Why Does Light Diffract?

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In summary, diffraction effects occur when waves travel through a small hole or slid. They diffract, or spread out, and their intensity varies depending on the size and spacing of the holes. This is because interference between waves occurs, and the width of the slit doesn't affect the diffraction. The Heisenberg uncertainty principle is responsible for the diffraction effects.
  • #1
scilover89
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Why waves, such as light, will diffract when they pass through a small hole or slid?
 
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  • #2
scilover89 said:
Why waves, such as light, will diffract when they pass through a small hole or slid?

It's because of interference between the light that goes through different parts of the slit or hole.

Remember two-slit interference? You add two waves (with different phases) together and they interfere. Now imagine dividing a single slit into lots of tiny "sub-slits," and add up the waves coming from each of those sub-slits. Each of those waves has a slightly different phase. Take the limit as the number of sub-slits becomes infinite, and the width of each slit becomes infinitesimally tiny. You end up with an integral that gives you the overall intensity.

Note that when we do two-slit interference, we usually start by assuming that the slits are very narrow so that their width doesn't matter, only the spacing between them. When we do take the slit width into account, we get a combination of the single-slit and double-slit interference patterns: the maxima in the double-slit pattern have different intensities, in a pattern that matches the single-slit pattern.
 
  • #3
jtbell said:
It's because of interference between the light that goes through different parts of the slit or hole.

Remember two-slit interference? You add two waves (with different phases) together and they interfere. Now imagine dividing a single slit into lots of tiny "sub-slits," and add up the waves coming from each of those sub-slits. Each of those waves has a slightly different phase. Take the limit as the number of sub-slits becomes infinite, and the width of each slit becomes infinitesimally tiny. You end up with an integral that gives you the overall intensity.

Note that when we do two-slit interference, we usually start by assuming that the slits are very narrow so that their width doesn't matter, only the spacing between them. When we do take the slit width into account, we get a combination of the single-slit and double-slit interference patterns: the maxima in the double-slit pattern have different intensities, in a pattern that matches the single-slit pattern.

Actually, this doesn't answer the question. If what you said is true, then how come you don't see such effects when the slit is WIDER? After all, why can't you still divide the slit into many, many more smaller slits? Why would the diffraction effects go away?

Note that a diffraction effect is really the "broadening" of the beam path. Within such broadening, you can get diffraction patterns. I've explained why this really is a manifestation of the uncertainty principle! Rather than repeat everything that I've written, I will simply refer to my journal entry below:

[11-15-2004 09:26 AM] - Misconception of the Heisenberg Uncertainty Principle

So diffraction is caused by the HUP. What causes the HUP? The 1st Quantization rule of QM that deals with non-commuting operators. What causes the 1st Quantization rule of QM? That, we don't quite know, and part of QM's postulate.

Zz.
 
  • #4
Well Zz,how come the classical theory of LIGHT DIFFRACTION (you know,the one made by Fresnel,Fraunhofer,Kirchhoff and Sommerfeld) is both correct & elegant,while the one proposed by Feynman (with those photons taking a bunch of paths) is horrible and noninteligeable...?

I'll say diffraction is a wave-like phenomenon and applying to it the famous duality,though correct,would complicate things beyond one's power of comprehension.It would be interesting to view water waves' diffraction through the eyes of the HUP...

The diffraction is so nice with Fresnel integrals and Bessel functions,why bring in operators and propagators...?

Daniel.
 
  • #5
dextercioby said:
Well Zz,how come the classical theory of LIGHT DIFFRACTION (you know,the one made by Fresnel,Fraunhofer,Kirchhoff and Sommerfeld) is both correct & elegant,while the one proposed by Feynman (with those photons taking a bunch of paths) is horrible and noninteligeable...?

I'll say diffraction is a wave-like phenomenon and applying to it the famous duality,though correct,would complicate things beyond one's power of comprehension.It would be interesting to view water waves' diffraction through the eyes of the HUP...

The diffraction is so nice with Fresnel integrals and Bessel functions,why bring in operators and propagators...?

Daniel.

