A) 750 m
First of all, let's find the wavelength of the microwave. We have
[tex]f=12GHz=12\cdot 10^9 Hz[/tex] is the frequency
[tex]c=3.0\cdot 10^8 m/s[/tex] is the speed of light
So the wavelength of the beam is
[tex]\lambda=\frac{c}{f}=\frac{3\cdot 10^8 m/s}{12\cdot 10^9 Hz}=0.025 m[/tex]
Now we can use the formula of the single-slit diffraction to find the radius of aperture of the beam:
[tex]y=\frac{m\lambda D}{a}[/tex]
where
m = 1 since we are interested only in the central fringe
D = 30 km = 30,000 m
a = 2.0 m is the aperture of the antenna (which corresponds to the width of the slit)
Substituting, we find
[tex]y=\frac{(1)(0.025 m)(30000 m)}{2.0 m}=375 m[/tex]
and so, the diameter is
[tex]d=2y = 750 m[/tex]
B) 0.23 W/m^2
First we calculate the area of the surface of the microwave at a distance of 30 km. Since the diameter of the circle is 750 m, the radius is
[tex]r=\frac{750 m}{2}=375 m[/tex]
So the area is
[tex]A=\pi r^2 = \pi (375 m)^2=4.42\cdot 10^5 m^2[/tex]
And since the power is
[tex]P=100 kW = 1\cdot 10^5 W[/tex]
The average intensity is
[tex]I=\frac{P}{A}=\frac{1\cdot 10^5 W}{4.42\cdot 10^5 m^2}=0.23 W/m^2[/tex]
