An object 5 cm in size is placed at a distance of 20 cm from the concave mirror of focal length 15 cm. At what distance from the mirror should a screen be placed to get a sharp image also find the nature of image

Object distance, $\text{u}=-\text{20}\text{.0cm}$

focal length, $\text{f}=-15.0\text{cm}$

Image distance, $\text{V}=\text{?}$

Image size, ${\text{h}}^1=?$

We know the mirror formula, $\frac{\text{1}}{\text{v}}+\frac{\text{1}}{\text{u}}=\frac{\text{1}}{\text{f}}$

$\frac{\text{1}}{\text{v}}=\frac{\text{1}}{\text{f}}-\frac{\text{1}}{\text{u}}$

$=\left(-\frac{1}{15}\right)+\left(\frac{1}{20}\right)=\frac{-1}{60}=-60\text{cm}$

The screen should be placed at 60 cm from the mirror. The image is real

magnification

$\text{m}=\frac{{\text{h}}^{\text{1}}}{\text{h}}=\frac{-\text{v}}{\text{u}}=\frac{60}{-20}=-3$

Height of the image, ${\text{h}}^{\text{1}}=\text{mh}=\left(-3\right)\times 5=-15\text{cm}$

The image is inverted and enlarged.