Question

The radar system of a navy cruiser transmits at a wavelength of 1.6 cm, from a circular antenna with a diameter of 2.3 m. At a range of 6.2 km, what is the smallest distance that two speedboats can be from each other and still be resolved as two separate objects by the radar system?

Short answer

The smallest distance between the boats after rounding off to two significant digits is $53m$

Step-by-step solution

Introduction

Radar (radio detection and ranging) is a detection system that uses radio waves to determine the distance (ranging), angle, and radial velocity of objects relative to the site. It can be used to detect aircraft, ships, spacecraft, guided missiles, motor vehicles, weather formations, and terrain

Concept

Rayleigh criterion states that the minimum angular separation for resolvability of two objects is given by the following expression:

${\theta }_R=\frac{1.22\lambda }{d}$

Here, $\lambda$is wavelength; d is the diameter of the aperture.

For the smallest distance (D) between the boats they can be still be resolved as to separate objects by the radar system is,$D=L{\theta }_R$

Here, L is the range and ${\theta }_R$is minimum angular separation.

Determine the radar system of a navy cruiser transmits

Convert range from kilometres to meters

$$\begin{gathered} L=6.2\text{km} \\ \text{=}\left(6.2\text{km}\right)\left(\frac{{10}^3\text{m}}{1\text{km}}\right) \\ =6.2\times {10}^3\text{m} \end{gathered}$$

Convert wavelength from centimetre to meters:

$$\begin{gathered} \lambda =1.6cm \\ =\left(1.6\text{cm}\right)\left(\frac{1\text{m}}{100\text{cm}}\right) \\ =1.6\times {10}^{-2}\text{m} \end{gathered}$$

To calculate the smallest distance between the boats,

Substitute ${\theta }_R=\frac{1.22\lambda }{d}$in the electron, $D=L{\theta }_R$

$D=L\left(\frac{1.22\lambda }{d}\right)$

Substitute $6.2\times {10}^3\text{m}$ for $1.6\times {10}^{-2}\text{m}$for $\lambda$, and 2.3 m for the diameter of the antenna (d)

$$\begin{gathered} D=\left(6.2\times {10}^3\text{m}\right)\left(\frac{1.22\left(1.6\times {10}^{-2}\text{m}\right)}{2.3\text{m}}\right) \\ =52.6\text{m} \end{gathered}$$

Hence, the smallest distance between the boats after rounding off to two significant digits is$53m$.