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When an object is placed at a distance of `60 cm` from a convex spherical mirror, the magnification produced is `1//2`. Where should the object be placed to get a magnification of `1//3` ? |
Answer» Here, `u_(1) = -60 cm, m_(1) = (1)/(2)` `u_(2) = ?, m_(2) = (1)/(3)` As `m_(1) = -(v_(1))/(u_(1)) :. (1)/(2) = -(v_(1))/(-(-60)), v_(1) = 30 cm` From `(1)/(f) = (1)/(v_(1)) + (1)/(u_(1)) = (1)/(30) - (1)/(60) = (1)/(60)` `f = 60 cm` Again, `m_(2) = -(v_(2))/(u_(2)) = (1)/(3), v_(2) = -(u_(2))/(3)` From `(1)/(v_(2)) + (1)/(u_(2)) = (1)/(f)` `(1)/(u_(2)) = (1)/(f) - (1)/(v_(2)) = (1)/(60) + (3)/(u_(2))` or `(1)/(u_(2)) - (3)/(u_(2)) = (1)/(60), u_(2) = -120 cm` |
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