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.
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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