A radar for tracking aircraft broadcasts at 12 GHz microwave beam from a 2.0-m-diameter circular radar antenna. From a wave perspective the antenna is a circular aperture through which the microwaves diffracts.
A. what is the diameter of the radar beam at a distance of 30 km?
B. if the antenna emits 100 kW of power what is the average microwave intensity at 30 km?
Answer
A. The-wavelength-of-the-wave-is,\\
\lambda =\frac{c}{f} \\
=\frac{3 \times 10^{8} \mathrm{~m} / \mathrm{s}}{12 \times 10^{2} \mathrm{~Hz}} \\ =0.025 \mathrm{~m} \\
=25 \mathrm{~mm}\\
The-angular-resolution-of-the-system-is,\\
\sin \theta=1.22\left(\frac{\lambda}{D}\right)\\
=1.22\left(\frac{25 \mathrm{~mm}}{2 \mathrm{~m}}\right)\\
\theta=\sin ^{-1}(0.01525)\\
=15.25 \mathrm{mrad}\\
The-diameter-of-the-radar-beam-from-the-half-angle-is,\\
d=2(15.25 \mathrm{mrad})\left(30 \mathrm{~km}\left(\frac{1000 \mathrm{~m}}{1 \mathrm{~km}}\right)\right)\\
=915 \mathrm{~m}\\
\\
B.\\
The-average-microwave-intensity-is,\\
I =\frac{P}{A} \\
=\frac{P}{\pi r^{2}} \\
=\frac{4 P}{\pi d^{2}} \\
=\frac{4\left(100 \times 10^{3} \mathrm{~W}\right)}{\pi(915 \mathrm{~m})^{2}} \\
=0.152 \mathrm{~W} / \mathrm{m}^{2}