Calculating wavelength using Lloyd mirror

[henk jan]

For an experiment I have to calculate de wavelength of a sound using a Lloyd mirror, the reciever and transmitter are fixed and the sound mirror is moving towards the optical axis. I know i have te work with fringe spacing or something like that but I dont know how exactly

[swansont]

Do you understand why there are fringes?

How does sound enter into this?

[henk jan]

1 hour ago, swansont said:

Do you understand why there are fringes?

How does sound enter into this?

Yes, fringes form because of destructive and constructive interference, and I understand that if the waves are in the same phase its constructive, but I dont know ow to calculate the wavelength if you know the length between fringes

[exchemist]

47 minutes ago, henk jan said:

Yes, fringes form because of destructive and constructive interference, and I understand that if the waves are in the same phase its constructive, but I dont know ow to calculate the wavelength if you know the length between fringes

I'm not expert on this but don't you need to do some geometry to work out how the path length changes with the off-axis distance of the sound reflector. As the reflector moves in towards the axis and this distance decreases, the difference in path length will decrease so your receiver will encounter successive maxima and minima in the sound, as the direct and reflected beams reinforce or cancel. (The reflected sound will be phase-shifted by π as well, I think.)  

Unless I've misunderstood the scenario.......

 

Edited May 26, 2021 by exchemist

[studiot]

2 hours ago, henk jan said:

For an experiment I have to calculate de wavelength of a sound using a Lloyd mirror

2 hours ago, swansont said:

How does sound enter into this?

 

My question as well.

As far as I know lloyd's mirror was an optical device to reduce the number of mirrors in fresnel's experiment from 1 to 2. Edit  from 2 to 1

Here is a description of a similar acoustic one that may answer your question.

Taken from The Student's Physics, volume ii  -  Acoustics.

Edited May 26, 2021 by studiot

[swansont]

58 minutes ago, henk jan said:

Yes, fringes form because of destructive and constructive interference, and I understand that if the waves are in the same phase its constructive, but I dont know ow to calculate the wavelength if you know the length between fringes

It’s a matter of geometry. The optical path length differs, and so there is a phase difference between the two beams. You need to calculate the path length difference and express it in terms of wavelengths

10 minutes ago, studiot said:

As far as I know lloyd's mirror was an optical device to reduce the number of mirrors in fresnel's experiment from 1 to 2.

Also a device that eliminates the single-slit diffraction effects of a Young’s double-slit apparatus.

[studiot]

4 minutes ago, swansont said:

15 minutes ago, studiot said:

As far as I know lloyd's mirror was an optical device to reduce the number of mirrors in fresnel's experiment from 1 to 2.

Also a device that eliminates the single-slit diffraction effects of a Young’s double-slit apparatus.

OOps

As far as I know lloyd's mirror was an optical device to reduce the number of mirrors in fresnel's experiment from 1 to 2.  from 2 to 1.

[henk jan]

Ok so now I have calculated the path length difference between two fringes, that should give the wavelength, right? But it gives answers that are wrong, a speed of of sound of roughly 250 m/s where it should give about 343 m/s

[exchemist]

8 minutes ago, henk jan said:

Ok so now I have calculated the path length difference between two fringes, that should give the wavelength, right? But it gives answers that are wrong, a speed of of sound of roughly 250 m/s where it should give about 343 m/s

What about the π phase shift of the reflected wave. Have you taken that into consideration when working out where the maxima and minima will be? 

[studiot]

20 minutes ago, henk jan said:

Ok so now I have calculated the path length difference between two fringes, that should give the wavelength, right? But it gives answers that are wrong, a speed of of sound of roughly 250 m/s where it should give about 343 m/s

A proper (but short) description of your experiment and calculation would go a long way to explaining the issue.

Until that happens we are just guessing blind.

[henk jan]

21 minutes ago, exchemist said:

What about the π phase shift of the reflected wave. Have you taken that into consideration when working out where the maxima and minima will be? 

But a pi phase shift will only mean half a wavelength, so multiplying by two would give me a speed of sound of 500m/s right

9 minutes ago, studiot said:

A proper (but short) description of your experiment and calculation would go a long way to explaining the issue.

