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if we take an very small point source of EM wave, we will find that the EM wave is polarized. How does the polarization vary depending on which point of the sphere ? Polarization is always shown along one axis, but in a real 3D spherical wave it's impossible to keep a transversal EM field vector without changing its direction EDIT: My question comes from optics, if the image of a point is formed from constructive interference, then for such interference to occur, the polarization must have a specific configuration in 3D

Manu de Hanoi
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1 Answers1

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The E and B vectors of a spherical EM wave lie in a plane normal to the direction of the propagation of the wave and their magnitude and direction (within that plane), at any point in space, are randomly changing with time.

The source of such wave could be considered a point source only in a sense that it is small relative to the distances at which the wave is evaluated. The source has to contain a large number of incoherent radiators to ensure a uniform distribution of radiation in all directions.

We can also loosely say that the EM spherical wave is unpolarized, such as EM radiation from a light bulb or a star, keeping in mind that it is in fact always polarized, but the polarization is randomly changing.

You can find more details on this topic in this post.

V.F.
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  • Thanks, let's assume it's not perfectly spherical, say, just half a sphere. Let's freeze time, how does polarization change around that half sphere ? see my EDIT on the question – Manu de Hanoi May 13 '18 at 16:31
  • Yes, the image of a point, is indeed is formed by interference of multiple wavelets and, since the interference could be constructive or destructive, the image will look like a series of rings (Airy pattern). All the wavelets forming such image will be coming from the same point of the source, i.e., the point will radiate light in all directions, some segment of that light will pass through a lens and converge on the image plane. In other words, all these wavelets will be coherent and have the same polarization. But the wavelets forming other points could have different polarization. – V.F. May 13 '18 at 20:25
  • how could they all be polarized the same, when, as you said, the E field is normal to the direction of the propagation and the direction of the propagation is different for every single ray; You can do this if all the rays are in a plane but not in 3d – Manu de Hanoi May 14 '18 at 06:33
  • For small apertures, like the pupil in a human eye, that difference in polarization won't be significant and won't cause any significant image distortion. – V.F. May 14 '18 at 12:22
  • im trying to understand how the polarization is distributed in order to research contrast enhancing microscopy techniques – Manu de Hanoi May 14 '18 at 17:59
  • The answer is that polarization is normal to the direction of propagation and random in magnitude and direction within the plane of the wave. Such distribution does not prevent constructive interference for individual image points for the reasons listed in my comments. Do you have other questions? – V.F. May 14 '18 at 18:17
  • thanks for your answers but you havent told me how the polarization is distributed along the wavefront of a point source. You said "polarization is normal to the direction of propagation and random in magnitude and direction(I add as time varies, but not a given time) within the plane of the wave."

    In the sentence above, if we understand the "direction" as a 3d vector, you will see that the wavefront from a point source (a curved surface) cannot have a single 3d vector although light is supposedly polarized in a single direction . continued ->

     – Manu de Hanoi May 14 '18 at 19:44
  • Therefore I'd like to know how that 3d direction vector varies as we move on a single polarized wavefront . If we take a hemishere as a model for the wavefront, you will see the tangeant to that hemisphere (= the polarization normal to the propagation direction) cannot all be parrallel – Manu de Hanoi May 14 '18 at 19:53
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    A single point won't produce a spherical or hemospherical wave. It could possibly be modeled as a dipole with its doughnut-shaped radiation pattern. E-vector of a dipole will be in the same plane as the dipole, will be getting smaller in magnitude as the altitude angle increases and will be normal to the direction of the wave propagation or to the line connecting the mid-point of the dipole with the point of interest. – V.F. May 14 '18 at 22:25