No, no. You read way too much into what I just said. I said it can be EXPLAINED by the HUP. I didn't say anyone in their right mind would USE it. It is the same with the 2-slit expt. I can show you how to do that using photons, but why do that when the wave picture is so much more convenient for most purposes?

This is why we continue to treat EM radiation as waves in many cases. But the diffraction from a single slit is the CLEAREST example of the HUP, something that many people simply do not realize. Such effects are not just restricted to microscopic, atomic scale.

Zz.
 
  • #6
A wave (light, sound, water, etc.) can always be thought of as constantly "regenerating" itself from every point.

If at some instant, the wave front is a plane, then Drawing an sphere out from each point with a fixed radius (representing the distance light would travel from that point) gives an "envelope" which is another plane parallel to the original one- the wave front at that later time.

If the wave is constrained to go through a pin pointopening (or a very small opening compared to the wave-length), the wave front will become a sphere centerd on that point.

If a plane wave meets a barrier with two pinpoint openings, the wave front on the other side of the barrier will be two spherical waves centered on the two points which can interfere with one another.

I remember doing that with water waves in high school physics.
 
  • #7
Yes,Halls,that picture of Huygens' principle is really nice.Though i'd still vote for Kirchhoff's integral...:wink:

Daniel.
 
  • #8
ZapperZ said:
Actually, this doesn't answer the question. If what you said is true, then how come you don't see such effects when the slit is WIDER? After all, why can't you still divide the slit into many, many more smaller slits?

You still can!

Why would the diffraction effects go away?

As the slit becomes wider and wider, the "diffraction effects" become smaller and smaller relative to the overall intensity at each point on the screen. But they're still there, in principle. Try putting a point source behind a large aperture, and a screen close to the aperture. Look closely and you'll see diffraction effects at the edge of the shadow of the aperture's edge. They still exist in the middle of the "beam", but they're imperceptibly tiny.

You have to be careful when doing the integral in such a situation, though. When we derive the standard single-slit diffraction formula, we use the Fraunhofer approximation, which assumes that the rays from various points in the slit to the image point on the screen are practically parallel, and that all points in the slit are equidistant from a point source. If the slit is wide enough, this approximation breaks down.

We can do better with the Fresnel approximation. Look in an optics book or on the Web, and you can probably find some nifty diffraction patterns generated that way. Try searching for "Cornu spiral", which is a graphical device to aid in generating such patterns. Once upon a time, I did some that recognizably approach the "no-diffaction" regime.

And finally for the ultimate accuracy and generality, do the integral numerically, customized for the specific source, slit and screen configuration.
 
  • #9
For large apertures it's almost the same as "edge diffraction" (semiinfinite plane,check chapterin Born & Wolf).

Daniel.
 
  • #10
jtbell said:
You still can!



As the slit becomes wider and wider, the "diffraction effects" become smaller and smaller relative to the overall intensity at each point on the screen. But they're still there, in principle. Try putting a point source behind a large aperture, and a screen close to the aperture. Look closely and you'll see diffraction effects at the edge of the shadow of the aperture's edge. They still exist in the middle of the "beam", but they're imperceptibly tiny.

You have to be careful when doing the integral in such a situation, though. When we derive the standard single-slit diffraction formula, we use the Fraunhofer approximation, which assumes that the rays from various points in the slit to the image point on the screen are practically parallel, and that all points in the slit are equidistant from a point source. If the slit is wide enough, this approximation breaks down.

We can do better with the Fresnel approximation. Look in an optics book or on the Web, and you can probably find some nifty diffraction patterns generated that way. Try searching for "Cornu spiral", which is a graphical device to aid in generating such patterns. Once upon a time, I did some that recognizably approach the "no-diffaction" regime.

And finally for the ultimate accuracy and generality, do the integral numerically, customized for the specific source, slit and screen configuration.

But I think that's the point I was trying to make. We can't just leave the explanation as "many point sources in the slit". This by itself cannot explain why the diffraction pattern gets wider as we make the slit smaller. Unfortunately, once we try to go beyond that, then we run into the things you mentioned and the explanation gets highly mathematical. I don't see a simple and intuitive way of doing that.

Zz.
 
  • #11
ZapperZ said:
We can't just leave the explanation as "many point sources in the slit".

