Q83 / 120 · NEET 2011 (Mains) · Physics
A conversing beam of rays is incident on a diverging lens. Having passed though the lens the rays intersect at a point $15 \mathrm{~cm}$ from the lens on the opposite side. If the lens is removed the point where the rays meets will move $5 \mathrm{~cm}$ closer to the lens. The focal length of the lens is
- A$-10 \mathrm{~cm}$
- B$20 \mathrm{~cm}$
- Correct answerC$-30 \mathrm{~cm}$
- D$5 \mathrm{~cm}$
Solution
Given $u=10 \mathrm{~cm}, y=15 \mathrm{~cm}$
$$
\begin{aligned}
& \frac{1}{f}=\frac{1}{y}-\frac{1}{u} \\
& \frac{1}{f}=\frac{1}{15}-\frac{1}{10} \\
& \frac{1}{f}=\frac{10-15}{150} \\
& \frac{1}{f}=\frac{-5}{150} \\
& f=-30 \mathrm{~cm}
\end{aligned}
$$
Hence, the correct answer is option: 'C'.