[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 T5_/*`F  
function sjr=nfdre(~) <WiyM[ep  
WvoJ^{\4N*  
%系统焦距及各镜间距输入，间距取负正负 O`^dy7>{U  
;u?L>(b  
f=input('f:'); Yn
_v'Os2  
d1=input('d1:'); <~bvfA=  
d2=input('d2:'); 6no&2a|D  
d3=input('d3:'); HkQ rij6  
pwg\b  
A=f^2/(d3*d2)-f/d1; Vr7L9%/wg  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); xFScj0Y  
C=d3/d2-f/d1; c }7gHud  
zHX7%x,Cq  
a1=(-B+sqrt(B^2-4*A*C))/(2*A);%α1 +C36OcmT~  
a2=d3/(a1*f);%α2 Z [YSET  
b2=a1*(1-a2)*f/d2;%β2 Tr.u'b(  
b1=(1-a1)*f/(d1*b2);%β1 {+^&7JX  
`]I p`_{  
s$Vz1B  
%曲率半径 STL+tLJ  
3rj7]:Vr  
R1=2*f/(b1*b2) W|L#Q/ RX  
R2=2*a1*f/(b2*(1+b1)) C <d]0)  
R3=2*a1*a2*f/(1+b2) -KL5sK  
AIG5a$}&  
A1=b2^3*(a1-1)*(1+b1)^3; 1wy?<B.f
 
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; T(=Z0M  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; $}t;c62  
AQGl}%k_  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); b?j\YX[e  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); v=~+o[  
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); f+%s.[;A  

YJF|J2u  
CB=[C1 B1;C2 B2]; a] P0PH~  
AB=[A1 B1;A2 B2]; WCP2x.gb5  
AC=[A1 C1;A2 C2]; 97pfMk1_  
GGU>={D)  
%非球面系数 /[I#3|  
k2=-(det(CB)/det(AB)); 2f5YkmGc";  
k3=-(det(AC)/det(AB)); W.c>("gC  
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 9*S9~  
k2=k2 ODxCD%L  
k3=k3 e3k58  
&<EixDi4q  
end 1oI2  
hO2W!68
 
%有中间像，焦距输入为正数 hf:\^w  
@2~;)*  
function sjr=yfdre(~) {Fvl7Sh  
x~E\zw  
f=input('f:'); '3=@UBs  
d1=input('d1:'); m#^;V  
d2=input('d2:'); AEd9H
+I  
d3=input('d3:'); 0|| 5r#  
~?Zm3zOCc2  
A=f^2/(d3*d2)-f/d1; fLZ99?J  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); 9vBW CCf
 
C=d3/d2-f/d1; dN@C)5pm5`  
tu^C<MV  
a1=(-B-sqrt(B^2-4*A*C))/(2*A); \;1nEjIA  
a2=d3/(a1*f); 0py29>"t  
b2=a1*(1-a2)*f/d2; j/F:j5O*  
b1=(1-a1)*f/(d1*b2); h\4enu9[RL  
U U3o(Yq  
%曲率半径 _
q}^#-  
JvF0s}#4  
R1=2*f/(b1*b2) w&*oWI$i  
R2=2*a1*f/(b2*(1+b1)) zFr#j~L"  
R3=2*a1*a2*f/(1+b2) _adW>-wQ!d  
+o]J0Gu  
A1=b2^3*(a1-1)*(1+b1)^3; P}w0=  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; 
oK3PA  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; d wku6lCk  
 63VgQ  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); s8;*Wt  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); $XcuU sG  
C2=b2*(a1-1)^2*(1+b1)*(1-b1)^2/(4*a1*b1^2)-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)*(1-b2)^2/(4*a1*a2*b1^2*b2^2)-b2*(a1-1)*(1-b1)*(1+b1)/(a1*b1)-(a2*(a1-1)+b1*(1-a2))*(1-b2)*(1+b2)/(a1*a2*b1*b2)-b1*b2+b2*(1+b1)/a1-(1+b2)/(a1*a2); Pk&$#J_  
_e "  
CB=[C1 B1;C2 B2]; /}k?Tg/  
AB=[A1 B1;A2 B2]; h*?]A  
AC=[A1 C1;A2 C2]; q!WiX|P  
cvC 7#i[G  
%二次系数 KB$Y8[  

C_&ZQlgQ  
k2=-(det(CB)/det(AB)); QO %;%p*  
k3=-(det(AC)/det(AB)); cRWYS[O?-  
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 \CBL[X5tr  
k2=k2 qmtH0I7)  
k3=k3 _$yS4=.  
'jYKfq~_cJ  
end

谢谢分享，学习一下 p+CK+m