An object is placed at 5 cm in front of a concave mirror with a radius of curvature of 20 cm. The distance of the image from mirror and its nature are

10 cm, virtual and erect

The correct answer is Option (1) → 10 cm, virtual and erect

For a concave mirror,

$R = 20 \text{ cm}$

So the focal length is

$f = \frac{R}{2} = 10 \text{ cm}$

Using Cartesian sign convention for mirrors:

$f = -10 \text{ cm}, \quad u = -5 \text{ cm}$

Mirror formula:

$\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$

Substituting values:

$\frac{1}{-10} = \frac{1}{v} + \frac{1}{-5}$

$\frac{1}{v} = -\frac{1}{10} + \frac{1}{5}$

$\frac{1}{v} = \frac{1}{10}$

$v = 10 \text{ cm}$

Positive $v$ means the image is formed behind the mirror, hence it is virtual and erect.