Polarization experiment with mircowaves

The metal grid acts as an polarizer for the microwave.

If the metal grid is at an angle of $\alpha = 0^\circ$ to the x-axis, the electric field of the wave and the grid bars are perpendicular to each other, so that no current flow is caused in the metal. In this case, the the wave passes undisturbed through the grating and an intensity of $I = I_0$ is measured by the receiver.

However, if the angle is $\alpha = 90^\circ$ , electrical voltage is caused along the metal wires by the electric field of the microwave. This leads to an alternating current in the metal whose current strength just compensates for the external voltage. The electric field on the side of the receiver therefore yields zero, so that the grating here acts as a mirror, at which the wave is completely reflected. Therefore, the measured intensity results $I = 0$ .

Otherwise, the electric field can be separated in components parallel and perpendicular to the grid bars: $\vec E = E_\parallel \vec n + E_\perp \vec t = E \sin \alpha \cdot \vec n + E \cos \alpha \cdot \vec t$ with unit vectors $\vec n$ and $\vec t$ with $|\vec n| = |\vec t| = 1$ , $\vec n \perp \vec t$ . Only the tangential component of the wave passes through the grid, so that the electric field behind the grid results $\vec E' = E \cos \alpha \cdot \vec t$ The receiver also acts as a polarizer and registers only the component parallel to the z-axis. The electric field is separated into $\vec E' = E_x' \vec e_x + E_z' \vec e_z = E' \sin \alpha \cdot \vec e_x + E' \cos \alpha \cdot \vec e_z$ The received electrical field therefore gives $\vec E'' = E' \cos \alpha \cdot \vec e_z = E (\cos \alpha)^2 \vec e_z$ Therefore, the received intensity results $\begin{aligned} I &= \frac{1}{2} c \varepsilon_0 |E''|^2 \propto |E''|^2 \\ \Rightarrow \quad I &= (\cos \alpha)^4 I_0 = \frac{1}{2^4} I_0 = 0.125 \cdot I_0 \end{aligned}$