| 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 [!{*)4$6
function sjr=nfdre(~) !@-j!Ub
KdF QlQaj
%系统焦距及各镜间距输入，间距取负正负 #{(?a.:
wP1dPl_j:0
f=input('f:'); b@N|sXt&C
d1=input('d1:'); '6{q;Bxo
d2=input('d2:'); D0PP
d3=input('d3:'); 4UoUuKzt
)8&Q.? T
A=f^2/(d3*d2)-f/d1; @{.rDz
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); 6KhH S@Z
C=d3/d2-f/d1; \~xsBPX+x
xXZ$#z\ Z,
a1=(-B+sqrt(B^2-4*A*C))/(2*A);%α1 uf`o\wqU
a2=d3/(a1*f);%α2 8x'rNb
b2=a1*(1-a2)*f/d2;%β2 -w 2 !k
b1=(1-a1)*f/(d1*b2);%β1 133lIX+(k
dk({J
}*$-rieg
%曲率半径 Y,WcHE
3z:
rUhA
R1=2*f/(b1*b2) I J q$GR
R2=2*a1*f/(b2*(1+b1)) [x!T<jJ
R3=2*a1*a2*f/(1+b2) u4$d# 0sA
] V]~I.
A1=b2^3*(a1-1)*(1+b1)^3; M O* m@
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; 6luCi$bL
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; "eI-Y`O,
dz5bW>
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); O j MDxG
w
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); 3-32q)8
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); uVDB; 6
`;*=2M<c
CB=[C1 B1;C2 B2]; / c/!13|
AB=[A1 B1;A2 B2]; &z{oVU+mA
AC=[A1 C1;A2 C2]; LLgN%! &
1$@k@* u\
%非球面系数 dSI Mwu6u
k2=-(det(CB)/det(AB)); ]|Vm!Q
k3=-(det(AC)/det(AB)); "5XD+qi
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 ; {I{X}b
k2=k2 g IX"W;
k3=k3 iv #9{T
J;m[1Mae&
end 3m7$$ N|
j(nPWEyJM
%有中间像，焦距输入为正数 v .r$]O
[p4a\ Qg0
function sjr=yfdre(~) .v Q2w
qcQ`WU {
f=input('f:'); 7jts;H=
d1=input('d1:'); n{4&('NRFP
d2=input('d2:'); e;rs!I !Yw
d3=input('d3:'); c ty~dzX^
%l: %c
A=f^2/(d3*d2)-f/d1; E6)FYz7x
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); 1%EY!14G+
C=d3/d2-f/d1; Bu!Gy8 \
n )`*{uv$
a1=(-B-sqrt(B^2-4*A*C))/(2*A); WHE*NWz>q
a2=d3/(a1*f); we bT
b2=a1*(1-a2)*f/d2; A|RAMO@le
b1=(1-a1)*f/(d1*b2); 0C3Yina9
*
H7qda' %>
%曲率半径 Mv4JF(,S
=N7N=xY
R1=2*f/(b1*b2) X$JKEW;0BP
R2=2*a1*f/(b2*(1+b1)) ^o?.Rph|i]
R3=2*a1*a2*f/(1+b2) #B+2qD>E
/ d6mlQS
A1=b2^3*(a1-1)*(1+b1)^3; kP8Ypw&
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; 5^*
d4[&+
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; db#y ]>^l
lZn <v'y
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); Grjm 9tbX}
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); Q~-g tEv+&
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"U\;Mw?
l[\[)X3$
CB=[C1 B1;C2 B2]; zKiKda%)
AB=[A1 B1;A2 B2]; (x.K%QC)
AC=[A1 C1;A2 C2]; 2d$hgR#v
TC R(
%二次系数 Z71"d"
W&bh&KzCW
k2=-(det(CB)/det(AB)); 4|++0=#D$
k3=-(det(AC)/det(AB)); R
)?8A\<E
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 H3a}`3}U
k2=k2 ?'h@!F%R'
k3=k3 N s1u0$fg
'{OZ[$E
end
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