Sign of magnification $=$ Positive Value of magnification $=$ Greater than 1 Sign of magnification $=$ Positive Value of magnification $=$ Greater than 1ORGiven, size of object, $H_o = 4\,\mathrm{cm}$ Object distance, $u = -25\,\mathrm{cm}$ Focal length of concave mirror, $f = -15\,\mathrm{cm}$(i) As per formula,$\;\frac{1}{f}\;=\;\frac{1}{v}+\;\frac{1}{u}$$\Rightarrow\;\; \;\frac{1}{-15}\;=\;\frac{1}{v}+\;\frac{1}{-25}$$\Rightarrow\;\; \;\frac{1}{v}\;=\;\frac{1}{-15}-\;\frac{1}{-25}=\;\frac{1}{-15}+\;\frac{1}{25}$$\;=\;\frac{-25+15}{375}=\;\frac{-10}{375}$$\;=\;\frac{1}{-37.5}$$\Rightarrow\;\; \;\frac{1}{v}\;=\;\frac{1}{-37.5}$$\Rightarrow\;\; v\;=-37.5\,\mathrm{cm}$$\therefore$ Image is formed on object side at 37.5 $\mathrm{cm}$ from pole of concave mirror.(ii) The formula for the size of an image$\frac{H_i}{H_O} = -\frac{v}{u}$Where $H_i$ is the size of image.Now,$\;\frac{H_i}{4\,\mathrm{cm}}\;=\;\frac{-(-37.5\,\mathrm{cm})}{(-25\,\mathrm{cm})}$$H_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$\frac{H_i}{H_O} = -\frac{v}{u}$Where $H_i$ is the size of image.Now,$\;\frac{H_i}{4\,\mathrm{cm}}\;=\;\frac{-(-37.5\,\mathrm{cm})}{(-25\,\mathrm{cm})}$$H_i\;=\;\frac{-4 \times 37.5}{25\,\mathrm{cm}} = -6\,\mathrm{cm}$$\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.