Given data

• Convex mirror radius of curvature: $R = 30\text{ cm}$
• So focal length: $f = \frac{R}{2} = 15\text{ cm}$

For a convex mirror, using Cartesian sign convention: $f = +15\text{ cm}$

The image is half the size of the object, so magnification is: $m = \frac{1}{2}$

Use mirror magnification formula For mirrors, $m = -\frac{v}{u}$ Hence, $\frac{1}{2} = -\frac{v}{u}$ $v = -\frac{u}{2}$

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

Substitute $f=15$ cm and $v=-u/2$: $\frac{1}{15} = \frac{1}{-u/2} + \frac{1}{u}$ $\frac{1}{15} = -\frac{2}{u} + \frac{1}{u}$ $\frac{1}{15} = -\frac{1}{u}$

Therefore, $u = -15\text{ cm}$

Check the result

• Negative object distance is correct for a real object placed in front of the mirror.
• This matches the physical behavior of a convex mirror.

Option matching $u = -15\text{ cm}$ So the correct option is B.