Biomedical Engineering Reference
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varying reconstruction distance unwrapping in
Figure 7.10F and G
, although it still fails to
unwrap the very noisy areas, accurately unwraps the rest of the image.
This approach can be useful even in a general case when the phase images from multiple
reconstruction distances are not available. Instead of varying the reconstruction distance,
one can simply add a constant to the single phase image followed by the modulo of 2
π
and
repeat all the same steps as if they were different reconstruction distance phase images.
This method will lose the advantage of multiple phase plane reconstructions, and therefore
the noisy areas in one image will remain noisy in the others, but it can still produce a phase
unwrapped result. When it was applied to the hologram in
Figure 7.10A
, it still resulted in a
better (compared to path following method) reconstruction, although not as good as the
varying reconstruction distance unwrapping (see
Figure 7.10H and I)
.
Finally, sudden vertical changes in height of an object which correspond to the phase
change greater than 2
π
will not be correctly unwrapped using this method alone. In such a
case, dual-wavelength phase imaging must be used to further increase the axial range at
which an unambiguous phase imaging can be performed.
7.4 Dual-Wavelength Phase Imaging
7.4.1 Synthetic Wavelength
In general, a multiple wavelength phase imaging technique that removes the 2
π
discontinuities is based on the comparison of phase maps acquired at different wavelengths,
as in this case the discontinuities, while being present in all phase maps, will occur at
different places for each wavelength
[9
12]
. For dual-wavelength phase imaging, the
object's height as a function of coordinate can be expressed as:
hðx; yÞ
5
λ
1
2
π
ϕ
1
ðx; yÞ
1
λ
1
m
1
ðx; yÞ
5
λ
2
π
ϕ
2
ðx; yÞ
1
λ
2
m
2
ðx; yÞ
(7.10)
2
where
m
1
and
m
2
are the unknown non-negative integer numbers of wavelengths. Let us
assume that
λ
2
is bigger than
λ
1
, in which case
m
1
cannot be smaller than
m
2
. Rearranging
Eq. (7.10)
, we obtain:
hðx; yÞ
5
λ
2
λ
1
λ
2
2
λ
1
U
ϕ
1
ð
x
;
y
Þ
2
ϕ
2
ð
x
;
y
Þ
5
Λ
12
2
π
ðϕ
1
ðx; yÞ
2
ϕ
2
ðx; yÞÞ
1
Λ
12
M
12
(7.11)
Here, the term
Λ
12
5
λ
2
λ
1
/
λ
2
2
λ
1
is known as the synthetic (“beat”) wavelength. Since the
height must be a positive number, the term
M
12
5
m
1
(
x
,
y
)
2
m
2
(
x
,
y
) must be non-negative.
Moreover, if
1
m
1
ðx; yÞ
2
m
2
ðx; yÞ
2
π
Λ
12
is
bigger than the entire height of the object, then
M
12
cannot exceed 1, so then
M
12
is either
ϕ
1
(
x
,
y
)
2
ϕ
2
(
x
,
y
)
,
0, then
M
12
$
1. Note that if we assume that
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