I'm new here and I'm wondering something about diffraction of waves

  • Thread starter Byrgg
  • Start date
  • #51
Hootenanny
Staff Emeritus
Science Advisor
Gold Member
9,622
6
Yes, sorry Byrgg I sould have stated that earlier any kind of diffraction through an appature (single / double / multiple slits, diffraction grating etc)has a maximum angle of diffraction of 90o, due to a physical barrier. I will emphises again (as jtbell has done) that [itex]\theta[/itex] in the discussed formula gives the angle at which a minimum occurs due to interference of diffrated waves. Waves are diffracted at all angles between 0 and the maximum, the discussed formulae give the angles at which two diffracted waves interefere distructively, i.e. when then have a phase diffrence of [itex]\frac{n\lambda}{2}[/itex] where n is an odd postive integer. As for diffraction of angles greater than 900, I have only ever seen these treated qualatively never quantatively, the example I remember was light incident on a sphere of very small radius. I hope this clears things up for you.

As Doc Al said, I am very impressed with the effort you have put into learning this material and you are obviously highly motivated with a kean interest in physics. Keep it going :smile:
 
Last edited:
  • #52
335
0
Ok so then would be accurate to say that diffraction doesn't have any one specific angle? It occurs in the areas between minima(the maxima)?
 
  • #53
Hootenanny
Staff Emeritus
Science Advisor
Gold Member
9,622
6
Byrgg said:
Ok so then would be accurate to say that diffraction doesn't have any one specific angle? It occurs in the areas between minima(the maxima)?
Yes, it occurs at all angles full stop. As I said in my previous post;
Waves are diffracted at all angles between 0 and the maximum...
If you like each ray is diffracted at a different angle, resuting in a different path lengths. It is when the diffrence in pathlengths is equal to [itex]\frac{n\lambda}{2}[/itex] where n is an odd positive integer that destructive interference occurs.
 
Last edited:
  • #54
335
0
Ok, I think I'm understanding most of this a little better now, but I have a few more questions, they're sort of geometric in nature...

Let's say the gap is much wider than than the wavelength, the wave coming from the gap is linear in the middle, with curvature around the edges, right? That is the appearance of wave I believe, as could easily be seen in a ripple tank. Geometrically speaking, can you roughly estimate where the first minima occur? Or, do you have to calculate the angle, then use a protractor to accurately draw the locations of the minima?

Also, in another example, say the wall was sort of v-shaped, like this: / \

and the wave came from below, would the edges of the wave extend around and past 90 degrees to reach the wall? Would that also allow the minima the occur at an angle greater than 90 degrees?

If I need to, I can try to make some detailed diagrams in ms paint and put them here, I can do that right?
 
  • #55
Hootenanny
Staff Emeritus
Science Advisor
Gold Member
9,622
6
Okay, assuming we are talking about single slit Fraunhofer diffraction, then the vertical displacement (y) of the first minima from the centre line of the slit (I presume that is what you are talking about) on a screen placed d meters away, is given by;

[tex]y = \tan\theta d[/tex]

Now, if [itex]a>>\lambda[/itex], then [itex]\theta[/itex] will be small. Therefore, we can use the small angle approximation [itex]\tan\theta\approx\sin\theta\approx\theta[/itex] (think about drawing a triangle with a very small angle, the opposite and adjacent sides will be approximatly the same length). Now, taking Fraunhofer's single slit relationship;

[tex]\sin\theta = \frac{m\lambda}{a}[/tex]

and the previous equation;

[tex]y = \tan\theta d \Leftrightarrow \tan\theta = \frac{y}{d}[/tex]

And applying the approximation [itex]\sin\theta \approx \tan\theta[/tex], we obtain;

[tex]\frac{y}{d} = \tan\theta \approx \frac{m\lambda}{a}[/tex]

Thus we obtain an expression for finding the vertical displacement of a minima from the centre line on a screen placed d metres away from the appature;

[tex]y \approx \frac{md\lambda}{a}[/tex]

Note that this only holds when the angle of diffraction is small, i.e. when [itex]a>>\lambda[/itex]
 
  • #56
335
0
Ok but what about my v-shaped question? If allowed(there being no physical barrier), would the waves be able diffract more(beyond 90 degrees)?

