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A converging beam of rays is incident on a diverging lens. Having passed through the lens the rays intersect at a point 15 cm across the lens. If the lens is removed the point where the rays meet will move 5 cm closer to the mounting that holds the lens. Find focal length (in cm) of the lens.

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Detailed Solution

For concave lens, $u = + 10 c m (v i r t u a l o b j e c t) a n d v = + 15 c m$
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ , $\frac{1}{+ 15} - \frac{1}{+ 10} = \frac{1}{f}$
$∴ \frac{1}{v} - \frac{1.5}{- 3} = \frac{1 - 1.5}{- 5} \Rightarrow v = - 2.5 c m$

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A converging beam of rays is incident on a diverging lens. Having passed through the lens the rays intersect at a point 15 cm across the lens. If the lens is removed the point where the rays meet will move 5 cm closer to the mounting that holds the lens. Find focal length (in cm) of the lens.