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I Abbe theory: diffraction pattern to primary image

  1. Mar 26, 2017 #1
    I have been attempting to unravel Abbe’s theory, of the role of diffraction in microscopic vision, in a ‘nuts and bolts’ sort of way, meaning in terms that I can understand: diffraction, interference and so on. While Fourier transforms are clearly at the heart of the process the assertion that an object undergoes a Fourier transformation into components (a diffraction pattern) which then undergo a ‘reverse’ transformation into an image is just an assertion, to me. I only feel I understand what is going on when I learn how the processes of diffraction and interference work to produce that pattern. All this is in widely available texts (I use Jenkins & White, Fundamentals of Optics) but I am struggling with the last bit of the process: how the diffraction pattern is re-transformed into the image.
    https://lh3.googleusercontent.com/y32sRf0V8G0x19IlSiwuIscME3xdInGRCVbr_CQa9p6U-0h9pwqrG24PnT_ss2KDBY1hIR_9o-5qF652HhnuU7pzbNQbVo_6tIFnxhCxh1ewcwFJIZ1Bq4jZPHGLr2fNHqDPC8Ej6wfverxufuX-cTDLM7Bv-g_fCnRsbhkCtEiDdce896TW24feo1TG6TT2AQyVAozFlrZkaGcFFK8UYEQ52ccwZeAF5zdaxyA4ISEzNUdZp_jkIw4aQ1BqN2s2tfCt0ACeBOHPi6CPqkQewrExkpsgYUUGRdeOPkg9q9gaoN4Qw7r2YXTIbdCrGlCbFE9o6cUIeaJWAwcOtQ1xBVDRnv2VM8oHwuJ9GdXI2bmp8Ot1FqKjqSbwkicUCzVGofzfjYqIyiq0NMe_pNcQhG2pqU4zO04aZRl6yb379F2dGwCR2QBXrnziCFAhb-VAnpXGZv5OfGQmjy79Z5cE8iub2LGKEfIrBuO15JwoLl0ExvoEFxZ4NNPEx_0gZhwlVJcp66SV90vbGC_wAY7pJ_dL9Ixcdn_bMZS7a3itjJd5QRfBKn-JGrkLdf9ftzPTf-QkTNNtH5PEVaOH4dqwhjfUuEglBonMb94wKUDqwHifNZoAo7BV=w1794-h972-no

    Figure 1 is my version of the conventional summary of the process. The object is a grating with a 1:1 ‘mark space’ ratio, or equal opaque and transparent elements. This gives rise to zero order light of amplitude >+A and sine elements of frequency F (same as the grating, but magnified x M) and amplitude -A (from the 1st maximum), also of 3F with amplitude +A/3 (from the 3rd maximum), 5F at –A/5 (from the 5th maximum) and so on: all the ‘odd’ orders alternating in phase and antiphase and diminishing in amplitude. These are represented in the figure as b, c, d and e. The brightfield image is all these combined in g, and the darkfield image is all except the zero order, in f.

    So how does this happen: how are the elements of the diffraction pattern ‘written’ into the primary image plane as sinusoidal ‘standing waves’ of illumination of the appropriate periodicity? I cannot find anything, for instance in Geoffrey Brooker’s Modern Classical Optics, except the familiar assertion above: a reverse Fourier transform.

