Fraunhofer Diffraction Doubt


by GPhab
Tags: diffraction, doubt, fraunhofer
GPhab
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#1
Sep7-08, 01:38 AM
P: 25
The relation between the angles for which the minima occurs for a Fraunhofer diffraction(parallel rays, large separation between screen and slit) is derived by using the destructive interference criterion. Why isn't there any mention of the criterion for maxima?

The results obtained from the below equation and the ones above are the same(E standing for amplitude).

[tex]E = E_{0} (sin\beta)/\beta[/tex] where
[tex]\beta=(\pi/\lambda)b sin\theta [/tex]
Can anybody make things clear for me?

One more question.

For Fraunhofer Diffraction in a circular aperture, the first dark ring is formed by rays which diffract at an angle [tex]\theta[/tex] where

[tex]sin \theta \approx 1.22\lambda /b[/tex] , b is the radius of the aperture

What if [tex]\lambda / b = 1.1/1.22[/tex]. That will imply [tex] sin \theta > 1[/tex]. It will still allow the light to go through. But where is the first dark ring formed?
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jtbell
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#2
Sep7-08, 02:50 AM
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Quote Quote by GPhab View Post
The relation between the angles for which the minima occurs for a Fraunhofer diffraction(parallel rays, large separation between screen and slit) is derived by using the destructive interference criterion. Why isn't there any mention of the criterion for maxima?
Take the derivative of [itex]E^2[/itex] (for the intensity) with respect to [itex]\beta[/itex] and set it to zero to find the maxima. The result isn't very simple!

For Fraunhofer Diffraction in a circular aperture, [...] What if [tex]\lambda / b = 1.1/1.22[/tex]. That will imply [tex] sin \theta > 1[/tex]. It will still allow the light to go through. But where is the first dark ring formed?
There is no dark ring, that is, there is no [itex]\theta[/itex] for which the intensity is zero. The intensity is maximum at [itex]\theta = 0[/itex], and it decreases as [itex]\theta[/itex] increases towards 90 degrees, but it never reaches zero.
Antenna Guy
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#3
Sep7-08, 12:54 PM
P: 303
Quote Quote by GPhab View Post
The relation between the angles for which the minima occurs for a Fraunhofer diffraction(parallel rays, large separation between screen and slit) is derived by using the destructive interference criterion. Why isn't there any mention of the criterion for maxima?

The results obtained from the below equation and the ones above are the same(E standing for amplitude).

[tex]E = E_{0} (sin\beta)/\beta[/tex] where
[tex]\beta=(\pi/\lambda)b sin\theta [/tex]
Can anybody make things clear for me?
First off, the designation "Fraunhofer" implies a far-field relation (i.e. where [itex]r>\frac{2D^2}{\lambda}[/itex]), which is why there is no "r" in the expression for field strength as a function of theta. I think you will find it a little more intuitive to express [itex]\beta[/itex] as follows:

[itex]\beta=\frac{2\pi D}{\lambda} sin\theta [/itex], where D is the distance between the two slits.

Expressed in this way, you can choose either of the two slits as your phase reference, and [itex]\beta[/itex] is now the relative phase of the other slit as a function of theta. As jtbell pointed out, the maxima will occur at [itex]\theta=0[/itex] (where [itex]\beta =0 [/itex]), and will alternately pass through minima and maxima every [itex]\pi[/itex] radians in [itex]\beta[/itex].

Regards,

Bill

Antenna Guy
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#4
Sep7-08, 01:34 PM
P: 303

Fraunhofer Diffraction Doubt


Quote Quote by GPhab View Post
One more question.

For Fraunhofer Diffraction in a circular aperture, the first dark ring is formed by rays which diffract at an angle [tex]\theta[/tex] where

[tex]sin \theta \approx 1.22\lambda /b[/tex] , b is the radius of the aperture

What if [tex]\lambda / b = 1.1/1.22[/tex]. That will imply [tex] sin \theta > 1[/tex]. It will still allow the light to go through. But where is the first dark ring formed?
If "the slit" is a single circle, the [itex] \beta[/itex] term would be a rotationally symmetric (about [itex]\theta=0[/itex]) version of what was described above. However, to make the pattern function rotationally symmetric, the wave incident upon the circular slit would have to be circularly polarized (to pass an equal amount of energy at any [itex]\phi[/itex]).

