**5. Fourier holograms**

36 Holograms – Recording Materials and Applications

Fig. 17. (a). Plot of the Diffraction efficiency to certain temperature during the recording process.

Fig. 17. (b). Temperature behavior during the recording process in the 17 μm thickness cell

The temperature changes in the cases showed in Figs 15, 16, and 17 are due that we do not have temperature control in our laboratory. It is important to say; the gratings were recorded along different days. We can conclude that the temperature most be constant during the

recording process about 25ºC to obtain a softly behavior and the highest DE values.

using the C3 concentration of two holographic gratings.

Several types of holograms exist; these are of transmission, amplitude, phase, reflection, computer generated holograms, Fourier holograms, etc., (Smith, 1975).

A hologram can be done registering the interference pattern intensity between two beams, called a reference and an object beam, in a photosensitive material or holographic film as is shown in Fig. 18.

Fig. 18. Basic setup to hologram recording process, the beam reflected by the object is called object beam and the reflected beam by the mirror is the reference beam.

In particular, recording the interference pattern between the reference beam and the Fourier transform of an object produces the Fourier holograms. One of the main characteristics of this type of holograms is that the necessary area to record is small compared with other types of holograms. The Fig. 19 we depicted the scheme to record Fourier holograms.

Fig. 19. Schematically representation for Fourier holograms recording. BS: beam splitter, BE beam expander, CL: Collimating lens, TL: transforming lens, PM: photosensitive material, M: mirror.

Norland Optical Adhesive 65® as Holographic Material 39

( ) ( ) ( )

<sup>λ</sup> <sup>=</sup> <sup>=</sup> <sup>f</sup> x, <sup>y</sup> <sup>e</sup> dxdy <sup>f</sup> <sup>1</sup> <sup>O</sup> <sup>O</sup> x,y i2 <sup>x</sup> <sup>y</sup>

− π ν +μ

O = F{ } f( ) x,y (7)

<sup>2</sup> <sup>I</sup> <sup>=</sup> <sup>R</sup> <sup>+</sup> O (8)

<sup>2</sup> <sup>2</sup> \* \* I = R + O + R O + RO (9)

<sup>1</sup> <sup>2</sup> <sup>3</sup> <sup>4</sup> I = I + I + I + I (11)

<sup>2</sup> I1 <sup>=</sup> A (12)

(6)

∞

∞

−∞

−∞

where the Fourier transform is denoted by F.

Fig. 21. Schematically representation of the Fourier holograms.

which can be expressed as

can obtain

where

The photosensitive material, PM, registers the interference pattern intensity given by

where the symbol \* denotes the complex conjugate. Substituting the eqs. (5) and (7) in (9) we

<sup>I</sup> <sup>A</sup> <sup>F</sup>{ } <sup>f</sup>( ) x, <sup>y</sup> Ae <sup>F</sup>{ } () <sup>f</sup>( ) x, <sup>y</sup> Ae <sup>F</sup> { } <sup>f</sup> x, <sup>y</sup> <sup>2</sup> <sup>2</sup> <sup>−</sup>i2πν<sup>a</sup> i2πν<sup>a</sup> \* <sup>=</sup> <sup>+</sup> <sup>+</sup> <sup>+</sup> (10)

or

In the scheme of Fig. 19 a laser beam is expanded by microscope objective and a pinhole (BE) and collimated by the lens (CL) to produce a plane wave that illuminates the object f(x, y); the converging lens (TL) obtains the Fourier transform of the object called object beam (O) and directs it into the photosensitive material (PM) where it interferes with the reference beam (R) coming from a mirror and the pattern is recorded.

Once the hologram was recorded, we use the same reference beam R to reconstruct the images as is depicted in Fig. 20. The reconstruction of the hologram produces two images, namely, real and conjugated images.

Fig. 20. Reconstruction of the real image of a hologram.

