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Self-mixing interference effects with a folding feedback cavity in Zeeman-birefringence dual frequency laser

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Abstract

The self-mixing interference effects with a folding feedback cavity in a Zeeman-birefringence dual frequency laser have been investigated theoretically and experimentally. The fringe frequency of the self-mixing interference system can be doubled due to the hollow cube corner prism, with which a folding cavity is formed. The intensities of the two frequencies are changed periodically in the modulation of the external cavity length. When the phase difference between the two frequencies equals π/2, the intensity modulation curves can be divided into four zones with equal width in a period. Each zone corresponds to one polarization state. Based on the experimental results, a novel displacement sensor with a high resolution of λ/16, as well as functions of direction discrimination, is discussed.

©2006 Optical Society of America

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Figures (6)

Fig. 1.
Fig. 1. Experimental setup.
Fig. 2.
Fig. 2. (a) The configuration of M2; (b) The configuration of HCCP (unit: mm).
Fig. 3.
Fig. 3. Simulation of the laser intensity versus the output voltage of the D/A card. (a) normal intensity modulation frequency; (b) doubled intensity modulation frequency with HCCP.
Fig. 4.
Fig. 4. Simulation of the laser intensity variations of o-light and e-light. (a) without a threshold intensity; (b) with a threshold intensity
Fig. 5.
Fig. 5. Observation of the laser intensity versus the output voltage of the D/A card. (a) normal intensity modulation frequency; (b) doubled intensity modulation frequency with HCCP.
Fig. 6.
Fig. 6. Observation of the laser intensity variations of o-light and e-light.

Equations (17)

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E ( t ) = r 1 r 2 exp ( j 4 πv nL c + gL ) E 0 ( t ) + r 1 r 2 r 3 ξ exp ( j 4 πv nL + l + Δ l c + gL ) E 0 ( t ) ,
r 1 r 2 exp ( j 4 πv nL c + gL ) [ 1 + t 2 r 3 ζ r 2 exp ( j 4 πv l c + j 4 πv Δ 1 c ) ] = 1 .
r 1 r 2 exp ( gL ) { [ 1 + α cos ( φ + δ l ) ] 2 + [ α sin ( φ + δ l ) ] 2 } ½ exp [ j ( 4 πv nL c + θ ) ] = 1 ,
tan θ = α sin ( φ + δ l ) 1 + α cos ( φ + δ l ) .
r 1 r 2 exp ( gL ) [ 1 + α cos ( φ + δ l ) ] exp [ j ( 4 πv nL c + θ ) ] = 1 .
g = 1 L [ In ( r 1 r 2 ) + α cos ( φ + δ l ) ] .
4 πv nL c + θ = 2 .
g 0 = 1 L In ( r 1 r 2 ) .
Δ g = g g 0 = α L cos ( φ + δ l ) .
I = I 0 ( 1 K Δ g ) ,
I 1 = I 0 [ 1 + α K L cos ( 2 φ + δ l ) ] ,
φ H = 4 πv∙ 2 Δ l c = 2 φ .
I 2 = I 0 [ 1 + α K L cos ( 2 φ + δ l ) ] .
Δ g 0 = Δ g e .
I o = I 0 o [ 1 + α K L cos ( 2 φ + δ lo ) ] ,
I e = I 0 e [ 1 α K L cos ( 2 φ + δ le ) ] ,
δ = 4 π Δ v l c ,
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