Why Does Diffraction Occur Even When Wavelength Exceeds Aperture Width?

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Diffraction occurs even when the wavelength is smaller than the aperture width, contrary to the common misconception that it only happens in such cases. The equation sin(theta) = wavelength/width opening is used to determine the angles of dark fringes in a single-slit diffraction pattern, but it becomes invalid when the wavelength is larger than the slit width, as the central maxima fills the entire area. As the width of the slit decreases or the wavelength increases, diffraction effects become more pronounced, leading to a broader spread of light. The phase differences in light waves traveling through the slit determine constructive and destructive interference, with significant diffraction observed even at small angles. Understanding these principles clarifies why diffraction patterns can still be observed under various conditions.
  • #31
Ok so here's an example, if the wavelength is less than the width opening(in a tank let's say), would you be able to see the curvature of the waves? Also, after that, let's say the wavelength was larger than(or equal to) the width of the opening, what would the curves look like then?
 
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  • #32
Byrgg said:
Ok so here's an example, if the wavelength is less than the width opening(in a tank let's say), would you be able to see the curvature of the waves? Also, after that, let's say the wavelength was larger than(or equal to) the width of the opening, what would the curves look like then?
http://www.ngsir.netfirms.com/englishhtm/Diffraction.htm" is a nice applet that illustrates the phenomenon.
 
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  • #33
The applet isn't working for some reason.
 
  • #34
Byrgg said:
The applet isn't working for some reason.
I think you need Java installed. To asnwer you question: If the wavelength was significantly less than the size of the opening then very little diffraction would occur, you may see some small curvature about the edges of the opening. If the wavelength was greater than or equal to the slit, you would see significant diffraction, the smaller the opening when compared with the wavelength the smaller the radius of the curve.
 
  • #35
Ok, just to make sure I understand this well enough, could you tell me if I'm right?

When the wavelength is less than the width of the opening, the pattern of bright/dark areas is formed, and the wave appears to be(if it can be seen clearly) a straight line with curved edges. When the wavelength is equal to or greater than the width of the opening, there is no pattern of bright/dark areas, and the wave is basically just a curve?

Is that right, or did I say something incorrect?
 
  • #36
Byrgg said:
Ok, just to make sure I understand this well enough, could you tell me if I'm right?

When the wavelength is less than the width of the opening, the pattern of bright/dark areas is formed, and the wave appears to be(if it can be seen clearly) a straight line with curved edges. When the wavelength is equal to or greater than the width of the opening, there is no pattern of bright/dark areas, and the wave is basically just a curve?

Is that right, or did I say something incorrect?
Sounds like a good summary to me.
 
  • #37
Ok, if the wavelength is less than the width of the opening, then will the edges always be curved, with the middle being a straight line? Or when the wavelength is almost as large as the width of the opening, will it be just one curve?

Also, do the curves always extend all the way back to the wall from the line bisecting the opening(forming a quarter of a circle)?
 
  • #38
Please someone answer.
 
  • #39
Byrgg said:
Ok, if the wavelength is less than the width of the opening, then will the edges always be curved, with the middle being a straight line? Or when the wavelength is almost as large as the width of the opening, will it be just one curve?
Yes, roughly speaking this is the case, there may still be some diffraction so that the 'centre' wavefronts are not completely linear and may have a slight curve but it if the gap was significantly bigger, then the centre wavefronts would be linear.

If \lambda\approx a, then yes, the wavefront would appear circular.
Byrgg said:
Also, do the curves always extend all the way back to the wall from the line bisecting the opening(forming a quarter of a circle)?
Yes, this is correct. The waves would appear semi-circular if \lambda\approx a with the centre of the circule located at the centre of a. Quite a good reference for diffraction as explained by Hugens pricinple using wavefronts is available http://physics.tamuk.edu/~suson/html/4323/diffract.html"
 
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  • #40
Ok, so then if the gap is significantly bigger than the wavelength, do the curves still make the quarter of a circle at the edges?

Basically, if the wave is linear, with the edges curved, do those edges reach back to the wall making the mentioned quarter of a circle?
 
  • #41
Byrgg said:
Ok, so then if the gap is significantly bigger than the wavelength, do the curves still make the quarter of a circle at the edges?

Basically, if the wave is linear, with the edges curved, do those edges reach back to the wall making the mentioned quarter of a circle?
Yes, the curves are approximatly circular and extend all the way back to the 'wall' in which the slit is cut. Perhaps these images may help (taken from the same site as the applet);
Slit width is smaller than or comparable to wavelength
http://www.ngsir.netfirms.com/applets/diffraction/diffraction2.png

Slit width is significantly geater than wavelength
http://www.ngsir.netfirms.com/applets/diffraction/diffraction1.png
 
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  • #42
Alright, just another confirmation, let's say the wavelength is larger than the slit, the diffraction is 90 degrees right? This means the waves cannot be diffracted any more, correct? So does increasing the wavelength after this point have any other effect? I think you may have mentioned this before, but could you tell me what happens in this case?
 
  • #43
Byrgg said:
Alright, just another confirmation, let's say the wavelength is larger than the slit, the diffraction is 90 degrees right? This means the waves cannot be diffracted any more, correct? So does increasing the wavelength after this point have any other effect? I think you may have mentioned this before, but could you tell me what happens in this case?
I think you may be confusing diffraction and interfernce due to diffraction here, probably my fault for not clarifying the matter earlier. The angle we were referring to previously (using Fraunhofer defraction) is the angle at which the minima occur due to the interence of diffracted waves. Now, that maxium angle this can [theoretically] occur at is 90o. It is possible however, that waves can be diffracted by an angle greater than 90o. I would recommed reading around hugene's principle.
 
