Past exam of the optics and electrodynamics course of the University of Cambridge / 2022 / Section A / 3

Each mutually incoherent patch of an extended star produces a displaced double-slit interference pattern. The detector adds their light intensities. For two apertures separated by a projected interferometric baseline $D$, a source direction offset by a small angle ${\vartheta }_x$ changes their phase difference by $2\pi D{\vartheta }_x/\lambda$. The normalized sum over the source's angular brightness distribution is its first-order optical coherence, whose magnitude is the interferometric visibility for equally bright channels.

For a uniform circular stellar disk of full angular diameter $\alpha$, the normalized angular Fourier transform gives the uniform-disk stellar visibilityHere $J_1$ is the order-one Bessel function of the first kind. This follows by integrating $e^{i(2\pi D/\lambda ){\vartheta }_x}$ over the disk and dividing by its area. In particular, its power series gives $2J_1(x)/x=1-x^2/8+O(x^4)$.

For an ordinary slit separation $d=1\mathrm{mm}$, using the paper's $\alpha =22.6\times 10^{-8}\mathrm{rad}$ and observing wavelength $\lambda =570\mathrm{nm}$,Even $d=10\mathrm{mm}$ gives a visibility loss of only $1.94\times 10^{-5}$. Thus millimetre separation leaves the interference fringes almost indistinguishable from those of a point source; it gives very little sensitivity to the diameter.

The movable mirrors in the Michelson stellar interferometer collect light at a large adjustable separation $D$ and redirect it into closely spaced output beams. They increase the sampling interferometric baseline independently of the fixed output slit spacing $d$: $D$ sets sensitivity to source structure, whereas the output optics set the screen's interference fringe spacing. The first Bessel function zero is $x_1=3.83170597$, henceThe interference fringes disappear at this baseline even though each beam separately remains bright.

The value $3.08\mathrm{m}$ assumes a uniform disk. The original question supplies a diameter without a surface brightness law; real limb darkening changes the precise visibility zero. A one-dimensional uniformly bright strip of the same full angular width instead has a sinc function visibility and first zero $D=\lambda /\alpha =2.522\mathrm{m}$. That is a different source model, not the circular-disk result. The robust conclusion without a surface brightness model is that a baseline of a few metres, rather than millimetres, is required.