Microwave standing wave calculations (max/min Intensity, Wavelength)

In summary: If we use the idea that the difference in distance between a maximum and minimum is ##\frac{\lambda}{4}##, then we have ##2\sqrt{0.5^2 + (0.119)^2} - 2\sqrt{0.5^2+0.084^2}## ##= \frac{\lambda}{4}## which solving gives ##\lambda = 0.056m##. If we work out the difference in distance between two maxima with the same method, we get ##0.282m##, which is consistent as this should be half the wavelength (Therefore I think that 0.056 is the correct answer).However I
  • #1
nav888
38
2
Homework Statement
Explain the presence of maximum and minimum intensities and determine the wavelength.
Relevant Equations
I = K A ^2
V = f Lambda
16849544414217637062667709172220.jpg

I have tried to subtract the two values of y for the minimum intensities to find half the wavelength but I am really stuck.
Any help would be appreciated thanks
 
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  • #3
kuruman said:
To receive help you must tell us how you approached this question and also what you tried and what you got.

Please read, understand and follow our homework guidelines, especially item 4, here
https://www.physicsforums.com/threads/homework-help-guidelines-for-students-and-helpers.686781/
Alright. So I understand that the wave will be reflected and will superimpose with the first wave leading to a standing wave with nodes and antinodes. The problem I'm having is I don't know how to find the wavelength.
 
  • #4
What do you think is going on here? Why should the intensity at the detector vary from a maximum to a minimum and then back to a maximum as ##y## is varied?
 
  • #5
The nodes and antinodes?
 
  • #6
And why do you get nodes and antinodes? What is going on here?
 
  • #7
The waves "add" together while moving towards one another.
Principle of Superposition
 
  • #8
nav888 said:
The waves "add" together while moving towards one another.
Principle of Superposition
They are not moving towards one another. The transmitter at point T emits waves in all directions. At D there is a detector that detects the superposition of two waves, one that goes directly from T to D and one bounces off the metal plate at R and then reaches D. What phenomenon is this? Hint: It starts with "I".
 
  • #9
kuruman said:
They are not moving towards one another. The transmitter at point T emits waves in all directions. At D there is a detector that detects the superposition of two waves, one that goes directly from T to D and one bounces off the metal plate at R and then reaches D. What phenomenon is this? Hint: It starts with "I".
Interference
 
  • #10
nav888 said:
Interference
The part that I am really stuck on is the wavelength bit.
 
  • #11
Forget the wavelength bit for the time being and it will sort itself out. Consider what must be true to have a maximum and a minimum. Think path length difference.
 
  • #12
At constructive interference we have maxima where path difference is of the form n lambda , for destructive we have n+1/2 lambda.
 
  • #13
nav888 said:
At constructive interference we have maxima where path difference is of the form n lambda , for destructive we have n+1/2 lambda.
Right. What does this suggest that you should do that would bring the wavelength into the picture?
 
  • #14
find the path difference, but once again i am stuck :(
 
  • #15
do i need to use pythagoras?
 
  • #16
Distance from node to anti node=1/4 lambda
 
  • #17
I have found the path difference for the first max but how do I know if this is lambda or not? I am unsure because when I divide the path difference of the minima by this path difference (which should be the wavelength as the first one is 1 lambda), I expect to get a number in the form n + 1/2 but I dont, can someone help/
n+1/2 (lambda) / lambda = n + 1/2 which cant be an integer but i get an integer
 
  • #18
nav888 said:
do i need to use pythagoras?
Yes you do and please post your results. It would hep if you used LaTeX. Its easy to learn. Click on the link "LaTeX Guide", lower left, to learn how.
 
  • #19
nav888 said:
At constructive interference we have maxima where path difference is of the form n lambda , for destructive we have n+1/2 lambda.
I don't think that you have taken into account the statement underlined below.

1684959793166.png


I believe this is why the ratios of the path differences that you mention in post #16 are not coming out as you expect.
 
  • #20
TSny said:
I don't think that you have taken into account the statement underlined below.