Also, could you clear this up for me? I'm thinking that in a typical single-slit situation, the actual diffraction would be 90 degrees no matter what. The varying angle we think of is only the angle at which the first minima occurs correct? In a sense we percieve the extent of diffraction as the angle at which the first minima appears. When a < wavelength, we see it as completely bending around the object, when in reality, the waves always bend around the object even when a > wavelength(I believe it was mentioned earlier that the curves of the wave always reaches back to the barrier). We just think of it differently when a < wavelength because there is no minima - a geometrical shadow as I read it was called I think.

So yeah, if you could answer the first part and let me know if my long "theory" sort of thing is correct I'd be very appreciative.
 
  • #57
335
0
Oh and I should mention that my "theory" comes from the fact the waves still pass into the "dark" areas, they just destructively interfere at those points.
 
  • #58
Hootenanny
Staff Emeritus
Science Advisor
Gold Member
9,622
6
Byrgg said:
Ok but what about my v-shaped question? If allowed(there being no physical barrier), would the waves be able diffract more(beyond 90 degrees)?
I cannot say for definate, but I see no reason why they shouldn't, like I said before, I have only seen this type of diffraction treated qualatively.
Byrgg said:
Also, could you clear this up for me? I'm thinking that in a typical single-slit situation, the actual diffraction would be 90 degrees no matter what. The varying angle we think of is only the angle at which the first minima occurs correct? In a sense we percieve the extent of diffraction as the angle at which the first minima appears. When a < wavelength, we see it as completely bending around the object, when in reality, the waves always bend around the object even when a > wavelength(I believe it was mentioned earlier that the curves of the wave always reaches back to the barrier). We just think of it differently when a < wavelength because there is no minima - a geometrical shadow as I read it was called I think.

So yeah, if you could answer the first part and let me know if my long "theory" sort of thing is correct I'd be very appreciative.
Byrgg said:
Oh and I should mention that my "theory" comes from the fact the waves still pass into the "dark" areas, they just destructively interfere at those points
That is most definatly correct, I have been trying to explain this all they way through (obviously not very well:blushing:). You are absolutely correct.
 
  • #59
335
0
Wow! Seriously? Is that all really correct? You're not joking?

One more question, in multiple-slit diffraction, do the waves from one slit ever diffract into another one of the other slits?

Oh, and how do you use the greek letters and such? I think it would help me a bit if I knew how to use them.
 
Last edited:
  • #61
335
0
Ok thanks for the link, but anyone have an answer(s) to my last questions(in the same post)?
 
  • #62
335
0
Anyone? Please?
 
  • #63
335
0
Please? I think this is my last question.
 
  • #64
Hootenanny
Staff Emeritus
Science Advisor
Gold Member
9,622
6
Byrgg said:
Wow! Seriously? Is that all really correct? You're not joking?
No, I'm not joking :biggrin:
Byrgg said:
One more question, in multiple-slit diffraction, do the waves from one slit ever diffract into another one of the other slits?
Byrgg said:
but anyone have an answer(s) to my last questions(in the same post)??
Byrgg said:
Anyone? Please?
Byrgg said:
Please? I think this is my last question.
Patients if a virtue my friend, remember some of us are on different time zones :tongue2:. No, as said previously, with any appature (including double slit), the maximum angle of diffraction is 90o, so when the wavefront reaches the other slit it is travelling perpendicular to the slit, so no diffraction will occur at that point and the waves are not diffracted back into the slit. However, the waves do interfere in the area between the slits and between the slits and the screen.
 
