| 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 C:C}5<fk x
function sjr=nfdre(~) au=o6WRa
U4-g^S[
%系统焦距及各镜间距输入，间距取负正负 }w< 7.I
()+ <)hg}2
f=input('f:'); vUU9 $x
d1=input('d1:'); Q/_f
zg
d2=input('d2:'); @ &:ar
d3=input('d3:'); >>o dZL
B$!)YD;
A=f^2/(d3*d2)-f/d1; uv(Sdiir8
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); -~ Mb
C=d3/d2-f/d1; `[)YEg s
>JCM.I0_|
a1=(-B+sqrt(B^2-4*A*C))/(2*A);%α1 %$ Z7x\_
a2=d3/(a1*f);%α2 2hkRd>)&5
b2=a1*(1-a2)*f/d2;%β2 % !>I*H
b1=(1-a1)*f/(d1*b2);%β1 WKIoS"?-F
H {k^S\K
H_o x_
u}
%曲率半径 "zRoU$X
`gb5 "`EZ
R1=2*f/(b1*b2) H77"
R2=2*a1*f/(b2*(1+b1)) jvFTR'R)=
R3=2*a1*a2*f/(1+b2) YmgL zGk`
^8Q62
A1=b2^3*(a1-1)*(1+b1)^3; ;)e2 @'Agl
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; 9 ;Ox;;w
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; [4C: r!
!@'6)/
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); %r6y
;vAf
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); B'EKM)dA
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); rZ^v?4Z\
^__Dd)(
CB=[C1 B1;C2 B2]; ~UjGSO)z}
AB=[A1 B1;A2 B2]; $8[r9L!
AC=[A1 C1;A2 C2]; 78OIUNm`
WjwLM2<nK7
%非球面系数 fasgmi}
k2=-(det(CB)/det(AB)); GF%314Xu
k3=-(det(AC)/det(AB)); EEZw_ 1
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 !M]\I &
k2=k2 XW s"jt
k3=k3 xz{IH,?IG
p4i]7o@
end ez!C?
5
Ho^N1q
%有中间像，焦距输入为正数 z;w ELz1L{
pL%r,Y_^\x
function sjr=yfdre(~) _({A\}Q|
?6jkI2w
f=input('f:'); ~\3kx]^10
d1=input('d1:'); (B-43!C
d2=input('d2:'); *@ )O7vB
d3=input('d3:'); s)2fG\1
9n5<]Q (
A=f^2/(d3*d2)-f/d1; `zt_7MD
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); g
HbxgeL
C=d3/d2-f/d1; `z ) N,fF
%T9 sz4V
a1=(-B-sqrt(B^2-4*A*C))/(2*A); 1`9xIm*9w
a2=d3/(a1*f); k`p74MWu
b2=a1*(1-a2)*f/d2; #\ n8M
b1=(1-a1)*f/(d1*b2); 'fNKlPMv4D
B8%{ }[q
%曲率半径 GS Q /NYK
-yg?V2
R1=2*f/(b1*b2) xOHgp=#D
R2=2*a1*f/(b2*(1+b1)) Css l{B
R3=2*a1*a2*f/(1+b2) N**g]T
0`
$gM8{.!
A1=b2^3*(a1-1)*(1+b1)^3; A4?+T+#d
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; mA@Me7m}
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; .rJi yED?!
~ Yngkt
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); v[n7"
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); ?k|H3;\
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); T,OwM\`.X{
OOz[-j>'Y+
CB=[C1 B1;C2 B2]; 0W()lQ
AB=[A1 B1;A2 B2]; 3#4 5m+D
AC=[A1 C1;A2 C2]; F5qFYL;
~ E^, =4
%二次系数 {Pu\?Cq
.ol'.t , S
k2=-(det(CB)/det(AB)); Z0>DNmH*
k3=-(det(AC)/det(AB)); a9?y`{% L
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 \ S)2
k2=k2 I;?X f
k3=k3 )
(Tom9 ^
VCcr3Dx()F
end
|
|