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
  • #36
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.
You told me those were the rules for no phase change.
This is really stressing me out now and I'd like some more guidance please I've got 4 different answers 😭
 
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  • #37
nav888 said:
You told me those were the rules for no phase change.
This is really stressing me out now and I'd like some more guidance please I've got 4 different answers 😭
OMG is the answer 28cm
 
  • #38
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.
You are right. The first maximum in the table is the first one with non-zero phase difference.
 
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  • #39
nasu said:
You are right. The first maximum in the table is the first one with non-zero phase difference.
Is the answer 28cm
 
  • #40
nav888 said:
This was never taught to me
Then let me teach it to you in an intuitively obvious way by using some imagery. Imagine that you ride on a crest of the wave in a special vehicle. In front of you there is a meter with a single needle that looks like a clock but has divisions in degrees. As you ride along with the wave, the needle turns at a constant rate.

Rule: Every time the needle makes one complete revolution, the vehicle has advanced the distance of one wavelength and the phase of the wave, expressed as a sine or cosine, has advanced by ##2\pi.##

Now imagine two observers, one on each wave, in two separate vehicles. They start together with their wave-o-meters synchronized, go their separate ways and eventually meet again and compare the readings on their meters.
  • If the two needles point at the same value, you have constructive interference
  • If the two needles point in exactly antiparallel directions (not necessarily 0 and 180 degrees), you have destructive interference
  • If the needles point in none-of-the-above directions, you have in between interference.
In this problem, when the wave is reflected off the metal plate the needle attached to that wave gets bumped ahead by 180 degrees and which means that the path is bumped by half a wavelength.

Does this help?
 
  • #41
nav888 said:
OMG is the answer 28cm
Have you tried to calculate the path differences corresponding to the values of y given in the table? What is the order of magnitude of these? What is the change in path difference from one maxima to the next? Are these compatible with a wavelength of 28 cm?
 
  • #42
nasu said:
Have you tried to calculate the path differences corresponding to the values of y given in the table? What is the order of magnitude of these? What is the change in path difference from one maxima to the next? Are these compatible with a wavelength of 28 cm?
I'm really struggling with this and would just like to know if the answer is 29 I've got a bunch of messy working out please
 
  • #43
nav888 said:
I'm really struggling with this and would just like to know if the answer is 29 I've got a bunch of messy working out please
IS IT 2.8
 
  • #44
nav888 said:
IS IT 2.8
Please show your work. If you made a mistake somewhere, we will be happy to show you where. If you started from a wrong assumption, we will point it out. You cannot solve a physics problem by asking is it this or is it that? Not a single post from you shows how you proceeded to work out the answers that you ask about. Besides "2.8" without units is meaningless.
 
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  • #45
kuruman said:
Then let me teach it to you in an intuitively obvious way by using some imagery. Imagine that you ride on a crest of the wave in a special vehicle. In front of you there is a meter with a single needle that looks like a clock but has divisions in degrees. As you ride along with the wave, the needle turns at a constant rate.

Rule: Every time the needle makes one complete revolution, the vehicle has advanced the distance of one wavelength and the phase of the wave, expressed as a sine or cosine, has advanced by ##2\pi.##

Now imagine two observers, one on each wave, in two separate vehicles. They start together with their wave-o-meters synchronized, go their separate ways and eventually meet again and compare the readings on their meters.
  • If the two needles point at the same value, you have constructive interference
  • If the two needles point in exactly antiparallel directions (not necessarily 0 and 180 degrees), you have destructive interference
  • If the needles point in none-of-the-above directions, you have in between interference.
In this problem, when the wave is reflected off the metal plate the needle attached to that wave gets bumped ahead by 180 degrees and which means that the path is bumped by half a wavelength.

Does this help?
I hate to revive this thread after seeing the tragedy before. I’m also stuck on this question and read through this thread a million times to try to understand the answer. How does the reflection knocking the reflected waves another 1/2 lambda mean you have to multiply the length by 4, surely it would just be 3?
 
  • #46
aidandv said:
How does the reflection knocking the reflected waves another 1/2 lambda mean you have to multiply the length by 4, surely it would just be 3?
I don't understand what you are asking. What length needs to by multiplied by 3 instead of 4 and why? Please show your calculations and explain where they are coming from.
 
  • #47
image.jpg

The answer is 2.8 which is double the value I calculated for wavelength. How does the fact it gets knocked 180 degrees out of phase cause my calculation or be off by a factor of 2?
 
  • #48
What do you think are these values calculated on your scratch paper (which is not with the right orientation)?
 
  • #49
i believe it is one wavelength due to the fact that the distance between adjacent maxima are one wavelength apart. Ive realsied ive only done from the transmitter to the reflector which is only half the distance so that is why i am off by a factor of two
 
  • #50
Would you be able to find the wavelength if all that was given to you were the first minimum at 8.4 cm? The answer is yes. Read post #28 for a hint. Each of the four given values for ##y## can be used to find the wavelength independently of the others. Figuring out how to do it will deepen your understanding of interference.
 
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  • #51
kuruman said:
Would you be able to find the wavelength if all that was given to you were the first minimum at 8.4 cm? The answer is yes. Read post #28 for a hint. Each of the four given values for ##y## can be used to find the wavelength independently of the others. Figuring out how to do it will deepen your understanding of interference.
Yes I understand how you only require one as long as you know the order. Thanks for helpjng
 
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  • #52
aidandv said:
i believe it is one wavelength due to the fact that the distance between adjacent maxima are one wavelength apart. Ive realsied ive only done from the transmitter to the reflector which is only half the distance so that is why i am off by a factor of two
Yes, you calculated just half of the path difference. This is why you got the wrong value for wavelength.
 

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