View attachment 327014

I believe this is why the ratios of the path differences that you mention in post #16 are not coming out as you expect.
If we use the idea that the difference in distance between a maximum and minimum is ##\frac{\lambda}{4}##, then we have ##2\sqrt{0.5^2 + (0.119)^2} - 2\sqrt{0.5^2+0.084^2}## ##= \frac{\lambda}{4}## which solving gives ##\lambda = 0.056m##. If we work out the difference in distance between two maxima with the same method, we get ##0.282m##, which is consistent as this should be half the wavelength (Therefore I think that 0.056 is the correct answer).
However I would like to ask why does it not work when I try and use path differences.
If I work out the path difference at the first maxima, we have ##2\sqrt{0.5^2 + 0.084^2} - 1##, which equals 0.014m, and as this is constructive interference, should be of the form ##n\lambda##, where ##n## is an integer however you can see its not, its actually a fourth of the wavelength which is super odd as it should be an integer multiple.
Could I get some help please?
Thanks
 
  • #21
nav888 said:
difference in distance between a maximum and minimum is ##\lambda/4##
In post #12 you wrote ##\lambda/2## and @kuruman confirmed that. As do I.
nav888 said:
If I work out the path difference at the first maxima, we have ##2\sqrt{0.5^2+0.084^2}−1##,
That would be true if there were no phase shift at the reflection.
DeBangis21 said:
Distance from node to anti node=1/4 lambda
Despite the title, it doesn’t look to me like the question has anything to do with standing waves.
 
  • #22
haruspex said:
In post #12 you wrote ##\lambda/2## and @kuruman confirmed that. As do I.

That would be true if there were no phase shift at the reflection.

Despite the title, it doesn’t look to me like the question has anything to do with standing waves
By minima, we mean a point of 0 displacement, or a point of "maximum" negative displacement? I think this is the root of my confusion.
 
  • #23
nav888 said:
By minima, we mean a point of 0 displacement, or a point of "maximum" negative displacement? I think this is the root of my confusion.
I got 29cm as my answer can someone check please.
 
  • #24
nav888 said:
I got 29cm as my answer can someone check please.
Seems too much. Verging on radio wave territory. Please post your working.
 
  • #25
nav888 said:
I got 29cm as my answer can someone check please.
This is incorrect. Here are some recommendations to troubleshoot it.
1. Use symbols instead of numbers. The path length difference is
##\Delta L=2\sqrt{L^2+y^2}-2L## where ##L =~##50 cm. Since ##y## is also given in centimeters, there is no reason to convert to meters. That avoids conversion errors.
2. For each value of ##y## calculate a value for the wavelength. Don't forget the phase shift upon reflection. This will give you 4 separate values. I recommend using a spreadsheet.
3. Take an average of the four values.

Below I show a plot of the 4 values as a function of ##y##. I removed the numbers from the vertical axis because it is against our rules to give the answer away. However, I can say that the origin of the vertical axis is at zero wavelength which indicates that there is little variation from one value to another.

Interference.png
 
  • #26
kuruman said:
This is incorrect. Here are some recommendations to troubleshoot it.
1. Use symbols instead of numbers. The path length difference is
##\Delta L=2\sqrt{L^2+y^2}-2L## where ##L =~##50 cm. Since ##y## is also given in centimeters, there is no reason to convert to meters. That avoids conversion errors.
2. For each value of ##y## calculate a value for the wavelength. Don't forget the phase shift upon reflection. This will give you 4 separate values. I recommend using a spreadsheet.
3. Take an average of the four values.

Below I show a plot of the 4 values as a function of ##y##. I removed the numbers from the vertical axis because it is against our rules to give the answer away. However, I can say that the origin of the vertical axis is at zero wavelength which indicates that there is little variation from one value to another.

View attachment 327054
I genuinely don't know what to do with the path difference because it's not of the form n lambda and this has confused me so much. Could I just get someone to give me a worked solution please
 
  • #27
nav888 said:
I genuinely don't know what to do with the path difference because it's not of the form n lambda and this has confused me so much. Could I just get someone to give me a worked solution please
Someone on stack exchange told me the answer is 29 which doesn't help my confusion
 
  • #28
nav888 said:
I genuinely don't know what to do with the path difference because it's not of the form n lambda and this has confused me so much. Could I just get someone to give me a worked solution please
Consider two separate paths (in meters) ##L_1## and ##L_2## that converge and are combined. The difference (also in meters) is ##\Delta L=L_2-L_1##. Assume that these are direct paths, i.e. no reflections or other phase shifts in between. Here are the rules in this case
  1. If ##\Delta L## is an even number of half-wavelengths, i.e. ##\Delta L= 2\frac{\lambda}{2},4\frac{\lambda}{2},\dots,2n\frac{\lambda}{2}\dots##, (##n=0,1,2,\dots~##), you have constructive interference.
  2. If ##\Delta L## is an odd number of half-wavelengths, i.e. ##\Delta L= 1\frac{\lambda}{2},3\frac{\lambda}{2},\dots,(2n+1)\frac{\lambda}{2}\dots##, you have destructive interference.
In this particular problem you have a reflection at point R that shifts the phase of the reflected wave by 180°. This translates into adding ##\frac{\lambda}{2}## to the reflected path ##L_2## but not to the direct path ##L_1##.

