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Reconstruction of two stacked fourier holograms
Volume 27, number 3
OPTICS COMMUNICATIONS
December 1978
RECONSTRUCTION OF TWO STACKED FOURIER HOLOGRAMS K. CHALASINSKA-MACUKOW, B. KARCZEWSKI and T. SZOPLIK Warsaw University, Institute of ExperimentaI Physics, 00-681 Warsaw,Hoza 69, Poland
Received 7 September 1978 The simplest case of the stack formed by two amplitude Fourier holograms linearly recorded is considered. Assuming that all beams in the system propagate paraxially the image field distributions for both holograms are calculated and compared.
1. Introduction The stacked hologram system was proposed by Caulfield, McMahon and Soref in 1972 [1]. The aim of this method was to obtain higher information capacity of holographic memory system. The method was designed especially for the purposes of archival read-only and read-mostly Storage with a short access time. In general the above mentioned method was also discussed 'by Lehman from the point of view of its applications to associative memory [2]. The stack of holograms was realised with volume [1,3] and plane holograms [4]. It seems to be interesting to compare the quality of reconstructions from the successive stacked holograms. Each next hologram is then reconstructed with a beam disturbed by the preceding hologram or holograms. The simplest case of the stack formed by two amplitude Fourier holograms will be discussed. Preliminary results of the present work were reported in [5].
2. Hologram recording
Two amplitude Fourier holograms of the circular hole are recorded in the focal plane of the transforming lens (fig. 1). However the linear recording of the holograms is assumed. The object being a circular hole is normally illuminated by a plane wave of an amplitude A. The plane reference wave illuminates the hologram at an angle 0 in the xz plane. The field distribution in the hologram plane is then
x
• !
o
Ix !
Z t~
Fig. 1. Recording the Fourier holograms O-object, L - lens of the focal length f, H - hologram, 0 - angle between the wave vector of the reference wave and the z axis. 311
Volume 27, number 3
OPTICS COMMUNICATIONS
December 1978
Ad S l (27rdp 1/;V') UH(Xl'Yl)=Bexp(ik sinOXl)+ i Pl
(1)
where x I = p 1 COS/31,J1 = p 1 Sin/~I are coordinates in the hologram plane, k is a wavenumber, 2d is a diameter of the object,f is a focal length of the transforming lens and B is the amplitude of reference wave. Due to relatively large aperture of the transforming lens its influence on the object wave propagation can be neglected. The recorded hologram transmittance is:
[AdJl (2rrdPl ~f)] 2 t(xl'Yl)=Co+CllUH(Xl'Yl )12=c2 +Cl t Pl _1 Ad Jl (2ndp 1/Xf) + ClB exp(-ik sin 0 x ) Ad J1 (2ndp 1/~f) Pl - clB exp(ik sin 0 Xl) i Pl ' 1
(2)
where c o is transmission coefficient of an unexposed film after developing, c 1 is the product of a slope of the Hurter-Driffield curve at an operating point and the exposure time; c 2 = co + ClB2.
3. R e c o n s t r u c t i o n f r o m t h e f i r s t h o l o g r a m
The first hologram is read out with the plane wave of an amplitude C conjugate to the previously used reference wave (fig. 2). This beam interacts with the last term of the transmittance (2) to form an image. The image wave is then transformed by a lens of the focal length f 1 . In the plane parallel to the hologram plane and at a distance 2f 1 from it the image field distribution is:
2~ABCcld U(Xpl 'YPl 'Zl + 2fl) =
~'fl
0J J0 (2~rplop 1/;kfl ) J1 (2~dp 1/~¢) do 1,
(3)
where p 1 is the polar coordinate in the first hologram plane, xpl = Pp i cos 71,yp i = Pp 1 sin ")'1 with pp ~ and 71 being polar coordinates in the image plane; 2t is the hologram diameter. The influence of a transforming lens on the image wave propagation is here ignored again because of its relatively large diameter. X1
X I Xp
H,
L
]
iI • f, L f~ I xp2
] 312
Fig. 2. X Z plane of the readout system, r 0 - distance between centers of holograms HI and H2 ; a - displacement of the second hologram from the z axis; x 1 = P 1 cos ~1, Y 1 = P l sin/31 and Xpl = PPl cos 3'1 'YPl = PPl sin ")'1 are the polar coordinates in the first hologram and image planes; x~ = p~ cos/32 = x2 +a,Y'2 = P'2 sin/32 =Y2 and Xp2 = PP2 c°s ~r2'YP2 = aP2 sin "Y2 are respectively the polar coordinates in the second hologram and image plane.
