The total magnification produced by a compound microscope is 20. The magnification produced by the eye piece is 5. The microscope is focussed on a certain object. The distance between the objective and eyepiece is observed to be 14 cm. If least distance of distinct vision is 20 cm, calculate the focal length of the objective and the eye piece..

Here is the answer:- Hope it helps! SJR

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You are given two converging lens of focal length 2.5cm and 8cm to design a compound microscope. if it is designes to have a magnification of 40.find out the sepration between the objective and eyepeice
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I had the same doubt, thanks for asking and solving this question meritnation community
 
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me = ve/ue ue =ve/me -20/5 = -4 cm 1/fe=1/ve - 1/ue 1/fe = -1/20 + 1/4 fe = 5 cm m = me x mo -20 = 5 x mo mo = - 4 vo + ue = 14 vo = 14 - 4 = 10 cm mo = 1 - (vo / fo) - 4 = 1 - 10 / fo fo = 2 cm
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