I've already claimed that the first minima should occur where [itex]\beta=\pi[/itex], but the above approximation appears to differ by a suspicious amount:

[tex]sin \theta \approx \frac{1.22\lambda}{b} \text{(1)}[/tex]

versus:

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

[tex]sin \theta = \frac{\lambda}{4b} \text{(2)}[/tex]

[tex]RHS (1) - RHS (2) \approx \frac{\lambda}{b}[/tex]

Which I think corresponds to [itex]\frac{\beta}{2} \text{ at } \theta = \frac{\pi}{2} [/itex]. [edit: should be pi/2]

Regards,

Bill
Andy Resnick
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#5
Sep7-08, 08:41 PM
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P: 5,468
Quote Quote by GPhab View Post
The relation between the angles for which the minima occurs for a Fraunhofer diffraction(parallel rays, large separation between screen and slit) is derived by using the destructive interference criterion. Why isn't there any mention of the criterion for maxima?

The results obtained from the below equation and the ones above are the same(E standing for amplitude).

[tex]E = E_{0} (sin\beta)/\beta[/tex] where
[tex]\beta=(\pi/\lambda)b sin\theta [/tex]
Can anybody make things clear for me?
The first equation is very recognizable to me- it's the Fourier transform of a 1-D slit of width [itex]\beta[/itex]. That makes sense for Fraunhofer diffraction: the far-field scattering amplitude of a hole in an opaque screen is the Fourier Transform of the shape of the hole. The second equation is a particular way of defining the width of the slit, and it looks slightly unfamiliar: what is the definition of b (hopefully the physical width of the slit) and [itex]\theta[/itex]? Using angular coordinates [itex]\beta[/itex] and [itex]\theta[/itex] rather than linear coordinates makes me think that this is a non-imaging system- the image/diffraction pattern is in fact, at infinity and not at an intermediate plane that we can observe in the lab.

This observation (hopefully correct) helps me understand the question about minima vs. maxima. Because in the context above, maxima correspond to *caustics*, and in fact, (formally) have infinite amplitude. But the book (or whatever) you are reading probably doesn't mean that- caustics are a fairly advanced topic.

So, to answer the second question, let's just go with the key fact that "the far field diffraction pattern (Frauhofer diffraction) of a hole is the Fourier Transform of the shape of the hole".

Quote Quote by GPhab View Post

One more question.

For Fraunhofer Diffraction in a circular aperture, the first dark ring is formed by rays which diffract at an angle [tex]\theta[/tex] where

[tex]sin \theta \approx 1.22\lambda /b[/tex] , b is the radius of the aperture

What if [tex]\lambda / b = 1.1/1.22[/tex]. That will imply [tex] sin \theta > 1[/tex]. It will still allow the light to go through. But where is the first dark ring formed?
If the Fourier transform of a rectangle of height 'h' and width 'w' is a product of Sinc functions (Sinc (x) = (sin x)/x): 1/(hw) Sinc(h)*Sinc(w), the Fourier transform of a circle is analogously a 'sombrero' function (Somb(x) = J0(x)/x, where J0 is the Bessel function of order 0). The first zero of this function is at 1.22/b, where b is the physical radius of the circular hole. But since you are working with these unfamiliar (to me) angular coordinates, I'm not sure what it means to vary [itex] \theta[/itex]: is it a scattering angle with respect to the surface normal, or the angle of incidence with respect to the surface normal, or some other angular measure?

