Diffraction of Light: Explaining Interference & Huygens' Principle

[brainyman89]

diffraction takes place when the wavelets are a source of more wavelets that expand in all directions, and the shape of the new wave front is curved. The wavelets of these diffracted, or bent, waves can now travel different paths and subsequently interfere with each other, producing interference patterns.
Then why it is crucial to have size of the opening comparable to or smaller than the wavelength of the incoming wave to get diffraction? what does make wavelets expand in all directions, and the shape of the new wave front is curved? Do Huygens’s Principle explain why these wavelets expand in all directions which makes the shape of the new wave front curved?

Huygens’s Principle states that every point on a wave front of light acts as a source of secondary waves that radiate from their centers. Then why interference does not occur when light is propagating in a straight line thus causing diffraction?

thanks in advance

 

[RedX]

The opening needs to be smaller than the wavelength so as to not get destructive interference in directions lateral to the direction that the light approaches the slit.

Imagine a line of sources that is several wavelengths long. The top of the line, and a distance half a wavelength below the top, interfere destructively in directions along the line. Similarly, 1/100 of a wavelength below the top of the line and half a wavelength+1/100 of a wavelength below the top, interfere destructively in directions along the line. So the length of the line needs to be smaller than a wavelength so this type of destructive interference can't occur: you don't want to be able to find pairs always half a wavelength apart.

If the length is longer than several wavelengths, then only in the direction of the incoming light will you not get destructive interference - you will get destructive interference in lateral directions. But that means light just goes in a straight line, so ray optics works and you don't have to consider diffraction.

The most ideal situation to see diffraction is for the width of the slit to be infinitisemal. That way there can be no destructive interference, since only one source can fit in the slit, and one source can't interfere with anyone else, so you get light going out in all directions, including the direction lateral to the incoming light.

 

[Andy Resnick]

Then why it is crucial to have size of the opening comparable to or smaller than the wavelength of the incoming wave to get diffraction?

Diffraction always occurs when the wavefront is finite is size- diffractive effects scale something like z/D, where D is the diameter of the aperture and z the propagation distance.

 

[jtbell]

Then why it is crucial to have size of the opening comparable to or smaller than the wavelength of the incoming wave to get diffraction?

It's not. See for example edge diffraction. (picture here)

 

[RedX]

It's not. See for example edge diffraction. (picture here)

What are the units on the x-axis in the graph of the link? I'm assuming the units are in wavelengths of the light, but does each tick mark mean 1/100 of a wavelength, 1/10 of a wavelength, etc?

Also, can one extend this to a finite plane to get the diffraction pattern around a barrier? For example in the link, the half-plane is in the negative y-axis. If you put the half-plane on the positive y-axis instead, and superpose the result with the half-plane in the negative y-axis, then would this be the diffraction pattern of an infinitisemal point situated at the origin?

Also, I'm not understanding the photograph of the diffraction pattern of the laser in your second link. It looks like the screen was photographed parallel to the screen, so that the pattern is on the horizon. Is this the case? So by machined metal edge, I assume this is the half-plane of a metal sheet, and black band on the bottom of the picture is the geometric shadow region that's blocked by the half-plane, and the top of the picture will be very bright as it's the geometric light region? Also, why does it matter that it's metal?

 

[jtbell]

What are the units on the x-axis in the graph of the link? I'm assuming the units are in wavelengths of the light, but does each tick mark mean 1/100 of a wavelength, 1/10 of a wavelength, etc?

Dealing with Fresnel diffraction and the Cornu spiral is a bit messy. It's been a while since I worked through an example in detail myself. Maybe the following pages that are linked to that page will help:

http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/fresgeo.html#c1
http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/cornu.html#c1

The treatment on Hyperphysics is pretty basic and doesn't have very many examples. If you Google on "Fresnel diffraction" you should be able to find more.

Also, I'm not understanding the photograph of the diffraction pattern of the laser in your second link.

It matches the diagram at the upper right corner of the first link. The screen is the vertical dotted line at the right side of that diagram. The source is on the left side.

why does it matter that it's metal?

It's easier to get a sharp straight edge with metal than with (say) cardboard. Brand-new razor blades are good for this.

 

[brainyman89]

The opening needs to be smaller than the wavelength so as to not get destructive interference in directions lateral to the direction that the light approaches the slit.

Imagine a line of sources that is several wavelengths long. The top of the line, and a distance half a wavelength below the top, interfere destructively in directions along the line. Similarly, 1/100 of a wavelength below the top of the line and half a wavelength+1/100 of a wavelength below the top, interfere destructively in directions along the line. So the length of the line needs to be smaller than a wavelength so this type of destructive interference can't occur: you don't want to be able to find pairs always half a wavelength apart.