To explain mathematically the Fourier holograms we use the Fig. 21, where a point source is located at coordinates x=-a and y=0. The divergent spherical beams is then transformed at plane wave for the lens and is used as the reference beam R, given by

$$\mathbf{R} = \mathbf{R}(\mathbf{x}, \mathbf{y}) = \frac{1}{\lambda \mathbf{f}} \int\_{-\infty}^{\infty} \int\_{-\infty}^{\infty} \mathbf{\hat{\mathbf{\hat{s}}}(\mathbf{x} + \mathbf{a})} \mathbf{\hat{\mathbf{\hat{s}}}(\mathbf{y})} e^{-i2\pi(\mathbf{v}\mathbf{x} + \mu \mathbf{y})} \mathbf{dx} d\mathbf{y} \tag{4}$$

where λ is the wavelength, f is the focal length of the lens, ν= x/λf and μ=y/λf are spatial frequencies. As is indicated in eq.\*, the reference beam is a plane wave expressed by the Fourier transform of the point source δ(x+a) δ(y) located at x=-a. Evaluating the integral in eq. (4), we can express the reference beam as:

$$\mathbf{R} = \mathbf{A}\mathbf{e}^{\mathbf{i}2\pi\mathbf{v}\mathbf{a}}\tag{5}$$

where **A** is the wave amplitude.

The object beam O is the Fourier transform of the object f(x, y) and is given by

$$\mathbf{O} = \mathbf{O}(\mathbf{x}, \mathbf{y}) = \frac{1}{\lambda \mathbf{f}} \prod\_{-\alpha - \alpha}^{\alpha} \oint \mathbf{f}(\mathbf{x}, \mathbf{y}) \mathbf{e}^{-i2\pi (\mathbf{v}x + \mu y)} \mathbf{dx} dy \tag{6}$$

or

38 Holograms – Recording Materials and Applications

In the scheme of Fig. 19 a laser beam is expanded by microscope objective and a pinhole (BE) and collimated by the lens (CL) to produce a plane wave that illuminates the object f(x, y); the converging lens (TL) obtains the Fourier transform of the object called object beam (O) and directs it into the photosensitive material (PM) where it interferes with the reference

Once the hologram was recorded, we use the same reference beam R to reconstruct the images as is depicted in Fig. 20. The reconstruction of the hologram produces two images,

To explain mathematically the Fourier holograms we use the Fig. 21, where a point source is located at coordinates x=-a and y=0. The divergent spherical beams is then transformed at

where λ is the wavelength, f is the focal length of the lens, ν= x/λf and μ=y/λf are spatial frequencies. As is indicated in eq.\*, the reference beam is a plane wave expressed by the Fourier transform of the point source δ(x+a) δ(y) located at x=-a. Evaluating the integral in

i2 <sup>a</sup> R Ae πν = (5)

<sup>1</sup> <sup>R</sup> R(x,y) <sup>−</sup>i2<sup>π</sup> <sup>ν</sup>x+μ<sup>y</sup> ∞

∞

−∞

The object beam O is the Fourier transform of the object f(x, y) and is given by

−∞

( ) ( ) ( ) <sup>x</sup> <sup>a</sup> <sup>y</sup> <sup>e</sup> dxdy <sup>f</sup>

<sup>δ</sup> <sup>+</sup> <sup>δ</sup> <sup>λ</sup> <sup>=</sup> <sup>=</sup> (4)

beam (R) coming from a mirror and the pattern is recorded.

Fig. 20. Reconstruction of the real image of a hologram.

eq. (4), we can express the reference beam as:

where **A** is the wave amplitude.

plane wave for the lens and is used as the reference beam R, given by

namely, real and conjugated images.

$$\mathbf{O} = \mathbf{F} \{ \mathbf{f}(\mathbf{x}, \mathbf{y}) \} \tag{7}$$

where the Fourier transform is denoted by F.

Fig. 21. Schematically representation of the Fourier holograms.

