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Science · 2023 · 3 marks
CBSE 2023 · Region 4 · Set 1 · Q30
Define the following terms in the context of a diverging mirror:(i)Principal focus(ii)Focal length Draw a labelled ray diagram to illustrate your answer.An object of height $\displaystyle 10$ cm is placed $\displaystyle 25$ cm away from the optical centre of a converging lens of focal length $\displaystyle 15$ cm. Calculate the image-distance and height of the image formed.
Define the following terms in the context of a diverging mirror:
(i)
Principal focus
(ii)
Focal length Draw a labelled ray diagram to illustrate your answer.
An object of height $\displaystyle 10$ cm is placed $\displaystyle 25$ cm away from the optical centre of a converging lens of focal length $\displaystyle 15$ cm. Calculate the image-distance and height of the image formed.
Marking-scheme solution
(i) It is a point on the principal axis of a diverging mirror from where the rays parallel to principal axis appear to diverge after reflection.
(ii) The distance between the pole and the principal focus of a mirror.
OR
(B) \(\displaystyle \frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}}\)
\[\frac{1}{v}=\frac{1}{f}+\frac{1}{u}
\]
\[\begin{aligned}
& f=15 \mathrm{~cm}, \mathrm{u}=-25 \mathrm{~cm}, \mathrm{~h}=10 \mathrm{~cm} \\
& \frac{1}{\mathrm{v}}=\frac{1}{15 \mathrm{~cm}}+\frac{1}{-25 \mathrm{~cm}}=\frac{2}{75}=+\frac{1}{37 \cdot 5} \\
& v=37 \cdot 5 \mathrm{~cm}
\end{aligned}
\]
\[\begin{aligned}
& \text { height of the image }=\frac{v}{u} \times \text { height of the object } \\
& =\frac{37 \cdot 5}{-25 \mathrm{~cm}} \times 10 \mathrm{~cm}
\end{aligned}
\]
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CBSE Class 10 Science past-paper question from the 2023board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.