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A compound microscope has a magnifying power `30`. The focal length of its eye-piece is `5cm`. Assuming the final to be at the least distance of distinct vision `(25 cm)`, calculate the magnification produced by objective. |
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Answer» `m =- 30, f_(e) = 5cm` `m_(e) = 1+(D)/(f_(e)) = 1+(25)/(5) = 6` `m = m_(0)xxm_(e)` `-30 = m_(0) xx 6 rArr m_(0) =-0 5` |
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