If the length is longer than several wavelengths, then only in the direction of the incoming light will you not get destructive interference - you will get destructive interference in lateral directions. But that means light just goes in a straight line, so ray optics works and you don't have to consider diffraction.

The most ideal situation to see diffraction is for the width of the slit to be infinitisemal. That way there can be no destructive interference, since only one source can fit in the slit, and one source can't interfere with anyone else, so you get light going out in all directions, including the direction lateral to the incoming light.

I've got the idea, but can u give me a link for an animation or video that shows the interference a line of sources.

 

[RedX]

It matches the diagram at the upper right corner of the first link. The screen is the vertical dotted line at the right side of that diagram. The source is on the left side.

Thanks. I think the fact that the oscillations get smaller as you go up the screen, makes it look like the bands are equal-sized but the camera is shooting parallel (and not perpendicular) to the screen, so that bands at the very top become foreshortened as the mind interprets them as close to the horizon where everything is spaced closer together due to foreshortening.

To me the picture initially looked just like that, with the very top bands serving as sort of a horizon. But the top bands really are smaller and this is a two dimensional photo and not a 3 dimensional one. I think they could have prevented this by having the half-plane the other way, on the top instead of the bottom.

I understand the picture in your link, and also this one:

http://www.sciencephoto.com/images/download_lo_res.html?id=652050119

and this one:

http://www.physics.umd.edu/lecdem/services/demos/demosm5/m5-12.htm

However, I do not understand this picture:

http://www.physics.montana.edu/demonstrations/video/6_optics/demos/knifeedgediffraction.html

It makes no sense to me! How can it be symmetrical? Is it because the razor blade has a notch in it, and they are shining the laser through the hole of the razor, so that to each side of the hole is a half plane?

 

[brainyman89]

The opening needs to be smaller than the wavelength so as to not get destructive interference in directions lateral to the direction that the light approaches the slit.

Imagine a line of sources that is several wavelengths long. The top of the line, and a distance half a wavelength below the top, interfere destructively in directions along the line. Similarly, 1/100 of a wavelength below the top of the line and half a wavelength+1/100 of a wavelength below the top, interfere destructively in directions along the line. So the length of the line needs to be smaller than a wavelength so this type of destructive interference can't occur: you don't want to be able to find pairs always half a wavelength apart.

If the length is longer than several wavelengths, then only in the direction of the incoming light will you not get destructive interference - you will get destructive interference in lateral directions. But that means light just goes in a straight line, so ray optics works and you don't have to consider diffraction.

The most ideal situation to see diffraction is for the width of the slit to be infinitisemal. That way there can be no destructive interference, since only one source can fit in the slit, and one source can't interfere with anyone else, so you get light going out in all directions, including the direction lateral to the incoming light.

this explanation illustrates diffraction when light fall on a small aperture, what about diffraction caused when light falls on a sharp edge?

 

[brainyman89]

when the size of the opening is greater than the wavelength of the incoming wave, will diffraction slightly occur or will not exist at all?

according to the previous illustration given by RedXi, it is written that the line of sources(according to Huygens) should be greater than the wavelength of wave in order for the light to propagate in straight line, does this mean that the width of the light should be greater than it wavelength to propagate in straight line?

 

[jtbell]

For a very wide aperture, you get two edge-diffraction patterns like the ones illustrated by the links in my previous posts, facing each other so to speak.

As the aperture becomes narrower, the two edge-diffraction patterns overlap and the result becomes complicated. You can't simply add the patterns because they "interfere" with each other.

As the aperture becomes very narrow you get the simple Fraunhofer single-slit pattern.

This page: http://www.falstad.com/diffraction/ has an applet that illustrates Fresnel diffraction. Choose "Aperture: slit". You can change the width of the slit by dragging one of the vertical red lines. You can zoom in and out, adjust the brightness etc. by using the sliders. To see the details of the pattern "inside" a wide slit you need to turn the brightness way down, and you might need to increase the resolution. Your screen resolution tends to "interfere" with the details in the pattern if they're too small.

 

[RedX]

when the size of the opening is greater than the wavelength of the incoming wave, will diffraction slightly occur or will not exist at all?

