Question

Millimetre-wave radar generates a narrower beam than conventional microwave radar, making it less vulnerable to antiradar missiles than conventional radar. (a) Calculate the angular width $\mathbf{2}\mathit{\theta }$of the central maximum, from first minimum to first minimum, produced by a 220 GHz radar beam emitted by a 55.0-cm diameter circular antenna. (The frequency is chosen to coincide with a low-absorption atmospheric “window.”) (b) What is $\mathbf{2}\mathit{\theta }$ for a more conventional circular antenna that has a diameter of 2.3 m and emits at wavelength 1.6 cm?

Short answer

(a) The angular width$2\theta$ of the central maximum is$0.346°$ .

(b) The value of$2\theta$ for a more conventional circular antenna is $0.97°$.

Step-by-step solution

Concept/Significance of first minima in diffraction pattern

The angle corresponding to the first minima in diffraction pattern is given by,

$$\begin{gathered} \mathit{s}\mathit{i}\mathit{n}\mathit{\theta }\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{22}\frac{\mathbf{\lambda }}{\mathbf{d}} \\ \mathit{\theta }\mathbf{=}\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\left(1.22\frac{\lambda }{d}\right) \end{gathered}$$ ….. (1)

(a) Find the angular width 2θ of the central maximum

Find the wavelength as follows.

$$\begin{gathered} \lambda =\frac{c}{f} \\ =\frac{3\times {10}^8}{220\times {10}^9} \\ =1.36\times {10}^{-3}\text{m} \end{gathered}$$

Substitute $0.55\text{m}$for d and$1.36\times {10}^{-3}\text{m}$ for $\lambda$in equation (1).

$$\begin{gathered} \theta ={\sin }^{-1}\left(\frac{1.22\left(1.36\times {10}^{-3}\text{m}\right)}{0.55\text{m}}\right) \\ \theta ={\sin }^{-1}\left(3.0167\times {10}^{-3}\right) \\ \theta =0.1728 \\ 2\theta \approx 0.346° \end{gathered}$$

Therefore, the angular width $2\theta$of the central maximum is $0.346°$.

(b) Find the value of for a more conventional circular antenna

Substitute$2.3\text{m}$ for d and $1.6\times {10}^{-2}\text{m}$for $\lambda$in equation (1).

$$\begin{gathered} \theta ={\sin }^{-1}\left(\frac{1.22\left(1.6\times {10}^{-2}\text{m}\right)}{2.3\text{m}}\right) \\ \theta ={\sin }^{-1}\left(8.487\times {10}^{-3}\right) \\ \theta =0.486° \\ 2\theta \approx 0.97° \end{gathered}$$

Therefore, the value of $2\theta$for a more conventional circular antenna is $0.97°$.