# A mysterious phase difference

Today my teacher was discussing the Poisson spot and gave a simple explanation for why there must be a bright spot on the axis of the disc when illuminated with parallel monochromatic light.

What he said was:

Say we instead have a circular aperture in an infinite plane, we know there must be a bright spot at the centre/axis (The Airy Disk pattern). Now, instead of making a circular hole in the plane, we remove the plane itself to get an opaque disk.

So, we have 2 systems one of an opaque disk and one of a circular aperture of the same dimensions in an infinite plane. If we superimpose both these systems, the light ceases to exist as there is no longer an opening. And since only bright can cancel bright, the center of our first system (opaque disk) must be brightly illuminated to cancel the airy disk pattern.

Now my issue is, how did a phase difference of $$\pi$$ come about between the two systems to induce a destructive interference? Is it just because of complementarity or is there something fundamental going on.

There is a path difference between them for sure, but how are we certain that it is $$(n+\frac{1}{2}) \lambda$$. Also, how are the intensities same.

• Apr 29, 2022 at 4:52
• That doesn't answer my question of why there is a phase difference between the two systems. Apr 29, 2022 at 4:58

Your teacher was referring to Babinet's principle. It is often a good idea to fix your ideas on actual computation. You have an incident light field $$\phi_i$$ on the plane. As it crosses the plane, it either gets multiplied by $$h_a=1{[r\leq R]}$$ in the case of the aperture of radius $$R$$ (origin at the center of the aperture), either multiplied by $$h_d=1-h_a$$ in the case of the disk due to the complementary nature.

The resulting diffraction pattern you are interested in is mathematically described by Faunhofer diffraction, which amounts to a Fourier transform, which is crucially linear. This means that your final field will be $$\phi_f=\mathcal F (h\phi_i)$$, so you see that for the disk: $$\phi_f=\mathcal F (\phi_i)-\mathcal F (h_a\phi_i)$$ The first term is what you would get without the plane, the overall forward beam, while the second term is the field diffracted by the aperture with an opposing phase, which came from the complementarity. For example, in the case of a monochromatic plane wave of normal incidence, $$\phi_i$$ is constant, the first term would be a Dirac peak at the origin and the second term your usual Airy diffraction pattern.

Hope this helps and tell me if you need more details.

• Hi, so my question was they have opposing phases only due to complementarity? That doesn't seem too fundamental. I was curious how it got an opposing phase, where is that (n+1/2)$\lambda$ path difference coming from. Apr 29, 2022 at 10:17
• Surprising as it may be, yes it is solely due to complementarity, and it is pretty fundamental because the arguments are easily generalized hence Babinet's principle. The minus sign can be traced back simply to the linearity of the Fourier transform and the definitions of the filters.
– lpz
Apr 29, 2022 at 10:25
• Careful, there is no path difference here, we are dealing with diffraction, not interference, there are an infinite number of paths over which you need to calculate the phase difference and sum up, which is captured by the FT. Thankfully, you don't need to do the actual computation and these arguments allow you to obtain the final result directly.
– lpz
Apr 29, 2022 at 10:25