  • #44
When the wavelength is larger than the slit, the "central maxima" of the intensity pattern extends to the limit, which is 90 degrees* (measured from the direction that the wave is traveling). There is essentially no dark fringe, but the intensity does vary. (It's a bit involved, but here's a link if you want to get into the nitty gritty: http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/sinint.html#c1)

90 degrees is the limit, in the case of single slit diffraction, for a very simple reason: past 90 degrees there's a barrier that blocks the wave! :smile:
 
  • #45
"90 degrees is the limit, in the case of single slit diffraction, for a very simple reason: past 90 degrees there's a barrier that blocks the wave!"

Ok that's what I thought, but what about when hootenanny said this:

"It is possible however, that waves can be diffracted by an angle greater than 90o. I would recommed reading around hugene's principle."

So in a single slit, could I assume that the maximum diffraction is 90 degrees? Then would hootenanny's post apply to non-single slit diffraction cases only?

Also, since I'm only in gr.11 if calculating the diffraction in these other cases isn't overly complicated, could someone show me how, or at least give a rough summary? Oh and, I should be able to understand up to gr. 12material, seeing as gr. 11 is basically done and I have a good mark.
 
  • #46
Someone? Please?
 
  • #47
Oh, and in addition, is 90 degrees the max angle for the minima? I'm sure this is true for single-slit, but does this change for non-single slit?
 
  • #48
Byrgg said:
So in a single slit, could I assume that the maximum diffraction is 90 degrees?
Since the material out of which the slit has been cut extends to each side, how could any light be found at an angle greater than 90 degrees? Yes, the maximum angle at which light could possibly be diffracted is 90 degrees for that simple reason.
Then would hootenanny's post apply to non-single slit diffraction cases only?
Right.

Also, since I'm only in gr.11 if calculating the diffraction in these other cases isn't overly complicated, could someone show me how, or at least give a rough summary? Oh and, I should be able to understand up to gr. 12material, seeing as gr. 11 is basically done and I have a good mark.
I applaud your moxie, but until you learn to actually calculate the single slit case--which is not so easy!--I wouldn't worry about other diffraction geometries. (By all means read up on those diffraction patterns--slit, edge, circular hole, disk--but don't think you'll be able to actually calculate those patterns without a lot of study.) The hyperphysics site has some good material. (Check the links I gave before and poke around.)

My recommendations:
(1) Heed Hoot's advice to learn about Huygen's principle.
(2) Study Young's double slit experiment (if you haven't already)--much easier to analyze and it will give you a good understanding of interference. And that will give you a head start in understanding single slit diffraction.
(3) Learn how that single slit equation in your first post is derived.
(4) If you still want more, grab yourself a copy of a decent college physics text. (Halliday and Resnick is standard.) You'll soon see that even at the (first year) college level that single slit diffraction is only handled up to a certain point. For example, understanding the derivation of that equation from your first post in this thread is usually as far as you get in some courses. Calculating the intensity pattern--via the method of phasors--is usually done only in higher level courses. (Depends on the instructor and the text.)
 
  • #49
Ok, well I don't know if I'll go around looking all for answers to everything right now, but what about my last question?

In a non-single slit case, is the maximum angle for the minima still 90 degrees?

Oh yeah, and in a single-slit scenario, is the angle of diffraction the same as the angle of the first minima? Is it ever the same as the first minima?
 
  • #50
Byrgg said:
In a non-single slit case, is the maximum angle for the minima still 90 degrees?

When you have diffraction through an aperture, or a set of apertures in a plane barrier, you're usually interested in the light that makes it through to the other side. In this situation, the maximum angle of either a minimum or maximum must be 90 degrees. Greater than 90 degrees, and you're dealing with waves that have been reflected back on the same side of the barrier as the source.

In principle, I suppose one could analyze this situation using similar mathematical tools as with transmission of light through a barrier, but introductory textbooks don't cover this. In fact, I don't think I've seen it in an intermediate-level textbook, either.

One situation where people do discuss "backscattering" is with things like shining light on small targets such as dust grains or molecules. This is a rather advanced topic, though.

Oh yeah, and in a single-slit scenario, is the angle of diffraction the same as the angle of the first minima? Is it ever the same as the first minima?

The usual elementary formula for single-slit diffraction, n \lambda = a \sin \theta, where n is an integer > 0, gives the angles of minimum intensity of the diffracted light. If n = 1 it gives you the first minimum on either side of the center.
 
  • #51
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 \theta 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 \frac{n\lambda}{2} 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:
 
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  • #52
Ok so then would be accurate to say that diffraction doesn't have anyone specific angle? It occurs in the areas between minima(the maxima)?
 
  • #53
Byrgg said:
Ok so then would be accurate to say that diffraction doesn't have anyone 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 \frac{n\lambda}{2} where n is an odd positive integer that destructive interference occurs.
 
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  • #54
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
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;

y = \tan\theta d

Now, if a>>\lambda, then \theta will be small. Therefore, we can use the small angle approximation \tan\theta\approx\sin\theta\approx\theta (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;

\sin\theta = \frac{m\lambda}{a}

and the previous equation;

y = \tan\theta d \Leftrightarrow \tan\theta = \frac{y}{d}

And applying the approximation \sin\theta \approx \tan\theta[/tex], we obtain;<br /> <br /> \frac{y}{d} = \tan\theta \approx \frac{m\lambda}{a}<br /> <br /> 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;<br /> <br /> y \approx \frac{md\lambda}{a}<br /> <br /> Note that this only holds when the angle of diffraction is small, i.e. when a&amp;gt;&amp;gt;\lambda
 
  • #56
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
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
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
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.
 
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