Question

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 38 x 104 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 and (ii) the final image (virtual) at 25 cm from the eye.​

Answer 1

Answer:

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Explanation:

f0=1m

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

=3.8×105km=3.8×108m

image distance from the objective = focal length of the objective=1 m

image size = image diameter= 0.92×10−2m

object size = object diameter ie diameter of moon=?

We know thatobject diameterImage diameter=Object distanceImage distance

∴ Diameter of moon= 3.8×108× Image distance

=3.8×108×0.92×10−2m=3.946×106m=3496km

For normal adjustment, the distance between the two lenses f0+fe=100+5=105cm

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

f0+(DfeD+fe)=100+(25×525+5)=104.2cm