Babinet's Principle

Suppose that a given aperture is characterized by the aperture function $f(u,v)$. Consider a so-called complementary aperture whose aperture function, $\bar{f}(u,v)$, satisfies

$$f(u,v)+\bar{f}(u,v) = 1$$ (10.111)

for all $u$ and $v$. In other words, the complementary aperture is transparent everywhere where the original aperture is opaque, and vice versa. Let

$$\bar{f}_c(u',v') =\frac{1}{2}\int_{-\infty}^\infty \bar{f}(u,v)\,\cos\left[\frac{\pi}{2}\,(u-u')^{\,2} +\frac{\pi}{2}\,(v-v')^{\,2}\right] du\,dv,$$ (10.112)

$$\bar{f}_s(u',v') =\frac{1}{2}\int_{-\infty}^\infty \bar{f}(u,v)\,\sin\left[\frac{\pi}{2}\,(u-u')^{\,2} +\frac{\pi}{2}\,(v-v')^{\,2}\right] du\,dv.$$ (10.113)

It is clear from the analysis in the previous section that

$$f_c(u',v')+\bar{f}_c(u',v') =0,$$ (10.114)

$$f_s(u',v')+\bar{f}_s(u',v') = 1.$$ (10.115)

Let $\bar{\psi}(u',v',t)$ denote the wave amplitude at the projection screen when the complementary aperture is illuminated. By analogy with Equation (10.102),

$$\bar{\psi}(u',v',t)\simeq -\psi_0\,\sin(\omega\,t-k\,R-\phi)\,\bar{f}_c(u',v') + \psi_0\,\cos(\omega\,t-k\,R-\phi)\,\bar{f}_s(u',v').$$ (10.116)

Thus, it follows from Equation (10.102), and the previous three equations, that

$$\psi(u',v',t)+\bar{\psi}(u',v',t) = \psi_0\,\cos(\omega\,t-k\,R-\phi).$$ (10.117)

In other words, the sum of the wave amplitude at the projection screen seen when the original aperture is illuminated and the wave amplitude seen when the complementary aperture is illuminated is equal to the wave amplitude seen when the aperture is completely transparent. This result is known as Babinet's principle.