Concave mirror of radius of curvature $20\text{cm}$ has object placed $15\text{cm}$ from its pole on the principle axis. Which of the following correctly determine the position of the image, magnification and nature of image.

The correct option is B $-30\text{cm}$, $-2$, Real
Given,

Radius of curvature, $R=-20\text{cm}$

Object distance, $u$ = $-15\text{cm}$

focal length, $f=\frac{R}{2}=-10\text{cm}$


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

$\Rightarrow \frac{1}{v}=\frac{1}{-10}-\frac{1}{-15}$

$\Rightarrow v=-30\text{cm}$

So, the image formed is at distance $30\text{cm}$ left of the pole.
As $v$ is negative, so the image formed is real and formed in front of the mirror.

We know that the magnification formula is,

$m=\frac{-v}{u}$

$\Rightarrow m=\frac{-(-30)}{-15}$

$\therefore m=-2$ So, the image formed is also magnified and inverted.

Hence, option (b) is the correct answer.

Why this Question: To understand the concept of mirror formula and image formation by spherical mirrors.