Two objects are placed on the principal axis of a thin converging lens. One is $10 c m$ from the lens and the other is on the other side of the lens at a distance of $40 c m$ from the lens. The image of both objects are in the same place. The focal length of the lens is (in $c m$ ).

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Solution

In one case image is virtual ( $u = - 10 c m$ )
In another case image is real ( $u = - 40 c m$ ) because rays are coming from the direction of object. So $u$ in this case will also be negative.
$v_{1} = u f / (u + f) = (- 10 f) / (- 10 + f) v_{2} = (- 40) / (- 40 + f)$
In both situations, sign convention is opposite.
$\Rightarrow v_{2} = - v_{1} \Rightarrow (- 40 f) / (- 40 + f) = (10 f) / (- 10 + f)$ Gives us, $f = 16 c m$

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