Find the position and nature of the image of an object 5cm high and 10cm in front of a convex lens of focal length 6cm.
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$h_1=5 cm u =-10 cm f =6 cm$
$\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}$
$\frac{1}{\text{v}}-\frac{1}{-10}=\frac{1}{6}$
$\frac{1}{\text{v}}=\frac{1}{6}-\frac{1}{10}$
$\frac{1}{\text{v}}=\frac{2}{30}=\frac{1}{15}$
$\text{v}=15\text{cm}$
Image is formed 15cm behind the convex lens and it is real and inverted.
$\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}$
$\frac{1}{\text{v}}-\frac{1}{-10}=\frac{1}{6}$
$\frac{1}{\text{v}}=\frac{1}{6}-\frac{1}{10}$
$\frac{1}{\text{v}}=\frac{2}{30}=\frac{1}{15}$
$\text{v}=15\text{cm}$
Image is formed 15cm behind the convex lens and it is real and inverted.
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