The photosensitive material, PM, registers the interference pattern intensity given by

$$\mathbf{I} = \left| \mathbf{R} + \mathbf{O} \right|^2 \tag{8}$$

which can be expressed as

$$\mathbf{I} = \left| \mathbf{R} \right|^2 + \left| \mathbf{O} \right|^2 + \mathbf{R}^\star \mathbf{O} + \mathbf{RO} \mathbf{O}^\star \tag{9}$$

where the symbol \* denotes the complex conjugate. Substituting the eqs. (5) and (7) in (9) we can obtain

$$\mathbf{I} = \mathbf{A}^2 + \left| \mathbf{F} \{ \mathbf{f}(\mathbf{x}, \mathbf{y}) \right|^2 + \mathbf{A} \mathbf{e}^{-\mathrm{i}2\pi \mathbf{v} \mathbf{a}} \mathbf{F} \{ \mathbf{f}(\mathbf{x}, \mathbf{y}) \} + \mathbf{A} \mathbf{e}^{\mathrm{i}2\pi \mathbf{v} \mathbf{a}} \mathbf{F}^\star \{ \mathbf{f}(\mathbf{x}, \mathbf{y}) \} \tag{10}$$

$$\mathbf{I} = \mathbf{I}\_1 + \mathbf{I}\_2 + \mathbf{I}\_3 + \mathbf{I}\_4 \tag{11}$$

where

$$\mathbf{I}\_1 = \mathbf{A}^2 \tag{12}$$

Norland Optical Adhesive 65® as Holographic Material 41

The term represented by eq. (23) is know as the conjugated image of the object f(x, y) and is

According to above results, we use the concentration C3, the sample thickness 110 mm, intensity beams ratio 1:1 and 5 degrees between the interference beam to record Fourier holograms. In Fig. 19 we showed the representation of the setup employed, where we use a He-Ne laser at 594 nm wavelength as recording beam and a laser at 544 nm as reading beam. In Fig. 22 we show the

exposed area in the cell and, as can be observed, this area is 1mm2 approximately.

Fig. 22. Photography of the exposed area, the white point is the area where the Fourier

Fig. 24. Real images of the Fourier holograms recorded in our photosensitive material.

reconstructed images are shown in Fig. 24 where only are the real images.

Fig. 23. Binary objects employed to Fourier holograms recording.

In Fig. 23 we show the two binary objects used to make Fourier holograms. The objects are the negative of the Universidad Michoacana de San Nicolás de Hidalgo logo and a text; the objects were photographed with a Nikon camera using a kodalith film and developed in a dark room following the Kodak instructions. After the objects were taken, we put it in the experimental setup and illuminates with a plane wave as is indicated in Fig 19. The

RI Af ( ) <sup>x</sup> a, <sup>y</sup> \* <sup>4</sup> <sup>=</sup> <sup>−</sup> <sup>−</sup> <sup>−</sup> <sup>−</sup> (24)

Equation (22) can be reduced as following

the -1 diffracted order.

hologram was recorded.

$$\mathbf{I}\_2 = \left| \mathbf{F} \{ \mathbf{f}(\mathbf{x}, \mathbf{y}) \} \right|^2 \tag{13}$$

$$\mathbf{I}\_3 = \mathbf{A}\mathbf{e}^{-i2\pi \mathbf{v}a} \mathbf{F}[\mathbf{f}(\mathbf{x}, \mathbf{y})] \tag{14}$$

$$\mathbf{I}\_4 = \mathbf{A}\mathbf{e}^{\mathbf{i}2\pi\mathbf{v}a}\mathbf{F}^\*\left[\mathbf{f}(\mathbf{x},\mathbf{y})\right] \tag{15}$$

In order to realize the reconstruction of the hologram, the reference beam R is used to illuminate the exposed film, so we can obtain

$$\text{RI} = \text{RI}\_1 + \text{RI}\_2 + \text{RI}\_3 + \text{RI}\_4 \tag{16}$$

The first two terms in right hand side of equation (15) are constants. The third term is expressed as

$$\text{RI}\_3 = \text{RAe}^{-i2\pi \text{va}} \text{F} \{ \mathbf{f}(\mathbf{x}, \mathbf{y}) \} \tag{17}$$

and substituting eq. (5) in eq. (16) we obtain

( ) ( ) ( ) { } ( ) ∞ −∞ ∞ −∞ <sup>−</sup> <sup>π</sup> <sup>ν</sup> +μ <sup>−</sup> <sup>π</sup> RI <sup>=</sup> <sup>A</sup> <sup>δ</sup> <sup>x</sup> <sup>+</sup> <sup>a</sup> <sup>δ</sup> <sup>y</sup> <sup>e</sup> <sup>e</sup> <sup>F</sup> <sup>f</sup> x, <sup>y</sup> dxdy i2 <sup>x</sup> <sup>y</sup> i2 <sup>a</sup> <sup>3</sup> (18)

