[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 3Ul*QN{6  
function sjr=nfdre(~) R5D1
w+  
8Wx=p#_  
%系统焦距及各镜间距输入，间距取负正负  DrR@n~  
r" y.KD^  
f=input('f:'); L#J1b!D&<6  
d1=input('d1:'); >
j/w@Fj  
d2=input('d2:'); paK2xX
8E  
d3=input('d3:'); ]`K2N  
2 nCA<&  
A=f^2/(d3*d2)-f/d1; wz%-%39q%  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); rH-23S  
C=d3/d2-f/d1; \85i+q:LuA  
 )2.Si#  
a1=(-B+sqrt(B^2-4*A*C))/(2*A);%α1 N['.BN  
a2=d3/(a1*f);%α2 yA
t^;  
b2=a1*(1-a2)*f/d2;%β2 f8~_E  
b1=(1-a1)*f/(d1*b2);%β1 ,prf;|e?  
Xhm c6?  
M >u_4AY  
%曲率半径 \dVO
wr  
CrLrw T  
R1=2*f/(b1*b2) HtFDlvdy]  
R2=2*a1*f/(b2*(1+b1)) DVA:Cmh\  
R3=2*a1*a2*f/(1+b2) ~Y;*u]^  
&8H'eAA  
A1=b2^3*(a1-1)*(1+b1)^3; Cye.gsCT  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; 6Oq7#3]  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; )e{aN+  
Da|z"I x  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); AH^/V}9H  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); 80I#TA6C  
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); rp$'L7lrX  
Y4-t7UlS;  
CB=[C1 B1;C2 B2]; hb$Ce'}N  
AB=[A1 B1;A2 B2]; qPNR`%}Q  
AC=[A1 C1;A2 C2]; _f83-':W6  
Nb\4 /;#  
%非球面系数 V0@=^Bls  
k2=-(det(CB)/det(AB)); 2G7Wi!J  
k3=-(det(AC)/det(AB)); .A
|udZ,  
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 'L'R9&o<X  
k2=k2 qGo.WZ$  
k3=k3 W1~0_;  
c{|p.hd  
end M%HU4pTW#o  
23PGq%R  
%有中间像，焦距输入为正数 dPlV>IM$z  
@JMiO^  
function sjr=yfdre(~) 3fj4%P"  
?,mmYW6TjB  
f=input('f:'); o-5TC  
d1=input('d1:'); b6bHTH0  
d2=input('d2:'); /v{I  
d3=input('d3:'); pBH
Ra?Y5  
01]f2.5  
A=f^2/(d3*d2)-f/d1; _6Sp QW  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); j#|ZP-=1_  
C=d3/d2-f/d1; 4?kcv59  
K;?+8(H  
a1=(-B-sqrt(B^2-4*A*C))/(2*A); pK*TE5]  
a2=d3/(a1*f); N~Jda o  
b2=a1*(1-a2)*f/d2; Txu/{M,  
b1=(1-a1)*f/(d1*b2); $Sq:q0  
|yCMt:Hk  
%曲率半径 -?\D\\+t  
O
-wzz  
R1=2*f/(b1*b2) )=+|i3]U  
R2=2*a1*f/(b2*(1+b1)) Gc?a+T  
R3=2*a1*a2*f/(1+b2) i-1op> Y  
MgZ/(X E  
A1=b2^3*(a1-1)*(1+b1)^3; 1MFbQs^  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3;  wwqEl(  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; 81F9uM0  
=;L|gtH"  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); Z,gk|M3.  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); M\j.8jG  
C2=b2*(a1-1)^2*(1+b1)*(1-b1)^2/(4*a1*b1^2)-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)*(1-b2)^2/(4*a1*a2*b1^2*b2^2)-b2*(a1-1)*(1-b1)*(1+b1)/(a1*b1)-(a2*(a1-1)+b1*(1-a2))*(1-b2)*(1+b2)/(a1*a2*b1*b2)-b1*b2+b2*(1+b1)/a1-(1+b2)/(a1*a2); /vt3>d%B;  
z{q`GwW  
CB=[C1 B1;C2 B2]; zLQx%Yg!  
AB=[A1 B1;A2 B2]; *. t^MP  
AC=[A1 C1;A2 C2]; ~%oR[B7=|  
k$VlfQ'+  
%二次系数 7V>M]  
kh<2BOV  
k2=-(det(CB)/det(AB)); C!gZN9-  
k3=-(det(AC)/det(AB)); kJU2C=m@e2  
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 P}iE+Z3  
k2=k2 !WlH'y-I  
k3=k3
*CMx-_  
;uW FHc5@B  
end

谢谢分享，学习一下 x-c"%Z|