A concave lens of focal length f produces an image $\left(\text{1/n}\right)$ times the size of the object. The distance of the object from the lens is:

A concave lens produce always virtual images given that

Focal length of concave lens $=-\text{f}$

Image size $=\frac{1}{\text{n}}\text{object}\text{size}$

$\text{m}=\frac{\text{image}\text{size}}{\text{object}\text{size}}=\frac{\text{V}}{\text{U}}=\frac{1}{\text{n}}$

Magnification s always$+\text{ve}$

$\text{V}=\frac{\text{u}}{\text{n}}\text{u}=?,\text{f}=-\text{f}$

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

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

$-\left(\text{n}-1\right)\text{f}=\text{u}$

$\therefore$ Distance of the object is $\left(\text{n}-1\right)\text{f}$