A double size real image is produced by a convex lens when an...

For a convex lens, the magnification $m$ in terms of focal length $f$ and object distance $u$ is given by $m = \dfrac{f}{f + u}$.

For a real image, the magnification is negative.

In the first case, a double size real image is formed when the object is placed $18 \text{ cm}$ away.

$m_1 = -2$ and $u_1 = -18 \text{ cm}$.

$-2 = \dfrac{f}{f - 18}$

$-2f + 36 = f$

$3f = 36 \Rightarrow f = 12 \text{ cm}$

In the second case, a triple size real image is required.

$m_2 = -3$

$-3 = \dfrac{12}{12 + u_2}$

$-36 - 3u_2 = 12$

$-3u_2 = 48 \Rightarrow u_2 = -16 \text{ cm}$

The object should be placed $16 \text{ cm}$ away from the lens.