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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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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 |
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