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References

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  1. H. Z. Hu, “Polarization Heterodyne Interferometry Using a Simple Rotating Analyzer. 1: Theory and Error Analysis,” Appl. Opt. 22, 2052 (1983).
    [CrossRef] [PubMed]
  2. This expression is given only up to first order in Ref. 1.
  3. N. A. Massie, R. D. Nelson, S. Holly, “High Performance Real-Time Heterodyne Interferometry,” Appl. Opt. 18, 1797 (1979).
    [CrossRef] [PubMed]

1983 (1)

1979 (1)

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Figures (1)

Fig. 1
Fig. 1

Rotating analyzer heterodyne interferometer: L, light source; PBS, polarizing beam splitter; MR, reference surface; TO, test object; C1, C2, and C3, λ/4 plates; RA, rotating analyzer.

Equations (29)

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I = a 2 + b 2 2 + a b sin ( 2 ω t α ) ,
Δ α = r cos α cot 2 β r 2 4 sin 2 α ,
β = tan 1 ( a / b ) .
a exp ( i α ) [ i sin δ / 2 cos δ / 2 ] ,
b [ cos δ / 2 i sin δ / 2 ] .
[ 1 0 0 0 ] [ i sin δ / 2 cos δ / 2 ] a exp ( i α ) = i a sin δ / 2 [ 1 0 ] exp ( i α ) ,
[ 1 0 0 0 ] [ cos δ / 2 i sin δ / 2 ] b = i b sin δ / 2 [ 0 1 ] .
[ a 0 ] exp ( i α ) and [ 0 b ] ,
cot 2 β = ( b 2 a 2 ) / 2 a b .
δ 1 = π / 2 + r 1 ,
δ 2 = π / 2 + r 2 .
cot 2 β = b 2 a 2 2 a b ( 1 + r 1 2 / 2 + r 2 2 / 2 ) ( b 2 r 2 2 b 2 r 1 2 ) ( 1 + r 1 2 / 2 + r 2 2 / 2 ) 2 a b .
Δ α = 2 ε ( 1 + sin α cot 2 β ) + 4 ε 2 cos α cot 2 β ε 2 sin 2 α .
a sin ( π / 2 + 2 ε 1 ) [ 1 0 ] exp ( i α ) ,
b sin ( π / 2 + 2 ε 2 ) [ 0 1 ] .
PBS ( T ) = [ 1 0 0 ρ 1 exp i ρ 3 ] .
PBS ( R ) = [ ρ 2 exp i ρ 4 0 0 1 ] ,
E 1 = b [ ρ 2 exp ( i ρ 4 ) 0 0 1 ] [ 0 1 1 0 ] [ 1 0 0 ρ 1 exp ( i ρ 3 ) ] [ 1 1 ]
= b [ ρ 1 ρ 2 exp i ( ρ 3 + ρ 4 ) 1 ] .
E 2 = [ 1 ρ 1 ρ 2 exp i ( ρ 3 + ρ 4 ) ] a exp ( i α ) .
E 1 + E 2 = [ a exp ( i α ) + b ρ 1 ρ 2 exp i ( ρ 3 + ρ 4 ) ρ 1 ρ 2 exp i ( ρ 3 + ρ 4 ) a exp ( i α ) + b ]
= [ X 1 exp i Y 1 X 2 exp i Y 2 ] ,
X 1 = [ a 2 + 2 a b ρ 1 ρ 2 cos ( ρ 3 + ρ 4 α ) + b 2 ρ 1 2 ρ 2 2 ] 1 / 2 ,
X 2 = [ b 2 + 2 a b ρ 1 ρ 2 cos ( α + ρ 3 + ρ 4 ) + a 2 ρ 1 2 ρ 2 2 ] 1 / 2 ,
Y 1 = α + b a ρ 1 ρ 2 sin ( ρ 3 + ρ 4 α ) ,
Y 2 = a b ρ 1 ρ 2 sin ( α + ρ 3 + ρ 4 ) .
E 1 + E 2 = [ a exp i α b ] .
Φ = α + ρ 1 ρ 2 a b [ b 2 sin ( ρ 3 + ρ 4 α ) a 2 sin ( ρ 3 + ρ 4 + α ) ] .
Δ α = ρ 1 ρ 2 a b [ b 2 sin ( ρ 3 + ρ 4 α ) a 2 sin ( ρ 3 + ρ 4 + α ) ] ,

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