This question asks us to find the distance of the object from a concave mirror when it forms a virtual image that is n times the size of the object. Let's break down the problem using the principles of optics.

Understanding Concave Mirror Properties and Virtual Images

A concave mirror is a converging mirror. It can form both real and virtual images depending on the position of the object. To form a virtual image of a real object, the object must be placed between the pole (P) and the principal focus (F) of the mirror. Virtual images formed by concave mirrors are:

• Erect (oriented the same way as the object)
• Magnified (larger than the object)
• Located behind the mirror

Applying Sign Conventions for Mirrors

To solve this problem accurately, we must use the standard sign conventions for mirrors:

• Distances measured in the direction of incident light are positive; distances measured against the direction of incident light are negative.
• The pole (P) is the origin.
• Object distance ($u$) is measured from the pole. For a real object, it is always negative ($u < 0$).
• Image distance ($v$) is measured from the pole. For a virtual image formed by a concave mirror, it is positive ($v > 0$).
• Focal length ($f$) is measured from the pole. For a concave mirror, the focal length is negative ($f < 0$). The question provides 'focal length $f$', implying $f$ is the magnitude, so the actual focal length used in formulas is $-f$.
• Magnification ($m$) is positive for erect images (virtual) and negative for inverted images (real).

Magnification Formula and Virtual Images

The magnification ($m$) produced by a mirror is defined as the ratio of the image distance ($v$) to the object distance ($u$), with a negative sign:

$m = -\frac{v}{u}$

We are given that the image is n times the size of the object. Since the image is virtual, it is erect, meaning the magnification is positive. Therefore:

$m = +n$

Equating the two expressions for magnification:

$n = -\frac{v}{u}$

We can express the image distance ($v$) in terms of the object distance ($u$):

$v = -nu$

Since $u$ is negative (real object) and $n$ is positive (magnification for virtual image, $n>1$), $v = -n(\text{negative value})$ results in $v$ being positive, which is consistent with a virtual image.

Applying the Mirror Formula

The mirror formula relates the object distance ($u$), image distance ($v$), and focal length ($f$) of a spherical mirror:

$\frac{1}{f_{actual}} = \frac{1}{u} + \frac{1}{v}$

Using our sign conventions, the actual focal length of the concave mirror is $f_{actual} = -f$. Substituting this and $v = -nu$ into the mirror formula:

$\frac{1}{-f} = \frac{1}{u} + \frac{1}{-nu}$

This can be written as:

$-\frac{1}{f} = \frac{1}{u} - \frac{1}{nu}$

Calculating Object Distance

Now, we need to solve this equation for the object distance ($u$). Let's combine the terms on the right side:

$-\frac{1}{f} = \frac{n}{nu} - \frac{1}{nu}$

$-\frac{1}{f} = \frac{n-1}{nu}$

To find $u$, we can rearrange the equation:

$nu = -f(n-1)$

$u = -\frac{f(n-1)}{n}$

The question asks for the distance of the object from the mirror. Distance is a scalar quantity and is typically represented as a positive value (the magnitude). Therefore, we take the absolute value of $u$:

Object Distance $= |u| = |-\frac{f(n-1)}{n}|$

Since $f$ represents the magnitude of the focal length, $f > 0$. For a concave mirror to form a virtual, magnified image, the object must lie within the focal length ($0 < |u| < f$), which implies the magnification $n$ must be greater than 1 ($n>1$). If $n>1$, then $(n-1)$ is positive.

Therefore, $\frac{f(n-1)}{n}$ is a positive value.

Object Distance $= \frac{f(n-1)}{n}$

This result matches one of the options provided.