    https://lh3.googleusercontent.com/e9uyMuCC14-XfNcGXr3VWx4I7nzA54Pi9YU2RKK0cVdDOv9xTT7Mugw3P6gkJMSOYx7_7ddlnPRO0ND8uQd0lSJdbCeu9jsS2jEWmJbO2o8bm9NZ1SLB_os3yo0KNTmz-LO0XN_vAgRriKoYZaobF5i8NBoXQkO_jw_cs0MxFYF6ai7YL3-dFND53CKsNVo2LBmuLbtpWgj1bBtJK2f6LxOOdqphbzUQ6qOwlkSsZJdirSrvIWKGhkcXLyeEsk2ecH2U9_hjT1hGLQEYFQ4A6Q3Z1TUrywPnR96RyHTU58uYfU6OfW8WSEyCQ80KC9iKzNQVlFmRoataO7L8fC8V87KzqTChBeRiva8eiKWvOOWxU5hwsLjjPhE-nHrbU3I3vLyHI0QufZ0_wqRQrvjngTdghdCcFjoXLuEMz9Hf8RNAH5WCohlCUBB09bxMh6CnXGGw7NXOz5_LTiWUgvdmLV-pEnuEaqBhUw838av5C8JnqUJEyfORtkDnpsbHzTZAskwN_4Kzb7HS3bcdQeNO49wKwE_bF-0KefzG0x8ZoneA2JbsHCSGSHzMrykCpqcu2NFqmbdvh1-joWICS58klpC2QpsMDQpWhIcLM86b8c_q01tu=w1233-h996-no
    Figure 2 is my guess. 2B follows rays from both edges of a slit object to their positions in the image plane, via a rather dumpy objective represented by its two equivalent planes. 2A notes that there is a path difference Pq between the rays from each pair of higher order diffraction maxima, in the interference plane (aka objective rear focal plane) and the image plane. The bit of elementary trigonometry below shows that this path difference is equal to the path difference between rays from opposite edges of the slit at the angle Theta. The trig. looks pretty outrageous but angle mnq is <2 degrees with a x40 objective so the small angle approximation should hold well. Supposing the outer rays to be from 3rd order maxima the phase difference across the slit will be 3L/2 (L being wavelength) so the rays mn and qn uniting in the image plane also have a phase difference of 3L/2. They will thus interfere destructively giving a minimum of intensity at this point. Rays from the centre of the object slit will unit in the image on axis, a will be zero, there will be no phase difference between them, and they will interfere constructively giving a maximum. Between centre and edge the phase difference will change from 0 to 3L/2 ‘writing in’ 1 ½ periods of a sine wave. The same thing happens in the other half of the image so a total of 3 waves will be ‘written in’ as required by theory (for the 3rd order maximum).

    All of which is great, but they are the wrong way up. See fig. 1. As will be the 1st order rays, here in antiphase to the zero order. If they interfere constructively on axis then there would be a dark spot in the brightfield image where a bright spot should be. What please is wrong with my picture? (fig. 2) What have I ignored / misunderstood / got mixed up? Or, is fig. 2 right and there is something wrong with the account in Fig. 1? I don’t think there is: I am a long time microscope user and it seems to ‘add up’.

    Any help very much appreciated. I can’t stop trying to figure this out and get on with the rest of my life… Thanks for reading this far, John SW.
     
    Last edited: Mar 26, 2017
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  3. Mar 26, 2017 #2

    Charles Link

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    Perhaps a simpler approach would be to just consider the focusing to a single point (small spot) in the focal plane (essentially at the focal point) for parallel rays from infinity for a paraboidal mirror. For spherical lenses and mirrors and working with finite object and image distances, the principles (diffraction theory) are similar, but in some ways, not quite as simple mathematically because the functions, (the focused diffraction patterns), will be more complex. In any case, you might find an Insights article that I recently authored, of interest. Here is a link to that article: https://www.physicsforums.com/insights/diffraction-limited-spot-size-perfect-focusing/ ## \\ ## Editing... In the article, I mainly focused on energy conservation, and I did not write out the diffraction integral with the ## e^{ik(x^2+y^2)/(2f)} ## phase term for the curved paraboidal surface. (The equation of the paraboloid is ## x^2+y^2=4fz ##). If I recall correctly (it's been a few years since I computed it), the integral for the intensity in the focal plane is most easily done by evaluating it in the ## dxdy ## rectangular form. It can readily be shown that the ## \Delta x ## and ## \Delta y ## of the intensity pattern in the focal plane are inversely proportional to the aperture size ## 2a ## in the x direction and ## 2b ## in the y direction. Thereby, the focused spot size area is inversely proportional to the area of the optical beam. ## \\ ## Additional item: The far field (Fraunhofer) diffraction pattern can be observed in the focal plane of a focusing optic (at close range), because parallel rays at a given angle come to a focus at a location in the focal plane that depends upon that angle.(So instead of going to that location in the far-field, you can simply look at the corresponding location in the focal plane. A simple way of observing the far-field diffraction pattern is to simply use a focusing optic and you can observe the far-field pattern in the focal plane.) In general, although the focal plane is said to contain the F.T. of the incident beam pattern, and the Fraunhofer diffraction equations contain mathematics that is essentially a Fourier transform, for most applications, I haven't found it of much use to emphasize the Fourier transform relationship. e.g. in evaluating most integrals that appear in diffraction theory, it is not necessary to consult a handbook of Fourier transforms.
     
