Register to reply

Single-Slit Diffraction

by Lil_Aziz1
Tags: diffraction, singleslit
Share this thread:
Lil_Aziz1
#1
May2-10, 04:06 PM
P: 21
Hi everyone. I need help comprehending single slit diffraction. I am using the textbook: College Physics by Serway / Vuille 8th Edition. This is what it says:

"To analyze the diffraction pattern, it's convenient to divide the slit into halves, as in Figure 24.17 (picture below). All the waves that originate at the slit are in phase. Consider waves 1 and 3, which originate at the bottom and center of the slit, respectively. Wave 1 travels farther than wave 3 by an amount equal to the path difference [tex] \frac{a}{2}\sin\theta[/tex], where a is the width of the slit. Similarly, the path difference between waves 3 and 5 is [tex] \frac{a}{2}\sin\theta[/tex]. If this path difference is exactly half of a wavelength, the two waves cancel each other and destructive interference results. Therefore, waves from the upper half of the slit interfere destructively with waves from the lower half of the slit when

[tex] \frac{a}{2}\sin\theta = \frac{\lambda}{2} [/tex]

The question is: why is a/2 the hypotenuse? why not just use a? I realize the first sentence says, "it's convenient to divide the slit into halves," but why?

Phys.Org News Partner Physics news on Phys.org
Engineers develop new sensor to detect tiny individual nanoparticles
Tiny particles have big potential in debate over nuclear proliferation
Ray tracing and beyond
Lil_Aziz1
#2
May4-10, 04:01 PM
P: 21
Though I'd bump this thread since I've gotten no reply in two days.
Doc Al
#3
May4-10, 05:01 PM
Mentor
Doc Al's Avatar
P: 41,477
Quote Quote by Lil_Aziz1 View Post
The question is: why is a/2 the hypotenuse? why not just use a? I realize the first sentence says, "it's convenient to divide the slit into halves," but why?
The point is that the first dark fringe is obtained when light from the top segment of the slit is exactly out of phase with the light from the middle of the slit. This implies that all the light "cancels out", since the light from any point in the top half is exactly out of phase with the light from the corresponding point in the bottom half. So it's natural to look at half the slit (a/2) and set the phase difference equal to λ/2.

Andy Resnick
#4
May4-10, 06:12 PM
Sci Advisor
P: 5,524
Single-Slit Diffraction

I agree with Doc Al, but from my perspective, this is a case of trying to make a concept overly simplified.

The far-field diffraction pattern from an aperture illuminated by a monochromatic plane wave is simply the Fourier transform of the aperture. For a single slit, this pattern is sin(ka)/ka, where k is 2*pi/l (l is the wavelength of illumination)

http://cnyack.homestead.com/files/af...iffraction.htm

The usual results follow.
cesiumfrog
#5
May4-10, 06:43 PM
P: 2,051
Quote Quote by Andy Resnick View Post
from my perspective, this is a case of trying to make a concept overly simplified.
Seconded. Technically it's valid in this context (but what good is that unless the reasons for it have been properly explained, and evidently this is not quite the case). Conceptually I think it lends itself to a misleading picture of the general process.

If they wanted a simple example, they could have just used two narrow slits. (If they want to go beyond that, why not proceed straight to full generality?)
sophiecentaur
#6
May5-10, 05:57 AM
Sci Advisor
Thanks
PF Gold
sophiecentaur's Avatar
P: 12,186
Not everyone finds the Fourier transform cuddly and familiar. Whilst Fourier is, undoubtedly, the way to go in the end, the fact is that you can extend the idea of two slit interference to two halves of a wide slit without Fourier being invoked. He's introduced / hinted at the idea of infinitessimals which is a step forward.
The nice thing is that the two approaches give the same result for the null directions - good reinforcement for the student who is fresh to these matters, I'd have thought.
I've seen worse!
Doc Al
#7
May5-10, 06:22 AM
Mentor
Doc Al's Avatar
P: 41,477
Quote Quote by Andy Resnick View Post
The far-field diffraction pattern from an aperture illuminated by a monochromatic plane wave is simply the Fourier transform of the aperture. For a single slit, this pattern is sin(ka)/ka, where k is 2*pi/l (l is the wavelength of illumination)
True, but you are forgetting the context here which is freshman physics. I don't think anyone's going to be doing Fourier transforms in physics 101. The analysis in the quoted text, as far as it goes, is perfectly legitimate and appears in just about every freshman physics textbook. (Even in my ancient edition of Halliday and Resnick.)
Andy Resnick
#8
May5-10, 07:38 AM
Sci Advisor
P: 5,524
Quote Quote by Doc Al View Post
True, but you are forgetting the context here which is freshman physics. I don't think anyone's going to be doing Fourier transforms in physics 101. The analysis in the quoted text, as far as it goes, is perfectly legitimate and appears in just about every freshman physics textbook. (Even in my ancient edition of Halliday and Resnick.)
Of course not- but that's no reason to generate a confusing alternate derivation. Most high school students are already familiar with the Fourier transform- they just call it an "equalizer" on iTunes or the stereo, or whatever.

