The image of an object place at 60cm in front of a lens is obtained on a screen at a distance of 120cm from it. Find the focal length of the lens. What would be the lens? What would be the height of the image? If the object is 5cm high?
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Object distance u $=-60\text{cm}$ Image distance v $=+120\text{cm}$ Using lens formula, $\frac{1}{\text{f}}=\frac{1}{\text{v}}-\frac{1}{\text{u}}$$\frac{1}{\text{f}}=\frac{1}{120}-\Big(-\frac{1}{60}\Big)$
$=\frac{1}{120}+\frac{1}{60}$
$=\frac{3}{120}=40$
$\text{f}=+40\text{cm}$
Therefore Focal length of lens is 40cm.$\frac{\text{H(image)}}{\text{H(object)}}=\frac{\text{v}}{\text{u}}$
$\text{H}_\text{i}=-\frac{120}{60}\times5$
$=-10\text{cm}$
Height of image is 10cm. Negative sign means image is real and inverted.
$=\frac{1}{120}+\frac{1}{60}$
$=\frac{3}{120}=40$
$\text{f}=+40\text{cm}$
Therefore Focal length of lens is 40cm.$\frac{\text{H(image)}}{\text{H(object)}}=\frac{\text{v}}{\text{u}}$
$\text{H}_\text{i}=-\frac{120}{60}\times5$
$=-10\text{cm}$
Height of image is 10cm. Negative sign means image is real and inverted.
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