Question

An astronaut in a space shuttle claims she can just barely resolve two point sources on Earth's surface, 240 km below. Calculate their (a) angular and (b) linear separation in meters, assuming ideal conditions. Take ? = 538 nm and the pupil diameter of the astronaut's eye to be 5.0 mm.

Answer 1

Given : $\lambda$ = 538 ×10^-9m ; d = 5×10^-3m ; D =240×10³ m

Solution:

(a)

Angular separation can be calculated as :

sin $\theta$ = 1.22 $\lambda$ /d

sin $\theta$ = (1.22)(538×10^-9)/(5×10^-3)= 1.313×10^-4

for small angles , sin $\theta$ ~ $\theta$

Hence angular separation, $\theta$ = 1.313×10^-4 rad.

(b)

Linear separation (S) can be calculated as :

S/D = sin $\theta$

S/(240×10³)= 1.313×10^-4

i.e. S = 31.512 m