Ray optics - telescope
A small telescope has an objective lens of focal length 150cm and eye piece of focal length 5cm. If this telescope is used to view a 100 m high tower 3 km away, find the height of the final image when it is formed 25cm away from eyepiece.
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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
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