Antennas causing destructive interference

[PirateFan308]

Homework Statement

Two radio antennas radiating in phase are located at points A and B, 200m apart. The radio waves have a frequency of 5.80MHz. A radio receiver is moved out from point B along a line perpendicular to the line connecting A and B.

At what distances from B will there be destructive interference? (Note: The distance of the receiver from the sources is not large in comparison to the separation of the sources, so equation [itex]dsin\theta = (m+0.5)λ[/itex] does not apply.)

The Attempt at a Solution

I tried looking at the difference in path lengths separately, because I know that for destructive interference to occur, (m+1/2)λ=Δray length.

I drew a right angle triangle, where one side was 200m, the other side was r2 and the hypotenuse was r1. I said that θ was the angle between the 200m and r1. So then:
[itex]r_1=200tanθ,~~ r_2=\frac{200}{cosθ}

~~~~~Δr = 200 \frac{sinθ-1}{cosθ} = (m+1/2)λ[/itex]

But I can't figure out how to solve this equation, nor can I see another way to solve this problem. Thanks!

 

[PeterO]

Homework Statement

Two radio antennas radiating in phase are located at points A and B, 200m apart. The radio waves have a frequency of 5.80MHz. A radio receiver is moved out from point B along a line perpendicular to the line connecting A and B.

At what distances from B will there be destructive interference? (Note: The distance of the receiver from the sources is not large in comparison to the separation of the sources, so equation [itex]dsin\theta = (m+0.5)λ[/itex] does not apply.)

The Attempt at a Solution

I tried looking at the difference in path lengths separately, because I know that for destructive interference to occur, (m+1/2)λ=Δray length.

I drew a right angle triangle, where one side was 200m, the other side was r2 and the hypotenuse was r1. I said that θ was the angle between the 200m and r1. So then:
[itex]r_1=200tanθ,~~ r_2=\frac{200}{cosθ}

~~~~~Δr = 200 \frac{sinθ-1}{cosθ} = (m+1/2)λ[/itex]

But I can't figure out how to solve this equation, nor can I see another way to solve this problem. Thanks!

The first thing I would have calculated is the wavelength of these radio waves, and thus how many wavelengths the separation (200m) represents. That is the maximum path difference possible.
That will show you how many points exist where there is destructive interference.

 

[PirateFan308]

[itex]\lambda = \frac{c}{f} = \frac{3.0*10^8}{5.80*10^6} = 51.7m[/itex]
So the separation represents 3.86 wavelengths, meaning there will be a maximum of 3 path differences.

As I was thinking further, I drew another right triangle with the distance between the sources being 200m, the hypotenuse being r1 and the other side being r2. If I put r2=x, then [itex] r_1 = \sqrt{x^2+200^2} [/itex] so i could solve as follows:

[itex] \sqrt{x^2+200^2}-x=(m+1/2)λ[/itex]

[itex]x^2+40000 = (m+1/2)^2(1/2*51.7m)^2+(m+1/2)(51.7m)x + x^2[/itex]

[itex]x=\frac{40000-(m+1/2)^2(25.86)^2}{(m+1/2)(51.7)}[/itex]

for m=0,

[itex]x=\frac{40000-(0+1/2)^2(25.86)^2}{(0+1/2)(51.7)} = 1540m[/itex]

for m=1,

[itex]x=\frac{40000-(1+1/2)^2(25.86)^2}{(1+1/2)(51.7)} = 496.4m[/itex]

for m=2,

[itex]x=\frac{40000-(2+1/2)^2(25.86)^2}{(2+1/2)(51.7)} = 277.1m[/itex]

for m=3,

[itex]x=\frac{40000-(3+1/2)^2(25.86)^2}{(3+1/2)(51.7)} = 175.78m[/itex]

form m=4, x=113.7. For m=5, x=69.53. For m=6, x=34.95. For m=7, x=6.15. For m=8, x=-18.9.

So the idea that I would only have 3 destructive interference spots doesn't line up with this. But I'm not sure where I made the mistake.

 

[PeterO]

[itex]\lambda = \frac{c}{f} = \frac{3.0*10^8}{5.80*10^6} = 51.7m[/itex]
So the separation represents 3.86 wavelengths, meaning there will be a maximum of 3 path differences.

