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.

Answer

`30cm`

Explanation

`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`