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Science · 2025 · 3 marks
CBSE 2025 · Region 2 · Set 1 · Q31
A convex lens forms an 8.0 cm long image of a 2.0 cm long object which is kept at a distance of 6.0 cm from the optical centre of the lens. If the…
A convex lens forms an $\displaystyle 8.0$ cm long image of a $\displaystyle 2.0$ cm long object which is kept at a distance of $\displaystyle 6.0$ cm from the optical centre of the lens. If the object and the image are on the same side of the lens, find (i) the nature of the image, (ii) the position of the image, and (iii) the focal length of the lens.
Marking-scheme solution
(i)
Nature: Virtual and erect
(ii)
Given \(\displaystyle \mathrm{h}^{\prime}=+8.0 \mathrm{~cm}, \mathrm{~h}=+2.0 \mathrm{~cm}, \mathrm{u}=-6 \mathrm{~cm}\)
\[\begin{aligned}
& \mathrm{m}=\frac{\mathrm{h}^{\prime}}{\mathrm{h}}=\frac{\mathrm{v}}{\mathrm{u}} \\
& \qquad \begin{aligned}
& =\frac{8.0 \mathrm{~cm}}{2.0 \mathrm{~cm}}=\frac{\mathrm{v}}{-6 \mathrm{~cm}} \\
\text { or } \mathrm{v} & =-24 \mathrm{~cm}
\end{aligned}
\end{aligned}
\]
Thus, the image is at a distance of $\displaystyle 24$ cm from the lens.
(iii)
Lens formula \(\displaystyle \frac{1}{v}-\frac{1}{u}=\frac{1}{f}\)
\[\begin{aligned}
\frac{1}{-24}-\frac{1}{-6} & =\frac{1}{f} \\
\frac{-1}{24}+\frac{1}{6} & =\frac{1}{f} \\
\frac{1}{8} & =\frac{1}{f} \\
\mathrm{f} & =8 \mathrm{~cm}
\end{aligned}
\]
Thus the focal length of the lens \(\displaystyle =8 \mathrm{~cm}\)
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CBSE Class 10 Science past-paper question from the 2025 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.