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 × 10^{5} 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.

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_{o}= 1 m

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

= 3.8 × 10^{5} km =3.8 × 10^{8} 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 × 10^{8})/1

Diameter of moon = 3.8 × 10^{8} × Image diameter

= 3.8 × 10^{8} × 0.92 × 10^{-2} m =3.946 × 10^{6} m =3496 km

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

f_{0} = f_{e} = 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