Q: A converging beam of rays is incident on a diverging lens. Having passed through the lens the rays intersect at a point 15 cm from the lens on the opposite side. If the lens is removed the point where the rays meet will move 5 cm closer to the lens. The focal length of the lens is

(a) – 10 cm

(b) 20 cm

(c) – 30

(d) 5 cm

Ans: (c)

Solution: Here, O acts as a virtual object and I acts as final image of the converging beam of light passing through a diverging lens.

Here, u = + 10 cm, v = +15 cm , f = ?

By using Lens formula ,

$\displaystyle \frac{1}{v} – \frac{1}{u} = \frac{1}{f} $

$\displaystyle \frac{1}{15} – \frac{1}{10} = \frac{1}{f} $

$\displaystyle \frac{1}{15} – \frac{1}{10} = \frac{1}{f} $

f = – 30 cm