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A compound microscope consists of an objective lens of focal length 2.00 cm and an eyepiece of focal length 6.25 cm separated by a distance of 15.0 cm. How far (in cm) from the objective lens should an object be placed in order to obtain the final image at the least distance of distinct vision (25.0 cm)? |
Answer»  For the eyepiece: `(1)/(f_(e)) = (1)/(v_(e)) -(1)/(u_(e)), (1)/(u_(e))= (1)/(v_(e))- (1)/(f_(e)) = (1)/(-25) - (1)/(6.25) (" as " u_(e)= -D= -25)` `(1)/(u_(e))= - (1)/(5) or u_(e)= - 5cm` or As `v_(0) +|u_(e)|=L= 15CM, v_(0) = 15-|u_(e)\=15-5= 10cm` For the OBJECTIVE. `(1)/(f_(0))= (1)/(v_(0)) - (1)/(u_(0))` or `(1)/(2)= (1)/(10) - (1)/(u_(0))` or `(1)/(u_(0))= (1)/(10) -(1)/(2)= -(4)/(10)` or `u_(0)= -2.5cm` Thus, the object should be PLACED at a distance 2.50 CM in front of the objective. |
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