A convex lens of focal length 16 cm forms a virtual image of double the size of the object. The distance of the object from the lens is

$\text{f}=16\text{cm}$

$\text{m}=\frac{\text{v}}{\text{u}}=\frac{\text{size\hspace{0.17em}of\hspace{0.17em}image}}{\text{size\hspace{0.17em}of\hspace{0.17em}object}}=2$ (for virtual image)

$\frac{\text{v}}{\text{u}}=\frac{2\left(\text{size\hspace{0.17em}of\hspace{0.17em}object}\right)}{\text{size\hspace{0.17em}of\hspace{0.17em}object}}=2$

$\Rightarrow \text{v}=2\text{u,\hspace{0.17em}f\hspace{0.17em}is\hspace{0.17em}+ve\hspace{0.17em}\hspace{0.17em}i.e.,\hspace{0.17em}f}=16,\text{\hspace{0.17em}\hspace{0.17em}u}=?$ 

We know$\frac{\text{1}}{\text{v}}-\frac{\text{1}}{\text{u}}=\frac{1}{\text{f}}$

$\frac{\text{1}}{\text{2u}}-\frac{\text{1}}{\text{u}}=\frac{1}{\text{f}}$

$\frac{-\text{1}}{\text{2u}}=\frac{1}{\text{16}}$

$\therefore \text{u}=-8\text{\hspace{0.17em}cm}$

$\therefore$The distance of object = 8 cm