A mirror is used to produce an image with magnification of $$\frac{1}{4}$$. If the distance between object and its image is 40 cm, then the focal length of the mirror is :

For a spherical mirror the transverse magnification is defined as
$$m \;=\; \frac{h_i}{h_o}\;=\;-\frac{v}{u} \quad -(1)$$
where $$u$$ is the object distance and $$v$$ is the image distance, both measured from the pole using the Cartesian sign convention.

The given magnitude of magnification is $$\frac{1}{4}$$.
Because the value is positive and less than $$1$$, the image is erect and virtual. Hence the mirror must be convex, so we take
$$m = +\frac{1}{4} \quad\Longrightarrow\quad \frac{1}{4}= -\frac{v}{u} \quad -(2)$$

From $$(2)$$,
$$v = -\frac{u}{4} \quad -(3)$$

With the sign convention, the object lies in front of the mirror, so $$u$$ is negative. Equation $$(3)$$ therefore makes $$v$$ positive, placing the virtual image behind the mirror, as expected for a convex mirror.

The distance between the object and its image is given to be $$40\ \text{cm}$$. Along the principal axis these two points are on opposite sides of the pole, so their separation equals the sum of their magnitudes:
$$|u| + |v| = 40 \quad -(4)$$

Put $$|v| = \frac{|u|}{4}$$ from $$(3)$$ into $$(4)$$:
$$|u| + \frac{|u|}{4} = 40 \;\Longrightarrow\; \frac{5|u|}{4} = 40 \;\Longrightarrow\; |u| = 32\ \text{cm}$$

Thus,
$$u = -32\ \text{cm}, \qquad v = +8\ \text{cm}$$

Apply the mirror formula
$$\frac{1}{f} = \frac{1}{v} + \frac{1}{u} \quad -(5)$$

Substituting $$v = 8\ \text{cm}$$ and $$u = -32\ \text{cm}$$ into $$(5)$$ gives
$$\frac{1}{f} = \frac{1}{8} - \frac{1}{32} = \frac{4 - 1}{32} = \frac{3}{32}$$

Therefore
$$f = \frac{32}{3}\ \text{cm} \approx 10.7\ \text{cm}$$

The focal length is positive, confirming that the mirror is indeed convex. Hence the correct option is Option C (10.7 cm).