Sign of magnification $=$ Positive
Value of magnification $=$ Greater than 1
Sign of magnification $=$ Positive
Value of magnification $=$ Greater than 1OR
Given, size of object, ${\mathrm{H}}_{\mathrm{o}}=4\mathrm{cm}$
Object distance, $\mathit{u}=-25\mathrm{cm}$
Focal length of concave mirror, $\mathit{f}=-15\mathrm{cm}$
(i) As per formula,
$\frac{1}{\mathit{f}}=\frac{1}{\mathit{v}}+\frac{1}{\mathit{u}}$
$\Rightarrow \frac{1}{-15}=\frac{1}{\mathit{v}}+\frac{1}{-25}$
$\Rightarrow \frac{1}{\mathit{v}}=\frac{1}{-15}-\frac{1}{-25}=\frac{1}{-15}+\frac{1}{25}$
$=\frac{\left(-25+15\right)}{375}=\frac{\left(-10\right)}{375}$
$=\frac{1}{-37.5}$
$\Rightarrow \frac{1}{\mathit{v}}=\frac{1}{-37.5}$
$\Rightarrow \mathit{v}=-37.5\mathrm{cm}$
$\therefore$ Image is formed on object side at 37.5 $\mathrm{cm}$ from pole of concave mirror.
(ii) The formula for the size of an image
${\mathrm{H}}_{\mathit{i}}∕{\mathrm{H}}_{\mathrm{O}}=-\frac{\mathit{v}}{\mathit{u}}$
Where ${\mathrm{H}}_{\mathit{i}}$ is the size of image.
Now,
$\frac{{\mathrm{H}}_{\mathit{i}}}{4\mathrm{cm}}=\frac{-\left(-37.5\mathrm{cm}\right)}{\left(-25\mathrm{cm}\right)}$
${\mathrm{H}}_{\mathit{i}}=\frac{-4\times 37.5}{25\mathrm{cm}}=-6\mathrm{cm}$
$\therefore$ Image is formed on object side at 37.5 $\mathrm{cm}$ from pole of concave mirror.
(ii) The formula for the size of an image
${\mathrm{H}}_{\mathit{i}}∕{\mathrm{H}}_{\mathrm{O}}=-\frac{\mathit{v}}{\mathit{u}}$
Where ${\mathrm{H}}_{\mathit{i}}$ is the size of image.
Now,
$\frac{{\mathrm{H}}_{\mathit{i}}}{4\mathrm{cm}}=\frac{-\left(-37.5\mathrm{cm}\right)}{\left(-25\mathrm{cm}\right)}$
${\mathrm{H}}_{\mathit{i}}=\frac{-4\times 37.5}{25\mathrm{cm}}=-6\mathrm{cm}$
$\therefore$ Size of image $=6\mathrm{cm}$
Hence, screen should place $37.5\mathrm{cm}$ left side of mirror and size of image is $6\mathrm{cm}$. It is inverted and larger than object's size.