Question

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?

Answer 1

The Rayleigh criterion for diffraction allows finding that the answer for the separation of the telescopes is:

          D = 6.54 10⁶ m

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 $\frac{\lambda}{D}$

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 $\frac{\lambda}{\theta}$

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 $\frac{1.3 \ 10^{-3} }{242.4 10^{-12}}$  

        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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