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; 4v#3UG  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; ebF},Q(48  
8YI.f  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); V0p@wG3  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); (0%0+vY  
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); GvQ|+vC  
aO@zeKg  
CB=[C1 B1;C2 B2]; GnbXS>  
AB=[A1 B1;A2 B2]; w}Q|*!?_  
AC=[A1 C1;A2 C2]; F8 4LMk?U  
m9^?p  
%二次系数 #+Lo&%p#3  
h[d|y_)f  
k2=-(det(CB)/det(AB)); 
C3`2{1  
k3=-(det(AC)/det(AB)); 38P_wf~\  
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 @vf{_g<  
k2=k2 Gq5)>'D?  
k3=k3 eW*nRha  
Mnpb".VU#T  
end

谢谢分享，学习一下 >u5}5OP7