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Science · 2023 · 3 marks
CBSE 2023 · Region 5 · Set 1 · Q30
(a) (i) An object of 5 cm height is placed at a distance of 20 cm from the optical centre of a concave lens of focal length 18 cm. Calculate (1)…
(i)
An object of $\displaystyle 5$ cm height is placed at a distance of $\displaystyle 20$ cm from the optical centre of a concave lens of focal length $\displaystyle 18$ cm. Calculate ($\displaystyle 1$) image distance and ($\displaystyle 2$) the magnification in this case.
(ii)
Compare the values of magnification obtained by a concave lens and a convex lens when both the lenses form virtual images.
A convex lens can form a (i) real, inverted and magnified image as well as (ii) virtual, erect and magnified image of an object. If the focal length of the lens is $\displaystyle 10$ cm, what should be the range of the object distance in both cases ? Draw ray diagrams to justify your answer.
Marking-scheme solution
(a) Given, Height of object \(\displaystyle (\mathrm{h})=5 \mathrm{~cm}\) Object distance \(\displaystyle (\mathrm{u})=-20 \mathrm{~cm}\) Focal length \(\displaystyle (\mathrm{f})=-18 \mathrm{~cm}\) Image distance (v) = ?
(i)
(1) \(\displaystyle \frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}}\)
\[\begin{aligned}
& \frac{1}{v}=\frac{1}{u}+\frac{1}{f}=\frac{1}{-20}+\frac{1}{-18}=-\left[\frac{18+20}{360}\right] \\
& =\frac{-38}{360}
\end{aligned}
\]
\[\Rightarrow v=\frac{-360}{38}=-9.47 \mathrm{~cm}
\]
(2) \(\displaystyle \mathrm{m}=\frac{\mathrm{v}}{\mathrm{u}}=\frac{-9.47}{-20}=0.47\)
(ii)
For convex lens : m > $\displaystyle 1$, for concave lens m < $\displaystyle 1$
(3)
(i) $\displaystyle 20$ cm > u > $\displaystyle 10$ cm / Between $\displaystyle 10$ cm and $\displaystyle 20$ cm
(ii)
Object distance less than $\displaystyle 10$ cm / $\displaystyle 10$ > u > $\displaystyle 0$
(i)
(ii)
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CBSE Class 10 Science past-paper question from the 2023 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.