There's no need to *derive* the sinc function, but in the spirit of plug-and-chug, surely providing a table of simple transform pairs is within the ability of a Freshman to use.
sophiecentaur
#9
May5-10, 09:37 AM
Sci Advisor
Thanks
PF Gold
sophiecentaur's Avatar
P: 12,186
Quote Quote by Andy Resnick View Post
Of course not- but that's no reason to generate a confusing alternate derivation. Most high school students are already familiar with the Fourier transform- they just call it an "equalizer" on iTunes or the stereo, or whatever.

There's no need to *derive* the sinc function, but in the spirit of plug-and-chug, surely providing a table of simple transform pairs is within the ability of a Freshman to use.
1. It's not a confusing treatment - it makes sense at all levels (as a nice extension to the Young's slits approach) and doesn't clash with Fourier when he arrives.
2. As for providing students with a set of transform pairs and telling them that they just happen to apply in diffraction; can that really be the way for them to get to understand what's going on? Fourier transforms to go between frequency and time domains make it far more approachable as a start (as you say, you can show it and listen to the effects through a loudspeaker but there is a distinction between the Fourier Transform and Fourier Analysis which can be missed if not careful). I can't think of an equivalent way to show the same thing with diffraction - apart from the effect of blurring through small apertures.
Peoples' brains don't always find plug and chug at all straightforward. You may be lucky enough not to have a problem but we're not all gifted in that way; it's a matter of confidence.

If you are hell bent on having a go at over-simplification of Physics teaching there are many more deserving targets - like why people insist on trying to explain simply every phenomenon in terms of particles (e.g. Convection, for Chrissake)
cesiumfrog
#10
May5-10, 06:57 PM
P: 2,051
Quote Quote by sophiecentaur View Post
As for providing students with a set of transform pairs and telling them that they just happen to apply in diffraction; can that really be the way for them to get to understand what's going on?
Isn't it already common practice to include a table of selected integral-antiderivative pairs along with textbook physics exercises? So as to say, "we expect you to set up the problem in full detail from first principles.. but when you do so you'll obtain expression X, which happens to be mathematically equivalent to expression Y.. knowing that fact is necessary to answer the question, but knowing the mathematical techniques of the proof for why X=Y isn't of as high importance and anyway it is above the level we expect from you in this unit."

Heck, you don't necessarily need to mention Fourier transforms (nor their "grammar"), just hand them the standard integral that happens to correspond to the particular Fourier transform they are performing, that way it's just as easy as any other problem they've done but it also lets them see a way to approach broader questions.

(Edit: Having learnt the technique in the OP's text, how then might the student begin if asked to obtain the intensity pattern of a triple-slit?)

Quote Quote by sophiecentaur View Post
why people insist on trying to explain simply every phenomenon in terms of particles (e.g. Convection, for Chrissake)
Yuck. Who?
Andy Resnick
#11
May5-10, 09:36 PM
Sci Advisor
P: 5,524
Quote Quote by sophiecentaur View Post
<snip>

If you are hell bent on having a go at over-simplification of Physics teaching
<snip>
I've only just finished my first year, but I've learned the essential difference between a good teacher and a bad one. A good teacher tries new ways to teach *every single time*. S/he keeps what works, and tosses what doesn't. A bad teacher is the opposite of that.

As for introducing the Fourier transform in an introductory course, I do exactly that (algebra-based, Giancoli). Was it successful? Honestly, I don't know yet- I have only one data point, so to speak. But I was able to present a unified treatment of sound and optics, among other things. Some of the students seemed to 'get it', some didn't- but I have information about how to adapt the presentation.

My relevant test had questions about resolution limits of 'perfect' vision v. Rayleigh limit, matching the pixel size of a camera to the magnification of the optical system, and comparison shopping for a telescope. Definitely not plug-and-chug.


Register to reply

Related Discussions
Single-slit diffraction General Physics 2
Single slit diffraction Introductory Physics Homework 4
Single-slit diffraction Introductory Physics Homework 0
Single Slit Diffraction General Physics 1
Help Single slit diffraction Introductory Physics Homework 20