Question

The Hubble Space Telescope has an aperture of 2.4 m and focuses visible light (400-700 nm). The Arecibo radio telescope in Puerto Rico is 305 m (1000 ft) in diameter (it is built in a mountain valley) and focuses radio waves of wavelength 75 cm.
Under optimal viewing conditions, what is the smallest crater that each of these telescopes could resolve on our moon?

Answer 1

Answer:

$y_{hubble} = 77\ \ m$

$y_{aceribo} = 1.1*10^6 \ \ m$

Explanation:

what is the smallest crater that each of these telescopes could resolve on our moon?

For moon ;

s = 3.8 × 10 ⁸ m

y = 1.22 λs/D

where;

λ = 400 nm = 400× 10 ⁻⁹

D = 2.4 m

The smallest crater for the hubble space is calculated as follows:

$y_{hubble} = 1.22*400*10^{-9}*3.8*10^8/2.4$

$y_{hubble} = 77\ \ m$

For Aceribo ;

y = 1.22 λs/D

where :

λ = 75 cm = 0.75 m

D = 305 m

$y_{acerbo} = 1.22*0.75 *3.8*10^8/305$

$y_{aceribo} = 1.1*10^6 \ \ m$