A radio-frequency antenna (such as a TV antenna) may be thought of semiquantitatively as an aperture with dimensions comparable to the physical dimensions (usually of the order of a wavelength). How far away from the antenna, expressed in wavelengths, must one be for the Fraunhofer case to be valid? Compare with the case for a pinhole of diameter \(0.1 \mathrm{~mm}\) for visible wavelengths.

We are given that the dimensions of the radio-frequency antenna are of the order of a wavelength, so let's denote that as \(D = \lambda\). Using the Fraunhofer condition, we have: $$ \frac{D^2}{L \lambda} \ll 1 \Rightarrow \frac{\lambda^2}{L \lambda} \ll 1. $$ Now, we can simplify and find \(L\) in terms of \(\lambda\): $$ \frac{\lambda}{L} \ll 1 \Rightarrow L \gg \lambda. $$ So, the distance for the Fraunhofer case to be valid for a radio-frequency antenna is much greater than its dimensions or wavelengths \(L \gg \lambda\).

Now, let's consider a pinhole of diameter \(0.1 \mathrm{~mm}\) for visible wavelengths. A typical visible wavelength \(\lambda\) lies roughly within the \(400\mathrm{~nm}\) to \(700\mathrm{~nm}\) range. So, given this information and the diameter of the pinhole is \(D = 0.1 \mathrm{~mm}\), we can use the Fraunhofer condition to calculate the distance \(L\) for the pinhole. $$ \frac{D^2}{L \lambda} \ll 1 \Rightarrow \frac{(0.1 \cdot 10^{-3})^2}{L \lambda} \ll 1. $$ We will consider the wavelength as \(\lambda = 500\mathrm{~nm}\) (a mid-value for visible light). In that case, we need to solve for \(L\): $$ \frac{(0.1 \cdot 10^{-3})^2}{L \cdot 500 \cdot 10^{-9}} \ll 1. $$

For the radio-frequency antenna, we found that \(L \gg \lambda\), which means the distance required for the Fraunhofer case to be valid is much larger than the wavelength. In the case of the pinhole, the distance \(L\) depends on the specific value of the wavelength (visible light) and the diameter \(D\) of the pinhole, which we can express as: $$ L \gg \frac{(0.1 \cdot 10^{-3})^2}{\lambda}. $$ From these results, we can see that the distance required for the Fraunhofer case to be valid varies for different apertures, and it heavily depends on the dimensions and wavelengths being considered. For radio-frequency antennas, the required distance is much larger than the wavelength, whereas for pinholes working with visible light, it depends on the ratio of the pinhole's diameter squared and the light's wavelength.