At a fundamental level, at least in terms of the Huygens-Fresnel diffraction theory, that's really all it is. More precisely, if the slit is illuminated by a point source, the basic idea is "many paths from the source to the point P on the screen," each path connecting the source and one sub-slit with a straight line, and the sub-slit and point P with another straight line. Actually, we should use the term "sub-aperture" rather than sub-slit, because this formulation applies also to two-dimensional apertures.

For example, see Hecht's Optics, which starts its discussion of the rectangular aperture on p. 497 with the contribution to the field at P, from a single sub-aperture:

[tex]dE_P = \frac {K(\theta) E_0} {\rho r \lambda} \cos [k(\rho + r) - \omega t] dS[/tex]

where E_0/\lambda is a constant related to the intensity of the source, \rho is the distance from the source to the sub-aperture, r is the distance from the sub-aperture to P, and K(\theta) is Kirchhoff's obliquity factor

[tex]K(\theta) = (1+\cos \theta) / 2[/tex]

where \theta is the angle between the two lines described above, which prevents the aperture from producing a wave that goes backwards towards the source. The \rho and r in the denominator produce an inverse-square falloff of intensity with distance, going from the source to the sub-aperture, and from the sub-aperture to P, as we should expect.

This is equivalent to having the sub-apertures as our "original" sources, producing waves with different amplitudes and phases, from different locations within the aperture.

Starting from here, the only difference between different situations (single slit, double slit, circular aperture, rectangular aperture, whatever) is the geometry of integration and the approximations that one uses to make solving the integral less complicated.

At this level, the only 'ad-hoc' thing in the expression above is the obliquity factor. But it emerges naturally when you do a rigorous derivation starting from the differential wave equation.
 
  • #12
jtbell said:
At a fundamental level, at least in terms of the Huygens-Fresnel diffraction theory, that's really all it is. More precisely, if the slit is illuminated by a point source, the basic idea is "many paths from the source to the point P on the screen," each path connecting the source and one sub-slit with a straight line, and the sub-slit and point P with another straight line. Actually, we should use the term "sub-aperture" rather than sub-slit, because this formulation applies also to two-dimensional apertures.

For example, see Hecht's Optics, which starts its discussion of the rectangular aperture on p. 497 with the contribution to the field at P, from a single sub-aperture:

[tex]dE_P = \frac {K(\theta) E_0} {\rho r \lambda} \cos [k(\rho + r) - \omega t] dS[/tex]

where E_0/\lambda is a constant related to the intensity of the source, \rho is the distance from the source to the sub-aperture, r is the distance from the sub-aperture to P, and K(\theta) is Kirchhoff's obliquity factor

[tex]K(\theta) = (1+\cos \theta) / 2[/tex]

where \theta is the angle between the two lines described above, which prevents the aperture from producing a wave that goes backwards towards the source. The \rho and r in the denominator produce an inverse-square falloff of intensity with distance, going from the source to the sub-aperture, and from the sub-aperture to P, as we should expect.

This is equivalent to having the sub-apertures as our "original" sources, producing waves with different amplitudes and phases, from different locations within the aperture.

Starting from here, the only difference between different situations (single slit, double slit, circular aperture, rectangular aperture, whatever) is the geometry of integration and the approximations that one uses to make solving the integral less complicated.

At this level, the only 'ad-hoc' thing in the expression above is the obliquity factor. But it emerges naturally when you do a rigorous derivation starting from the differential wave equation.

Unfortunately, I don't think I'm getting through, here.

You'll notice that, as I've said in the previous posting, that as soon as you try to explain this, you cannot avoid but invoke a complete mathematical explanation - which is fine and which I understand completely (I did, after all, used Hecht and Zajak for undergrad optics course eons ago). However, it is difficult to explain conceptually why the diverging beam that goes way beyond the size of the slit as the slit gets smaller and smaller, and yet, it is about the same size as the slit when it gets larger and larger (for plane waves impinging on the slit).

I could easily explain this mathematically by saying "Look, take the Fourier transform of the slit (i.e. assume it is a square pulse, the width of which equals to the width of the slit), and you'll see the Fourier transform of this (in k space) corresponds to the actual diffraction pattern that you get. This is as accurate as any. But it says nothing conceptually.

