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3 Holographic Interferometry for Studying …
of object waves, and their phases relative to the reconstructing beam are equal to the
phases which had object beams relative to corresponding to them reference beams
during the recording process. Therefore, the phase difference between the waves
reconstructed by the hologram structure, which was formed by the nth and the n +
1th wave pairs, is
δϕ = |ϕ n − ϕ n+1 | ≈
4ππλ
λ 2
(3.2)
where λ is the central wavelength of the dye laser spectrum and λ = λ n –λ n−1 .
Then, in each point, we have interference of four waves:
E = E 0
3
n=0
exp
i
4π zλn
λ 2
,
(3.3)
For full light intensity in the point of an image, we will get the following formula:
I = I 0
3
n=0
exp
i
4π zλn
λ 2
2
.
(3.4)
Equation (3.4) can be rewritten in a simpler way:
I = I 0
sin
2
4
δϕ
2
sin
2
δϕ
2
(3.5)
Provided that all the generation lines had the same intensity, the maximal intensity
in the point of the image is reached when:
δϕ = 2π k,
(3.6)
where k is the integral number.
These image points are connected by the contours, which are the intersection of
object surfaces with the family of surfaces specified by:
δϕ = 2π k, z =
kλ
2
2λ
(3.7)
In such a case, the distance along the normal between the neighboring contour
lines is
l =
λ
2
2λ
(3.8)
3 Holographic Interferometry for Studying …
of object waves, and their phases relative to the reconstructing beam are equal to the
phases which had object beams relative to corresponding to them reference beams
during the recording process. Therefore, the phase difference between the waves
reconstructed by the hologram structure, which was formed by the nth and the n +
1th wave pairs, is
δϕ = |ϕ n − ϕ n+1 | ≈
4ππλ
λ 2
(3.2)
where λ is the central wavelength of the dye laser spectrum and λ = λ n –λ n−1 .
Then, in each point, we have interference of four waves:
E = E 0
3
n=0
exp
i
4π zλn
λ 2
,
(3.3)
For full light intensity in the point of an image, we will get the following formula:
I = I 0
3
n=0
exp
i
4π zλn
λ 2
2
.
(3.4)
Equation (3.4) can be rewritten in a simpler way:
I = I 0
sin
2
4
δϕ
2
sin
2
δϕ
2
(3.5)
Provided that all the generation lines had the same intensity, the maximal intensity
in the point of the image is reached when:
δϕ = 2π k,
(3.6)
where k is the integral number.
These image points are connected by the contours, which are the intersection of
object surfaces with the family of surfaces specified by:
δϕ = 2π k, z =
kλ
2
2λ
(3.7)
In such a case, the distance along the normal between the neighboring contour
lines is
l =
λ
2
2λ
(3.8)