Question

What diameter telescope is needed to resolve the separation between an Earth-like planet and its star at 550 nm if the linear separation between them is 1 AU and the star system is 3 pc from Earth

Answer 1

The Rayleigh criterion allows finding the result for the diameter of the telescope that allows solving the separation of the star and the planet is:

• The diameter of the telescope is D = 0.415 m

The Rayleigh criterion is used to find the separation of two points, it is based on the fact that the diffraction maximuum pattern of the first object coincides with the first minimum of the second object.

By entering in the diffraction ratio for slits you will find.

           sin θ  = $\frac{\lambda}{a}$  

In general in diffraction experiments the angles are very small,

           $tan \theta = \frac{y}{x} = \frac{sin \theta}{cos \theta} \\sin \theta = \frac{y}{x}$

 

For the case of circular apertures, when solving in polar coordinates, a constant appears.

 

        $\frac{y}{x} = 1.22 \frac{\lambda}{D}$

       $D = 1.22 \frac{\lambda \ x}{y}$

Where λ is the wavelength of light and D is the diameter of the aperture.

They indicate that the separation between the star and the planet is 1 AU and the distance from the system to the Earth is 3 parce.

Let's reduce the parce to astronomical units

       x = 3 pc (  $\frac{206264 AU}{1 pc}$ )

       x = 6.18 10⁵ AU

Let's calculate

          D = $D = 1.22 \ \frac{550 \ 10^{-9 } \ 6.18 \ 10^5 }{1}$  

          D = 0.415  m

In conclusion, using the Rayleigh criterion we can find the result for the diameter of the telescope that allows solving the separation of the star and the planet is:

•  The diameter of the telescope is D = 0.415 m

Learn more about the Rayleigh Criterion here:  brainly.com/question/20113743