When an object is placed at a distance of 60cm from a convex mirror the magnification produced is half Where should the object placed to get a magnification of 1 3?

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Text Solution

Solution : Here, `u_(1) = -60 cm, m_(1) = (1)/(2)` <br> `u_(2) = ?, m_(2) = (1)/(3)` <br> As `m_(1) = -(v_(1))/(u_(1)) :. (1)/(2) = -(v_(1))/(-(-60)), v_(1) = 30 cm` <br> From `(1)/(f) = (1)/(v_(1)) + (1)/(u_(1)) = (1)/(30) - (1)/(60) = (1)/(60)` <br> `f = 60 cm` <br> Again, `m_(2) = -(v_(2))/(u_(2)) = (1)/(3), v_(2) = -(u_(2))/(3)` <br> From `(1)/(v_(2)) + (1)/(u_(2)) = (1)/(f)` <br> `(1)/(u_(2)) = (1)/(f) - (1)/(v_(2)) = (1)/(60) + (3)/(u_(2))` <br> or `(1)/(u_(2)) - (3)/(u_(2)) = (1)/(60), u_(2) = -120 cm`