| 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 6zIgQ4Bp24
function sjr=nfdre(~) w:o-klKXY
X E}H 3/2
%系统焦距及各镜间距输入，间距取负正负 ^*R (!P^
4z:#I ;
f=input('f:'); rZ_>`}O2
d1=input('d1:'); gQ~5M'#
d2=input('d2:'); IfDx@ ?OB
d3=input('d3:'); 8 _d -81Dd
[ neXFp}S
A=f^2/(d3*d2)-f/d1; <-UOISyf
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); # $\fh;!W
C=d3/d2-f/d1; r^HA a GpC
1 7hTr
a1=(-B+sqrt(B^2-4*A*C))/(2*A);%α1 ID1?PM
a2=d3/(a1*f);%α2 *.f2VQ~H
b2=a1*(1-a2)*f/d2;%β2 v_PdOp[
k
b1=(1-a1)*f/(d1*b2);%β1 <Y'>F!?#
}tST)=M`
SC#sax4N!=
%曲率半径 (}!C4S3#
2aNT#J"_
R1=2*f/(b1*b2) _-c K{
R2=2*a1*f/(b2*(1+b1)) {bF1\ S]2
R3=2*a1*a2*f/(1+b2) ++[5q+b
xPmN},i'R$
A1=b2^3*(a1-1)*(1+b1)^3; Cd#E"dY6
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; qH%")7>
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; #~?kYCtC)
-ewQp9)G
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); (UEXxUdQ_Q
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); NB8&
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); FE5Q?*Ea
H D /5 !d
CB=[C1 B1;C2 B2]; w,.qCp T$_
AB=[A1 B1;A2 B2]; O>DNC-m)i{
AC=[A1 C1;A2 C2]; i4Z4xTn
`T!#@&+
%非球面系数 {LE&ylE
k2=-(det(CB)/det(AB)); xc'vS >&
k3=-(det(AC)/det(AB)); ?~ULIO '
k1=(k2*a1*b2^3*(1+b1)^3-k3*a1*a2*(1+b2)^3+a1*b2^3*(1+b1)*(1-b1)^2-a1*a2*(1+b2)*(1-b2)^2)/(b1^3*b2^3)-1 X=p"5hhfn
k2=k2 j033%p+Xc
k3=k3 Ef7:y |?
]j.k?P$U}
end &m{'nRU}c
PE@+w#i7*
%有中间像，焦距输入为正数 4 \ $3
B 0mLI%B
function sjr=yfdre(~) OOy}]uYF`
k{Lv37H
f=input('f:'); s_wUM)!
d1=input('d1:'); @YL}km&Fw
d2=input('d2:'); KS>$`ax,
d3=input('d3:'); )3E,D~1e%
/NBTvTI
A=f^2/(d3*d2)-f/d1; X:q _c =X
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); #n})X,ip2
C=d3/d2-f/d1; gT 1P*N;v
6* rcR]
a1=(-B-sqrt(B^2-4*A*C))/(2*A); Px4/O~bLk
a2=d3/(a1*f); ,jh~;, w2
b2=a1*(1-a2)*f/d2; RtF8A5ys
b1=(1-a1)*f/(d1*b2); we]>(|
$iMC/Kym
%曲率半径 o)]FtL:mm
.)PqN s:
R1=2*f/(b1*b2) !L\'Mk/=A
R2=2*a1*f/(b2*(1+b1)) $-G`&oT
R3=2*a1*a2*f/(1+b2) -z C]^Ho@
O*7i } \{
A1=b2^3*(a1-1)*(1+b1)^3; IcoowZZ
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; Gt^d;7x]
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; 79ZxqvB\
ju{%'D!d9
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); H7e /
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); wJIB$3OT
C2=b2*(a1-1)^2*(1+b1)*(1-b1)^2/(4*a1*b1^2)-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)*(1-b2)^2/(4*a1*a2*b1^2*b2^2)-b2*(a1-1)*(1-b1)*(1+b1)/(a1*b1)-(a2*(a1-1)+b1*(1-a2))*(1-b2)*(1+b2)/(a1*a2*b1*b2)-b1*b2+b2*(1+b1)/a1-(1+b2)/(a1*a2); xh'^c^1
O~'FR[J
CB=[C1 B1;C2 B2]; %Y',|+A rx
AB=[A1 B1;A2 B2]; @ ^. *$E5
AC=[A1 C1;A2 C2]; fi=?n{e'
mTjm92
%二次系数 sc xLB;
s v=^k(d3
k2=-(det(CB)/det(AB)); 76MsrOv55
k3=-(det(AC)/det(AB)); ( p D7
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 )Qr6/c 8}
k2=k2 @36S}5Oa
k3=k3 e[4V%h
SfDQ;1 ?
end
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