Volume 27, number 3
OPTICS COMMUNICATIONS
December 1978
4. The zero order beam propagation The zero order beam serves as a readout wave for the second hologram (see fig. 2). This beam propagates along the z' axis making an angle 0 with z axis. It is now our aim to derive the form of that wave in the plane of the second hologram. In the x2Y 2 plane the field distribution is given by the well known integral
tZ(x2,y2,z2) =
1
e ikr
f f u(xl,Yl,Zl) -7- cos(F, F) dS,
(4)
HI
where
U(x 1,Yl ,z l)
exp(-ik sin 0 x 1) [c 2 + c 1 A2d 2 j2
= C
(21rdPl/~f)"
L
4
is the field distribution of the zero order beam just behind the first hologram. The integration in the formula (4) is extended over the first hologram. Since the zero order beam propagates paraxially with respect to the z' axis r can be expanded (see fig. 2) into:
'=r°+
x,22 +y,2 x~ +y2
x,2x 1 + y l y 1
ro
+xlnO-x'2O"
For the same reason cos(F, #) ~ cos 0. Substituting the above formulae in (4) we obtain two integrals. The first one connected with the unmodulated part of hologram transmittance is
, c2Ck cos 0 UI(x'2 'Y2' z2) = ir0 exp(ikro) exp(ik/322/2/'0) exp(-ik sin O cos ~2 P2) t
PI .IkPp2
.
X 0/trexp" ~ ]2J 0 ~ r ' - ~ ]pl"Ol"
(5)
The second integral connected with the autocorrelation of the object function has the form:
A2d2Cl Ckc°sO UII(x2'Y2' z2) =
X0
ir 0
~ro)JOt--~O)
exp(ikr0)
~1
exp(ik p22 ~ 2r 0 ] exp(-ik sin 0 cos/32 P2)
Pl dOl"
(6)
The integral appearing in (6) may be approximately evaluated if we replace J21(21rdp1/~f)/p21 by exp(-p 12/o2), provided that the relation O~ J21(2~'dP 1FA.{~ • ~1 d P l = 0/ exp(-p~/o2)dPl holds. From this relation we obtain the unknown parameter o characteristic of the gaussian profile: o = 16 ~f/3rr5/2d. Now our problem becomes mathematically very similar to the difraction problem already investigated by Schell and Tyras [6]. In both integrals (eqs. (5) and (6)) we use now the expansion of Bessel function of zero order [7]. 313
Volume 27, number 3
OPTICS COMMUNICATIONS
December 1978
Then the equations (5) and (6) can be combined to give:
U(x'2'Y'2 z2) - Ckircos 0 exp(ikro ) exp 0
ik
'
× ~(-1)nikP'2
n=O(n!)2 \2-~0
c2 f0e x p
2"70
exp(-ik sin 0 cos ~2 P2)
P~^2n+ldp 1 +A2d2Clfexp(-~Tp~)v^2n+l 1 dp 1
2r0I Vl
,
(7)
0
where 7/= -ik/2r 0 + 1/o2. After substituting p~ = x in the first integral of (7) we have [8] : f t exp (ik ~rO#~p2n+ 1l]
dp 1 =½t2n+2P(1)P(n+ + 2) l!M(n+ l'n+2'ikt2/2ro
0 where M is the Kummer's function. For large values of l ikt 2/2r01 the function M can be expanded into the confluent hy~ergeometric series [9]. It is worthy to notice that for k = 6 × 10-4 mm and t ~ 0.5 mm from the condition kt'>> 2r0 one can derive the upper limit for r 0 equal to 102 mm. This requirement may be easily met in practice. After expanding the Kummer's function we obtain (see (7)):
~ (-l)n(k~r20)2nfexp(ikO~)p~n+I
n=0 (n!) 2
0
~00
dPl
(8)
ro { ~ -l[-ikP22~n~(n+l)ll-ikt21-' + ~
=--
n--o ,~-'~-~ro I t--o - - Y - . ~ 2ro I
N
[ikt2'~ ~
( - 1 ) n ( ~ ) 2n
t}
(-1) / likt2 -I ~ - 2r0] 1=0 (n - / ) [ \ 2r 0 1
exp - -
.=o - ~ .
