Here, magnification in area `= 6.25 :.` Linear magnification `m = sqrt(6.25) = 2.5`
As `m = (v)/(u) or v = mu = 2.5 u`
As `(1)/(v) - (1)/(u) = (1)/(f) :. (1)/(2.5u) - (1)/(u) = (1)/(10)`
`(1 - 2.5u) - (1)/(u) = (1)/(10)`
`2.5 u = - 15, or u = - 6cm :. v = 2.5 u = 2.5(-6) = - 15 cm`
As the virtual image is at `15 cm` , whereas distance of distnct vision is `25 cm`, therefore, the image cannot be seen distnctly by the eye.