Until that happens we are just guessing blind.

Sorry, we have a sound transmitter with a frequency of 40 kHz  and a sound receiver , with 40 cm between them. We have a sound mirror (just a steel plate) on a moving axis perpendicular to the axis of the transmitter and receiver (optical axis). As the mirror moves towards optical axis, the receiver will take measurements and show these on a graph. This gives a graph with nodes and antinodes. Now we have to calculate the speed of sound using this graph. We tried calculating the path length difference between two antinodes, but this gives the wrong result

I have put a picture of the setup, here bron means source and ontvanger means receiver

Edited June 4, 2021 by henk jan

[swansont]

Show your calculation, please

[studiot]

53 minutes ago, henk jan said:

Sorry, we have a sound transmitter with a frequency of 40 kHz  and a sound receiver , with 40 cm between them. We have a sound mirror (just a steel plate) on a moving axis perpendicular to the axis of the transmitter and receiver (optical axis). As the mirror moves towards optical axis, the receiver will take measurements and show these on a graph. This gives a graph with nodes and antinodes. Now we have to calculate the speed of sound using this graph. We tried calculating the path length difference between two antinodes, but this gives the wrong result

I have put a picture of the setup, here bron means source and ontvanger means receiver

Excellent, thank you.  +1

I suggest that you use 2d for the distance between Bron and Ontvanger as this makes the maths simpler.

What formula are you using for path length difference that is what is the relation between x and d and path length ?

and what does the plot of intensity against x  from your machine show (can you show a picture ?)

[henk jan]

9 minutes ago, swansont said:

Show your calculation, please

2*sqrt((d/2)^2+a^2)-2*sqrt((d/2)^2+b^2) . With d the distance between transmitter and receiver, a is x at an antinode and b is x at the next antinode

5 minutes ago, studiot said:

Excellent, thank you.  +1

I suggest that you use 2d for the distance between Bron and Ontvanger as this makes the maths simpler.

What formula are you using for path length difference that is what is the relation between x and d and path length ?

and what does the plot of intensity against x  from your machine show (can you show a picture ?)

I am using the formula as stated above and this is the graph that we got, but I am primarily using the numbers from the data set, instead of the actual graph

[henk jan]

28 minutes ago, swansont said:

How are you getting from this to the speed of sound?

I thought the path length difference had to be 1 wave length so I multiplied this with the frequency of 40 kHz, this gave me an average speed of sound of around 250 m/s

[studiot]

2 hours ago, henk jan said:

2*sqrt((d/2)^2+a^2)-2*sqrt((d/2)^2+b^2) . With d the distance between transmitter and receiver, a is x at an antinode and b is x at the next antinode

How is this correct ?

 

What two waves are interfering to produce the nodes and antinodes ?

The sound mirror is at position a or position b, but not both at the same time so how can these interfere ?

Surely the difference in path length is between the direct wave ie d (if you must but I still say 2d is easier) and the indirect wave which is reflected off the sound mirror.

Now exchemist has asked a very good question.

What is the length of the indirect wave ?

Well is it just the distance from Bron to the reflector plus the distance from the reflector to Ontvanger ?

Or does any phase change at the solid boundary (metal sheet) make a difference ?

The article I posted for you describes this as does the sound wave section in Wikipedia, which you should be able to get in Dutch.

https://en.wikipedia.org/wiki/Reflection_phase_change

If so what difference ?

Bear in mind that a phase change can be related to a distance change via the wavelength.

Because you are measuring distances you should be working in terms of wavelengths.

Remember that a phase change of π (distance = λ/2) represents destructive interference ie a minimum and a phase change of 2π represents constructive interference (distance  = λ)

 

[exchemist]

3 minutes ago, studiot said:

How is this correct ?

 

What two waves are interfering to produce the nodes and antinodes ?

The sound mirror is at position a or position b, but not both at the same time so how can these interfere ?

Surely the difference in path length is between the direct wave ie d (if you must but I still say 2d is easier) and the indirect wave which is reflected off the sound mirror.

Now exchemist has asked a very good question.

What is the length of the indirect wave ?

Well is it just the distance from Bron to the reflector plus the distance from the reflector to Ontvanger ?

