Solving Double Slit Homework Qs: Phase Difference & Destructive Interference

[MartinFreeman]

Homework Statement

Two antennas located at points A and B are broadcasting radio waves of frequency 98.0 MHz, perfectly in phase with each other. The two antennas are separated by a distance d= 6.20 m. An observer, P, is located on the x axis, a distance x= 60.0 m from antenna A, so that APB forms a right triangle with PB as hypotenuse.

Homework Equations

phase difference / 2 pi = r2-r1 / lambda
d(y/L)= n(lamdba) if constructive
d(y/L) = (n+1/2)(lambda) if deconstructive.

The Attempt at a Solution

There are three questions, I have managed to solve the first one , which is :

What is the phase difference between the waves arriving at P from antennas A and B?

I have found, that the solution is : phase = (2(pi ) sqrt( (6.2)^2 + (60)^2) - 60) / 3.0612m - > 0.6557 rad.

the two following parts are where i become confused , I have searched online for advice, but I cannot seem to solve the following parts; here they are.

Now observer P walks along the x-axis toward antenna A. What is P's distance from A when he first observes fully destructive interference between the two waves?

and finally, the third part :

If observer P continues walking until he reaches antenna A, at how many places along the x-axis (including the place you found in the previous problem) will he detect minima in the radio signal, due to destructive interference?note: If you do help, I very much appreciate it! However, If you could be as clear as possible. Thanks in advance!

 

[Doc Al]

I have found, that the solution is : phase = (2(pi ) sqrt( (6.2)^2 + (60)^2) - 60) / 3.0612m - > 0.6557 rad.

Do you understand this solution? The same thinking is used for all parts of the problem.

the two following parts are where i become confused , I have searched online for advice, but I cannot seem to solve the following parts; here they are.

Now observer P walks along the x-axis toward antenna A. What is P's distance from A when he first observes fully destructive interference between the two waves?

Ask yourself: What's the path length difference between AP and BP? What is the value at the initial point? How many wavelengths is that? As the distance AP decreases, what must the path length difference equal to have destructive interference? (Represent the distance AP as "x" in your equations, since you'll be solving for it.)

 

[MartinFreeman]

thanks for responding Doc, however I must admit that I am still no further in understanding how to get to the first destructive point.
Going from your quick series of questions , this is what I have come up with ( and I have a feeling I've made a mistake )

The path length between AP and BP is sqrt(6.2^2 +60^2) - 60 = 0.319482m

how many wavelengths is that? 0.319482m/3.0612m = 0.104 of a lamda, (10.4%) ( this i become uncertain of what to do )

AP must decrease such that the pathlength between AP and BP should cause BP's wavelength to be out of phase, which is pi / 180 degrees (?)

so... ( this is a guess ) BP-AP must = 1/2 lamda (?) such that there would be a destructive point,

is AP = 60-x and BP = to sqrt( (6.2^2)+ (X^2)) ? I don't know what to do I'm sorry.
~ AP= X
~ X = 60 - x

 

[Doc Al]

You're on the right track. Yes, the path length difference must equal λ/2. Call the distance AP = x. Now write a general expression for the path length difference in terms of x. (I think you've almost got it.) Then set it equal to λ/2 and solve for x. (It will be a quadratic equation.)

 

[MartinFreeman]

You're on the right track. Yes, the path length difference must equal λ/2. Call the distance AP = x. Now write a general expression for the path length difference in terms of x. (I think you've almost got it.) Then set it equal to λ/2 and solve for x. (It will be a quadratic equation.)

thanks again for responding, I am not to sure how I'm getting a quadratic,

The relationship I've come up with is this , r2-r1 = lamda /2
r2= BP = sqrt( 6.2^2 +x^2)
r1 = AP = x

path difference = 3.06/2
~sqrt ( 6.2^2 =x^2) = 3.06/ 2
~38.44 +x^2 - 3600 = (3.06/2)^2
x^2 = 3563.9
x = + 59.69

I've spent a few hours on this question and I think I'm losing my mind on this. Do you mind sharing the solution with me? Thanks a lot for your patience and help!

 

[Doc Al]

The relationship I've come up with is this , r2-r1 = lamda /2
r2= BP = sqrt( 6.2^2 +x^2)
r1 = AP = x

Good.

path difference = 3.06/2

Good.

~sqrt ( 6.2^2 =x^2) = 3.06/ 2

What happened to r1? (The approximation is not justified.)

 

[MartinFreeman]

Good.

Good.

What happened to r1? (The approximation is not justified.)

is it possibly, sqrt( 6.2^2 + x^2) - x = 1.53m? now I'm stuck for sure.

 

[Doc Al]

is it possibly, sqrt( 6.2^2 + x^2) - x = 1.53m? now I'm stuck for sure.

Solve for x.

 

[MartinFreeman]

Solve for x.

X cancels itself though ,

38.44 +x^2 - x^2 = 2.3409m
38.44= 2.3409 . .

 

[MartinFreeman]

Solve for x.

Sorry for wasting your time , Il just skip it

 

[MartinFreeman]

Solve for x.

Can i simply ask you how to arrange the equation?

 

[Doc Al]

X cancels itself though ,

No it doesn't.
Do this:
sqrt( 6.2^2 + x^2) - x = 1.53
sqrt( 6.2^2 + x^2) = 1.53 + x

Now square both sides to get rid of that square root.

 

[MartinFreeman]

No it doesn't.
Do this:
sqrt( 6.2^2 + x^2) - x = 1.53
sqrt( 6.2^2 + x^2) = 1.53 + x

Now square both sides to get rid of that square root.

Thank you very much for helping me Doc I've managed to get the solution

 

[Doc Al]

Thank you very much for helping me Doc I've managed to get the solution