| 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 ^*ez j1
function sjr=nfdre(~) (Cd {#j<
jy@i(@Z
%系统焦距及各镜间距输入，间距取负正负 EQOP?>mWx!
R{KIkv
f=input('f:'); \ j X N*A
d1=input('d1:'); +ls * //R
d2=input('d2:'); 7O9hn2?e
d3=input('d3:'); n ,: .]3v%
[x p,&
A=f^2/(d3*d2)-f/d1; %x 8`fm
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); a(DZGQ-as
C=d3/d2-f/d1; u#@{%kPW
hd
;S>K/C
a1=(-B+sqrt(B^2-4*A*C))/(2*A);%α1 >aC\_Mc
a2=d3/(a1*f);%α2 ; zE5(3x
b2=a1*(1-a2)*f/d2;%β2 Z t+FRR=
b1=(1-a1)*f/(d1*b2);%β1 N| O]z
^N2M/B|0
sqpOS!]
%曲率半径 ) !}-\5F
Ao, ! z
R1=2*f/(b1*b2) 1H,tP|s
R2=2*a1*f/(b2*(1+b1)) .i&ZT}v3
R3=2*a1*a2*f/(1+b2) $7DcQ b9
pz35trW
A1=b2^3*(a1-1)*(1+b1)^3; Ag4Ga?&8ec
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; 6+B{4OY
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; H~dHVQtJZ
hKZ`DB4
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); KA-/k@1&
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); vG<pc_ak
C2=b2*(a1-1)^2*(1+b1)*(1-b1)^2/(4*a1*b1^2)-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)*(1-b2)^2/(4*a1*a2*b1^2*b2^2)-b2*(a1-1)*(1-b1)*(1+b1)/(a1*b1)-(a2*(a1-1)+b1*(1-a2))*(1-b2)*(1+b2)/(a1*a2*b1*b2)-b1*b2+b2*(1+b1)/a1-(1+b2)/(a1*a2); [&pW&>p3
c6Aut`dK
CB=[C1 B1;C2 B2]; %X"m/4c8}
AB=[A1 B1;A2 B2]; lHKf#|
AC=[A1 C1;A2 C2]; ZAX0n!db3
4o4 =
%非球面系数 W\qLZuQ
k2=-(det(CB)/det(AB)); Z 5)_B,E:X
k3=-(det(AC)/det(AB)); 'LbeL1ca
k1=(k2*a1*b2^3*(1+b1)^3-k3*a1*a2*(1+b2)^3+a1*b2^3*(1+b1)*(1-b1)^2-a1*a2*(1+b2)*(1-b2)^2)/(b1^3*b2^3)-1
GKyG
#Fl
k2=k2 B&i0j5L
k3=k3 Q-8'?S
t.E4Tqzc>
end w &|R5Q
9XoQO 9*Q
%有中间像，焦距输入为正数 8L -4}!~C
JpxbB)/
function sjr=yfdre(~) W[>iJJwz
R{)
Q1~H=q
f=input('f:'); /j' B\,
d1=input('d1:'); Wyq~:vU.S
d2=input('d2:'); ran^te^Ks(
d3=input('d3:'); J}(6>iuQY?
{+"g':><
A=f^2/(d3*d2)-f/d1; sp=OT-Pfp
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); AU xM)H
C=d3/d2-f/d1; )>y
k-
N~):c2Kp<9
a1=(-B-sqrt(B^2-4*A*C))/(2*A); iIsEQ h
a2=d3/(a1*f); \+iu@C
b2=a1*(1-a2)*f/d2; ms}f>f=
b1=(1-a1)*f/(d1*b2); @q&|MML t
=9pw uH
%曲率半径 l@N;sI<O-
3a#PA4Ql
R1=2*f/(b1*b2) [6; N3?+
R2=2*a1*f/(b2*(1+b1)) ]am~aJ|L
R3=2*a1*a2*f/(1+b2) pd4cg?K
&:c:9w
A1=b2^3*(a1-1)*(1+b1)^3; tx0Go'{
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; Mny'9h sl
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; #M5_e m4kN
IV{FH&t^T"
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); wfxOx$]z K
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); ge0's+E+1
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); EJZ@p7*Oj
xMD rE?
CB=[C1 B1;C2 B2]; z wL3,!t
AB=[A1 B1;A2 B2]; ,AH0* L
AC=[A1 C1;A2 C2]; a`H\- G
N%9 h~G
%二次系数 z,{e]MB)M
JbE?a[Eg?
k2=-(det(CB)/det(AB)); d/XlV]#2x\
k3=-(det(AC)/det(AB)); lD W!Fg
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 #B @X
k2=k2 gbm 0H-A:*
k3=k3 *Ph]F$ZP
`gBD_0<T7
end
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