The aperture diameter of a telescope is 5 m. The separation between the moon and the ear this 4 × 105 km. With light of wavelength of 5500 mathringA, the minimum separation between objects on the surface of moon, so that they are just resolved, is close to :

Question

The aperture diameter of a telescope is 5 m. The separation between the moon and the ear this 4 × 105 km. With light of wavelength of 5500 mathringA, the minimum separation between objects on the surface of moon, so that they are just resolved, is close to : (A) 600 m (B) 20 m (C) 60 m (D) 200 m

Tardigrade Admin · Accepted Answer

Let distance is x then dθ=(1.22λ/D) (D = diameter) (x/d)=(1.22λ/D) (d = distance between earth moon) x=(1.22×(5500×10-10)×(4×108)/5)=53.68 m most appropriate is 60m.

Q. The aperture diameter of a telescope is 5 m. The separation between the moon and the ear this $4 \times 1 0^{5}$ km. With light of wavelength of 5500 $\overset{˚}{A}$ , the minimum separation between objects on the surface of moon, so that they are just resolved, is close to :

600 m

20 m

60 m

200 m

Let distance is $x$ then
$d θ = \frac{1.22 λ}{D}$ ( $D =$ diameter)
$\frac{x}{d} = \frac{1.22 λ}{D}$ ( $d =$ distance between earth & moon)
$x = \frac{1.22 \times ( 5500 \times 1 0 ^{- 10} ) \times ( 4 \times 1 0 ^{8} )}{5} = 53.68 m$
most appropriate is $60 m .$

Q. The aperture diameter of a telescope is 5 m. The separation between the moon and the ear this 4×105 km. With light of wavelength of 5500 A˚, the minimum separation between objects on the surface of moon, so that they are just resolved, is close to :