Let's put it this way... if you had given your original answer, and included the additional explanation as you did in your 2nd response, I think I wouldn't have said anything, since that clearly indicated that to know what happens as the slit width gets larger, the superpostion of many, many sub-slits will conspire to produce a nasty mathematics that eventually will give us what we observe. I don't believe that there is (at least, I can't do it) a conceptually easier or hand-waving way to explain this.

Zz.
 
  • #13
dextercioby said:
Yes,Halls,that picture of Huygens' principle is really nice.Though i'd still vote for Kirchhoff's integral...:wink:

Daniel.

Kirchhoff's Integral is inexact. It is fine for far-field optics, but for near-field optics, it cannot be applied. Maxwell's equations are the only equations that can precisely predict any diffracted field, given an incident field and an aperture.

Claude.
 
  • #14
Kirchhoff's Integral is inexact,but not for the reason you mentioned(BTW,Fresnel's theory is based on this integral,too,check Jackson or,better,Born & Wolf).It's because it's SCALAR THEORY OF DIFFRACTION.The correct theory is the vector one...:wink:

Daniel.
 
  • #15
Sorry, I meant near-field, not in relation to Fresnel's theory, but in relation to sub-wavelength apertures and the field within one aperture length of the aperture (If that makes sense). Near-field is just one of those terms that can mean different things in a given context, I guess I got caught out (again!).

I seem to recall that there was another reason why Kirchhoff's Integral fails for sub-wavelength apertures in the near-field (not just because it is a scalar equation), something to do with the way it assumes the form of the field inside the aperture and so forth.

Claude.
 
  • #16
all


Why doesn't this happen with a laser (1 coherent freq.) through the slit?


TRoc
 
  • #18
T.Roc said:
all


Why doesn't this happen with a laser (1 coherent freq.) through the slit?


TRoc

What made you say that this doesn't happen with a laser? The ONLY way to see clear diffraction pattern IS to use a source with plane, monochromatic light.

Zz.
 
  • #19
Probably never walking into a lab...??:confused:

Though i have a hunch he's not exactly a theorist...

Daniel.
 
  • #20
dex, zz


I simply mean that I don't get the same results through the same slits when I use a laser, and when I use "white" light. What does this mean?

TRoc
 
  • #21
T.Roc said:
dex, zz


I simply mean that I don't get the same results through the same slits when I use a laser, and when I use "white" light. What does this mean?

TRoc

Your "white light" is not monochromatic. The diffracton patterns are washed out. So you should have asked, instead, on why you do not get a clear diffraction effect when you use white light.

Zz.
 
  • #22
Zz,

Actually, I started with white light, and produced the pattern predicted by the text. It was months later that I got my hands on a laser.

Do not let this throw the question. Your last post is what I want to learn about. Namely, what are the expected differences in refraction through 2 closley space slits based on monochromatic, dichromatic, and polychromatic light waves?

TRoc
 
  • #23
T.Roc said:
Zz,

Actually, I started with white light, and produced the pattern predicted by the text. It was months later that I got my hands on a laser.

Then (1) your "white light" isn't white, i.e. not a combination of many spectrum of visible light, or (2) there's something other weird things going on that you missed, such as a filter somewhere, or (3) who knows what.

Do not let this throw the question. Your last post is what I want to learn about. Namely, what are the expected differences in refraction through 2 closley space slits based on monochromatic, dichromatic, and polychromatic light waves?

TRoc

Since you were able to "produce the pattern predicted by the text", I would suggest you look closely at the derivation of those pattern and look at the fact that the wavelength (or frequency) used to derive the predicted textbook case does not consist or assume to be a range of value, but rather only one value at a time! If you have a source consisting of 1, 2, 3, 4, ... infinite range of wavelengths, then just plug it in!

Zz.
 
  • #24
I suggest that those of you having difficulties with jtbell's ideas read a freshman physics book. For example, you will find a very nice, physically based discussion of diffraction in Halliday and Resnick, Vol 2, Chapter 44 -- there are even pictures. And a clear explanation of why diffraction max and min tend to vanish as the slit width becomes greater than the wave length. (The slit becomes more and more infinite looking to the most of the wave front. See Resnick and Halliday for the details.)

