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Poisson's spot and the circularity of the obstacle

  • #1
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Homework Statement


For poisson's spot to be observed, how accurately circular must the stop be? How is the argument for the Poisson's spot changed if the light is either; white or spatially incoherent?

Homework Equations



The Cornu spiral for the circular aperture.

The Attempt at a Solution



Using the Cornu spiral I understand that the Poisson spot will be present, however I don't understand the diffraction theory enough to be able to deduce how the solution to the Fresnel integral is going to change as a result of the non perfect circularity of the obstacle.

The way I tried solving the problem was to think about the Fresnel half period zones. From the Cornu spiral it is clear that when the obstacle covers and odd number of the half period zones the Poisson spot will disappear. So if the defect in the obstacle is periodic (suppose like a sine wave) which is centred on the boundary between the ultimate and penultimate covered half zones, the if the sin has the amplitude equal to the width of the half plate zone the result will be the disappearance of the Poisson's spot.

The second part of the question gets a bit more complicated, namely; the width of the half period zones is wavelength dependent. However it seems to me that this should not prevent there existing a Poisson's spot for certain wavelengths for certain obstacle sizes and distances? However I don't know what would be the effect of the spatial incoherence?

Is the above reasoning correct? is there a way to make the explanation a bit less
hand-wavy without involving mathematics?
 

Answers and Replies

  • #2
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Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
 
  • #3
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No unfortunately, I did not get any further.
 
  • #4
I agree with your argument with regard to the accuracy of the stop's edge. As long as the edge corrugation of the disc is much smaller, eg 1/10, than the width of the adjacent Fresnel zone: w = \sqrt(R^2 + (lambda *g *b)/(g+b)), where g and b are the distances between source to disc and disc to detector, respectively, the Spot will not be notably dampened by the edge corrugation. R is the disc radius, and lambda the wavelength. Temporal coherence, i.e. the width of the wavelength distribution is not so important - the Poisson spot is white for a white light source (within the bound of the above equation, and that the intensity of the spot is roughly proportional to wavelength). However, spatial coherence, i.e. the size of the source is very important. If the transverse coherence length at the position of the disc is about equal to the diameter of the disc, the relative intensity of Poisson's spot will be 1, i.e. equal to the intensity of the unobstructed light field. For large sources of width w_s the intensity is approximately 2/pi * g/w_s * lambda/R. More details you can find in the following article: https://doi.org/10.1088/1367-2630/aa5e7f
 
  • #5
Charles Link
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The previous post probably answered most of the questions on this, but in any case, I do remember our Optics professor telling us that this is a difficult experiment to perform, apparently requiring a very precise alignment between the collimated source and the circular obstacle if a good bright spot is going to be observed.
 

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