Where should an object be placed in front of a convex lens of focal length 0.12 m to obtain a real image of size three times the size of the object, on the screen?

(i) As we know,

The formula for magnification of a lens is —

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

Given,

f = +0.12 m

Image is real and magnified, m = -3

Substituting the values in the formula we get,

$-3 = \dfrac{v}{u} \\[0.5em] \Rightarrow v = -3u \\[0.5em]$

Therefore, we get, v = -3u

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,

$\dfrac{1}{-3u} – \dfrac{1}{u} = \dfrac{1}{0.12} \\[0.5em] \dfrac{-1-3u}{3u} = \dfrac{1}{0.12} \\[0.5em] -\dfrac{4}{3u} = \dfrac{1}{0.12}\\[0.5em] u = - \dfrac{4 \times 0.12}{3} \\[0.5em] \Rightarrow u = - 0.16 m \\[0.5em]$

Therefore, object should be placed at a distance of 0.16 m infront of the lens.