Last edited:
  • #65
335
0
Ok, sorry about that, wasn't thinking of time zones. Is there any reason why it doesn't diffract through the other holes? Or does this not occur because the waves destructively interfere between gaps(do they destructively interfere there?). Even if there were multiple gaps, and only one had waves coming from it, it still wouldn't go through the other holes?
 
  • #66
Hootenanny
Staff Emeritus
Science Advisor
Gold Member
9,622
6
Byrgg said:
Ok, sorry about that, wasn't thinking of time zones. Is there any reason why it doesn't diffract through the other holes? Or does this not occur because the waves destructively interfere between gaps(do they destructively interfere there?). Even if there were multiple gaps, and only one had waves coming from it, it still wouldn't go through the other holes?
Take a look at http://micro.magnet.fsu.edu/primer/java/interference/doubleslit/doubleslitjavafigure1.jpg" [Broken], now the diffracted waves will br traveling either parallel to the slits or away from them. The waves do not diffract into the slits simply because there is nothing to diffracted through or around. Does that make sense?
 
Last edited by a moderator:
  • #67
335
0
Yeah, I think I get it now. They only diffract if they're going towards the gap or perpendicular right? Parallel wouldn't really do it I don't think... but then how exactly does diffraction around a corner work then? Say the waves were travelling along a barrier and came to a corner, that could be seen as a gap sort of couldn't it?, and don't the waves bend around corners?

Say a wave is moving forward, if it comes to a corner, then it continues forward, while diffracting around the corner as well right? The forward gap is perpendicular to the wave's direction, while the right/left gap is parallel. Is this explanation ok or should I try to make a diagram?
 
  • #68
Hootenanny
Staff Emeritus
Science Advisor
Gold Member
9,622
6
Byrgg said:
Yeah, I think I get it now. They only diffract if they're going towards the gap or perpendicular right? Parallel wouldn't really do it I don't think... but then how exactly does diffraction around a corner work then? Say the waves were travelling along a barrier and came to a corner, that could be seen as a gap sort of couldn't it?, and don't the waves bend around corners?

Say a wave is moving forward, if it comes to a corner, then it continues forward, while diffracting around the corner as well right? The forward gap is perpendicular to the wave's direction, while the right/left gap is parallel. Is this explanation ok or should I try to make a diagram?
When sound gets diffracted around a corner, a component of the wave will be travelling towards the slit, because the waves are radiating from a point source, they are not travelling completely perpendicular to the slit. http://hyperphysics.phy-astr.gsu.edu/hbase/sound/imgsou/difr2.gif" [Broken] Whereas in the double slit case, the source is in line with the slit and so the waves are either travelling away from the slit or perpendicular to it. Does that make sense?
 
Last edited by a moderator:
  • #69
335
0
Oh I think I see now, you said at a corner it still goes because it's partially going towards the slit right? You also mentioned that it's from a point source, that basically means it's radiating out and this is why it's partially aimed towards the slit right? If the point source was right in the wall(in theory) then would the waves still go around the corner?
 
Last edited:
  • #70
Hootenanny
Staff Emeritus
Science Advisor
Gold Member
9,622
6
Byrgg said:
Oh I think I see now, you said at a corner it still goes because it's partially going towards the slit right? You also mentioned that it's from a point source, that basically means it's radiating out and this is why it's partially aimed towards the slit right? If the point source was right in the wall(in theory) then would the waves still go around the corner?
Yeah thats it.
 
  • #71
335
0
Actually what I meant is if the source is right in the wall then isn't it sort of like already being diffracted through a single slit? The waves would be parallel(wavelength perpendicular) to the left/right slit. In this case the waves shouldn't pass through should they?
 

Related Threads on I'm new here and I'm wondering something about diffraction of waves

Replies
86
Views
7K
Replies
15
Views
2K
Replies
1
Views
3K
  • Last Post
Replies
6
Views
2K
  • Last Post
Replies
3
Views
658
Replies
4
Views
3K
  • Last Post
2
Replies
30
Views
4K
Top