A 5cm tall object is placed perpendicular to the principal axis of a convex lens of focal length 18cm at a distance of 12cm from it. Use lens formula to determine the position, size and nature to the image formed.
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$\text{h}_0=5\text{cm},\text{ u}=-30\text{cm},\text{ v}=?,\text{ f}=+10\text{cm}$$\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}$
$\frac{1}{\text{v}}-\frac{1}{(-30)}=\frac{1}{10}$
${\text{v}}=15\text{cm}$
$\text{m}=\frac{\text{v}}{\text{u}}$
$\text{m}=\frac{15}{(-30)}$
$\text{m}=-\frac{1}{2}$
$\text{m}=\frac{\text{v}}{\text{u}}=\frac{\text{h}_\text{i}}{\text{h}_0}$
$-\frac{1}{2}=\frac{\text{h}_\text{i}}{5}$
$\text{h}_\text{i}=-\frac{5}{2}$
$\text{h}_\text{i}=-2.5\text{cm}$
$\frac{1}{\text{v}}-\frac{1}{(-30)}=\frac{1}{10}$
${\text{v}}=15\text{cm}$
$\text{m}=\frac{\text{v}}{\text{u}}$
$\text{m}=\frac{15}{(-30)}$
$\text{m}=-\frac{1}{2}$
$\text{m}=\frac{\text{v}}{\text{u}}=\frac{\text{h}_\text{i}}{\text{h}_0}$
$-\frac{1}{2}=\frac{\text{h}_\text{i}}{5}$
$\text{h}_\text{i}=-\frac{5}{2}$
$\text{h}_\text{i}=-2.5\text{cm}$
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