CBSE 2025 · Region 2 · Set 2 · Q31 · 3 marks
An object is placed at a distance of $\displaystyle 20$ cm from the optical centre of a concave lens and its image is formed on the same side of the lens as the object. If the distance of the image from optical centre of the lens is $\displaystyle 10$ cm, use lens formula to determine (i) focal length, and (ii) power of the lens in new Cartesian sign conventions.
Marking-scheme solution
(i) \(\displaystyle \mathrm{u}=-20 \mathrm{~cm}, \mathrm{v}=-10 \mathrm{~cm}\)
\[\begin{gathered}
\frac{1}{v}-\frac{1}{u}=\frac{1}{f} \\
\frac{1}{-10}-\frac{1}{-20}=\frac{1}{f} \\
\frac{-1}{10}+\frac{1}{20}=\frac{1}{f} \\
\frac{-2+1}{20}=\frac{1}{f} \\
\frac{-1}{20}=\frac{1}{f} \\
\mathrm{f}=-20 \mathrm{~cm}
\end{gathered}
\]
(ii) \(\displaystyle \mathrm{P}=\frac{1}{f(m)}\)
\[=\frac{1}{-0.2}=-5 D
\]
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