Abstract

We describe a reciprocal-lattice vector method for analysis of the diffractive self-imaging (or Talbot effect) of a two-dimensional periodic object. Using this method we analyze the fractional Talbot effect of a hexagonal array and deduce a simple analytical expression for calculation of the complex amplitude distribution at any fractional Talbot plane. Based on this new formula, we design a hexagonal array illuminator (HTAI) with a high fractional parameter. A computer simulation for demonstration of the HTAI is also given.

© 2007 Optical Society of America

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References

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Figures (2)

Fig. 1
Fig. 1

Sketches of (a) a hexagonal lattice and (b) the corresponding reciprocal lattice.

Fig. 2
Fig. 2

Example of (a) the HTAI designed according to Eq. (17) with a fractional parameter of β = 16 and (b) the simulated output intensity distribution of the HTAI shown in (a) at the fractional Talbot plane of z β = z T 16 . In this example, the wavelength of the input beam is taken as λ = 0.6328 μ m , and the output hexagonal array is designed with a basis vector length of Δ = 256 μ m and a spot size of 16 μ m .

Equations (18)

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U ( r ) = u 0 ( r ) lattice ( r , R n ) ,
F { lattice ( r , R n ) } = 1 S n lattice ( ρ , K h ) , F 1 { lattice ( ρ , K h ) } = 1 S h lattice ( r , R n ) ,
b 1 = a 2 × a 0 a 0 ( a 1 × a 2 ) , b 2 = a 0 × a 1 a 0 ( a 1 × a 2 ) , a 1 = b 2 × b 0 b 0 ( b 1 × b 2 ) , a 2 = b 0 × b 1 b 0 ( b 1 × b 2 ) .
U z = 1 S n exp ( i 2 π z λ ) F 1 { G 0 ( ρ ) lattice ( ρ , K h ) exp ( i π λ z ρ 2 ) } ,
λ z K h 2 = 2 m ( m is an integer ) ,
U z = 1 S n exp ( i 2 π z λ ) F 1 { G 0 ( ρ ) lattice ( ρ , K h ) exp ( i 2 π m ) } = exp ( i k z ) U ( r ) .
a 1 = Δ i , a 2 = Δ 2 i + 3 Δ 2 j .
b 1 = 1 Δ i 1 ( 3 Δ ) j , b 2 = 2 ( 3 Δ ) j ,
4 ( h 1 2 + h 2 2 h 1 h 2 ) λ z 3 Δ 2 = m .
z T = 3 Δ 2 2 λ .
U z = 1 S n exp ( i 2 π z λ ) F 1 { G 0 ( ρ ) lattice ( ρ , K h ) exp [ i π 2 β ( h 1 2 + h 2 2 h 1 h 2 ) ] } ,
U z = 1 S n exp ( i 2 π z λ ) F 1 { G 0 ( ρ ) [ C ( ρ ) lattice ( ρ , β K h ) ] } ,
C ( ρ ) = h 2 = 0 β 1 h 1 = 0 β 1 δ ( ρ K h ) exp [ i 2 π β ( h 1 2 + h 2 2 h 1 h 2 ) ] .
U z = exp ( i 2 π z λ ) u 0 ( r ) [ c ( n 1 , n 2 , β ) lattice ( r , R n β ) ] .
c ( n 1 , n 2 , β ) = A ( n 1 , n 2 , β ) exp [ ϕ ( n 1 , n 2 , β ) ] .
A ( n 1 , n 2 , β ) = { 3 β , β = 3 L , n 1 n 2 = 3 m 0 , β = 3 L , n 1 n 2 3 m 1 β , β 3 L } ,
ϕ ( n 1 , n 2 , β ) = 2 π 3 ( 1 β + α ) ( n 1 2 + n 2 2 + n 1 n 2 ) 2 α π 2 ,
α = { 0 , β = 3 L , L = 1 , 2 , 3 1 , β = 3 L 1 , L = 1 , 2 , 3 2 , β = 3 L + 1 , L = 1 , 2 , 3 } .

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