A convex lens has a focal length of 10cm. At what distance from the lens should the object be placed so that it forms a real and inverted image 20cm away from the lens? What would be the size of the image formed if the object is 2cm high? With the help of a ray diagram show the formation of the image by the lens in this case.
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Focal length = f = 10cm Image distance = v = 20cm Using sign convention, F is positive, v is also positive Lens formula, $\frac{1}{\text{f}}=\frac{1}{\text{v}}-\frac{1}{\text{u}}$$\frac{1}{10}=\frac{1}{20}-\frac{1}{\text{u}}$
$\frac{1}{\text{u}}=\frac{1}{20}-\frac{1}{10}$
$\frac{1}{\text{u}}=\frac{(10-20)}{200}$
$\frac{1}{\text{u}}=\frac{-10}{200}$
${\text{u}}=-20$
Object distance $=20 cm$ (minus indicates that the lens is converging lens) $h_i=$ height of image and $h_0=$ height ofobject$\frac{\text{h}_\text{i}}{\text{h}_0}=\frac{-\text{v}}{\text{u}}$
$\frac{\text{h}_\text{i}}{2}=\frac{-20}{20}$
${\text{h}_\text{i}}=2\text{cm}$
$\frac{1}{\text{u}}=\frac{1}{20}-\frac{1}{10}$
$\frac{1}{\text{u}}=\frac{(10-20)}{200}$
$\frac{1}{\text{u}}=\frac{-10}{200}$
${\text{u}}=-20$
Object distance $=20 cm$ (minus indicates that the lens is converging lens) $h_i=$ height of image and $h_0=$ height ofobject$\frac{\text{h}_\text{i}}{\text{h}_0}=\frac{-\text{v}}{\text{u}}$
$\frac{\text{h}_\text{i}}{2}=\frac{-20}{20}$
${\text{h}_\text{i}}=2\text{cm}$
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