    Last edited: Mar 26, 2017
  4. Mar 28, 2017 #3
    Hi Chales Link

    Thank you very much for your thought provoking comments and link to the beautifully clear analysis of the point source ‘image’ formed by a parabolic mirror. If I understand it correctly this ‘image’ is what is often loosely described as the Point Spread Function (PSF) of the converging optical system, the parabolic mirror. It is interesting to contrast this case with what is occurring in a short focus, large Angular Aperture system such as a microscope.

    The key difference seems to be that, while both systems form an FT of the source ‘object’ in their posterior (downstream) focal plane, in one case this plane is also the image plane (for the parabolic mirror w. source distant, say case 1) while in the other case (large AA objective with source close, say case 2) it is the aperture plane, the image being formed at a distant conjugate focus.

    The spot size, case 1, is determined by conditions of interference between diffracted light spreading from the converging spherical wavefront aka image forming cone establishing the radius of the first minimum that radius being smaller as the conic angle increases, for a given wavelength. All light ‘on focus’ is in phase, only diffracted light ‘propagating’ off the conjugate focus becomes out of phase.

    In case 2 the ‘spot’ or image size is determined by the ‘wavelength’ of the various Fourier components of the original object ‘waveform’ written in to the image space, as per my fig. 1. This at any rate is the conventional account and I have no issues with it. The geometry of path difference in my fig.2 is again conventional and it seems at least plausible that it applies to this situation, the formation of the primary image from FT elements of the diffraction pattern. It accounts perfectly for the spatial ‘frequencies’ required and obtained in the image-component waveforms but not for their phase (I refer to the image component phases, not to the phase of the image forming light).

    The composite waveforms shown in fig.1 g & f are substantially correct, as I know from experience with the microscope. The grating spaces show light in a brightfield image and the bars show dark; in a darkfield image the opposite is the case. For this to occur the components c, d, and e must be as shown, as well as the zero order b. I we call the image forming light of b and d in phase, then that forming c and e is in antiphase. The grating space in g will only be light if there is a minimum (of anti phase light) due to interference at the centre point of ‘waveform’ c but the conditions of interference shown in fig. 2 suggest the opposite to be the case. Likewise components d and e must both have minima at the centre point to approximate to the ‘square wave’ form of the grating image, but again I cannot see how this occurs. This is my problem.
     
  5. Mar 28, 2017 #4

    Andy Resnick

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    There's a lot here, I can't immediately grasp all of it. One thing I notice is that you are confusing 'amplitude' and 'intensity'. Figs 1 b-g seem to be the intensity, which is the square of the field- the FT of the intensity is the autocorrelation of the transform of the field amplitude. I guessed this because your 'darkfield' 1f is incorrect- removing the DC component will enhance the edges of your square wave object.
     
  6. Mar 28, 2017 #5

    Charles Link

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    @JohnSW Your exploration of the case of the diffraction image formed by an illuminated scene in a microscope is very interesting. This case is one that I have yet to explore in detail.
     