which can be reduced as

$$\text{RI}\_3 = \text{A} \int\_{-\infty}^{\infty} \int\_{-\infty}^{\infty} \text{F}[\mathbf{f}(\mathbf{x} + \mathbf{a}, \mathbf{y})] \text{e}^{-i2\pi (\mathbf{v}(\mathbf{x} + \mathbf{a}) + \mu \mathbf{y})} \text{dxdy} \tag{19}$$

The result shown in eq. (18) can be interpreted as the double Fourier Transform of the object f(x, y), i. e.

$$\text{RI}\_3 = \text{F}[\text{F}[\text{f}(\text{x} + \text{a}, \text{y})]] = \text{Af}(-\text{x} - \text{a}, -\text{y}) \tag{20}$$

where the Fourier transform properties was employed. This term is know as the real image of the object f(x, y) located in the position x=a and correspond to the +1 diffracted order. Similarly, developing the fourth term RI4 we obtain

$$\text{RI}\_4 = \text{RAe}^{\text{i}2\pi \text{va}} \text{F}^\* \left[ \text{f}(\text{x}, \text{y}) \right] \tag{21}$$

and substituting eq. (4) in eq. (21) and using the relation F\*(f(x, y))=-F(f\*(x, y)) we obtain

$$RI\_4 = -A \int\int\int \delta(\mathbf{x} + a) \, \delta(y) e^{-i2\pi(\mathbf{v}\mathbf{x} + \mu y)} e^{i2\pi a} F\{f^\*(\mathbf{x}, y)\} dx dy \tag{22}$$

which can be reduced as

$$\mathrm{RI}\_4 = -\mathrm{A} \int\_{-\infty}^{\infty} \int\_{-\infty}^{\infty} \mathrm{F} \left[ \mathrm{f}^\*(\mathbf{x} + \mathbf{a}, \mathbf{y}) \right] \mathrm{e}^{-i2\pi (\mathbf{v}(\mathbf{x} - \mathbf{a}) + \mu \mathbf{y})} \mathrm{d}x \mathrm{d}y \tag{23}$$

Equation (22) can be reduced as following

40 Holograms – Recording Materials and Applications

In order to realize the reconstruction of the hologram, the reference beam R is used to

 RI <sup>=</sup> RI1 <sup>+</sup> RI2 <sup>+</sup> RI3 <sup>+</sup> RI4 (16) The first two terms in right hand side of equation (15) are constants. The third term is

( ) ( ) ( ) { } ( )

The result shown in eq. (18) can be interpreted as the double Fourier Transform of the object

where the Fourier transform properties was employed. This term is know as the real image of the object f(x, y) located in the position x=a and correspond to the +1 diffracted order.

RI RAe <sup>F</sup> { } <sup>f</sup>( ) x, <sup>y</sup> i2 <sup>a</sup> \* <sup>4</sup>

\* , *i xy i a RI A x a y e e F f x y dxdy* πν μ

<sup>−</sup> <sup>π</sup> <sup>ν</sup> <sup>−</sup> +μ RI <sup>=</sup> <sup>−</sup><sup>A</sup> <sup>F</sup> <sup>f</sup> <sup>x</sup> <sup>+</sup> a, <sup>y</sup> <sup>e</sup> dxdy \* i2 <sup>x</sup> <sup>a</sup> <sup>y</sup> <sup>4</sup>

( ) ( ) ( ) { } ( ) <sup>2</sup> <sup>2</sup>

{ } ( ) ( ) ( )

π

=− + (22)

and substituting eq. (4) in eq. (21) and using the relation F\*(f(x, y))=-F(f\*(x, y)) we obtain

 δ

∞

−∞

∞ ∞ − +

{ } ( ) ( ) ( )

