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A `2.5` dipote lens forms a virtual image which is 4 times the object placed perpendicually on the principle axis of the lens. Find the required distance of the object form the lens. |
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Answer» Correct Answer - `30cm` `f = (1)/(p) = (1)/(2.5)m = 40cm, m = (v)/(u) = 4, v = 4u` Using lens formula `(1)/(40) = (1)/(4u) - (1)/(u) = (1-4)/(4u), (1)/(40) = (-3)/(4u) rArr u = -30 cm` So, required distance `= 30 cm` |
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