Question

A radar for tracking aircraft broadcasts a $12\mathrm{GHz}$microwave beam from a $2.0-\mathrm{m}$-diameter circular radar antenna. From a wave perspective, the antenna is a circular aperture through which the microwaves diffract.

a. What is the diameter of the radar beam at a distance of $30\mathrm{km}$?

b. If the antenna emits $100\mathrm{kW}$of power, what is the average microwave intensity at $30\mathrm{km}$?

Short answer

a) The diameter of the beam at a distance of 30km is$915\mathrm{m}$

b) The average microwave intensity at 30km is$0.152\mathrm{W}/{\mathrm{m}}^2$

Step-by-step solution

Radar beam (part a)

a) When a simple right triangle is constructed, the hypotenuse is the ray leading to the limit of the central maximum, that is, the beginning of the first dark circle, the catheter in front of the small angle is clearly visible. It's also important to remember that when we refer to this little angle, we're referring to its value.

$\tan \left({\theta }_1\right)=\frac{R}{L}=\frac{\Phi }{2L}$

As a consequence, we can compute our circumference in relation of both the angle as follows:

$\mathrm{\Phi }=2L\tan \left({\theta }_1\right)$

Keep in mind that we can get the angle as from darker circles criteria as

${\theta }_1=\frac{1.22\lambda }{D}$

$$\begin{gathered} \mathrm{\Phi }=2L\tan \left(\frac{1.22\lambda }{D}\right) \\ =2L\tan \left(\frac{1.22c}{DV}\right) \end{gathered}$$ ($\lambda =\frac{c}{\nu }$)

$$\begin{gathered} \mathrm{\Phi }=2\times 3\times {10}^4\tan \left(\frac{1.22\times 3\times {10}^8}{2\times 1.2\times {10}^{10}\mathrm{V}}\right) \\ =915\mathrm{m} \end{gathered}$$

(Numerically)

Power and Diameter (part b)

b) We can determine the rate based on the power and diameter

$I=\frac{P}{A}=\frac{P}{\pi R^2}=\frac{4P}{\pi {\mathrm{\Phi }}^2}$

localid="1649147611891" $I=\frac{4\times 1\times {10}^5}{\pi \times {915}^2}=0.152\mathrm{W}/{\mathrm{m}}^2$(Numerically)