If you could define the terms in the formulas, I could better answer your questions- they are good questions, and show the real power of Fourier optics to model systems.
GPhab
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#6
Sep8-08, 01:23 AM
P: 25
The real problem is that I am still in high school. In the book which I am reading, the results are stated without proof(except the criterion for minima). Andy was right. The author was referring to the diffraction pattern obtained at infinity.
Antenna Guy
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#7
Sep8-08, 05:55 PM
P: 303
Quote Quote by Andy Resnick View Post
...the Fourier transform of a circle is analogously a 'sombrero' function (Somb(x) = J0(x)/x, where J0 is the Bessel function of order 0). The first zero of this function is at 1.22/b, where b is the physical radius of the circular hole.
Hi Andy,

I'm not sure why an ideal aperture pattern describes (even approximately) the first minima of a diffraction pattern. The diffraction effects come from the edges of the openning, in a manner consistent with what I described earlier. I assumed that the two slits I was working with were much narrower than a wavelength, and of negligible width relative to b - which should be analgous to the edge effects of the single open aperture of width 2b that you chose. If you superimpose the ideal pattern function of the open aperture with the diffraction effects, I think you will arrive at what the approximate relation attempts to describe (otherwise, I don't see why the relation would be approximate).

Regards,

Bill
Andy Resnick
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#8
Sep8-08, 11:07 PM
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Antenna Guy,

The relationship I described- the far-field diffraction pattern of an aperture- is a limit, not an exact solution. One crucial limit is that Fraunhofer diffraction is not applicable in the near field. That is, in the far-field, diffraction does not originate from the edges of the aperture, it's a result of interference between all the waves that pass through the aperture- the angular spectrum of plane waves that pass through the aperture.

Note I am talking about spatial spectra, not temporal spectra. Even though there is a plane wave with a single (spatial) frequency and wavevector incident upon the screen, because of the boundary conditions at the aperture plane which require the electromagnetic field in all of space to obey certain properties- continuity, for example- what comes out of the hole has a range of spatial wavevectors (propagation directions)- with definite phase relationships between the different wavevectors.

If the dimensions of the hole are much less than a wavelength, then the far-field diffraction pattern is zero- even though it is clearly not in the near field.

Honestly, when I first learned the basic equations of Fourier optics (and the Van Cittert-Zernike theorem), I felt like I had an epiphany. It's like magic.
Antenna Guy
Antenna Guy is offline
#9
Sep9-08, 07:15 AM
P: 303
Quote Quote by Andy Resnick View Post
The relationship I described- the far-field diffraction pattern of an aperture- is a limit, not an exact solution. One crucial limit is that Fraunhofer diffraction is not applicable in the near field. That is, in the far-field, diffraction does not originate from the edges of the aperture, it's a result of interference between all the waves that pass through the aperture- the angular spectrum of plane waves that pass through the aperture.
Plane waves are a theoretical limit too - you cannot generate one without an infinite size aperture having uniform amplitude and phase distribution. You must consider that an aperture can be a source of distant energy - not just a receiver. I do not dispute that a spectrum of plane waves can be used to characterize a far-field radiation pattern (in fact, that has been the standard method employed in any planar near-field measurement system I've seen). However, spherical waves are a much more intuitive means of characterization when one considers reciprocity.

Note I am talking about spatial spectra, not temporal spectra. Even though there is a plane wave with a single (spatial) frequency and wavevector incident upon the screen, because of the boundary conditions at the aperture plane which require the electromagnetic field in all of space to obey certain properties- continuity, for example- what comes out of the hole has a range of spatial wavevectors (propagation directions)- with definite phase relationships between the different wavevectors.
Spherical waves are spatial too - they characterize the amplitude (and phase) distribution over angular space at fixed radius.

If the dimensions of the hole are much less than a wavelength, then the far-field diffraction pattern is zero- even though it is clearly not in the near field.
This is not true. Non-propogating modes are an artifact of locality. Their existence does not imply that the energy they represent is not propogating, only that you are too close to the aperture to construct a wave spectrum that does not change with distance from the source.

Honestly, when I first learned the basic equations of Fourier optics (and the Van Cittert-Zernike theorem), I felt like I had an epiphany. It's like magic.
They don't call it "black magic" for nothing.

BTW - I realize Mr Zernike is more well known in optics circles, but did you know that the polynomials he developed for lenses are commonly used to optimize reflector surface shapes?

Regards,

Bill


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