Question

A small telescope has an objective lens of focal length 140 cm and an eyepiece of focal length 5 cm if this telescope is used to view a hundred metre tall Tower 3 kilometre away what is the height of the final image of the tower if it is formed at 25 CM

Answer 1

Answer:

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=−fefo(1+Dfe)

m=−5150(1+255)=−36

Also, m=tanαtanβ

tanα=H/u=100/3000=1/30

tanβ=−3036

tanβ=DH′

Height of the image of the tower, H′=−30−36×25=−30cm

Explanation:

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