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A compound microscope has an objective of focal length 2 cm ahd eye piece of focal length 5 cm. The distance between two lenses is 25 cm. If final image is at 25 cm from the eye - piece, find the magnifying power of the microscope : |
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Answer» 56.5 `(1)/(-25) - (1)/(-u_(e))=(1)/(5) or u_(e)=(25)/(6)` `L = v_(0) + u_(e) or 25 = v_(0) + (25)/(6)` or `v_(0) = (125)/(6) cm` For objective `(1)/(((125)/(6)))-(1)/(-u_(0))=(1)/(2)oru_(0)=(250)/(113)cm` `THEREFORE M=(v_(0))/(u_(0))[1+(D)/(f_(e))]=((125)/(6))/((259)/(113))[1+(25)/(5)]` |
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