Question

Please help!

A satellite has two bright lights at the end of its two solar panels to indicate its position. The distance between the two lights is 5.39 m. The wavelength of the light emitted by the indicator lights is 548 nm. If a telescope with an aperture of 68.9 cm can just resolve the two indicator lights, then how far is the satellite from the telescope? (Hint: Use the Rayleigh Criterion, and the small angle approximation: sin(θ) = tan(θ).)

Answer 1

Given,

y = 5.39 m ; $\lambda$ = 548 nm ; d = 69.9 cm

We know that for circular apperture:

$\theta$ = 1.22 $\lambda$/d = 1.22 x 548 x 10^-9 m / 69.9 x 10^-2 = 9.56 x 10^-7 rad

From small angle approximations, tan$\theta$ = sin$\theta$ = $\theta$ = y/D

9.56 x 10^-7 rad = 5.39/D

D = 5.39/9.56 x 10^-7 = 5.64 x 10⁶ m

Hence, distance of satellite from the telescope = D = 5.64 x 10⁶ m.