# Electromagnetic waves

• sapiental
In summary, we considered an AM radio station with two antennas separated by 80m and broadcasting at the frequency of 1500 kHz and wavelength of 200m. We found that the phase difference between the two antennas for a point north of them is 12.56 degrees or 0.213 radians. The ratio of intensity for a point north of the antennas to the intensity east of the midpoint between the antennas is 1. In order to have maximum intensity along a line running from North-East to South-West, we need to delay the signal from one antenna by a certain time phase shift, which can be calculated by comparing the extra distance one signal has to travel to a wavelength.

## Homework Statement

1. Tune the Antenna Consider an AM radio station broadcasting at the frequency 1500
kHz, and wavelength 200 m.. Suppose the station has two antennas separated by 80m, one north
of the other.
a) (2 pts.) If the two antennas broadcast their signals in phase with each other, what is the phase
difference (in radians) of the radio waves arriving at a point north of the two antennas?
b) (4 pts.) If the two antennas broadcast their signals in phase with each other, what is the ratio
of the intensity of the radio signal at a point north of the antennas to the intensity east of the
midpoint between the antennas? (Both at the same long distance from the center of the antennas.)
c) (4 pts.) Suppose we want the maximum intensity in the antenna radiation pattern to lie on the
line running frm North-East to South-West, 45 degrees from north. What must be the phase difference
between the signals broadcast from the two antennas?

a) phase difference = (2pi/wavelength)(r2-r1)

I picked the point 20m from s1 and 100m from s2 so

phase difference = (2pi/wavelength)(100m-80m) = 12.56deg or .213rad

b) I kept the same point for the one north of the antennas. since they say from the midpoint east; 40m and then I did 40m to the east or -x.

I = I_o cos^2 (phase dif/2)

North point intensity = I = I_o cos^2 (12.56/2)

East from mindpoint = I = I_o cos^2 (0)

so the ratio is 1?

You did not do exactly what you said you did in part a) Try that again. I assume you need to fix a) to do b), so take another look at that. For c) you need to delay the signal from one antenna so that the signals are in phase along a "front" that is perpendicular to the direction of maximum intensity. Find the extra distance one signal has to go and compare that to a wavelength (similar to part a). What time phase shift do you need to compensate for the path difference?

I would like to clarify a few things before providing a response to the content. Firstly, electromagnetic waves are a form of energy that do not require a medium to travel through. They can travel through vacuum and are characterized by their frequency, wavelength, and amplitude. Therefore, the statement "Suppose the station has two antennas separated by 80m" may be misleading as the distance between the antennas does not affect the frequency or wavelength of the electromagnetic waves being emitted.

Moving on to the questions, for part a) the phase difference between the signals at a point north of the two antennas would depend on the distance of the point from each antenna and the wavelength of the electromagnetic waves. In this case, if the point is 20m from antenna s1 and 100m from antenna s2, the phase difference would be 12.56 degrees or 0.213 radians, as you have calculated.

For part b), the ratio of the intensity of the radio signal at a point north of the antennas to the intensity east of the midpoint between the antennas would depend on the distance of the points from each antenna. However, since both points are at the same distance from the center of the antennas, the ratio would be 1. This is because the intensity of an electromagnetic wave decreases with distance from the source, and since both points are equidistant from the antennas, the intensity would be the same.

For part c), the phase difference required for the maximum intensity in the antenna radiation pattern to lie on the line running from North-East to South-West would depend on the distance of the point from each antenna, the wavelength of the electromagnetic waves, and the angle of the line from North-East to South-West. Without knowing these values, it is not possible to calculate the exact phase difference. However, in general, to achieve maximum intensity in a specific direction, the phase difference between the signals from the two antennas needs to be adjusted in such a way that the waves combine constructively in that direction. This can be achieved by changing the distance between the antennas, the wavelength of the waves, or by using a phase shifter.

## What are electromagnetic waves?

Electromagnetic waves are a form of energy that is created by the movement of electrically charged particles. They can travel through vacuum and do not require a medium to propagate, unlike mechanical waves.

## What are the different types of electromagnetic waves?

The different types of electromagnetic waves are radio waves, microwaves, infrared radiation, visible light, ultraviolet radiation, X-rays, and gamma rays. These waves differ in their wavelengths and frequencies, which determine their properties and uses.

## How do electromagnetic waves interact with matter?

Electromagnetic waves can be absorbed, reflected, or transmitted when they encounter matter. The type and amount of interaction depend on the properties of the matter and the wavelength of the wave. For example, visible light is mostly transmitted through air, but is absorbed by opaque objects like walls.

## What is the speed of electromagnetic waves?

The speed of electromagnetic waves in a vacuum is approximately 3 x 10^8 meters per second, also known as the speed of light. This speed is constant and is denoted by the letter "c" in scientific equations.

## What are the practical applications of electromagnetic waves?

Electromagnetic waves have a wide range of practical applications in our daily lives. Radio waves are used for broadcasting, microwaves are used for cooking, infrared radiation is used for thermal imaging, visible light allows us to see, ultraviolet radiation is used for disinfection, X-rays are used for medical imaging, and gamma rays are used for cancer treatment.