And do recall that Fresnel, Fraunhofer and others at the time did not know about the Uncertanty Principle -- nor did Airy in 1835 when he gave the first solution to diffraction from a circular aperature.

Regards,
Reilly Atkinson
 
  • #25
since lasers were not around when this experiment was first done, how was their "monochromatic" light made?

TRoc
 
  • #26
Read

T.Roc said:
since lasers were not around when this experiment was first done, how was their "monochromatic" light made?

TRoc

Read Halliday and Resnik or any book on optics.
Regards,
Reilly Atkinson
 
  • #27
T.Roc,have you heard of light filters...?

Daniel.

Daniel.
 
  • #28
dex,

how did you manage to memorize so much stuff, and not learn such basic concepts?

you need to spend some time outside of "the lab", and let some of the air out of your ego.

since you had never heard of additive theory for visible light mixing, it stands to reason that you don't know the subtractive rules. a filter can not absorb all but one frequency of light, so the result of filtered light can not be monochromatic.

here I will insert a quote from claude bile on the thread "photon interactions". I do not make lasers for a living, and am happy to refer to someone more experienced in that area.

"Lasers are monochromatic



This is incorrect, all lasers posess a finite linewidth, some lasers (supercontinuum lasers) have linewidths that can cover a large portion of the frequency spectrum."

so can you explain diffraction without saying you must start with monochromatic light?

and can we let a forum be a forum? I can find lists of books to read on the subject, I'm not posting for references, but open discussion. this is a fairly easy experiment to do at "home", maybe you should try it before you challenge someone else's ability to be a "theorist".


TRoc
 
  • #29
T.Roc said:
"Lasers are monochromatic



This is incorrect, all lasers posess a finite linewidth, some lasers (supercontinuum lasers) have linewidths that can cover a large portion of the frequency spectrum."

so can you explain diffraction without saying you must start with monochromatic light?

However, you also need to understand what is meant by "monochromatic" in the PRACTICAL sense. While it is true that ALL light source, even lasers have finite linewidth, the question is whether the light source is as BAD as an ordinary light bulb, or if it has a PREDOMINANTLY one single wavelength with an acceptable bandwidth. If you visit a synchrotron light source (after all, if you are insisting someone get out of a lab, it is fair that you are told to go visit one), the use of monochrometers, inteferometers, etc. allows you to select a particular frequency to use so that the light source you are using is predominantly of just one frequency with a very small linewidth. You will be laughed at if you go to these people and tell them that their light source cannot be thought of to be "monochromatic" just because their linewidth is not zero. Considering the resolution of a particular experiment, such a spread in frequency is often irrelevant.

Zz.
 
  • #30
Zz,

Now we're going to be practical?? I told you that I have done the experiment both ways, so I have a "practical" understanding. I came here to deepen this understanding.

You said, quote :

"The ONLY way to see clear diffraction pattern IS to use a source with plane, monochromatic light."

and :

"your "white light" isn't white, i.e. not a combination of many spectrum of visible light,.."

I don't have a problem with broadening the definition of "monochromatic" to allow for the limitations of lasers, synchrotrons, etc.
But you also are saying that I could only get the diffraction pattern from white light (light bulb) because it is the combination of MANY frequencies? This is too broad (not monochromatic) to be practical for me.

Then, for what reason did you change the order of the statements between Dex and myself?

He said:

"Probably never walking into a lab...??

Though i have a hunch he's not exactly a theorist..."

and I responded:

"you need to spend some time outside of "the lab", and let some of the air out of your ego."

So, who is not being fair? If you don't want to help, and you have nothing to learn, what are you doing here - trying to bust people's balls?

To theorise something, one must take the risk of being wrong. I don't mind being told I'm wrong, but I don't want to dance in circles with people who are not offering legitimate advice.

TRoc
 
  • #31
Look. You do get diffraction from everything. But to be able to measure or quantify it, the range of frequencies you use has to be very very narrow. Of course, it can't be at a single wavlength, like exactly 562.00000000... nm, but whatever equipment we use, there will usually be a small range. Bus as Zz says, depending on the resolution of the image formed etc. it doesn't matter too much.

