A conversing beam of rays is incident on a diverging lens. Having passed though the lens the rays intersect…

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
  1. $-10 \mathrm{~cm}$
  2. $20 \mathrm{~cm}$
  3. $-30 \mathrm{~cm}$
  4. $5 \mathrm{~cm}$

Solution

Given $u=10 \mathrm{~cm}, y=15 \mathrm{~cm}$ $\begin{aligned} & \frac{1}{f}=\frac{1}{y}-\frac{1}{u} \\ & \frac{1}{f}=\frac{1}{15}-\frac{1}{10} \\ & \frac{1}{f}=\frac{10-15}{150} \\ & \frac{1}{f}=\frac{-5}{150} \\ & f=-30 \mathrm{~cm} \end{aligned}$

Asked in: NEET 2011 (Mains)

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