9.4 Babinet's principle

Babinet's principle

Say you have a slit with light passing through it. You will get a diffraction pattern, lets call it ${\vec{E}}_s$ . If you cover the slit with a piece of material that fits just inside the slit,then there is no $\vec{E}$ field in the Fraunhofer limit. The is means that the $\vec{E}$ field of the blocker, lets call it $vec{E}_b$ , must exactly cancel ${\vec{E}}_s$ . The only way this can happen is if $${\vec{E}}_b=-{\vec{E}}_s$$ . Now if you take the slit away the $\vec{E}$ field of the blocker must still remain and then irradiance must be $$I_b={(-E_s)}^2=I_s$$ The interference pattern looks the same. You can verify this yourself by taking a strand of your hair and a laser pointer. Shine the laser pointer at awall and then put a strand of your hair in front of the light beam. The resulting interference pattern is the exact same as one would obtain from aslit with the same width as your hair.

We can use Babinet's Principle to solve complex problems. For example, say you have square aperture with sides of length $L$ . The the diffraction pattern for light passing through it is $$\vec{E}={\vec{\varepsilon }}_A{e}^{i(kr-\omega t)}\frac{L^2}{R}\left[\frac{\sin {\beta }_L}{{\beta }_L}\right]\left[\frac{\sin {\alpha }_L}{{\alpha }_L}\right]$$ where (assuming the aperture lies in the $y-z$ plane) $${\beta }_L=kLY/2R$$ $$\alpha =kLZ/2R\text{.}$$ Now put an opaque square of length $d$ in the middle of the aperture. Now the resulting $\vec{E}$ field is $$\vec{E}=\frac{{\vec{\varepsilon }}_A}{R}{e}^{i(kr-\omega t)}\left[L^2\frac{\sin {\beta }_L}{{\beta }_L}\frac{\sin {\alpha }_L}{{\alpha }_L}-d^2\frac{\sin {\beta }_d}{{\beta }_d}\frac{\sin {\alpha }_d}{{\alpha }_d}\right]$$ or $$I=I_0{\left[L^2\frac{\sin {\beta }_L}{{\beta }_L}\frac{\sin {\alpha }_L}{{\alpha }_L}-d^2\frac{\sin {\beta }_d}{{\beta }_d}\frac{\sin {\alpha }_d}{{\alpha }_d}\right]}^2$$ $${\beta }_L=kLY/2R$$ $${\alpha }_L=kLZ/2R\text{.}$$ $${\beta }_d=kdY/2R$$ $${\alpha }_d=kdZ/2R\text{.}$$