Distance between object and its image (magnified by $$-\dfrac{1}{3}$$) is 30 cm. The focal length of the mirror used is $$\left(\dfrac{x}{4}\right)$$ cm, where magnitude of value of x is ________.

For a spherical mirror the linear magnification is given by
$$m = -\dfrac{v}{u}$$
where $$u$$ is the object distance (from the pole) and $$v$$ is the image distance. Distances are measured from the pole, positive to the right (toward the reflecting side), negative to the left (toward the incident light).

The magnification in the question is $$m = -\dfrac{1}{3}$$. Using the formula:
$$-\dfrac{v}{u} = -\dfrac{1}{3}$$
gives
$$\dfrac{v}{u} = \dfrac{1}{3} \quad -(1)$$

This means the image distance is one-third of the object distance, with the same sign. Let the magnitude of the object distance be $$a$$ centimetres, so
$$u = -a \quad (\text{object is on the left})$$
Then from $$(1)$$:
$$v = -\dfrac{a}{3}$$

The actual separation between the object and its image is the difference of their magnitudes:
$$|\,u - v\,| = \bigl|\,(-a) - \!\left(-\dfrac{a}{3}\right)\bigr| = a - \dfrac{a}{3} = \dfrac{2a}{3}$$

This separation is given to be 30 cm, so
$$\dfrac{2a}{3} = 30 \;\Longrightarrow\; a = 45\text{ cm}$$
Hence
$$u = -45\text{ cm}, \qquad v = -15\text{ cm}$$

Apply the mirror equation
$$\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}$$
Substituting $$v = -15$$ cm and $$u = -45$$ cm:
$$\dfrac{1}{f} = \dfrac{1}{-15} + \dfrac{1}{-45} = -\dfrac{1}{15} - \dfrac{1}{45} = -\dfrac{3 + 1}{45} = -\dfrac{4}{45}$$
Therefore
$$f = -\dfrac{45}{4}\text{ cm}$$

The focal length is written in the form $$\dfrac{x}{4}\text{ cm}$$, so $$x = -45$$. The question asks for the magnitude of $$x$$, hence

$$|x| = 45$$

Answer: 45