
#1
Jul1109, 09:58 AM

P: 7

Yesterday,our lab members discussed about the diffraction of incoherent beam.
I argued that if the beam is completely incoherent, it will not make diffraction patterns that follows sinc or bessellike function. My argument is that the width of diffraction pattern is determined by light's coherence area. If coherence area is small, the diffraction pattern will have width wider than that of perfect coherent light. (We were talking about farfield diffraction using lens) If coherence area>0 , it will be spreaded all across the focal plane of the lens. I think such an effect is consistent with van CittertZernike theorem. However, other lab members did not agree with me. They say that such a conclustion is different from our experience. Their examples are sunlight and fluorescent tube light. If we focus such incoherent light, we will not see the effect I argued above. I calculated the effect quantitatively today. It seems that coherence area of sunlight(~0.02 mm ) is too large to see the effect I described above,if we use a lens with focal length 10 cm or so. It means that even if intensity distribution at focal plane(of such a lens) seem to be focused,it is consistent with my argument. Anyway,I'm not sure if my method to calculate the effect is correct. I modeled partially coherent light as many independant wave packets that has small coherence area. I thought that the coherence area is like the slit width. Diffraction pattern width will be deteremined by the coherence area. I don't exactly know how to calculate the pattern width for partially coherent light. I'm reading Born and Wolf for this subject, but still confused.. Is there any compact explanation about this kind of effect? 



#2
Jul1209, 02:02 PM

Sci Advisor
P: 5,468

There are two conceptual limits to any real beam: a fully coherent beam and a fully incoherent beam, and the diffraction pattern for these are fairly simple. The diffraction pattern of a partially coherent beam is significantly more complicated.
For an aperture (or exit pupil) illuminated by a fully coherent beam (i.e. one with a measurable phase everywhere, at all times), the farfield diffraction is the Fourier Transform of the aperture circular apertures have an Airy function pattern, square ones have sinc functions, etc. An aperture/exit pupil illuminated by a fully incoherent beam will have as the farfield diffraction pattern the square of the Fourier Transform of the pupil. A subtle thing to remember is that the Fourier transform is complex, and so it's the modulus squared. Similarly, for coherent illumination, the MTF is the pupil function, while for incoherent illumination, it's the autocorrelation of the pupil function. Goodman's book "Fourier Optics" has an excellent section on the difference between coherent and incoherent illumination, and why one is not really 'better' than another. Incoherent illumination has a higher cutoff frequency, for example. Now, the Van CittertZernike theorem states that the mutual coherence function in the observation plane is the Fourier transform of the intensity distribution in the source plane which can be used to measure the size of objects. Having two adjustable pinholes in the observation plane allows one to measure the mutual coherence function (by measuring the subsequent fringe visibility), which can then be used to infer the size of the original object (i.e. a star). I don't really see how it relates to your problem, tho. does this help? 



#3
Jul1209, 11:25 PM

P: 7

Yes, It helps.
Thanks.^^ However,I'm still a little bit confused. Van CittertZernike Theorem is mainly about propagation of coherence. Let's define f(x,x') as this coherence function. x and x' are postions on same plane. If we consider the case x=x', f(x,x) becomes Intensity at x. And according to VanCittert Zernike theorem, the intensity does not depend on the postion x because f(x,x') depends only on xx'. If we think about intensity of farfield diffraction pattern, an aperture illuminated by fully incoherent light must not make any significant pattern? Does my argument make sense? 



#4
Jul1309, 12:24 PM

Sci Advisor
P: 5,468

diffraction of incoherent beam
Hmmm... ok, so we have the mutual coherence function f(x,x') which can also give the intensity (f(x,x) = I(x) and f(x',x') = I(x').
The vCZ theorem states "The degree of mutual coherence between a point x and a variable point x' is equal to the (intensity, farfield) diffraction pattern of the source (replaced by a hole of the same size and shape), centered on point x' and measured at point x." That's different that wondering if the farfield diffraction pattern exists. Again, an aperture illuminated by incoherent light will still have a diffraction pattern the only difference is that for coherent imaging, we have: E(image) = h * E(source) Where E is the field amplitude, 'h' is the point spread function, and '*' is convolution. For incoherent imaging, we have: I(image) = h^2 * I(source) Where we have to work with the intensity, since the field is not measurable. 



#5
Jul1309, 04:24 PM

Sci Advisor
P: 1,563

There is no (completely) spatially incoherent light after you send it through an aperture. Compare this to the case of temporally incoherent light. The temporal autocorrelation is basically the fourier transform of the power spectrum. Therefore a spectrally broad source will give temporally incoherent light and vice versa. A high spread in the light frequencies will lead to incoherent light. If you use a spectral filter on broadband white light, its coherence time will increase. Loosely speaking the description is similar for spatial coherence. A high spread in the kvectors (corresponding to a high angular size of the light source) leads to low spatial coherence. If you put an aperture inside the beam, you cut off some of the light and thereby create a new "light source" with the size of the aperture. Therefore you now have a light source with smaller angular size and a smaller spread in kvectors. As a consequence you have in fact increased spatial coherence. This is why you use a single slit in front of the double slit when you use incoherent light for the double slit and this is also why light from distant stars is spatially very coherent (for example the original HBTexperiment relied on this). 



#6
Jul1409, 01:08 AM

P: 7

you wrote..
I(image) = h^2 * I(source) I think..below may be more exact expression. "I(x1)=Integral h*(x1,x')h(x1,x) f(x'x)dx dx' " If source is incoherent f(x'x) can easily be repaced to I(x) (f(x'x) may have form I(x)delta(xx')) Now, what is h(x1,x) for the far field diffraction? For the far field diffraction, it may have forms proportional to exp[i*b*x*x1] if we let b some coefficient to link x and x1. If we do the integration,h*(x1,x')h(x1,x) will cancel each other and we will get mere a constant.. (because of the delta function term.. If we replace f(x,x') to I(x)..the result is the same) So I(x1) is proportional to "Integral I(x) dx" since terms related to x1 elliminated, I(x1) does not depend on x1. It is same everywhere! Another approach is the incoherent sum. If we detect light at overlapping position of light from two incoherent sources, intenisity is proportional to I(s1)+I(s2). (if sourcedetector distance is same for the two sources) We don't need to care interference effect. (I don't know how to include a picture in the post) Same effect applies to far field diffraction. There will be no destructive inteference in the farfield because point sources on the source plane are incoherent. Of course, this not physically realizable. (It is the reason why I believe that completely incoherent source can not exist) However, as source becomes more incoherent,the diffraction pattern will be more like mere a constant distribution. Does my argument make sense? (Anyway,how can I include formulas written with MathType into my post?) 



#7
Jul1509, 08:00 AM

Sci Advisor
P: 5,468

For your post, I think you erred in writing down the point spread function when f(x,x') [tex] \propto \delta (xx')[/tex] then I(x1) =[tex]\int h^{*}(x1,x')h(x1,x)f(x',x)dx dx'[/tex] goes to [tex]I(x1) \propto \int h^{*}(x1,x')h(x1,x) \delta (xx')dx dx' \propto \int h(x1,x)^{2}dx[/tex] Also remember that even though the coherent case Ui = h * Uo, we measure the intensity U*U, and so there are interference terms which are not present in Ii = h^2 * Io Thus, although the intensity from two partially coherent sources is a simple sum, the field may not be there can be constructive, destructive, or partial intereference the Rayleigh criteria for two partially coherent sources is different than from two mutually coherent sources. Are we getting anywhere, or going around in circles? 


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