At PF we do not provide worked solutions to homework problems submitted by students. I have given you all you need to put this together.
 
  • #29
kuruman said:
Consider two separate paths (in meters) ##L_1## and ##L_2## that converge and are combined. The difference (also in meters) is ##\Delta L=L_2-L_1##. Assume that these are direct paths, i.e. no reflections or other phase shifts in between. Here are the rules in this case
  1. If ##\Delta L## is an even number of half-wavelengths, i.e. ##\Delta L= 2\frac{\lambda}{2},4\frac{\lambda}{2},\dots,2n\frac{\lambda}{2}\dots##, (##n=0,1,2,\dots~##), you have constructive interference.
  2. If ##\Delta L## is an odd number of half-wavelengths, i.e. ##\Delta L= 1\frac{\lambda}{2},3\frac{\lambda}{2},\dots,(2n+1)\frac{\lambda}{2}\dots##, you have destructive interference.
In this particular problem you have a reflection at point R that shifts the phase of the reflected wave by 180°. This translates into adding ##\frac{\lambda}{2}## to the reflected path ##L_2## but not to the direct path ##L_1##.

At PF we do not provide worked solutions to homework problems submitted by students. I have given you all you need to put this together.i

kuruman said:
Is the answer 56cm
 
  • #30
You need to use the fact that you are told that the values in the table are succesive maxima and minima. You cannot calculate the wavelength from any single value in the table, because you don't know the order of the maxima or minima (the value of n). There is no reason to assum that the first value in the table corresponds to a specific value of the path difference. You cannot calculate a wavelenth for each entry in the table unless you know the order of the maxima/minima.
But the difference in phase between any succesive maxima should be ##2\pi## and the path difference between two succesive maxima should be one wavelength.
 
  • #31
nasu said:
You need to use the fact that you are told that the values in the table are succesive maxima and minima. You cannot calculate the wavelength from any single value in the table, because you don't know the order of the maxima or minima (the value of n). There is no reason to assum that the first value in the table corresponds to a specific value of the path difference. You cannot calculate a wavelenth for each entry in the table unless you know the order of the maxima/minima.
But the difference in phase between any succesive maxima should be ##2\pi## and the path difference between two succesive maxima should be one wavelength.
Does Maxima mean the highest positive amplitude and minimal mean the lowest possible negative amplitude? Or is mimina 0 amplitude (in which case a lowest possible negative ampltude would be a Maxima too?)
 
  • #32
nasu said:
There is no reason to assum that the first value in the table corresponds to a specific value of the path difference.
I disagree. The problem clearly states, "The metal sheet is moved away from the line joining T and D so that y increases." It is reasonable to interpret this as saying that the metal plate starts at the line joining T and D and then moved away. Unless the author deliberately wants to misdirect the solver, the first value entered in the table must be ##y## for the first maximum.
 
  • Like
Likes nasu
  • #33
kuruman said:
I disagree. The problem clearly states, "The metal sheet is moved away from the line joining T and D so that y increases." It is reasonable to interpret this as saying that the metal plate starts at the line joining T and D and then moved away. Unless the author deliberately wants to misdirect the solver, the first value entered in the table must be ##y## for the first maximum.
I'm still confused on how to get from the path difference to the wavelength. This was never taught to me and I need to solve this problem
Thanks for your help so far
 
  • #34
nav888 said:
I'm still confused on how to get from the path difference to the wavelength. This was never taught to me and I need to solve this problem
Thanks for your help so far
I gave you two equations for the constructive and destructive interference conditions in post #28. What part about them is confusing you? You have ##\Delta L## on the left hand side and a quantity that is a number times half a wavelength on the right hand side. That's how you relate the path length difference to the wavelength.
 
  • #35
kuruman said:
I gave you two equations for the constructive and destructive interference conditions in post #28. What part about them is confusing you? You have ##\Delta L## on the left hand side and a quantity that is a number times half a wavelength on the right hand side. That's how you relate the path length difference to the wavelength
There is three unknowns.
both n values and lambda
 

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