Answer to Question #203688 in Astronomy | Astrophysics for Urain

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 Earth and Jupiter is 968 × 10^6 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 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

Answer:-

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