where "r = kP'2/r0 and M,N = 0,1,2 ..... Since kt 2 ~"2 r0 in the sum Zff--o('" ") we take into account only the term with l = 0. That term is equal to 1. Then the first double sum in (8) reduces to -exp(-ikP'22/2ro).In the second double sum appearing in (8) we employ the series expansion of Bessel function of the lth order [10] and this sum takes on the following final form: N exp(ikt2/2r0) ~
(1°2/it~Jl(Tt).
The second integral in (7) can be dealt with in the same way as the first one. For large values of [-r/t 21 the Kummer's function is similarly expanded and we obtain again two double sums. After analogical approximations we finally arrive at the formula for the field distribution in the second hologram plane:
U(x2'Y2'Z2) - C cos 0 exp(ikro) exp(ik 0~2/2r0) exp(-ik sin 0 cos/~2 P2) I
F
B
I
x/czLexpt-~o] -expt~00 ]~:-~t it! A2d2Clk I e x p (__72 N (~tt ) Jl(Tt)]}, - ~ )--exp(--r/t 2 ) ~ -'tl
4 ~ 2iro
314
(9)
Volume 27, number 3
OPTICS COMMUNICATIONS
December 1978
As it is seen from (9) the field distribution in plane of the second hologram consists of four components correspondN ")" The first coming respectively to exp(-ikP'22/2ro), exp(ikt2/2ro) ZI--0(" N • ")' exp(-72/417) and e x p ( - n t 2) ZI--0("" ponent is a part of the zero order beam undisturbed by the first hologram and is proportional to the hologram transmittance term connected with the reference wave• The second component is a diffractional correction originating in the diffraction from the aperture of the first hologram. The third component undisturbed by the first hologram is connected with the image wave recorded on the first hologram and the fourth component is a corresponding diffractional correction. This is the entire readout wave incident on the second hologram.
5. Reconstruction from the second hologram
Diffracted image wave from the second hologram propagating in the direction parallel to the z axis is transformed by a lens of focal length f I to form an image in the xp2yp2 plane (see fig. 2). The field distribution in that image plane is as follows:
U(Xp2'YP:'Z2+ 2 f l ) :
2nABCcld cos 0 Xf1
.
[
"
t
exp(ikro)JJo~--~l 0
t
[2rtdP2~exp(ik p~2 ]Jl['-~] x ~_oIS(P2)dP,
/2rtP2PP2~ -
(10)
where S(p~) denotes the whole parenthesis {. . . . . . . . } from eq. (9). Obviously the image wave is composed of four components (see (9)). However only the first component is focused in the back Fourier plane of a lens used in readout process. The remaining three components of the image wave are not focused in the image plane z = z 2 + 2]"1 (see fig. 2). They are focused in two different plane situated somewhat further. Both correction components are focused in the same plane.
