A telescope objective of focal length 1 m forms a real image of the moon 0.92 cm in diameter. Calculate the diameter of the moon taking its mean distance from the earth to be 3.8 ×10^5 km…

Q: A telescope objective of focal length 1 m forms a real image of the moon 0.92 cm in diameter. Calculate the diameter of the moon taking its mean distance from the earth to be 3.8 × 105 km.If the telescope uses an eyepiece of 5 cm focal length, what would be the distance between the two lenses for (i) the final image to be formed at infinity (ii) the final image (virtual) at 25 form eye.

Sol: fo = 1 m

Object distance from the objective = distance of the moon from the earth

= 3.8 × 105 km =3.8 × 108 m

Image distance from the objective = focal length of the objective = 1 m size = image size = image
diameter = 0.92 × 10-2 m

object size = object diameter

i.e. diameter or moon =?

We know that ,

(Object diameter )/(Image diameter) =(Object distance )/(Image distance )

(Diameter of moon )/(Image diameter )=(3.8 × 108)/1

Diameter of moon = 3.8 × 108 × Image diameter

= 3.8 × 108 × 0.92 × 10-2 m =3.946 × 106 m =3496 km

(i) For normal adjustment, the distance between the two lenses

f0 = fe = 100 + 5 = 105 cm

(ii) For the final image at 25 cm form the eye, the distance between the two lenses

$\large = f_o + \frac{D f_e}{D + f_e} $

$\large = 100 + \frac{25 \times 5}{25 + 5} $

= 104.2 cm