An interferometer observes a spectroscopic binary whose two stars sit too close…

PHYSICS · Multi-part · Fermi University Physics · Optics & Photonics · Coherence and statistical optics · Coherence and statistical optics

What an Instrument Can See, Ch Spatial coherence and the van Cittert-Zernike theorem

Problem

An interferometer observes a spectroscopic binary whose two stars sit too close together for any single telescope to resolve, at wavelength λ=550 nm\lambda = 550\ \mathrm{nm}. Each star is unresolved on its own, so it may be treated as a point (delta-function) incoherent source: a primary of flux I1I_1 and a secondary of flux I2=rI1I_2 = rI_1, with brightness ratio r≤1r \le 1, separated on the sky by a small angle α\alpha. The interferometer's baseline B\mathbf{B} is aligned with the line joining the two stars, so the problem is one-dimensional, and the pair sits at angular positions θ=∓α/2\theta = \mp\alpha/2 (primary, secondary) along that line. The van Cittert-Zernike theorem of the propagation-of-coherence section gives the complex coherence factor as the normalised Fourier transform of the source's angular brightness, μ12(B)=∫I0(θ) e−i2πθB/λ dθ∫I0(θ) dθ,\mu_{12}(B) = \frac{\displaystyle\int I_0(\theta)\,e^{-i2\pi\theta B/\lambda}\,d\theta}{\displaystyle\int I_0(\theta)\,d\theta}, evaluated at the baseline BB (the phase factor the theorem divides out is dropped throughout). ((a)) Write I0(θ)I_0(\theta) as two delta functions of weight I1I_1 and I2I_2 at θ=∓α/2\theta=\mp\alpha/2, and show that μ12(B)=I1 eiπαB/λ+I2 e−iπαB/λI1+I2.\mu_{12}(B) = \frac{I_1\,e^{i\pi\alpha B/\lambda} + I_2\,e^{-i\pi\alpha B/\lambda}}{I_1+I_2}. ((b)) Show that the squared fringe visibility is ∣μ12(B)∣2=1−4r(1+r)2sin⁡2 ⁣(παBλ),|\mu_{12}(B)|^2 = 1 - \frac{4r}{(1+r)^2}\sin^2\!\left(\frac{\pi\alpha B}{\lambda}\right), and check that setting r=1r=1 recovers the equal-brightness result ∣μ12(B)∣=∣cos⁡(παB/λ)∣|\mu_{12}(B)| = |\cos(\pi\alpha B/\lambda)|, which does reach zero. ((c)) The measured visibility first dips to a minimum ∣μ12∣=0.60|\mu_{12}| = 0.60 at a baseline B1=4.0 mB_1 = 4.0\ \mathrm{m}, and no shorter baseline shows a smaller value. Find the angular separation α\alpha of the two stars, in milliarcseconds. ((d)) Find the brightness ratio rr from that minimum, and explain why this binary's visibility curve never falls to zero on any baseline.

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