Calculating Maxima in Radio Wave Interference Around Two Towers

In summary, 20 maxima will be detected if a person walks around the radio towers in a circle of radius 10m.
  • #1
G01
Homework Helper
Gold Member
2,704
19
Two radio towers are positioned 10m apart. They emit waves of 750mHz. How many maxima will you detect if you walk around the towers in a circle of radius=10m?

I know that the path length difference, [tex]\Delta r [/tex], must = an integer number of wavelengths:

[tex] \Delta r = n\lambda [/tex]

Where I'm confused is how to find an expression using this information to solve for the number of maximas in a circle. Can anyone give me any hints? I'd show more work if I could think of anything else. The only other work I have is solving for the wavelength:

[tex] c = \lambda f [/tex]

[tex] \lambda = \frac{c}{f} = \frac{3X10^8m/s}{750000000Hz} = .4m [/tex]
 
Physics news on Phys.org
  • #2
It looks like 2, since
[tex]d \sin(\theta) = \pm \lambda[/tex]
[tex]\sin(\theta) = \pm 1[/tex]
with in [tex]0, 2\pi[/tex] interval
[tex]\theta = \pi, 3 \pi /2[/tex]
 
  • #3
The answe is 20, which is what is confusing me. I got 2 also, but at [tex] \pi/2 , 3\pi/2 [/tex] I don't know how I'm supposed to find the 20 spots. I know there has to be a way to derive a formula for it. Or something.
 
  • #4
What the heck I've done, the wavelength isn't 10m, it's 0.4m! Plus, 0 is also an integer, so there would be 4 solution (yours and mine should add, if d=wavelength!) Oh my! I was totally sleepy!

So then, let me try again. The difference in distance of waveves should be [tex]10 \sin(\theta)[/tex], and if this difference is equal to 0 or an integer multiple of wavelength, then there should be a maxima
[tex]25 \lambda \sin(\theta) = n \lambda[/tex]
[tex]\sin(\theta) = [-1,1] = n/25[/tex] where n is an integer.

Hmm.. I still don't have 20. I don't know where I got wrong, though.
I assumed center of the circle is in the middle of towers, BTW.
 
  • #5
Sorry I screwed up. The distance between the towers is 2 m not 10.
 
  • #6
Ok, I found that the angles, 0, [tex] \pi/2, \pi, 3\pi/2, [/tex] all have maximas so that gives me four. Now i need to find a way to find the maximas in one of the quadrants and I can multpily that by four and then add four to that and I should get 20. I am lost on how to find the maximas in between quadrant angles. How do you get the the distance between waves is [tex] 10\sin(\theta) [/tex]
 
  • #7
http://sol.sci.uop.edu/~jfalward/lightinterference/lightinterference.html
 
Last edited by a moderator:
  • #8
im sorry, i don't know which sectionyou want me to look at. I read the page and I have seen most of that stuff, but I don't see what section answers my question about why the distance bewteen the waves equals 10sin(x),
 
  • #9
"Constructive Inteference" section
 
  • #10
i got the answer thanks.
 

What is radio wave interference?

Radio wave interference occurs when multiple radio signals overlap and disrupt each other's transmission. This can result in distorted or weakened signals, causing communication issues.

What causes radio wave interference?

Radio wave interference can be caused by a variety of factors, including physical obstacles such as buildings or mountains, electronic devices emitting strong signals, and weather conditions such as lightning storms.

How does radio wave interference affect communication?

Radio wave interference can cause disruptions in communication, leading to distorted or lost signals. This can affect various forms of communication, such as radio broadcasting, cell phone calls, and Wi-Fi connections.

What can be done to reduce radio wave interference?

There are several strategies that can be used to reduce radio wave interference, such as using directional antennas to focus the signal, using different frequencies, and properly grounding electronic devices. Additionally, regulations and standards are in place to minimize interference from electronic devices.

What role does wavelength play in radio wave interference?

Wavelength is an important factor in radio wave interference, as signals with longer wavelengths are more likely to be affected by interference. This is because longer wavelengths are more easily disrupted by physical obstacles and atmospheric conditions.

Similar threads

  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
20
Views
2K
  • Introductory Physics Homework Help
Replies
10
Views
924
  • Introductory Physics Homework Help
Replies
19
Views
3K
  • Introductory Physics Homework Help
Replies
8
Views
2K
  • Introductory Physics Homework Help
Replies
8
Views
1K
Replies
4
Views
335
  • Introductory Physics Homework Help
Replies
6
Views
4K
  • Introductory Physics Homework Help
Replies
3
Views
9K
Back
Top