6. Conclusions
If the reference wave amplitude B is much larger than that of the object wave A, then on the basis of (9) and (10): 1° our assmption concerning the linear recording is fully justified, 2 ° the main contribution into the image field in the z 2 + 2f I plane comes from the first and second component (the first bracket in (9)), 3 ° the second component which is a diffractional correction is of course of less importance than the first one, 4 ° the last two components (the second bracket in (9)) can be interpreted as a noise. The image field reconstructed from the first hologram (see (3)) corresponds exactly to the first component of the image field obtained from the second hologram (see (9) and (10)). It is easy to predict that images from next holograms would be more noisy in geometrical progress. In view of the above considerations we may state that a stack of the amplitude Fourier holograms is not of particular interest in practice. For real memory system it would be better to use phase holograms. That would be more advantageous from the point of view of energy distribution in the process. However one can expect that signal to noise ratio in that case would not be improved essentially.
References
[1] H.J. Caulfied,D.H. MeMahonand R.A. Soref, US Patent 3635 (1972) 538. [2] M. Lehman, Preprints Intern. Conf. on Optical computing in research and development, Visegrad, 1977, Hungary, [3] J.B. Thaxter and M. Kestigian, Appl. Opt. 13 (1974) 913. [4] D. Pohl, Appl. Opt. 13 (1974) 341. 315
Volume 27, number 3
OPTICS COMMUNICATIONS
December 1978
[5] K. Chaiasifiska-Macukow,B. Karczewski and T. Szoplik, preprints Intern. School of Optical recording and processing of information, Primorsko, 1978, Bulgaria. [6] R.G. Schell and G. Tyras, J. Opt. Soc. Am. 61 (1971) 31. [7] A. Papoulis, Systems and transforms with applications in optics (McGraw Hill, 1968) p. 170, eq. (5-26). [8] M. Abramowitz and I.A. Stegun, eds., Handbook of mathematical functions (Dover, 1965) p. 505, eq. (13.2.1). [9] A. Erdelyi, ed., Higher transcendental functions, Vol. 1, (McGraw HiU, 1953) p. 278, eq. (2). [10] See [8], p. 360, eq. (9.1.10).
316
OPTICS COMMUNICATIONS
December 1978
RECONSTRUCTION OF TWO STACKED FOURIER HOLOGRAMS K. CHALASINSKA-MACUKOW, B. KARCZEWSKI and T. SZOPLIK Warsaw University, Institute of ExperimentaI Physics, 00-681 Warsaw,Hoza 69, Poland
Received 7 September 1978 The simplest case of the stack formed by two amplitude Fourier holograms linearly recorded is considered. Assuming that all beams in the system propagate paraxially the image field distributions for both holograms are calculated and compared.
1. Introduction The stacked hologram system was proposed by Caulfield, McMahon and Soref in 1972 [1]. The aim of this method was to obtain higher information capacity of holographic memory system. The method was designed especially for the purposes of archival read-only and read-mostly Storage with a short access time. In general the above mentioned method was also discussed 'by Lehman from the point of view of its applications to associative memory [2]. The stack of holograms was realised with volume [1,3] and plane holograms [4]. It seems to be interesting to compare the quality of reconstructions from the successive stacked holograms. Each next hologram is then reconstructed with a beam disturbed by the preceding hologram or holograms. The simplest case of the stack formed by two amplitude Fourier holograms will be discussed. Preliminary results of the present work were reported in [5].
2. Hologram recording
Two amplitude Fourier holograms of the circular hole are recorded in the focal plane of the transforming lens (fig. 1). However the linear recording of the holograms is assumed. The object being a circular hole is normally illuminated by a plane wave of an amplitude A. The plane reference wave illuminates the hologram at an angle 0 in the xz plane. The field distribution in the hologram plane is then
x
• !
o
Ix !