As I was thinking further, I drew another right triangle with the distance between the sources being 200m, the hypotenuse being r1 and the other side being r2. If I put r2=x, then [itex] r_1 = \sqrt{x^2+200^2} [/itex] so i could solve as follows:

[itex] \sqrt{x^2+200^2}-x=(m+1/2)λ[/itex]

[itex]x^2+40000 = (m+1/2)^2(1/2*51.7m)^2+(m+1/2)(51.7m)x + x^2[/itex]

[itex]x=\frac{40000-(m+1/2)^2(25.86)^2}{(m+1/2)(51.7)}[/itex]

for m=0,

[itex]x=\frac{40000-(0+1/2)^2(25.86)^2}{(0+1/2)(51.7)} = 1540m[/itex]

for m=1,

[itex]x=\frac{40000-(1+1/2)^2(25.86)^2}{(1+1/2)(51.7)} = 496.4m[/itex]

for m=2,

[itex]x=\frac{40000-(2+1/2)^2(25.86)^2}{(2+1/2)(51.7)} = 277.1m[/itex]

for m=3,

[itex]x=\frac{40000-(3+1/2)^2(25.86)^2}{(3+1/2)(51.7)} = 175.78m[/itex]

form m=4, x=113.7. For m=5, x=69.53. For m=6, x=34.95. For m=7, x=6.15. For m=8, x=-18.9.

So the idea that I would only have 3 destructive interference spots doesn't line up with this. But I'm not sure where I made the mistake.

Potentially, destructive interference occurs at (m+1/2)λ as you said

That predicts 0.5λ, 1.5λ, 2.5λ, 3.5λ, 4.5λ, 5.5λ, etc

You calculated the maximum path difference [separation] = 3.86λ so any from the list that are less than that are possible.

0.5λ, 1.5λ, 2.5λ, 3.5λ

So you should have expected 4 points, not just 3. I think your following mathematics prompted that too.

EDIT: sorry for the delay in response - time zone problems; I am in Australia.

 

[PirateFan308]

Potentially, destructive interference occurs at (m+1/2)λ as you said

That predicts 0.5λ, 1.5λ, 2.5λ, 3.5λ, 4.5λ, 5.5λ, etc

You calculated the maximum path difference [separation] = 3.86λ so any from the list that are less than that are possible.

0.5λ, 1.5λ, 2.5λ, 3.5λ

So you should have expected 4 points, not just 3. I think your following mathematics prompted that too.

EDIT: sorry for the delay in response - time zone problems; I am in Australia.

No worries - thank you for your help!

I first tried answering with 34.94,69.53,113.7,175.78 before I realized that was the distance less than 200m rather than the correct m≤3. Next, I tried answering 175.78,277.1,496.4,1540 and this was also wrong, but I can't find any errors in my computation.

 

[PeterO]

No worries - thank you for your help!

I first tried answering with 34.94,69.53,113.7,175.78 before I realized that was the distance less than 200m rather than the correct m≤3. Next, I tried answering 175.78,277.1,496.4,1540 and this was also wrong, but I can't find any errors in my computation.

In the second lne of your working (copied crudely here)

x2+40000=(m+1/2)2(1/2∗51.7m)2+(m+1/2)(51.7m)x+x2

You have halved the wavlength, while the formula you were using had the 1/2 with the m rather than with the λ. Perhaps I am mis-reading something?

 

[PeterO]

No worries - thank you for your help!

I first tried answering with 34.94,69.53,113.7,175.78 before I realized that was the distance less than 200m rather than the correct m≤3. Next, I tried answering 175.78,277.1,496.4,1540 and this was also wrong, but I can't find any errors in my computation.

An interesting way to get the answers would be:

Instead of starting with directions perpendicular to the line joining A & B, find the places on a circle created with A as the centre and radius 200m

Lets go for the point where m = 3, the path difference is 3.5 * 51.7 ≈ 181

The chord length from B must be 200 - 181 = 19

From that you can find the angle to the destructive interference point.

Now look at the perpendicular direction, and use trigonometry [tangent function actually], along with the angle you have calculate above, to compute the distance from B along the perpendicular.

Repeat for the other 3 path differences.

 

[PirateFan308]

In the second lne of your working (copied crudely here)

x2+40000=(m+1/2)2(1/2∗51.7m)2+(m+1/2)(51.7m)x+x2

You have halved the wavlength, while the formula you were using had the 1/2 with the m rather than with the λ. Perhaps I am mis-reading something?

You're right, for some reason I plugged in half λ. It should have been [itex]x^2+40000 = (m+1/2)^2λ^2+2(m+1/2)λx+x^2[/itex] which would reduce to [itex]x=\frac{40000-(m+1/2)^2λ^2}{2(m+1/2)λ}[/itex]. Thank you so much! You were a huge help!