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What is the difference in the construction of an astronomical telescope and a compound microscope? The focal lengths of the objective and eyepiece of a compound microscope are 1.25cm and 5.0cm, respectively. Find the position of the object relative to the objective in order to obtain an angular magnification of 30 when the final image is formed at the near point. |
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Answer» Solution :In an astronomical telescope we take a CONVEX lens of large focal LENGTH and large aperture as the objective lens and a convex lens of small focal length and SMALLER aperture as the eye PIECE. However, focal lengths as well as aperture of both objective and eye piece lenses is small in a compound microscope and focal length aperture of objective are even smaller than that of eyepiece. Since final image is formed at near point i.e., the least distance of distinct vision i.e., D = 25cm. In that case, angular magnification of eyepiece. `m_(e )=(1+(D)/(f_(e )))=1+(25)/(5)=6 "" [because f_(e )=+5cm]` As magnification of microscope m = 30 and `m=m_(0)xxm_(e )` `rArr m_(0)=(m)/(m_(e ))=(30)/(6)=5` `therefore m_(0)=(v_(0))/(u_(0))=5` `or v_(0)=5u_(0)` and as per sign convention`u_(0)` is -ve but `v_(0)` is +ve and `f_(0)=+1.25cm`. Hence, we have `(1)/(v_(0))-(1)/(u_(0))=(1)/(f_(0)) or (1)/(5u_(0)) -(1)/((-u_(0))) =(1)/(1.25) or (6)/(5u_(0))=(1)/(1.25)` `rArr u_(0)=1.5cm` |
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