Confused about interefence patterns

[sdfsfasdfasf]

Homework Statement

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Relevant Equations

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I am confued with what is meant when it says "about the line AB", does this mean horizontally or vertically? The question states that the sources produce vertically polarised microwaves (I know this is relavant) and I also know that waves moving in different planes cannot interfere. Does it mean that X is rotated "towards us" and then the full rotation has X facing backwards, "to the left"? If so, how can there be a pattern at 180 degrees as X isnt facing the detector?
Sorry if it sounds like im being stupid, I am really confused.

Thank you.

 

[kuruman]

I think it means that line AB and transmitter X stay fixed in the plane of the screen while detector Y is raised and rotated above the plane of the screen keeping the distance X-Y constant. At 90°, Y is directly above X and the line joining Y and X is perpendicular to the screen. At 180°, Y is back in the plane of the screen but on the other side of X (above the words "student connects" in the figure.)

 

[vela]

I'd take "rotated about the line AB" to mean that AB is the axis of rotation. You (the OP) are describing rotation about a line perpendicular to the plane of the page.

 

[sdfsfasdfasf]

I'd take "rotated about the line AB" to mean that AB is the axis of rotation. You (the OP) are describing rotation about a line perpendicular to the plane of the page.

Does your answer agree with kurumans answer?

 

[vela]

Obviously not. He's talking about moving transmitter (not detector) Y. I'm saying Y and D don't move and X rotates about its axis.

 

[sdfsfasdfasf]

Obviously not. He's talking about moving transmitter (not detector) Y. I'm saying Y and D don't move and X rotates about its axis.

I am struggling to visusalise what you mean when you say rotate about its axis, can you draw a picture please?

 

[haruspex]

I am struggling to visusalise what you mean when you say rotate about its axis, can you draw a picture please?

Suppose that at first the polarisation of waves from X is in the plane of the page.
Imagine an ant looking at X from B. In the diagram as shown, we are looking down on the top of the ant. The ant sees the polarisation as horizontal.
If we rotate the transmitter about line AB then the ant sees the polarisation as rotating. After 180° it is horizontal again. The face marked X is now at the bottom, facing away from us.
The transmitter points at B at all times.

 

[sdfsfasdfasf]

If im understanding correctly after 180 degrees the waves emitted are "upside down" so the maxima / minima are reversed on the intereference pattern?

 

[nasu]

After rotation by 180 degrees the plane of polarization is the same as before the rotation. There is no upside down. The electric field vector oscillates in the polarization plane, is not a constant vector. Only the orientation of the plane matters.

 

[sdfsfasdfasf]

After rotation by 180 degrees the plane of polarization is the same as before the rotation. There is no upside down. The electric field vector oscillates in the polarization plane, is not a constant vector. Only the orientation of the plane matters.

Ok, do the maxmima / minima flip? If so why?

 

[Steve4Physics]

Ok, do the maxmima / minima flip? If so why?

We often explain interference of two EM waves by considering vector addition of the waves’ electric fields (say $\vec {E_1}$ and $\vec {E_2}$) at the same time and place.

For vertically polarised waves, we get constructive interference where $\vec {E_1}$ and $\vec {E_2}$ always point in the same direction: both point up during the same half-cycle and both point down during the other half-cycle.

We get destructive interference at places where $\vec {E_1}$ and $\vec {E_2}$ always point in opposite directions: during the half-cylce when$\vec {E_1}$ points up, $\vec {E_2}$ points down and vice versa for the other half-cycle.

Rotating transmitter X about AB by $180^o$ (i.e. turning X upside down) will reverse its field direction. That means what would have been an upwards field at some moment in time will now be a downwards field, and vice versa.

So, can you put that all together and answer your question?

Minor edits to improve wording.

 

[kuruman]

Obviously not. He's talking about moving transmitter (not detector) Y. I'm saying Y and D don't move and X rotates about its axis.

Yes, I thought of this as a path-length difference experiment not as a polarization experiment. If it's a polarization experiment, one would have to assume that there is at least some signal in the starting configuration, i.e. there is no destructive interference to begin with. Then just rotating transmitter X about its axis leaving Y and D alone, would modulate the signal as explained by @Steve4Physics in post #11.

I think that rotating X and Y as one about AB (to avoid polarization effects) as described earlier and treating this as a path-length difference situation makes the answer to the question posed, "Describe and explain the observed patterns of maximum and minimum intensity", more interesting.

Below are plots of the path-length difference divided by the wavelength, $\Delta L/\lambda$ as a function of angle at fixed wavelength. Variable $y$ is the perpendicular distance of transmitter X to line PQ and $d$ is the constant separation between transmitters X and Y.

Interference is constructive at angles where $N=\Delta L/\lambda$ is an integer and destructive where it is a half-integer. The zeroth order maximum is at θ = 60° where it can be shown with a diagram and simple geometry that triangle XDY is isosceles.

 

[haruspex]

We often explain interference of two EM waves by considering vector addition of the waves’ electric fields (say $\vec {E_1}$ and $\vec {E_2}$) at the same time and place.

For vertically polarised waves, we get constructive interference where $\vec {E_1}$ and $\vec {E_2}$ always point in the same direction: both point up during the same half-cycle and both point down during the other half-cycle.

We get destructive interference at places where $\vec {E_1}$ and $\vec {E_2}$ always point in opposite directions: during the half-cylce when$\vec {E_1}$ points up, $\vec {E_2}$ points down and vice versa for the other half-cycle.

Rotating transmitter X about AB by $180^o$ (i.e. turning X upside down) will reverse its field direction. That means what would have been an upwards field at some moment in time will now be a downwards field, and vice versa.

So, can you put that all together and answer your question?

Minor edits to improve wording.

Am I misreading something? As I read posts #8, #9 and #11, @nasu says @sdfsfasdfasf is wong to say a 180° rotation flips maxima and minima and you are saying post #8 is right.

 

[Steve4Physics]

Am I misreading something? As I read posts #8, #9 and #11, @nasu says @sdfsfasdfasf is wong to say a 180° rotation flips maxima and minima and you are saying post #8 is right.

Yes, I think the suggestion in post #8 is correct.

Turning X upside down (i.e. a $180^o$ rotation about AB) effectively introduces a phase-change of $\pi$ between the two signals. (Not the best analogy, but compare it to reversing the polarity of one of a pair of stereo loudspeakers.)

After X’s $180^o$ rotation about AB, X’s plane of polarisation is restored, but the direction of X’s electric field in this plane has been reversed.

Edit - typo'.

 

[sdfsfasdfasf]

As an A Level student we don't learn about the actual fields that make up an electromagnetic wave, but this makes enough sense to me, the anaolgy of a phase shift of pi makes the most sense. Thank you all

 

[Steve4Physics]

As an A Level student we don't learn about the actual fields that make up an electromagnetic wave, but this makes enough sense to me, the anaolgy of a phase shift of pi makes the most sense. Thank you all

Make sure you have carefully read and thought about what the question is asking.

For another mark, I'd guess that you are required to say what happens to the interefernce pattern (along PQ) when the angle of rotation of X about AB is $90^0$ ...

 

[sdfsfasdfasf]

Make sure you have carefully read and thought about what the question is asking.

For another mark, I'd guess that you are required to say what happens to the interefernce pattern (along PQ) when the angle of rotation of X about AB is $90^0$ ...

No interference