
#1
Sep708, 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? 



#2
Sep708, 02:50 AM

Mentor
P: 11,231





#3
Sep708, 12:54 PM

P: 303

[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 



#4
Sep708, 01:34 PM

P: 303

Fraunhofer Diffraction DoubtI'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 



#5
Sep708, 08:41 PM

Sci Advisor
P: 5,468

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". 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. 



#6
Sep808, 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.




#7
Sep808, 05:55 PM

P: 303

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 



#8
Sep808, 11:07 PM

Sci Advisor
P: 5,468

Antenna Guy,
The relationship I described the farfield 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 farfield, 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 farfield 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 CittertZernike theorem), I felt like I had an epiphany. It's like magic. 



#9
Sep908, 07:15 AM

P: 303

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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