A half-dipole illuminates a parabolic mirror.Why are parabolic reflectors as part of antenna systems so common? The reason is that they provide a very high directivity. In this problem you will learn about this very basic concept of antenna theory to quantify emitters and receivers at the examples of a dipole, a "half-dipole" and a parabolic reflector.

Problem Statement

A half-dipole illuminates a parabolic mirror.Calculate the directivity of a dipole oscillating in \(z\)  direction. Then, assume the radiation propagating in negative \(x\) -direction gets absorbed in the farfield of the dipole. How does the directivity change?
Now we want to put this “half-dipole” in the focal point of a parabolic reflector with opening radius \(R\) and focal distance \(f\), see figure on the right. How can we calculate the directivity of this device approximately for a small opening angle \(\delta\theta \approx R/f\) Assume that the backreflected light is not influenced by the “half-dipole” itself and calculate \(D\) for a wavelength \(\lambda=500\,\)nm and an opening radius \(R=20\,\)cm.

 



Background: Directivity in Antenna Theory

The radiated power \(dP_{\mathrm{rad}}\) into a certain angle \(d\Omega=\sin\theta d\theta d\varphi\) is related to the radial component of the time-averaged Poynting vector \(S_{r}\left(r,\theta,\varphi\right)\),\[\begin{eqnarray*}  dP_{\mathrm{rad}}&=&r^{2}S_{r}\left(r,\theta,\varphi\right)d\Omega\equiv r^{2}S_{r,\mathrm{max}}\left(r\right)F\left(\theta,\varphi\right)d\Omega\ .  \end{eqnarray*}\]As defined above, one may also normalize \(S_{r}\) und use the normalized radiation intensity \(F\) instead. This quantity is used to define the directivity \(D\) of a certain antenna or source,\[\begin{eqnarray*}  D&=&\frac{4\pi}{\Omega_{p}\equiv\int_{0}^{2\pi}\int_{0}^{\pi}F\left(\theta,\varphi\right)\sin\theta d\theta d\varphi}\ .  \end{eqnarray*}\]

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