No one's trying to bust anyone's balls, but the reason this could cause frustration is that this is basic stuff, and trying to make a theory of light (or whatever), without knowing this sort of stuff, and hoping that it will excel quantum electrodynamics is being a bit hopeful, don't you think?
 
  • #32
T.Roc said:
Zz,

Now we're going to be practical?? I told you that I have done the experiment both ways, so I have a "practical" understanding. I came here to deepen this understanding.

You said, quote :

"The ONLY way to see clear diffraction pattern IS to use a source with plane, monochromatic light."

and :

"your "white light" isn't white, i.e. not a combination of many spectrum of visible light,.."

I don't have a problem with broadening the definition of "monochromatic" to allow for the limitations of lasers, synchrotrons, etc.
But you also are saying that I could only get the diffraction pattern from white light (light bulb) because it is the combination of MANY frequencies? This is too broad (not monochromatic) to be practical for me.

Then, for what reason did you change the order of the statements between Dex and myself?

He said:

"Probably never walking into a lab...??

Though i have a hunch he's not exactly a theorist..."

and I responded:

"you need to spend some time outside of "the lab", and let some of the air out of your ego."

So, who is not being fair? If you don't want to help, and you have nothing to learn, what are you doing here - trying to bust people's balls?

To theorise something, one must take the risk of being wrong. I don't mind being told I'm wrong, but I don't want to dance in circles with people who are not offering legitimate advice.

TRoc

Read my last posting. Are you saying that in the whole response that I posted, I had NO LEGITIMATE ADVICE?

Question: have you seen the spectrum from an incandescant light bulb using a spectrometer? If you have (and presumably, if you have done diffraction experiment, you'd see such a thing) then you would have notice that the diffraction and interference pattern is CONTINUOUS!

Notice that I said you DO get A diffraction pattern, but this is NOT the diffraction pattern described in standard text because those descriptions are based on a source having just one particular wavelength! All you need to do is plug in all the various wavelength and see for yourself that if you do not have just ONE wavelength, you get multiple patterns that will simply overlap each other! I could have sworn I've said this already.

But somehow, you are now insisting that light sources such as "lasers" and radiation from synchrotrons are not "exactly" monochromatic! What's the point? That such experiments then cannot or should not be done, or the results cannot be believed since there's some finite spread in frequency or wavelength? I will bet you money that you cannot design a typical single-slit experiment whose resolution is SMALLER than the linewidth of the Ti-Sapphire laser that I'm using now.

Moral of the story: In MOST instances, the resolution of your experiment is considerably WORSE than any finite linewidth of light sources such as this. This CLEARLY means that such a spread in wavelength might as well be non-existent. If this is not being "practical", I don't know what is.

Zz.
 
  • #33
Zz,

First, thanks for "hanging in there" with me. 2nd, no I didn't mean you had no legitimate advise to offer, or that everything you said was of no help. Just felt the conversation was starting to go in circles. Dex ruffled my feathers more than your post, you just were in the middle. Sorry.

I realize that you pros are constantly being asked to complete people's homework, or to be drawn into an endless debate with "kooks", and that's got to be frustrating. So is being neither of these, but trying to get answers to questions that will require thoughtful responses, and not reciteful ones.

I'm not saying that the experiments that have been done are not legit, or science is bogus, etc, etc.

My question is better said in 2 parts:

1. HAS an experiment been done to anyones' knowlegde that had a truly mono-chromatic wavelength through a slit? (hydrogen n=2, 658nm for instance)

2. IF this hasn't been done, and IF it were possible to devise a test (so we're only left with opinions) do you think we would get a "washed out" pattern, or just a slit shaped dot? (assume a 658nm slit)

thanks
TRoc
 
  • #34
T.Roc said:
Zz,

First, thanks for "hanging in there" with me. 2nd, no I didn't mean you had no legitimate advise to offer, or that everything you said was of no help. Just felt the conversation was starting to go in circles. Dex ruffled my feathers more than your post, you just were in the middle. Sorry.

