Solving Double Slit Homework Qs: Phase Difference & Destructive Interference

In summary,An observer P is located at point A on the x-axis, a distance x from antenna A of 60.0 m. P observes destructive interference between waves arriving from antennas A and B, with a phase difference of 0.6557 rad. If P walks the x-axis until he reaches antenna A, he will detect minima in the signal.
  • #1
MartinFreeman
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0

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!
 
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  • #2
MartinFreeman said:
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.
MartinFreeman said:
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.)
 
  • #3
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
 
  • #4
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.)
 
  • #5
Doc Al said:
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!
 
  • #6
MartinFreeman said:
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.
MartinFreeman said:
path difference = 3.06/2
Good.
MartinFreeman said:
~sqrt ( 6.2^2 =x^2) = 3.06/ 2
What happened to r1? (The approximation is not justified.)
 
  • #7
Doc Al said:
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.
 
  • #8
MartinFreeman said:
is it possibly, sqrt( 6.2^2 + x^2) - x = 1.53m? now I'm stuck for sure.
Solve for x.
 
  • #9
Doc Al said:
Solve for x.
X cancels itself though ,

38.44 +x^2 - x^2 = 2.3409m
38.44= 2.3409 . .
 
  • #10
Doc Al said:
Solve for x.
Sorry for wasting your time , Il just skip it
 
  • #11
Doc Al said:
Solve for x.
Can i simply ask you how to arrange the equation?
 
  • #12
MartinFreeman said:
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.
 
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  • #13
Doc Al said:
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
 
  • #14
MartinFreeman said:
Thank you very much for helping me Doc I've managed to get the solution
:thumbup:
 

1. What is the purpose of solving double slit homework questions?

Solving double slit homework questions is a way for students to understand the concepts of phase difference and destructive interference in wave mechanics. It allows them to apply these principles to real-life scenarios and develop problem-solving skills.

2. What is the difference between phase difference and destructive interference?

Phase difference refers to the difference in the phase of two waves, which can result in constructive or destructive interference. Destructive interference occurs when two waves with a phase difference of 180 degrees cancel each other out, resulting in a decrease in amplitude.

3. How do I calculate the phase difference between two waves?

The phase difference can be calculated by finding the number of wavelengths between the two waves and multiplying it by 360 degrees. This can also be expressed in radians by multiplying the number of wavelengths by 2π.

4. What factors affect the phase difference and destructive interference in double slit experiments?

The factors that affect phase difference and destructive interference include the distance between the two slits, the wavelength of the waves, and the angle at which the waves are incident on the slits.

5. How can I use the concept of phase difference and destructive interference to solve real-life problems?

The principles of phase difference and destructive interference can be applied to various situations, such as in acoustics, optics, and electronics. For example, it can be used to explain the phenomenon of interference in music or to design noise-cancelling headphones.

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