<sup>−</sup> <sup>π</sup> <sup>ν</sup> +μ <sup>−</sup> <sup>π</sup> RI <sup>=</sup> <sup>A</sup> <sup>δ</sup> <sup>x</sup> <sup>+</sup> <sup>a</sup> <sup>δ</sup> <sup>y</sup> <sup>e</sup> <sup>e</sup> <sup>F</sup> <sup>f</sup> x, <sup>y</sup> dxdy i2 <sup>x</sup> <sup>y</sup> i2 <sup>a</sup> <sup>3</sup> (18)

<sup>−</sup> <sup>π</sup> <sup>ν</sup> <sup>+</sup> +μ RI <sup>=</sup> <sup>A</sup> <sup>F</sup> <sup>f</sup> <sup>x</sup> <sup>+</sup> a, <sup>y</sup> <sup>e</sup> dxdy i2 <sup>x</sup> <sup>a</sup> <sup>y</sup> <sup>3</sup> (19)

RI3 = F{ }( ) F{ } f( ) x + a, y = Af − x − a,−y (20)

πν = (21)

(23)

<sup>I</sup> Ae <sup>F</sup> { } <sup>f</sup>( ) x,y i2 <sup>a</sup> \* <sup>4</sup>

illuminate the exposed film, so we can obtain

and substituting eq. (5) in eq. (16) we obtain

∞

∞

−∞

 ∞

∞

−∞

−∞

δ

 ∞

−∞

−∞ −∞

−∞

Similarly, developing the fourth term RI4 we obtain

4

which can be reduced as

expressed as

f(x, y), i. e.

which can be reduced as

{ } ( ) <sup>2</sup> I2 <sup>=</sup> <sup>F</sup> <sup>f</sup> x,y (13)

<sup>I</sup> Ae <sup>F</sup>{ } <sup>f</sup>( ) x,y i2 <sup>a</sup> <sup>3</sup> <sup>−</sup> πν <sup>=</sup> (14)

πν <sup>=</sup> (15)

RI RAe <sup>F</sup>{ } <sup>f</sup>( ) x,y i2 <sup>a</sup> <sup>3</sup> <sup>−</sup> πν <sup>=</sup> (17)

$$\text{RI}\_4 = -\text{Af}^\*( -\text{x} - \text{a}, -\text{y} )\tag{24}$$

The term represented by eq. (23) is know as the conjugated image of the object f(x, y) and is the -1 diffracted order.

According to above results, we use the concentration C3, the sample thickness 110 mm, intensity beams ratio 1:1 and 5 degrees between the interference beam to record Fourier holograms. In Fig. 19 we showed the representation of the setup employed, where we use a He-Ne laser at 594 nm wavelength as recording beam and a laser at 544 nm as reading beam. In Fig. 22 we show the exposed area in the cell and, as can be observed, this area is 1mm2 approximately.

Fig. 22. Photography of the exposed area, the white point is the area where the Fourier hologram was recorded.

In Fig. 23 we show the two binary objects used to make Fourier holograms. The objects are the negative of the Universidad Michoacana de San Nicolás de Hidalgo logo and a text; the objects were photographed with a Nikon camera using a kodalith film and developed in a dark room following the Kodak instructions. After the objects were taken, we put it in the experimental setup and illuminates with a plane wave as is indicated in Fig 19. The reconstructed images are shown in Fig. 24 where only are the real images.

Fig. 23. Binary objects employed to Fourier holograms recording.

Fig. 24. Real images of the Fourier holograms recorded in our photosensitive material.

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In order to compare our photosensitive material, we use the SO-253 film from Kodak to make the Fourier hologram of the text and the reconstructed real image is show in Fig 25 It is important to note, that the amount of text in Figs. 24 and 25 are different because we use 1 inch of diameter lens in one optical setup and 2 inch of diameter lens in the optical setup used in order to illuminate the photosensitive material. Another difference is that the SO-253 film has its sensibility at the line λ=633 nm so we record the Fourier hologram with this wavelength and we use the same wavelength to reconstruct the real image.

Fig. 25. Real image of hologram stored in SO 256 film from Kodak.

The advantage of Fourier holograms is that the storage area is small compared with Fresnel holograms and we can apply some multiplexing technique to optimizing the storage area. In table 6 we show the diffraction efficiencies of the holograms showed above measured for the +1-diffracted order.


Table 6. Diffraction efficiencies for hologram reconstruction.
