Distance of an object from the first focus of an equi-convex lens is 10 cm and the distance of its real image from second focus is 40 cm. The focal length of the lens is:

25 cm

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10 cm

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20 cm

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40 cm

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Solution

The correct option is C 20 cm

$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
$\frac{1}{40 + f} - \frac{1}{10 + f} = \frac{1}{f}$
on solving, we get
$f = 20 c m$

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Distance of an object from the first focus of an equi-convex lens is 10 cm and the distance of its real image from second focus is 40 cm. The focal length of the lens is:

The correct option is C 20 cm 1v–1u=1f 140+f−110+f=1f on solving, we get f=20cm

The correct option is C 20 cm

$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
$\frac{1}{40 + f} - \frac{1}{10 + f} = \frac{1}{f}$
on solving, we get
$f = 20 c m$