Z t~
Fig. 1. Recording the Fourier holograms O-object, L - lens of the focal length f, H - hologram, 0 - angle between the wave vector of the reference wave and the z axis. 311
Volume 27, number 3
OPTICS COMMUNICATIONS
December 1978
Ad S l (27rdp 1/;V') UH(Xl'Yl)=Bexp(ik sinOXl)+ i Pl
(1)
where x I = p 1 COS/31,J1 = p 1 Sin/~I are coordinates in the hologram plane, k is a wavenumber, 2d is a diameter of the object,f is a focal length of the transforming lens and B is the amplitude of reference wave. Due to relatively large aperture of the transforming lens its influence on the object wave propagation can be neglected. The recorded hologram transmittance is:
[AdJl (2rrdPl ~f)] 2 t(xl'Yl)=Co+CllUH(Xl'Yl )12=c2 +Cl t Pl _1 Ad Jl (2ndp 1/Xf) + ClB exp(-ik sin 0 x ) Ad J1 (2ndp 1/~f) Pl - clB exp(ik sin 0 Xl) i Pl ' 1
(2)
where c o is transmission coefficient of an unexposed film after developing, c 1 is the product of a slope of the Hurter-Driffield curve at an operating point and the exposure time; c 2 = co + ClB2.
3. R e c o n s t r u c t i o n f r o m t h e f i r s t h o l o g r a m
The first hologram is read out with the plane wave of an amplitude C conjugate to the previously used reference wave (fig. 2). This beam interacts with the last term of the transmittance (2) to form an image. The image wave is then transformed by a lens of the focal length f 1 . In the plane parallel to the hologram plane and at a distance 2f 1 from it the image field distribution is:
2~ABCcld U(Xpl 'YPl 'Zl + 2fl) =
~'fl
0J J0 (2~rplop 1/;kfl ) J1 (2~dp 1/~¢) do 1,
(3)
where p 1 is the polar coordinate in the first hologram plane, xpl = Pp i cos 71,yp i = Pp 1 sin ")'1 with pp ~ and 71 being polar coordinates in the image plane; 2t is the hologram diameter. The influence of a transforming lens on the image wave propagation is here ignored again because of its relatively large diameter. X1
X I Xp
H,
L
]
iI • f, L f~ I xp2
] 312
Fig. 2. X Z plane of the readout system, r 0 - distance between centers of holograms HI and H2 ; a - displacement of the second hologram from the z axis; x 1 = P 1 cos ~1, Y 1 = P l sin/31 and Xpl = PPl cos 3'1 'YPl = PPl sin ")'1 are the polar coordinates in the first hologram and image planes; x~ = p~ cos/32 = x2 +a,Y'2 = P'2 sin/32 =Y2 and Xp2 = PP2 c°s ~r2'YP2 = aP2 sin "Y2 are respectively the polar coordinates in the second hologram and image plane.
Volume 27, number 3
OPTICS COMMUNICATIONS
December 1978
4. The zero order beam propagation The zero order beam serves as a readout wave for the second hologram (see fig. 2). This beam propagates along the z' axis making an angle 0 with z axis. It is now our aim to derive the form of that wave in the plane of the second hologram. In the x2Y 2 plane the field distribution is given by the well known integral
tZ(x2,y2,z2) =
1
e ikr
f f u(xl,Yl,Zl) -7- cos(F, F) dS,
(4)
HI
where
U(x 1,Yl ,z l)
exp(-ik sin 0 x 1) [c 2 + c 1 A2d 2 j2
= C
(21rdPl/~f)"
L
4
is the field distribution of the zero order beam just behind the first hologram. The integration in the formula (4) is extended over the first hologram. Since the zero order beam propagates paraxially with respect to the z' axis r can be expanded (see fig. 2) into:
'=r°+
x,22 +y,2 x~ +y2
x,2x 1 + y l y 1
ro
+xlnO-x'2O"
For the same reason cos(F, #) ~ cos 0. Substituting the above formulae in (4) we obtain two integrals. The first one connected with the unmodulated part of hologram transmittance is
, c2Ck cos 0 UI(x'2 'Y2' z2) = ir0 exp(ikro) exp(ik/322/2/'0) exp(-ik sin O cos ~2 P2) t
PI .IkPp2
.
