A convex lens of focal length 6cm is held 4cm from a newspaper which has print 0.5cm high. By calculation, determine the size and nature of the image produced.
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f = 6cm u = -4cm $h _1=0.5 cm$ Lens formula, $\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}$$\frac{1}{\text{v}}-\frac{1}{-4}=\frac{1}{6}$
$\frac{1}{\text{v}}=\frac{1}{6}-\frac{1}{4}=\frac{2-3}{12}$
$\text{v}=-12\text{cm}$
$\text{m}=\frac{\text{v}}{\text{u}}=\frac{\text{h}_2}{\text{h}_1}$
$\frac{-12}{-4}=\frac{\text{h}_2}{0.5}$
$\text{h}_2=1.5\text{cm}$
Image is 1.5cm high, virtual, erect and magnified.
$\frac{1}{\text{v}}=\frac{1}{6}-\frac{1}{4}=\frac{2-3}{12}$
$\text{v}=-12\text{cm}$
$\text{m}=\frac{\text{v}}{\text{u}}=\frac{\text{h}_2}{\text{h}_1}$
$\frac{-12}{-4}=\frac{\text{h}_2}{0.5}$
$\text{h}_2=1.5\text{cm}$
Image is 1.5cm high, virtual, erect and magnified.
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