# Oxford PAT 2012, Question 19

## 13 thoughts on “Oxford PAT 2012, Question 19”

1. Laurie Smith says:

Wouldn’t there be maxima and minima after the microphone moves to the left of the Loudspeaker, so the answer is no, until M is to the right of L?

1. (You mean for part b). Yes you are right. Though actually, I suspect that the questioners didn’t intend you to consider that the microphone might move past the loudspeaker.

But when the microphone is between the loudspeaker and the screen then, if we let the distance between the microphone and the screen be D, we can see that the waves travel a distance x directly from the speaker to the microphone and a distance (x + 2D) from the speaker to the screen and back to the microphone. So again the distance that the microphone moves between maxima is λ/2.

2. George says:

Hey, for part b, I am not sure if it is correct to approach in this way: if microphone move, LSM tend to LSM-x , LM tend to LM-x. So we can say that path difference is 0 therefore, the answer will be yes?

I am using the same approach for part a: if L moves, LSM tend to LSM-x, LM tend to LM+x so the path difference is 2x therefore x=lambda /2 so the answer is yes

3. Max Leftley says:

I am still not sure I totally understand the question but I am going to kinda just rephrase what I think you said to check if I know it….. So each time the path difference between reflected waves and waves just emitted is n(lambda), there is constructive interference between them. This means it occurs for path differences of many different values of n, for example 1,2,3 etc… so 2d (distance travelled by reflected waves) = n = )1 or 2,3…. etc.)*(lambda) and d is half that, so the difference between any one of these is (lambda)/2?

Okay, that aside, I am not quite sure exactly how the microphone detects the maxima and when. The maximum is produced at the speaker isn’t it? Does this resultant wave travel to the microphone or not..? I have seen some answers in terms of standing waves which have really confused me. Yours has really helped but there are just a few oddities in my thinking. Also, is the significance of the speed of the wall being so slow that there is no noticeable Doppler shift that will prevent the waves from being cohesive?

1. Sorry my description here is a bit terse, so I’m going to flesh it out a little bit. The loudspeaker transmits sound ‘in all directions’. Most of this sound never reaches the microphone, but sound that travels in two directions does reach the microphone: first, the sound that goes straight there; second, the sound that bounces off the screen. These two waves interfere like all waves do. If the microphone is a distance D from the speaker and the screen is a distance d from the speaker, then the direct and reflected waves travel a distance D and (D+2d) respectively to reach the microphone. The waves constructively interfere when they are in phase, i.e. when 2d = nλ. It’s as simple as that.

The question says that the screen moves ‘slowly to the right’, indicating that speed is effectively zero, so you should ignore the Doppler effect. I don’t find it interesting or useful to think about standing waves to solve this question, but maybe some other analysis does…

1. Max Leftley says:

Ahh…. Thank you so much. That is the best explanation of the question I have seen.

4. ian says:

hi , for this question i know there is a formula of x= lambda (d)/D, where x is the distance between bright fringe, but how do u get 2d =n lambda???

1. ian says:

is it because the the path difference is 2d?

1. Yes exactly, the path difference is 2d and therefore there is constructive interference when 2d = nλ, where n is n integer. That is, when the reflected wave has travelled a whole number of wavelengths.

5. Jacob says:

Hello, as I study optics, I am confused by the idea of calculating the maximum intensity of the fringes based on the relative phases of the waves. I understand constructive and destructive interference in the general sense, but how do you find when the wave is in phase and when they are out of phase?

In my textbook, it says that the difference in path lengths (delta l)=mlambda for constructive (m is an integer). and (delta l)=(m+1/2)lambda for destructive. I am not really sure what these mean in terms of relating the numerics to interference and how to utilize it for out of phase vs in phase…

Sorry for the rather general question! Thank you so much for this website 🙂

1. Hi Jacob. It’s quite hard to describe this in words, without getting quite technical. MAybe it’s worth you looking at a video like this one: https://www.youtube.com/watch?v=CAe3lkYNKt8 (about 1 minute in). This has some explanation of what phase is and how it relates to interference.

6. Lee says:

Hi i was wondering for the second part, i can understand part a but i couldn’t get a grip on part b. I thought that there is a series of standing waves established behind the speaker, and thus when the microphone was moved along the x-axis, it would progress from an antinode to a node and back to the antinode again. Is it correct to say so? thks ><

1. Hi Lee, thanks for your question. Here is some more detail about the solution: I hope it helps.

We can assume that the screen is smooth and so reflects the sound such that the angle of incidence equals the angle of reflection. This means you can simplify your reasoning by observing that you only need to consider signals travelling along the line that joins M and L. So we can consider the reflection at the screen as being like a point source at the intersection of the screen and the line joining M and L. Since the loudspeaker is a point source too, the whole situation is equivalent to two point sources of sound on a line and a microphone on the same line, and this is equivalent to two co-located point sources that are out pf phase with each other (with the phase being decided by the distance between them).