X 0/trexp" ~ ]2J 0 ~ r ' - ~ ]pl"Ol"
(5)
The second integral connected with the autocorrelation of the object function has the form:
A2d2Cl Ckc°sO UII(x2'Y2' z2) =
X0
ir 0
~ro)JOt--~O)
exp(ikr0)
~1
exp(ik p22 ~ 2r 0 ] exp(-ik sin 0 cos/32 P2)
Pl dOl"
(6)
The integral appearing in (6) may be approximately evaluated if we replace J21(21rdp1/~f)/p21 by exp(-p 12/o2), provided that the relation O~ J21(2~'dP 1FA.{~ • ~1 d P l = 0/ exp(-p~/o2)dPl holds. From this relation we obtain the unknown parameter o characteristic of the gaussian profile: o = 16 ~f/3rr5/2d. Now our problem becomes mathematically very similar to the difraction problem already investigated by Schell and Tyras [6]. In both integrals (eqs. (5) and (6)) we use now the expansion of Bessel function of zero order [7]. 313
Volume 27, number 3
OPTICS COMMUNICATIONS
December 1978
Then the equations (5) and (6) can be combined to give:
U(x'2'Y'2 z2) - Ckircos 0 exp(ikro ) exp 0
ik
'
× ~(-1)nikP'2
n=O(n!)2 \2-~0
c2 f0e x p
2"70
exp(-ik sin 0 cos ~2 P2)
P~^2n+ldp 1 +A2d2Clfexp(-~Tp~)v^2n+l 1 dp 1
2r0I Vl
,
(7)
0
where 7/= -ik/2r 0 + 1/o2. After substituting p~ = x in the first integral of (7) we have [8] : f t exp (ik ~rO#~p2n+ 1l]
dp 1 =½t2n+2P(1)P(n+ + 2) l!M(n+ l'n+2'ikt2/2ro
0 where M is the Kummer's function. For large values of l ikt 2/2r01 the function M can be expanded into the confluent hy~ergeometric series [9]. It is worthy to notice that for k = 6 × 10-4 mm and t ~ 0.5 mm from the condition kt'>> 2r0 one can derive the upper limit for r 0 equal to 102 mm. This requirement may be easily met in practice. After expanding the Kummer's function we obtain (see (7)):
~ (-l)n(k~r20)2nfexp(ikO~)p~n+I
n=0 (n!) 2
0
~00
dPl
(8)
ro { ~ -l[-ikP22~n~(n+l)ll-ikt21-' + ~
=--
n--o ,~-'~-~ro I t--o - - Y - . ~ 2ro I
N
[ikt2'~ ~
( - 1 ) n ( ~ ) 2n
t}
(-1) / likt2 -I ~ - 2r0] 1=0 (n - / ) [ \ 2r 0 1
exp - -
.=o - ~ .
where "r = kP'2/r0 and M,N = 0,1,2 ..... Since kt 2 ~"2 r0 in the sum Zff--o('" ") we take into account only the term with l = 0. That term is equal to 1. Then the first double sum in (8) reduces to -exp(-ikP'22/2ro).In the second double sum appearing in (8) we employ the series expansion of Bessel function of the lth order [10] and this sum takes on the following final form: N exp(ikt2/2r0) ~
(1°2/it~Jl(Tt).
