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Science · 2026 · 3 marks
CBSE 2026 · Region 5 · Set 2 · Q36
An object is placed at a distance of 30 cm in front of a convex lens of focal length 15 cm. Use lens formula to determine the position of the image.…
An object is placed at a distance of $\displaystyle 30$ cm in front of a convex lens of focal length $\displaystyle 15$ cm. Use lens formula to determine the position of the image. What will be the size of the image in this case ?
Marking-scheme solution
\(\displaystyle u=-30 \mathrm{~cm}, f=15 \mathrm{~cm}\)
\[\begin{aligned}
& \frac{1}{\mathrm{v}}-\frac{1}{u}=\frac{1}{f} \\
& \frac{1}{\mathrm{v}}=\frac{1}{f}+\frac{1}{u} \\
& \frac{1}{\mathrm{v}}=\frac{1}{15}+\frac{1}{-30} \\
& \mathrm{v}=30 \mathrm{~cm}
\end{aligned}
\]
\(\displaystyle \mathrm{m}=\frac{v}{u}=+\frac{30}{-30}=-1\) Hence, size of image will be same as the object.
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Size of image will be same as size of object because when object is kept at \(\displaystyle 2 \mathrm{~F}_{1}\), image will be formed at \(\displaystyle 2 \mathrm{~F}_{2}\).
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