The focal length of a convex lens is 25 cm. At what distance from the optical centre of the lens an object be placed to obtain a virtual image of twice the size ?

As we know,

The formula for magnification of a lens is —

$m = \dfrac{v}{u} \\[0.5em]$

Given,

f = + 25 cm

As image is virtual and magnified, so m = + 2

Substituting the values in the formula we get,

$+2 = \dfrac{v}{u} \\[0.5em] \Rightarrow v = 2u \\[0.5em]$

Therefore, we get, v = 2u

Now as we know, the lens formula is —

$\dfrac{1}{v} – \dfrac{1}{u} = \dfrac{1}{f} \\[0.5em]$

Substituting the values in the formula, we get,

$\dfrac{1}{2u} – \dfrac{1}{u} = \dfrac{1}{25} \\[0.5em] \dfrac{1-2}{2u} = \dfrac{1}{25} \\[0.5em] -\dfrac{1}{2u} = \dfrac{1}{25} \\[0.5em] 2u = - 25 \\[0.5em] \Rightarrow u = - 12.5 cm \\[0.5em]$

Therefore, object should be placed at a distance of 12.5 cm infront of the lens.