Question

Public Radio station KXPR-FM in Sacramento broadcasts at 88.9 MHz. The radio waves pass between two tall skyscrapers that are 15.0 m apart along their closest walls. (a) At what horizontal angles, relative to the original direction of the waves, will a distant antenna is 3.50 W>m at the antenna, what is the intensity at

Answer 1

At horizontal angles, relative to the original direction of the waves, will be  1.4597 rad and the intensity will be 1.576 W/m².

What is a radar antenna?

A radar antenna is a device that sends out radio waves and listens for their reflections. The ability of an antenna to identify the exact direction in which an item is placed determines its performance.

The horizontal angle for the given condition is;

sinθ∝1

sinθ=(nλ)/a

n∝(λ/a)

$\rm n = \frac{88.9 \times 10^6 \times 15}{3 \times 10^8 } \\\\ n=4.4454$

$\rm n_{max}=4$

The maximum angle is;

$\rm \theta_4=\frac{4 \times 3 \times 10^8}{15 \times 88.9 \times 10^6 } \\\\ \rm \theta_4=64.14^0$

The value of the wavelength is;

$\rm \lambda= \frac{c}{v} \\\\ \rm \lambda= \frac{3 \times 10^8 }{88.9 \times 10^6} \\\\ \lambda= 3.3745 \ m$

The value of the intensity is;

$\rm I= I_{max}[\frac{sin \frac{\beta}{2}}{\frac{\beta}{2}} ]\\\\ \frac{\beta}{2} =\frac{\pi}{4} (a sin\theta)\\\\ \beta=\frac{\pi}{3.3745} (15 sin 6^0)\\\\ \beta=1.4597 \ rad$

$\rm I = 3.4[ \frac{sin 1.4597}{1.4597} ]^2\\\\ I=1.576 \ W/m^2$

Hence,at horizontal angles, relative to the original direction of the waves, will be  1.4597 rad and the intensity will be 1.576 W/m².

To learn more about the radar antenna, refer to the link;

https://brainly.com/question/24067190

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