  7. Mar 28, 2017 #6

    Andy Resnick

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    Goodman's book "Fourier Optics" is the go-to reference. It matters if the imaging system is coherent (amplitude detection) or incoherent (intensity detection), optical techniques like phase contrast (matched filters), DIC, etc are coherent processes. For coherent systems, spatial frequencies 'f_x' at the object plane end up as Bragg diffraction peaks in the exit pupil plane, at location x = λf_x*f, where f is the focal length of the lens. You don't generally image the exit pupil, but certain techniques do exactly that (static light scattering, for example) using a special "Bertrand lens" instead of a tube lens.

    https://www.researchgate.net/publication/238916989_Optical_properties_of_ordered_arrays_of_large_latex_particles/figures?lo=1[/URL]

    In addition to Bragg peaks, you can also obtain Kossel lines:

    [URL]http://www.nature.com/articles/srep44575[/URL]
     
    Last edited by a moderator: May 8, 2017
  8. Mar 28, 2017 #7

    Charles Link

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    @JohnSW This subject goes slightly beyond my expertise, but one question I have, particularly in looking at Fig.1g, might it be possible to get most of the structure of the resulting image from a point by point and/or point to spot transfer of diffraction pattern from object plane to image plane, or does the entire E-field in the object plane generate an interference/diffraction pattern that is substantially different from the point-to-point transfer? I'm on a learning curve here=I'm just trying to ask a question that might lead you more quickly to the correct result.
     
  9. Mar 29, 2017 #8
    Thank you both for your thoughts and (further) suggestions. I will order a copy of Goodman and try to work through the 'simple' case of the point object at the focus of a parabolic mirror which may indeed be a good place to start. I cannot answer the question about point to point vs. 'entire field' approaches but again this sounds like a profitable line of enquiry. When I say 'try to work through' it is because my maths is not the greatest. I have only got as far as I have on this subject with the aid of the 'vectoresque' graphical method of phasor addition. And I am encouraged by Goodman's reputation for clarity.
    With regard to intensity vs. amplitude I have tried to be clear and consistent on this. Some mix up between the two could indeed be the cause of my difficulties, essentially just inverting the answer I have arrived at. I chose amplitude for fig. 1 because it conveys more information: thus the in-phase (with zero order) components appear above their respective base lines while the antiphase components appear below theirs with + and - signs as appropriate. Is there something I have ignored here?
    I had imagined that the brightfield and darkfield curves would be essentially identical, with the exception of the flat, 'dc' component in the former. I believe that in the microscope image the appearance of sharpness in the DF image is in part an artefact due to the higher contrast and in part real but due to the fact that the objective is necessarily working at full aperture whereas in BF operation it is nearly always stopped down (to 3/4 or 2/3) for the sake of the dark-field-illumination.jpg bright-field-illumination.jpg improvement in contrast this gives. The rather clumsily inserted images from 'the web' look to me to be relatively free of post processing, probably with a low medium power (x10?) objective. (not sure if the images will come out: they dont show in the preview window but I have to go so will leave them as is)
    Thank you for your interest and encouragement and I will report back when I have tackled Goodman if not before.
     
  10. Mar 29, 2017 #9

    Charles Link

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    @JohnSW Good to hear from you. I think you might find, (I don't know for sure), that the answer to my question is that in cases such as the phase contrast microscope, the diffraction pattern from the beam across the entire plane can be considerably different from that of the point to point/spot transfer. I looked forward to hearing what your further exploration of the subject reveals.
     
  11. Mar 30, 2017 #10

    Andy Resnick

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    Let's start with something simple- something that for me, really helped understand what is going on.

    Compare the far-field diffraction pattern (which appears at the exit pupil, etc) of an incident plane wave that passes through two different apertures. One is an amplitude grating T = 1+cos(2πf0x), the other is a phase grating: T = e^[i(1+cos(2πf0x)].

    The far-field diffraction pattern is the (complex) Fourier Transform of the incident field amplitude, and then square it to obtain the intensity. Goodman works this out in detail, but as I said, working this out for myself was extremely informative.
     
  12. Mar 31, 2017 #11
    Hi Gentlemen. Nearly, but not quite, put together a response to your most welcome comments. Going away a few days will respond on return. JohnSW
     
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