Question

A small telescope has an objective lens of focal length 150 cm and eyepiece of focal length 5 cm

Answer 1

Answer :

Given that ,

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=−fe/fo(1+fe/D)

m=−150/5(1+5/25) = −36

tanβ=H′/D

Also, m=tanα/tanβ

tanα=H/u=100/3000=1/30

tanβ=−36/30

Height of the image of the tower, H′

= -(-36)× 25 / 30

= - 30cm

Hope this would help you