Answer to Question #201542 in Astronomy | Astrophysics for J17

Question

Telescopes are an essential tool for astronomers to study the universe. You plan to build your

own telescope that can resolve the Great Red Spot on the surface of Jupiter at a wavelength of

600 nm. The farthest distance between the Earth and Jupiter is 968 × 106 km and the Great Red

Spot has currently a diameter of 16,500 km.

(a) Use the Rayleigh criterion to determine the diameter of the lens’ aperture of your telescope

that is needed to resolve the Great Red Spot on Jupiter.

Impacts have formed many craters on the Moon’s surface. You would like to study some of the

craters with your new telescope. The distance between Moon and Earth is 384,400 km.

(b) What is the smallest possible size of the craters that your telescope can resolve?

Answer 1

Let us determine the angular size of the spot. We should divide the diameter of the spot by the distance, so we'll get the size in radians:

$\alpha = \dfrac{R}{d} = \dfrac{16500\,\mathrm{km}}{968\cdot10^6\,\mathrm{km}} \approx 1.7\cdot10^{-5}\,\mathrm{rad}.$

(a) The diameter of the lens will be smallest, if this angle is equal to the angular resolution of the telescope, or

$\alpha = 1.22\dfrac{\lambda}{D}$ . Therefore

$1.22\dfrac{\lambda}{D} = 1.7\cdot10^{-5},\;\; D= \dfrac{1.22\lambda}{1.7\cdot10^{-5}}, \\ D = \dfrac{1.22\cdot600\cdot10^{-9}\,\mathrm{m}}{1.7\cdot10^{-5}} = 0.04\,\mathrm{m}.$

(b) The smallest angular size of the crater should not be less than the angular resolution of the telescope. So let us consider an equality

$\alpha = 1.22\dfrac{\lambda}{D} = 1.7\cdot10^{-5} = \dfrac{d_c}{L},$ where $d_c$ is the diameter of the crater and $L$ is the distance from Earth to Moon.

Therefore, $d_c = 1.7\cdot10^{-5}\cdot L = 1.7\cdot10^{-5}\cdot 384400\,\mathrm{km}. = 6.6\,\mathrm{km}.$