Or does any phase change at the solid boundary (metal sheet) make a difference ?

The article I posted for you describes this as does the sound wave section in Wikipedia, which you should be able to get in Dutch.

https://en.wikipedia.org/wiki/Reflection_phase_change

If so what difference ?

Bear in mind that a phase change can be related to a distance change via the wavelength.

Because you are measuring distances you should be working in terms of wavelengths.

Remember that a phase change of π (distance = λ/2) represents destructive interference ie a minimum and a phase change of 2π represents constructive interference (distance  = λ)

 

Actually I have more or less persuaded myself that the phase change on reflection is irrelevant, after all.

If one is just measuring the change in path length between one peak (or trough) and the next, it doesn't seem to matter: the phase shift simply turns what would otherwise have been peaks into troughs and vice versa, but without altering their spacing relative to one another. 

 

[studiot]

8 minutes ago, exchemist said:

Actually I have more or less persuaded myself that the phase change on reflection is irrelevant, after all.

If one is just measuring the change in path length between one peak (or trough) and the next, it doesn't seem to matter: the phase shift simply turns what would otherwise have been peaks into troughs and vice versa, but without altering their spacing relative to one another. 

 

 

Is there a phase change for sound ??

Yes for light

Quote

Wikipedia

Sound waves

 

 

ound waves in a solid experience a phase reversal (a 180° change) when they reflect from a boundary with air.^[2] Sound waves in air do not experience a phase change when they reflect from a solid, but they do exhibit a 180° change when reflecting from a region with lower acoustic impedance. An example of this is when a sound wave in a hollow tube encounters the open end of the tube. The phase change on reflection is important in the physics of wind instruments.

 

[exchemist]

9 minutes ago, studiot said:

 

Is there a phase change for sound ??

Yes for light

 

Aha! I suppose that makes sense, when one thinks about what happens during reflection. Thanks for the correction. 

This problem is fun! But I must break off, as I am going to watch Carlo Rovellis' Oxford webinar on this history and philosophy of science....

[swansont]

1 hour ago, studiot said:

What two waves are interfering to produce the nodes and antinodes ?

The sound mirror is at position a or position b, but not both at the same time so how can these interfere ?

Surely the difference in path length is between the direct wave ie d (if you must but I still say 2d is easier) and the indirect wave which is reflected off the sound mirror.

Include a “d” term for both and it may make more sense. It will cancel, of course. The change in the path difference is entirely from the upper path.

3 hours ago, henk jan said:

I thought the path length difference had to be 1 wave length so I multiplied this with the frequency of 40 kHz, this gave me an average speed of sound of around 250 m/s

You could do this at 1 Hz and the path difference would be the same, wouldn’t it? I don’t see how the speed of the mirror ties in with the data you’ve presented.

[studiot]

15 hours ago, swansont said:

17 hours ago, studiot said:

 

Include a “d” term for both and it may make more sense. It will cancel, of course. The change in the path difference is entirely from the upper path.

Yes perhaps I was making things too complicated.

@henk jan

I really think your results are OK.

Have you counted peaks correctly ?

I make your estimate of the sound velocity 368 m/s

21 hours ago, henk jan said:

Ok so now I have calculated the path length difference between two fringes, that should give the wavelength, right? But it gives answers that are wrong, a speed of of sound of roughly 250 m/s where it should give about 343 m/s

If you look carefully, there is a big variation in apparent wavelength close to the mirror and a weak waveform at far distance as the pattern dies away.

So I have chosen the 13 minima from the middle section to provide an average over 12 wavelengths.

Also how accurate do you think the frequency of your sound is ?

Does this now help ?

 

[henk jan]

On 6/5/2021 at 10:33 AM, studiot said:

Yes perhaps I was making things too complicated.

@henk jan

I really think your results are OK.

Have you counted peaks correctly ?

I make your estimate of the sound velocity 368 m/s

If you look carefully, there is a big variation in apparent wavelength close to the mirror and a weak waveform at far distance as the pattern dies away.

So I have chosen the 13 minima from the middle section to provide an average over 12 wavelengths.

Also how accurate do you think the frequency of your sound is ?

Does this now help ?

 

Oh this helps a lot, thank you so much