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Experiments to Control Atom Number and Phase-Space Density in ...

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where<br />

is used.<br />

ˆǫ ′ <strong>in</strong>c ·ˆǫ ∗ 1 = 1 √ (ˆx<br />

2 ′ ∓iˆy ′ )· ˆx ′ = 1<br />

√<br />

2<br />

ˆǫ ′ <strong>in</strong>c ·ˆǫ ∗ 2 = 1 √ (ˆx<br />

2 ′ ∓iˆy ′ )· ˆy ′ = ∓ i<br />

√<br />

2<br />

The scattered field is then found by subtract<strong>in</strong>g the <strong>in</strong>cident field:<br />

Escat = Eobj −E<strong>in</strong>c<br />

= E0<br />

<br />

−nξ1σ0<br />

√2 exp<br />

2(1+δ 2 <br />

(1+iδ) ˆǫ1 ∓<br />

)<br />

iE0<br />

<br />

√2 ˆǫ2 −<br />

= E0<br />

<br />

−nξ1σ0<br />

√2 exp<br />

2(1+δ 2 <br />

(1+iδ) −1 ˆǫ1<br />

)<br />

1<br />

E0√<br />

(ˆǫ<br />

2 ′ 1 ∓iˆǫ ′ 2)<br />

<br />

(7.21)<br />

(7.22)<br />

(7.23)<br />

To derive the <strong>in</strong>tensity profile <strong>in</strong> the imag<strong>in</strong>g plane, the effect of the imag<strong>in</strong>g optics on<br />

the fields has <strong>to</strong> be taken <strong>in</strong><strong>to</strong> account. In the th<strong>in</strong> lens approximation [110] the scattered<br />

<strong>and</strong> <strong>in</strong>cident fields are thus given by<br />

Escat(x ′ ,y ′ ) = E0<br />

<br />

−nξ1σ0<br />

√2 exp<br />

2(1+δ 2 <br />

ik<br />

(1+iδ) −1 exp<br />

)<br />

<br />

<br />

ik<br />

E<strong>in</strong>c = E0 exp<br />

= E0 exp<br />

2f (x′2 +y ′2 )<br />

ik<br />

2f (x′2 +y ′2 )<br />

ˆǫ ′ <strong>in</strong>c = E0 exp<br />

1<br />

<br />

ˆǫ1 (7.24)<br />

2f (x′2 +y ′2 )<br />

<br />

ik<br />

2f (x′2 +y ′2 <br />

1√2<br />

) (ˆǫ ′ 1 ∓iˆǫ ′ 2)<br />

√2 (ˆx ′ ±iˆy ′ ). (7.25)<br />

The <strong>to</strong>tal electric field <strong>in</strong> the image plane is determ<strong>in</strong>ed by the sum of the scat-<br />

tered <strong>and</strong> the <strong>in</strong>cident field:<br />

E<strong>to</strong>t = Escat +E<strong>in</strong>c<br />

= E0<br />

<br />

−nξ1σ0<br />

√2 exp<br />

2(1+δ 2 <br />

ik<br />

(1+iδ) −1 exp<br />

)<br />

<br />

ik<br />

E0 exp<br />

2f (x′2 +y ′2 <br />

1√2<br />

) (ˆǫ ′ 1 ∓iˆǫ ′ 2)<br />

= E0<br />

1 −n2 √2 exp<br />

σ0<br />

2(1+δ 2 <br />

ik<br />

(1+iδ) −1 exp<br />

)<br />

<br />

1√2<br />

E0 exp (ˆx ′ ±iˆy ′ )<br />

ik<br />

2f (x′2 +y ′2 )<br />

145<br />

2f (x′2 +y ′2 )<br />

2f (x′2 +y ′2 )<br />

<br />

ˆǫ1 +<br />

<br />

ˆx ′ +

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