Superposition of waves, result visible?


by A Dhingra
Tags: result, superposition, visible, waves
A Dhingra
A Dhingra is offline
#1
Jan26-13, 03:35 AM
P: 196
hi...
Got to ask about superpostion of waves...
When two coherent light waves cross each at some point in space moving in different directions, do they superimpose?
If they do, do we need a screen to be able to see the resultant or we can see it directly.
Let's just assume we have a laser and with some sort of mirror arrangement we make the incident ray meet the reflected ray. The point where they meet can we see constructive or destructive interference (assuming 100% reflection and no losses)?
Thanks for any help!
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DaleSpam
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#2
Jan26-13, 07:00 AM
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P: 16,466
Quote Quote by A Dhingra View Post
hi...
Got to ask about superpostion of waves...
When two coherent light waves cross each at some point in space moving in different directions, do they superimpose?
Yes.

Quote Quote by A Dhingra View Post
If they do, do we need a screen to be able to see the resultant or we can see it directly.
Generally you need a screen, but you could also use something like dust.

Quote Quote by A Dhingra View Post
The point where they meet can we see constructive or destructive interference (assuming 100% reflection and no losses)?
You will see both constructive and destructive interference.
A Dhingra
A Dhingra is offline
#3
Jan27-13, 01:53 PM
P: 196
Thanks..
Looks like I got what I wanted.

A Dhingra
A Dhingra is offline
#4
Jan30-13, 11:19 AM
P: 196

Superposition of waves, result visible?


can i ask one more thing here?

If i have a laser pointing towards the mirror, there will be superposition of light after it retraces its path. If a glass plate of negligible thickness is placed in between obliquely, then can we obtain the result of superposition in the form of interference fringe pattern. In this if we have assumed the light wave to be continuous (which it is for classical purposes) this pattern should be obtained in for any distance between the plate and the mirror. But there comes coherence length into picture. If light is continuous how does the path difference affect the result in any way??
DaleSpam
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#5
Jan30-13, 01:26 PM
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P: 16,466
I am not sure that I understand what you are asking.

Are you asking about the length of different paths within the coherence length or are you asking about what happens after you are outside of the coherence length or are you asking about what happens as you move from within the coherence length to outside it?
mickybob
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#6
Jan31-13, 05:13 AM
P: 34
Quote Quote by A Dhingra View Post
can i ask one more thing here?

If i have a laser pointing towards the mirror, there will be superposition of light after it retraces its path. If a glass plate of negligible thickness is placed in between obliquely, then can we obtain the result of superposition in the form of interference fringe pattern. In this if we have assumed the light wave to be continuous (which it is for classical purposes) this pattern should be obtained in for any distance between the plate and the mirror. But there comes coherence length into picture. If light is continuous how does the path difference affect the result in any way??
I'm not sure quite what you mean by 'continuous'. If you means that the light could be considered to be a perfect sine wave of infinite length, then the coherence length would also be infinitely long, so you would always see interference. In reality you can't have an infinite coherence length, of course.

If you build an interferometer then you are splitting light into two arms and then recombining it. Roughly speaking, if the difference in the optical path lengths between the two arms is shorter than the coherence length then you will see interference - if it's longer then you won't. In practice it's not a sharp cut-off of course - it has a profile which is related to the Fourier transform of the source spectrum.

If you're asking why the coherence length affects whether or not you see interference then I'm happy to explain that, although it involves either some maths or a fairly conceptually difficult written explanation.
A Dhingra
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#7
Jan31-13, 08:04 AM
P: 196
Quote Quote by mickybob View Post
I'm not sure quite what you mean by 'continuous'. If you means that the light could be considered to be a perfect sine wave of infinite length, then the coherence length would also be infinitely long, so you would always see interference. In reality you can't have an infinite coherence length, of course.
I got this. And by continuous i had assumed light to be a perfect sine wave not at all affected by anything around it, so i got my answer.

Quote Quote by mickybob View Post
If you're asking why the coherence length affects whether or not you see interference then I'm happy to explain that, although it involves either some maths or a fairly conceptually difficult written explanation.
About this, if i have understood correctly, coherence length is that length for which the phase fluctuation of the source can be safely taken as a constant. In absence of such a coherence interference pattern will not be stationary with time, hence no evident interference pattern is found. Actually, i have taken this statement for granted and didn't question it! Is this it, or there is more to the theory? If the concepts can be extended any further, i would really like to see them.


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