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A compound microscope conists of an objective lens of focal length 2.0 cm and an eyepiece of focal length 6.25 cm separated by a distance of 15 cm. How far from the objective should an object be placed in order to obtain the final image at (a) the least distance of distinct vision (25cm), and (b) at infinity? What is the magnifying power of the microscope in each case? |
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Answer» Solution :Here `f_(0) = + 2.0`cm and `f_(e) = + 6.25 ` cm and distance between the lenses L = 15 cm (a) If the final image is formed at the least distance of distinct vision (i.e. `v_(e) = -D = -25 cm)` Then, `1/v_(e) - 1/u_(e) = 1/f_(e)` i.e., `1/(-25) -1/u_(e) =1/(6.25)` `therefore -1/u_(e) =1/6.25 + 1/25 = 1/5` or `u_(e) = -5 cm` `therefore v_(0) = L-|u_(e)| = 15-5 = 10 cm` Now applying lens formula for objective lens, we have `1/10 - 1/u_(0) = 1/2 rArr u_(0) = -10/4 = -2.5 cm` `therefore` Multiplying power of microscope `m=-v_(0)/u_(0) (1+D/f_(e)) = -10/2.5 (1+ 25/6.25) = 20` (b) When the final image is formed at infinity, `v_(e) = INFTY`. Hence, applying lens formula for EYE lens, we have `1/infty -1/mu_(e) = 1/(6.25) rArr u_(e) = -6.25 cm` `therefore v_(0) = L -|mu_(e)| = 15-6.25 = 8.75 cm` Using lens formula for objective lens, we GET `1/(+8.75) - 1/u_(0) =1/2` or `-1/u_(0) =1/2 -1/(8.75) = 6.75/17.5 rArr u_(0) = -(17.5)/(6.75) = -2.6 cm` `therefore` Magnifying power of microscope `m=-v_(0)/u_(0) (D/f_(e)) =-8.75/2.6 (25/6.25) = 13.5` |
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