A converging beam of rays is incident on a concave lens. After passing through the lens, the rays converge at a distance of 15 cm from the lens on the other side. If the lens is removed, the converging point of rays decreases by 5 cm. The focal length of lens is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → -30 cm
Without the lens the beam would meet at a point behind the lens; for the lens this point acts as a virtual object. Use the lens formula with the Cartesian sign convention (distances measured from the lens, positive in the direction of the incident light).
\( \text{Without the lens the rays meet at } 15 - 5 = 10\,\mathrm{cm} \text{ from the lens} \)
\( u = +10\,\mathrm{cm}\ (\text{virtual object}) \)
\( v = +15\,\mathrm{cm} \)
\( \displaystyle \frac{1}{f} = \frac{1}{v} - \frac{1}{u} = \frac{1}{15} - \frac{1}{10} \)
\( \displaystyle \frac{1}{f} = \frac{2 - 3}{30} = -\frac{1}{30} \;\Rightarrow\; f = -30\,\mathrm{cm} \)
The negative sign agrees with the lens being concave.