I realize that you pros are constantly being asked to complete people's homework, or to be drawn into an endless debate with "kooks", and that's got to be frustrating. So is being neither of these, but trying to get answers to questions that will require thoughtful responses, and not reciteful ones.

I'm not saying that the experiments that have been done are not legit, or science is bogus, etc, etc.

My question is better said in 2 parts:

1. HAS an experiment been done to anyones' knowlegde that had a truly mono-chromatic wavelength through a slit? (hydrogen n=2, 658nm for instance)

2. IF this hasn't been done, and IF it were possible to devise a test (so we're only left with opinions) do you think we would get a "washed out" pattern, or just a slit shaped dot? (assume a 658nm slit)

thanks
TRoc

In return, I will say that you have not offended me at all. I have much thicker skin than that after being on the 'net for more than a dozen years.

To answer your question, are you restricting a "monochromatic" source as being a H transition line? If it is, then I can tell you that even this isn't that good. There are line widths even for this, and depending on your source (arc lamp? discharge tube?), the line width can be quite worse than, for example, a YAG laser (we won't get even get to a mode-locked Ti-Sapphire laser). If this is all that you want (i.e. something as good or better than an H transition line), then yes, such an experiment has been done. In fact, I can easily stick a slit anywhere in the beam path and see the diffraction/interference pattern. I don't know why this would be new though...

What I don't understand is why would this be any different than what you get from, let's say, a He-Ne laser (which is very common), or the one from an ordinary laser pointer? Are you able to look at the diffraction pattern from such sources and actually SEE the finite linewidths from these sources? I would find that very hard to believe.

Zz.
 
  • #35
To the original poster:

Diffraction happens because wave amplitudes (of almost any type of wave)
can add or subtract (called superposition), and because waves tend to spread
out from a particular location as they travel.

These two features will cause diffraction to occurr whenever a wave of any
type meets an obstruction of any type.
 
<h2>1. Why does light diffract?</h2><p>Light diffracts because it behaves as a wave, and when it encounters an obstacle or passes through a narrow opening, it spreads out or bends around the edges of the obstacle or opening. This phenomenon is known as diffraction.</p><h2>2. What causes light to diffract?</h2><p>The diffraction of light is caused by the interference of light waves as they pass through or around an obstacle. This interference causes the light to bend or spread out, creating a diffraction pattern.</p><h2>3. How does the wavelength of light affect diffraction?</h2><p>The wavelength of light has a direct effect on the amount of diffraction that occurs. Light with a longer wavelength, such as red light, diffracts more than light with a shorter wavelength, such as blue light.</p><h2>4. Does the size of the obstacle or opening affect diffraction?</h2><p>Yes, the size of the obstacle or opening does affect diffraction. The smaller the opening or obstacle, the more pronounced the diffraction will be. This is because smaller openings or obstacles cause greater interference between the light waves.</p><h2>5. Can diffraction be observed in everyday life?</h2><p>Yes, diffraction can be observed in everyday life. For example, when you see a rainbow, the different colors are a result of diffraction. Also, the edges of shadows are often fuzzy due to diffraction of light around the edges of objects.</p>

1. Why does light diffract?

Light diffracts because it behaves as a wave, and when it encounters an obstacle or passes through a narrow opening, it spreads out or bends around the edges of the obstacle or opening. This phenomenon is known as diffraction.

2. What causes light to diffract?

The diffraction of light is caused by the interference of light waves as they pass through or around an obstacle. This interference causes the light to bend or spread out, creating a diffraction pattern.

3. How does the wavelength of light affect diffraction?

The wavelength of light has a direct effect on the amount of diffraction that occurs. Light with a longer wavelength, such as red light, diffracts more than light with a shorter wavelength, such as blue light.

4. Does the size of the obstacle or opening affect diffraction?

Yes, the size of the obstacle or opening does affect diffraction. The smaller the opening or obstacle, the more pronounced the diffraction will be. This is because smaller openings or obstacles cause greater interference between the light waves.

5. Can diffraction be observed in everyday life?

Yes, diffraction can be observed in everyday life. For example, when you see a rainbow, the different colors are a result of diffraction. Also, the edges of shadows are often fuzzy due to diffraction of light around the edges of objects.

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