The filament of a lamp is 80cm from a screen and a converging lens forms an image of it on a screen, magnified three times. Find the distance of the lens from the filament and the focal length of the lens.
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-u + v = 80cm ....(i) m = -3 (The image is real, Since it form on a screen)$\text{m}=\frac{\text{v}}{\text{u}}=-3$
$\text{v}=-3\text{u}$
Put in eq. (i) -u - 3u = 80 -4u = 80 u = -20cm Distance lens from filament is 20cm v = -3u = 60cm Lens formula, $\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}$$\frac{1}{60}-\frac{1}{-20}=\frac{1}{\text{f}}$
$\frac{1}{\text{f}}=\frac{1+3}{60}=\frac{4}{60}$
$\text{f}=15\text{cm}$
$\text{v}=-3\text{u}$
Put in eq. (i) -u - 3u = 80 -4u = 80 u = -20cm Distance lens from filament is 20cm v = -3u = 60cm Lens formula, $\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}$$\frac{1}{60}-\frac{1}{-20}=\frac{1}{\text{f}}$
$\frac{1}{\text{f}}=\frac{1+3}{60}=\frac{4}{60}$
$\text{f}=15\text{cm}$
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