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Question

A small telescope has an objective lens of focal length 150 cm and eyepiece of focal length 5 cm. What is the magnifying power of the telescope for viewing distant objects in normal adjustment?
If this telescope is used to view a 100 m tall tower 3 km away, what is the height of the image of the tower formed by the objective lens?

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Solution

Given, Focal length of objective, fo=150 cm

Focal length of eye-piece, fe=5 cm

Height of tower, H=100 m

Distance of tower, u=3 km

Magnification of telescope is given by:

m=fofe(1+feD)

m=1505(1+525)=36

Also, m=tanβtanα

tanα=H/u=100/3000=1/30

tanβ=3630

tanβ=HD

Height of the image of the tower, H=36×2530=30cm

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