When a concave mirror is placed facing the sun, the sun's rays converge to a point $10\text{cm}$ from the mirror. Now, an erect $2\text{cm}$ long pin is placed $15\text{cm}$ away on the principal axis of the mirror. A sharp image of this pin is obtained on a card. What would be the height of the image?

The correct option is A $4\text{cm}$
We know that the parallel rays from the sun converge at the focus of the concave mirror.

Given:
Object distance, $u=-15\text{cm}$
Focal length, $f=-10\text{cm}$
Object height, $h_o=+2\text{cm}$

From mirror formula,

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

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

$\Rightarrow \frac{1}{v}=-\frac{1}{30}$

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

Now, magnification,
$m=\frac{h_i}{h_o}=\frac{-v}{u}$

$\Rightarrow \frac{h_i}{+2}=\frac{-(-30)}{(-15)}$

$\Rightarrow h_i=-4\text{cm}$

Hence, the height of the image is $4\text{cm}$.