InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
A telescope has an objective of focal length 50 cm and an eye-piece of focal length 5 cm. The least distance of distinct vision is 25 cm. The telescope is focused for distinct vision on a scale 2 m away from the objective. Calculate (i) magnification produced and (ii) separation between objective and eyepiece. |
|
Answer» Given`f_(o)=50cm and f_(e)=5 cm` For objective,`1/(v_(o))-1/(-200)=1/50` `therefore v_(o)=200/3cm` `m_(o)=(v_(o))/(u_(o))=((200//3))/(-200)=-1/3` eye piece,`1/(-25)-1/(u_(e))=1/5` `therefore u_(e)=-25/6cmandm_(e)=(v_(e))/(u_(e))=(-25)/(-(25//6))=6` (i) Magnification,m=`m_(o)xxm_(e)=-2` (ii) Seperation between objective and eye piece, `L=v_(o)+abs(u_(e))=200/3+25/6=425/6=70.83 cm` Note Here, object is placed at finite distance from the objective. Hence, formulae derived for angular magnification M cannot be applied directly as they have been derived for the object to be at infinity. Here, it will be difficult to find angular magnification. So, only linear magnification can be obtained. |
|