I was referring to Fraunhofer diffraction. The intensity through a single slit is given by:

$$I(x)=\left(\frac{sin x}{x}\right)^2$$

where $$x=\frac{\pi L \sin \theta}{\lambda}$$

As x gets large, I(x) goes to zero. So if the width of the slit L is a lot greater than the wavelength lambda, you only get strong intensity at theta=0, which is a straight line through the slit.

jtbell is referring to the much more interesting case of Fresnel diffraction which is more accurate, and you get all those beautiful pictures. Great applet by the way!

As the aperture becomes very narrow you get the simple Fraunhofer single-slit pattern.

What I find interesting is that when you make the single slit very small in the applet, you get total darkness, whereas Fraunhofer diffraction predicts total brightness! Is this a contradiction? I always thought if the slits were infinitismal, that was the best case scenario?

 

[jtbell]

The intensity patterns that you usually see in textbooks show relative intensity, usually normalized to 1.0 at the central maximum. The actual overall brightness decreases as the slit gets smaller, both because less light from the source "hits" the slit in the first place, and because it gets spread out over a larger area on the screen as the central maximum becomes wider.

 

[brainyman89]

The opening needs to be smaller than the wavelength so as to not get destructive interference in directions lateral to the direction that the light approaches the slit.

Imagine a line of sources that is several wavelengths long. The top of the line, and a distance half a wavelength below the top, interfere destructively in directions along the line. Similarly, 1/100 of a wavelength below the top of the line and half a wavelength+1/100 of a wavelength below the top, interfere destructively in directions along the line. So the length of the line needs to be smaller than a wavelength so this type of destructive interference can't occur: you don't want to be able to find pairs always half a wavelength apart.

If the length is longer than several wavelengths, then only in the direction of the incoming light will you not get destructive interference - you will get destructive interference in lateral directions. But that means light just goes in a straight line, so ray optics works and you don't have to consider diffraction.

The most ideal situation to see diffraction is for the width of the slit to be infinitisemal. That way there can be no destructive interference, since only one source can fit in the slit, and one source can't interfere with anyone else, so you get light going out in all directions, including the direction lateral to the incoming light.

this explanation illustrates what causes diffraction when light fall on a small aperture
but what causes diffraction when light falls on a sharp edge?

 

[RedX]

this explanation illustrates what causes diffraction when light fall on a small aperture but what causes diffraction when light falls on a sharp edge?

Right, so the Fraunhofer formula:
$$I(x)=\left(\frac{sin x}{x}\right)^2$$ where $$x=\frac{\pi L \sin \theta}{\lambda}$$
predicts that when L goes to infinity, as would be the case for the (infinite) space above an infinite half-plane, then you only get intensity in the direction theta=0. So for an infinite half-plane there is only the geometric shadow. But that is what Fraunhofer diffraction predicts, and should only be true when the screen is infinitely far away. For a finite distance between screen and sharp-edge, you have to use Fresnel formulas. I can't really explain why there is diffraction in this case, but evidently if you do the math there is - in the Feynman Lectures on Physics Volume I, Feynman demonstrates graphically why you get diffraction at the edge of an infinite half-plane by drawing a picture (Cornu Spiral).

Anyways, so if you want diffraction around a line segment of finite size, then you would put an infinite line of sources both above and below the line segment, and add up the field contribution from all these sources (without using a first order approximation as you would with the Fraunhofer method), and then you should see diffraction around the line segment - the line segment doesn't quite produce a geometric shadow equal to its length.

The intensity patterns that you usually see in textbooks show relative intensity, usually normalized to 1.0 at the central maximum. The actual overall brightness decreases as the slit gets smaller, both because less light from the source "hits" the slit in the first place, and because it gets spread out over a larger area on the screen as the central maximum becomes wider.

I understand now. Thanks.

 

[jtbell]

I was referring to Fraunhofer diffraction. The intensity through a single slit is given by:

$$I(x)=\frac{sin x}{x}$$

where $$x=\frac{\pi L \sin \theta}{\lambda}$$

I just noticed an error here. Your I(x) is actually more nearly the amplitude of the electric field (or the magnetic field) in the wave. The intensity or "brightness" of the light is proportional to the square of the amplitude, so

$$I = I_0 \frac{\sin^2 x}{x^2}$$

where $I_0$ is the maximum intensity, at the center of the diffraction pattern.

 

[dlgoff]

Dear Professor jtbell,

I've always been interested in the part of light that is in the "geometric shadow" from a single edge.

In college, we had an advanced lab class and the instructor would let me use the equipment to test a curiosity about how polarization of the laser source "might" effect the outcome of the diffraction pattern. I was unable to really complete the experiment, but am still curious if there would be a difference due to the source polarization relative to a vertical edge. I'm not speculation any difference but I don't think I've ever seen an experiment about this. Probably since the math doesn't make a distinction?

