| songshaoman |
2020-05-25 15:25 |
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
%无中间像，焦距输入为负数 a L8Z| *
function sjr=nfdre(~) i \ )$
?{OU%usQwE
%系统焦距及各镜间距输入，间距取负正负 }~v0o#
I
T7 (U6yN
f=input('f:'); Z ..s /K {
d1=input('d1:'); V$ " ]f6
d2=input('d2:'); MX|@x~9W
d3=input('d3:'); OXV9D:bIa
;jmT5XzL
A=f^2/(d3*d2)-f/d1; u )Vn7zh
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); k !g%vx
C=d3/d2-f/d1; 2:1
kSR^Ky
6 _#C vQ
a1=(-B+sqrt(B^2-4*A*C))/(2*A);%α1 [-nPHmZV[
a2=d3/(a1*f);%α2 0)9n${P7d
b2=a1*(1-a2)*f/d2;%β2 4CxU
eq
b1=(1-a1)*f/(d1*b2);%β1 [qxDCuxq
)kE1g&
Lv- M.
%曲率半径 6^z):d #u
r <$"T
R1=2*f/(b1*b2) T ?{F7
R2=2*a1*f/(b2*(1+b1)) @:P:`Zk
R3=2*a1*a2*f/(1+b2) A/~^4DR
+ ;B K|([#
A1=b2^3*(a1-1)*(1+b1)^3; %)y-BdSp .
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; ] q|U0(q9
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; 3o_@ 3-Y%
*>jJ<8!
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); "]yfx@) _
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); 3Io7!:+
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); p$$0**p!`
({ h W
CB=[C1 B1;C2 B2]; ^:ehG9
AB=[A1 B1;A2 B2]; %p^`,b }
AC=[A1 C1;A2 C2];
^{64b
5Qxm\?0J
%非球面系数 1sXVuto
k2=-(det(CB)/det(AB)); P2
z~U
k3=-(det(AC)/det(AB)); `m-7L
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 W;.L N<bx
k2=k2 3/CKy##r%]
k3=k3 ]fU0;jzX
v@qVT'qlU
end >8gb/?z
}J_#N.y
%有中间像，焦距输入为正数 =u. hHkx
UQJ
function sjr=yfdre(~) P?<G:]W
`q7X(x
f=input('f:'); Dx G8`} +
d1=input('d1:'); ;sY n=r
d2=input('d2:'); [f`7+RHrd
d3=input('d3:'); r f
=Wq_
q) y<\cEO
A=f^2/(d3*d2)-f/d1; Uq( fk9`6
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); }i9VV+L#1
C=d3/d2-f/d1; 17!<8vIV$C
+w (B9rH
a1=(-B-sqrt(B^2-4*A*C))/(2*A); w!52DBOe+
a2=d3/(a1*f); G4J)o?:m@
b2=a1*(1-a2)*f/d2; +{s -F g
b1=(1-a1)*f/(d1*b2); 2h`Tn{&1/
eJ60@N\A
%曲率半径 jJe?pT]o
\mNN ) K@
R1=2*f/(b1*b2) 1"RC!
R2=2*a1*f/(b2*(1+b1)) :E2 ww`
R3=2*a1*a2*f/(1+b2) @ gjA8mL
T!t9`I0Zz
A1=b2^3*(a1-1)*(1+b1)^3; otdm r w|
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; S H6T\}X:
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; rWip[>^
NoT%z$ 1n
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); |6b&k hAM
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); %G'P!xQhy
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[7$F&&n
=8Gpov1!V~
CB=[C1 B1;C2 B2]; _''9 -t;n,
AB=[A1 B1;A2 B2]; /2: s g1
AC=[A1 C1;A2 C2]; =v=u+n O
|L(h+/>aWX
%二次系数 sQLjb8!7
sQMfU{S /
k2=-(det(CB)/det(AB)); C\}M_MD
k3=-(det(AC)/det(AB)); F? ?gVa aj
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 @$5= 4HA
k2=k2 [s~6,wz
k3=k3 E[ c6*I
E>bpq ^;r
end
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