An object is placed at a distance of 4cm from a concave lens of focal length 12cm. Find the position and nature of the image.
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u = -4cm f = -12cm Lens formula, $\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}$$\frac{1}{\text{v}}-\frac{1}{-4}=\frac{1}{-12}$
$\frac{1}{\text{v}}=-\frac{1}{12}-\frac{1}{4}$
$\frac{1}{\text{v}}=\frac{-4}{12}$
$\text{v}=-3\text{cm}$
Image is formed 3cm infront of the concave lens. Image is virtual and erect.
$\frac{1}{\text{v}}=-\frac{1}{12}-\frac{1}{4}$
$\frac{1}{\text{v}}=\frac{-4}{12}$
$\text{v}=-3\text{cm}$
Image is formed 3cm infront of the concave lens. Image is virtual and erect.
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