114
2 Holographic Microscopy of Phase and Diffuse Objects …
the object beam nerve fiber. In this case, initially on a hologram, a comparison wave
was recorded which went through all the elements of the optical scheme without the
object. Then, the nerve fiber was placed and was lighted up with a plane wave, which
transformed into a spherical one behind the lens. As a result of interference of the
recovered reconstructed spherical wave and wave having phase perturbation caused
by the object, an interference fringes system occurs. To produce finite width bands,
the incidence angle of the reconstructed wave on the hologram was modified. Taking
into account the fact that the distance between the centers of the spherical waves is
much smaller than their curvature radius in the interference region, then the bands
can be considered as almost straight.
If coordinate system is placed in such a way that z is an optical axis of the holographic interference microscope, and y is the object symmetry axis, then a hologram
turning around x-axis by a small angle shifts the center of the reconstructed spherical
wave along y-axis.
In this case, the dark band equation is the following [335]
δ(x, y)
2π
=
y − y m
T
,
(2.26)
where δ(x, y) is the phase difference between the object wave and the comparison
wave; T is the interference pattern period; y − y m is the m-order interference fringe
deviation.
The phase perturbation δ(x, y) in the plane directly behind the fiber is the following
δ(x, y) = δ(x) = 2k
√
1−x 2
0
n(x, z)dz,
(2.27)
where k = 2π /λ, λ is the wavelength.
The object is supposed to have axial symmetry and be lightened up by the plane
wave. A nerve fiber with a good degree of accuracy can be estimated as an axially
symmetric object.
On the interferogram, it is very easy to identify the fiber edge and its other areas
using the coordinates of its image for correspondence determination. In any plane z
= z p
δ
x p
= δ
−
f
z p
x
.
The multiplier −
f
z p
is the scale change in the observation plane z = z p concerning
the object plane. Normalizing the coordinates in the observation field in such a way
that the image radius was equal to 1, and making the replacement of the variable
x = r cos ϕ, z = r sin ϕ in (2.27) and considering the fact that n = n(r ) and
n(r ) = n(r ) − n 0 (where n(r ) is the nerve refraction index; n 0 is the physiological
2 Holographic Microscopy of Phase and Diffuse Objects …
the object beam nerve fiber. In this case, initially on a hologram, a comparison wave
was recorded which went through all the elements of the optical scheme without the
object. Then, the nerve fiber was placed and was lighted up with a plane wave, which
transformed into a spherical one behind the lens. As a result of interference of the
recovered reconstructed spherical wave and wave having phase perturbation caused
by the object, an interference fringes system occurs. To produce finite width bands,
the incidence angle of the reconstructed wave on the hologram was modified. Taking
into account the fact that the distance between the centers of the spherical waves is
much smaller than their curvature radius in the interference region, then the bands
can be considered as almost straight.
If coordinate system is placed in such a way that z is an optical axis of the holographic interference microscope, and y is the object symmetry axis, then a hologram
turning around x-axis by a small angle shifts the center of the reconstructed spherical
wave along y-axis.
In this case, the dark band equation is the following [335]
δ(x, y)
2π
=
y − y m
T
,
(2.26)
where δ(x, y) is the phase difference between the object wave and the comparison
wave; T is the interference pattern period; y − y m is the m-order interference fringe
deviation.
The phase perturbation δ(x, y) in the plane directly behind the fiber is the following
δ(x, y) = δ(x) = 2k
√
1−x 2
0
n(x, z)dz,
(2.27)
where k = 2π /λ, λ is the wavelength.
The object is supposed to have axial symmetry and be lightened up by the plane
wave. A nerve fiber with a good degree of accuracy can be estimated as an axially
symmetric object.
On the interferogram, it is very easy to identify the fiber edge and its other areas
using the coordinates of its image for correspondence determination. In any plane z
= z p
δ
x p
= δ
−
f
z p
x
.
The multiplier −
f
z p
is the scale change in the observation plane z = z p concerning
the object plane. Normalizing the coordinates in the observation field in such a way
that the image radius was equal to 1, and making the replacement of the variable
x = r cos ϕ, z = r sin ϕ in (2.27) and considering the fact that n = n(r ) and
n(r ) = n(r ) − n 0 (where n(r ) is the nerve refraction index; n 0 is the physiological