Introduction to Holography, 2nd Edition

Appendix A The Fresnel-Kirchoff Integral

Our task here is to find the intensity distribution in a plane beyond the region where diffraction has occurred. To do this we start a with a monochromatic wave

$$\tilde{\mathrm{E}}(x,y,z)\exp (-j\omega t)$$

which, when substituted in the wave equation (Equation (1.9))

$${\nabla }^2\tilde{E}\exp (-j\omega t)={\displaystyle \frac{1}{c^2}}{\displaystyle \frac{{\partial }^2\tilde{E}\exp (-j\omega t)}{\partial t^2}}$$

results in Helmholtz’s equation

(A.1)$${\nabla }^2\tilde{E}+k^2\tilde{E}=0$$

The function ${\displaystyle \frac{\exp (jk{r}^{\prime })}{r^{\prime }}}$ also satisfies Helmholtz’s equation since

(A.2)$${\nabla }^2\left({\displaystyle \frac{e^{jk{r}^{\prime }}}{r^{\prime }}}\right)+k^2\left({\displaystyle \frac{e^{jk{r}^{\prime }}}{r^{\prime }}}\right)=0$$

$$\begin{array}{l}\mathrm{Hence}\mathrm{\hspace{0.33em}}{\displaystyle \underset{V}{\iiint }\left\{\tilde{E}{\nabla }^2\left({\displaystyle \frac{e^{jk{r}^{\prime }}}{r^{\prime }}}\right)-\left({\displaystyle \frac{e^{jk{r}^{\prime }}}{r^{\prime }}}\right){\nabla }^2\tilde{E}\right\}}\mathrm{d}V\\ \mathrm{\hspace{1em}}={\displaystyle \underset{V}{\iiint }\left\{-k^2\left({\displaystyle \frac{e^{jk{r}^{\prime }}}{r^{\prime }}}\right)\tilde{E}+\left({\displaystyle \frac{e^{jk{r}^{\prime }}}{r^{\prime }}}\right)k^2\tilde{E}\right\}}\mathrm{d}V=0\end{array}$$

$$\begin{array}{l}\mathrm{By}\mathrm{\hspace{0.33em}}\mathrm{Green}’\mathrm{s}\mathrm{\hspace{0.33em}}\mathrm{theorem}{\displaystyle \underset{V}{\iiint }\left\{\tilde{E}{\nabla }^2\left({\displaystyle \frac{e^{jk{r}^{\prime }}}{r^{\prime }}}\right)-\left({\displaystyle \frac{e^{jk{r}^{\prime }}}{r^{\prime }}}\right){\nabla }^2\tilde{E}\right\}}\mathrm{dV}\\ \mathrm{\hspace{1em}}={\displaystyle \underset{S}{\iint }\left\{\tilde{E}\nabla \left({\displaystyle \frac{e^{jk{r}^{\prime }}}{r^{\prime }}}\right)-{\displaystyle \frac{e^{jk{r}^{\prime }}}{r^{\prime }}}\nabla \tilde{E}\right\}}\mathrm{d}S\end{array}$$

(A.3)$$\mathrm{so}\mathrm{\hspace{0.33em}}{\displaystyle \underset{S}{\iint }\left\{\tilde{E}\nabla \left({\displaystyle \frac{e^{jk{r}^{\prime }}}{r^{\prime }}}\right)-{\displaystyle \frac{e^{jk{r}^{\prime }}}{r^{\prime }}}\nabla \tilde{E}\right\}}\cdot \mathrm{d}S=0$$

The calculation of the left-hand integral over a surface, S₁, enclosing a point P from which r′ is measured, involves ...