Calculating Maxima in Radio Wave Interference Around Two Towers

[G01]

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, \Delta r, must = an integer number of wavelengths:

\Delta r = n\lambda

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:

c = \lambda f

\lambda = \frac{c}{f} = \frac{3X10^8m/s}{750000000Hz} = .4m

 

[gulsen]

It looks like 2, since
d \sin(\theta) = \pm \lambda
\sin(\theta) = \pm 1
with in 0, 2\pi interval
\theta = \pi, 3 \pi /2

 

[G01]

The answe is 20, which is what is confusing me. I got 2 also, but at \pi/2 , 3\pi/2 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.

 

[gulsen]

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 10 \sin(\theta), and if this difference is equal to 0 or an integer multiple of wavelength, then there should be a maxima
25 \lambda \sin(\theta) = n \lambda
\sin(\theta) = [-1,1] = n/25 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.

 

[G01]

Sorry I screwed up. The distance between the towers is 2 m not 10.

 

[G01]

Ok, I found that the angles, 0, \pi/2, \pi, 3\pi/2, 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 10\sin(\theta)

 

[gulsen]

http://sol.sci.uop.edu/~jfalward/lightinterference/lightinterference.html

 

[G01]

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),

 

[gulsen]

"Constructive Inteference" section

 

[G01]

i got the answer thanks.