Thanks for any information if there is any.

 

[RedX]

I just noticed an error here. Your I(x) is actually more nearly the amplitude of the electric field (or the magnetic field) in the wave. The intensity or "brightness" of the light is proportional to the square of the amplitude, so

$$I = I_0 \frac{\sin^2 x}{x^2}$$

where $I_0$ is the maximum intensity, at the center of the diffraction pattern.

I fixed the mistake and edited everything so it is now squared. I'll go the textbook route and have it so that straight through the slit the intensity is 1, and every other angle is a relative intensity. Thanks for pointing that out.

 

[jtbell]

am still curious if there would be a difference due to the source polarization relative to a vertical edge.

I've never seen any real discussion of this possibility, but my guess there would be either no difference or only a very small difference. All the treatments of diffraction that I've seen in intermediate-level optics textbooks (which is as far as I've studied it myself) don't assume anything about the polarization of the light or the interaction of the light with the edges of the aperture. The edge of the aperture simply serves as a go / no-go boundary for the wave.

It seems to me that any such interaction would directly affect only the light that is going right next to the edge of the aperture, whereas the diffraction pattern is the result of superposition of waves from all points where the light passes unobstructed through the aperture plane. In general, there are a lot more "source points" in the aperture plane that are relatively far from the edge than close to it, at least for things like edge diffraction or large apertures. Polarization might have an effect with very narrow apertures, made out of certain materials.

 

[dlgoff]

I've never seen any real discussion of this possibility, but my guess there would be either no difference or only a very small difference. All the treatments of diffraction that I've seen in intermediate-level optics textbooks (which is as far as I've studied it myself) don't assume anything about the polarization of the light or the interaction of the light with the edges of the aperture. The edge of the aperture simply serves as a go / no-go boundary for the wave.

It seems to me that any such interaction would directly affect only the light that is going right next to the edge of the aperture, whereas the diffraction pattern is the result of superposition of waves from all points where the light passes unobstructed through the aperture plane. In general, there are a lot more "source points" in the aperture plane that are relatively far from the edge than close to it, at least for things like edge diffraction or large apertures. Polarization might have an effect with very narrow apertures, made out of certain materials.

Thank you sir. You have answered a "long wondering" question. And your explanation makes sense to me.

Regards

 

[brainyman89]

the line of sources (according to Huygens) should be greater than the wavelength of wave in order for the light to propagate in straight line, does this mean that the width of the rays of light should be greater than it wavelength to propagate in straight line?

another question concerning edge diffraction, how comes diffraction occurs though light is propagating with width greater than lambda?

do radiations of large wavelength undergo diffraction easier as large openings could lead to diffraction?

visible light of wavelength between 400nm and 800nm undergoes diffraction when it falls on a slit of width smaller than 1mm, how could this happen though 1mm>>>400nm or 800nm??

 

[brainyman89]

Imagine a line of sources that is several wavelengths long. The top of the line, and a distance half a wavelength below the top, interfere destructively in directions along the line. Similarly, 1/100 of a wavelength below the top of the line and half a wavelength+1/100 of a wavelength below the top, interfere destructively in directions along the line. So the length of the line needs to be smaller than a wavelength so this type of destructive interference can't occur: you don't want to be able to find pairs always half a wavelength apart.

doesn't this case also occur when light falls on a edge, then why edge diffraction takes place?

 

[RedX]

the line of sources (according to Huygens) should be greater than the wavelength of wave in order for the light to propagate in straight line, does this mean that the width of the rays of light should be greater than it wavelength to propagate in straight line?

I think so. I don't think you can collimate light to a beam shorter than its wavelength, for otherwise diffraction would undo that.

another question concerning edge diffraction, how comes diffraction occurs though light is propagating with width greater than lambda?

As has been mentioned, it occurs but it's small. The best way is to probably do the math.

do radiations of large wavelength undergo diffraction easier as large openings could lead to diffraction?

Yes. Perhaps this website will be helpful:

http://esfscience.wordpress.com/2009/03/26/diffraction-wave-spreading-around-an-edge/

visible light of wavelength between 400nm and 800nm undergoes diffraction when it falls on a slit of width smaller than 1mm, how could this happen though 1mm>>>400nm or 800nm??

You can view diffraction because the screen is far away, of the order of meters, and the diffraction pattern is of size centimeters. So the angle where there is spreading is still very small, and it takes a great distance (meters) to view a little spread (centimeters).