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A telescope has an objective of focal length 50cm and an eyepiece of focal length 5cm. The least distance of distinct vision is 25cm. The telescope is focused for distinct vision on a scale 2m away from the objective. Calculate (a) magnification produced and (b) separation between objective and eye piece. |
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Answer» Solution :Given, `f_(o)=50cmandf_(e)=5CM` For objective: `1/v_(0)-1/(-200)=1/50." "thereforev_(0)=200/3cm` `m_(0)=v_(0)/u_(0)=((200//3))/(-200)=-1/3` For eyepiece : `1/(-25)-1/u_(e)=1/5," "thereforeu_(e)=-25/6cm` and `m_(e)=v_(e)/u_(e)=(-25)/(-(25//6))=6` (a) Magnification, `m=m_(o)xxm_(e)=-2` (B) Separation between objective and eyepiece, `L=v_(o)+|u_(e)|=200/3+25/6=425/6=70.83cm` |
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