The second integral in (7) can be dealt with in the same way as the first one. For large values of [-r/t 21 the Kummer's function is similarly expanded and we obtain again two double sums. After analogical approximations we finally arrive at the formula for the field distribution in the second hologram plane:
U(x2'Y2'Z2) - C cos 0 exp(ikro) exp(ik 0~2/2r0) exp(-ik sin 0 cos/~2 P2) I
F
B
I
x/czLexpt-~o] -expt~00 ]~:-~t it! A2d2Clk I e x p (__72 N (~tt ) Jl(Tt)]}, - ~ )--exp(--r/t 2 ) ~ -'tl
4 ~ 2iro
314
(9)
Volume 27, number 3
OPTICS COMMUNICATIONS
December 1978
As it is seen from (9) the field distribution in plane of the second hologram consists of four components correspondN ")" The first coming respectively to exp(-ikP'22/2ro), exp(ikt2/2ro) ZI--0(" N • ")' exp(-72/417) and e x p ( - n t 2) ZI--0("" ponent is a part of the zero order beam undisturbed by the first hologram and is proportional to the hologram transmittance term connected with the reference wave• The second component is a diffractional correction originating in the diffraction from the aperture of the first hologram. The third component undisturbed by the first hologram is connected with the image wave recorded on the first hologram and the fourth component is a corresponding diffractional correction. This is the entire readout wave incident on the second hologram.
5. Reconstruction from the second hologram
Diffracted image wave from the second hologram propagating in the direction parallel to the z axis is transformed by a lens of focal length f I to form an image in the xp2yp2 plane (see fig. 2). The field distribution in that image plane is as follows:
U(Xp2'YP:'Z2+ 2 f l ) :
2nABCcld cos 0 Xf1
.
[
"
t
exp(ikro)JJo~--~l 0
t
[2rtdP2~exp(ik p~2 ]Jl['-~] x ~_oIS(P2)dP,
/2rtP2PP2~ -
(10)
where S(p~) denotes the whole parenthesis {. . . . . . . . } from eq. (9). Obviously the image wave is composed of four components (see (9)). However only the first component is focused in the back Fourier plane of a lens used in readout process. The remaining three components of the image wave are not focused in the image plane z = z 2 + 2]"1 (see fig. 2). They are focused in two different plane situated somewhat further. Both correction components are focused in the same plane.
6. Conclusions
If the reference wave amplitude B is much larger than that of the object wave A, then on the basis of (9) and (10): 1° our assmption concerning the linear recording is fully justified, 2 ° the main contribution into the image field in the z 2 + 2f I plane comes from the first and second component (the first bracket in (9)), 3 ° the second component which is a diffractional correction is of course of less importance than the first one, 4 ° the last two components (the second bracket in (9)) can be interpreted as a noise. The image field reconstructed from the first hologram (see (3)) corresponds exactly to the first component of the image field obtained from the second hologram (see (9) and (10)). It is easy to predict that images from next holograms would be more noisy in geometrical progress. In view of the above considerations we may state that a stack of the amplitude Fourier holograms is not of particular interest in practice. For real memory system it would be better to use phase holograms. That would be more advantageous from the point of view of energy distribution in the process. However one can expect that signal to noise ratio in that case would not be improved essentially.
References
[1] H.J. Caulfied,D.H. MeMahonand R.A. Soref, US Patent 3635 (1972) 538. [2] M. Lehman, Preprints Intern. Conf. on Optical computing in research and development, Visegrad, 1977, Hungary, [3] J.B. Thaxter and M. Kestigian, Appl. Opt. 13 (1974) 913. [4] D. Pohl, Appl. Opt. 13 (1974) 341. 315
Volume 27, number 3
OPTICS COMMUNICATIONS
December 1978
[5] K. Chaiasifiska-Macukow,B. Karczewski and T. Szoplik, preprints Intern. School of Optical recording and processing of information, Primorsko, 1978, Bulgaria. [6] R.G. Schell and G. Tyras, J. Opt. Soc. Am. 61 (1971) 31. [7] A. Papoulis, Systems and transforms with applications in optics (McGraw Hill, 1968) p. 170, eq. (5-26). [8] M. Abramowitz and I.A. Stegun, eds., Handbook of mathematical functions (Dover, 1965) p. 505, eq. (13.2.1). [9] A. Erdelyi, ed., Higher transcendental functions, Vol. 1, (McGraw HiU, 1953) p. 278, eq. (2). [10] See [8], p. 360, eq. (9.1.10).
316