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a telescope has an objective lens of focal length 150cm and an eyepiece of focal length 5cm.if his telescope is used to view a 100m high tower 30km away,find the height of final image when it is formed 25cm away from the eyepiece
Asked by hibasherinichu | 23 Dec, 2023, 07:02: AM

Magnification of objective lens , mo = vo/uo

where vo and uo are  lens-to-image distance and  lens-to object distance  respectively.

Since lens-to-object distance uo = 30 km and uo .>> fo , where fo =0.15 m is focal length of objective lens,

lens-to- image distance vo = fo

Magnitude of magnification of objective lens , mo = fo/uo = 0.15 / ( 3 × 104 ) = 5 × 10-6

If tower height is 100 m , then image height = 100 × 5 × 10-6 = 5 × 10-4 m

Focal length of eye piece , fe = 5 cm

eye piece lens to image distance,  ve = 25 cm

In telescope image of objective lens formed very near to foacl point of eye piece .

Hence magnification of objective lense =  ue / ve = fe / ve = 25/5 = 5

Hence size of final image = 5 × 5 ×10-4 m = 2.5 × 10-3 m = 2.5 mm

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