The Event Horizon Telescope (EHT) recently observed the shadow of M 87* (the supermassive black hole, of about 6 billion solar masses, at the center of the giant elliptical galaxy M87, which is about 55 Mly away). If the observations were made at the radio wavelength of 1.3 mm & were at the diffraction limit for angular resolution at about 50 micro-arcseconds, about what was the distance between EHT’s component telescopes?

The Rayleigh criterion determines how close two light sources can be so that they can be seen separated, the Rayleigh criterion statement is that the first minimum of diffraction of a star coincides with the maximum of the other star, this method gives that the diffraction equation remains

θ = 1.22 [tex]\frac{\lambda}{D}[/tex]

Where θ is the diffraction limit, λ the wavelength and d the diameter of the detection system.

As the observation is taken by two telescopes, the diameter of the lens is the separation of the telescopes

D = 1.22 [tex]\frac{\lambda}{\theta}[/tex]

Let's reduce the magnitudes to the system intentional of measures

x = 55 106 ly () = 55 10¹¹ m

θ = 50 10-6 arcseconds ( ) = 242.4 10⁻¹² radians

λ = 1.3 mm ( ) = 1.3 10⁻³ m

Let's calculate

D = 1.22 [tex]\frac{1.3 \ 10^{-3} }{242.4 10^{-12}}[/tex]

D = 6.54 10⁶ m

In conclusion using the Rayleigh criterion for diffraction we can find the answer for the separation of the telescopes is:

D = 6.54 10⁶ m

Learn more about the Rayleigh criterion here:

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