Two identical objects are placed in front of convex mirror and concave mirror having same radii of curvature of 12 cm, at same distance of 18 cm from the respective mirrors. The ratio of the sizes of images formed by convex mirror and by concave mirror is :

Radius of curvature of each mirror is $$R = 12\ \text{cm}$$. For a spherical mirror the focal length is given by $$f = \frac{R}{2}$$.

Therefore $$f = 6\ \text{cm}$$ in magnitude.
  • For the concave mirror, the focus lies on the object side, so $$f = -6\ \text{cm}$$.
  • For the convex mirror, the focus lies behind the mirror, so $$f = +6\ \text{cm}$$.

The objects are placed at the same distance in front of each mirror: $$u = -18\ \text{cm}$$ (negative by the new Cartesian sign convention).

Mirror formula: $$\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$$.

Substituting $$f = -6$$ and $$u = -18$$:
$$\frac{1}{-6} = \frac{1}{v} + \frac{1}{-18}$$

$$\frac{1}{v} = -\frac{1}{6} + \frac{1}{18} = -\frac{3}{18} + \frac{1}{18} = -\frac{2}{18} = -\frac{1}{9}$$

$$v = -9\ \text{cm}$$ (image is real and formed in front of the mirror).

Linear magnification: $$m_c = -\frac{v}{u} = -\frac{-9}{-18} = -\frac{1}{2}$$

Magnitude of image size for concave mirror: $$|m_c| = \frac{1}{2}$$.

Using $$f = +6$$ and $$u = -18$$ in the mirror formula:
$$\frac{1}{6} = \frac{1}{v} + \frac{1}{-18}$$

$$\frac{1}{v} = \frac{1}{6} + \frac{1}{18} = \frac{3}{18} + \frac{1}{18} = \frac{4}{18} = \frac{2}{9}$$

$$v = \frac{9}{2}\ \text{cm} = 4.5\ \text{cm}$$ (image is virtual, behind the mirror).

Linear magnification: $$m_v = -\frac{v}{u} = -\frac{4.5}{-18} = \frac{4.5}{18} = \frac{1}{4}$$

Magnitude of image size for convex mirror: $$|m_v| = \frac{1}{4}$$.

Ratio of the sizes (convex : concave) is
$$\frac{|m_v|}{|m_c|} = \frac{\frac{1}{4}}{\frac{1}{2}} = \frac{1}{2}.$$

Hence the required ratio is $$\mathbf{\frac{1}{2}}$$, which corresponds to Option A.