Two satellites at an altitude of 1200 km are separated by 28 km. If they broadcast 3.6-cm microwaves, what minimum receiving-dish diameter is needed to resolve (by Rayleigh’s criterion) the two transmissions?

Respuesta :

as we know that

[tex]a \delta x = 1.22\lambda L[/tex]

here we know that

a = distance of two sources = 28 km

[tex]\Delta x [/tex] = diameter of dish

[tex]\lambda = 3.6 cm[/tex]

L = 1200 km

now from above equation

[tex]28 \delta x = 1.22 \times 0.036 \times 1200[/tex]

[tex]\delta x = \frac{1.22\times 0.036 \times 1200}{28}[/tex]

[tex]\delta x = 1.88 m[/tex]

so diameter will be 1.88 m