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Diffraction Efficiency

One of the main hologram characteristics is the diffraction efficiency $\eta$. In the general case, the diffraction efficiency is determined by the ratio of the power of the diffracted light beam $P_{\mathrm{diff}}$ to the incident power of the beam $P_{\mathrm{inc}}$ given by this relation:
\begin{displaymath}
\eta = \frac{P_{\mathrm{diff}}}{P_{\mathrm{inc}}}.
\end{displaymath} (23)

The power of the beam is determined by the integral of the light intensity $(I)$ over the surface of the detector $(S)$:
\begin{displaymath}
P = \int\!\!\!\int_{S} I \mathrm{d} S.
\end{displaymath} (24)

In the case of lasers with Gaussian light beams, the relation between the mean light intensity $\bar{I}$ and power $P$ is described by:
\begin{displaymath}
\bar{I} = \frac{P}{\pi r^{2}}.
\end{displaymath} (25)

where $r$ is the Gaussian radius of the beam. If the light intensity of the subject and reference beams are $I_{S}$ and $I_{R}$ and the diffracted beam intensities are $\acute{I_{S}}$, $\acute{I_{R}}$ then the diffraction efficiency is given by:
\begin{displaymath}
\eta = \frac{\acute{I}_{S(R)}}{I_{R(S)}}.
\end{displaymath} (26)

The diffraction efficiency depends on the wavelength as well as on the thickness of the recording medium. For different elementary holographic gratings the diffraction efficiency is summarized in Table 1. The diffraction efficiency depends on whether the hologram is 3-D or 2-D, phase or amplitude, and on the type of grating profile. Note that the brightness of the images made in amplitude holograms does not depend very strongly on the thickness of the medium, the maximum for 3-D holograms is 3.7%, which is significantly smaller than for 2-D, 6.25%, Table 1. Amplitude recordings have a much lower diffraction efficiency than phase holograms which, of course, makes them less attractive. The diffraction efficiency can be used to determine the type (amplitude/phase, thick/thin) of hologram being investigated.

Table 1: Diffraction efficiency for different transmission grating types [2]
$\eta$ diffraction efficiency; $t_0$ amplitude transmittance; $m$ modulation factor $(m = t_A/t_0)$; $ \kappa_A$ amplitude of modulated index of absorbtion; $\kappa_0$ mean index of absorbtion after exposure; $d$ thickness; $\theta , \theta_i$ angles of incidence in a vacuum and in a medium, respectively.
Dim Profile Equation for diffraction efficiency $\eta_{\mathrm{max}},\%$
Amplitude holograms
3-D sinusoidal $\eta =\exp{[(-2\kappa_{0}d/\cos{\theta_{i}})]\times
\sinh^{2}{[\kappa_{A}d / (4\cos{\theta_{i}})]}}$  
    $\eta \to \eta_{\mathrm{max}} \mathrm{at}\
\kappa_{A}d/\cos{\theta_{i}}
= 2\ln{3} \mathrm{and} \kappa_{A}=\kappa_{0}$ 3.7
2-D sinusoidal $\eta = (t_{0}^{2}m^{2}/4)\times\cos^{3}{\theta} $  
    $\eta \to \eta_{\mathrm{max}} \mathrm{at} t_{0} = 1/2 , m=1\
\mathrm{and} \theta \to 0$ 6.25
2-D rectangular $\eta = \pi^{-2}\cos^{3}{\theta} $  
    $\eta \to \eta_{\mathrm{max}} \mathrm{at} \theta \to 0$ 10.1
Phase holograms
3-D sinusoidal $\eta =\sin{^2}{(\pi n_A d /\lambda\cos{\theta_i})}$  
    $\eta \to \eta_{\mathrm{max}} \mathrm{at} n_A d/cos{\theta_{i}}=(\lambda/2)(2l+1)$ 100
$l=0,1,2,\ldots $  
2-D sinusoidal $\eta =J_1^2(2\pi n_A d /\lambda\cos{\theta_i} ) $  
    $\eta \to \eta_{\mathrm{max}} \mathrm{at} (2\pi n_A d
/\lambda\cos{\theta_i})\to 2$ 33.0
2-D rectangular $\eta_{\mathrm{max}} = (4\pi^2)\cos^{3}{\theta}$ 40.5


In general, however, the real time recording almost always leads to mixed amplitude/phase holograms. For high efficiency hologram recording the absorption of the medium (at the recording wavelength $\lambda_{1}$) must be sufficiently strong. The properties of the recorded hologram depend on the wavelength used during reconstruction, $\lambda_{2}$. Upon reconstruction with light of the same wavelength $(\lambda_{1}=\lambda_{2})$, the hologram is usually an amplitude and during readout the holographic properties can be changed by the photoactive light. Therefore, it is more common for the reconstruction light to be of longer wavelength $(\lambda_{2} > \lambda_{1})$. In such a case, the hologram is usually a phase hologram with a higher diffraction efficiency and the readout process would not change the properties of the recorded hologram (i.e., the light of $\lambda_{2}$ is not photoactive). However, for optimal reconstruction efficiency at $\lambda_{2}\neq \lambda_{1}$, the angle of incidence for the reference beam during readout must be changed from $\theta_{1}$ to $\theta_{2}$. According to (17), $\theta_{2}$ is given by:
\begin{displaymath}
\theta_{2} = \arcsin{ \frac {\lambda_{2}}{2\Lambda}}=
\arcsin{\left[\frac{\lambda_2}{\lambda_1}\sin{\theta_1}\right]}
\end{displaymath} (27)

Grating period $\Lambda$ is determined by $\lambda_1$ and $\theta_1$. For light sensitive $\mathrm{As_{2}S_{3}}$ films, the optimal wavelengths for recording and readout are $\lambda_1 =
514.5$ nm ($\mathrm{Ar^+}$ laser) and $\lambda_2 = 632.8$ nm (He-Ne laser). For grating periods of 0.5-1.0 $\mu$m and $\lambda_{1}\neq \lambda_{2}$, the angle correction $\Delta\theta
/\theta$ is approximately 20%. Under such conditions, in 10  $\mu$m thick films, diffraction efficiency of 80% was reached which is close to the theoretical limit (Table 1). The diffraction efficiency is decreased from the theoretical maximum (100% for 3-D phase holograms) when the amount of reflected light increases. Thus, materials with very high refractive indices like $\mathrm{As_{2}S_{3}}$ films $(n = 2.5)$, which makes a reflection coefficient of 18% must be coated with antireflection layers. This is particularly important for optoelectronic devices [10,6]. Note that $\eta$ is a function of the amplitude of the induced holographic grating $(\kappa_{A}, n_{A}$, Table 1, Fig. 8) which depends on the length of exposure and its intensity $I \times t$.
next up previous
Next: Holographic Optical Elements Up: Holography Previous: Hologram Classifications
root 2002-05-23