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Science · 2023 · 3 marks
CBSE 2023 · Region 5 · Set 3 · Q29
(a) A student has focussed the image of an object of height 3 cm on a white screen using a concave mirror of focal length 12 cm. If the distance of…
(a)
A student has focussed the image of an object of height $\displaystyle 3$ cm on a white screen using a concave mirror of focal length $\displaystyle 12$ cm. If the distance of the object from the mirror is $\displaystyle 18$ cm, find the values of the following :
(i)
Distance of the image from the mirror
(ii)
Height of the image
OR
(b)
Define power of a lens. The focal length of a lens is -$\displaystyle 10$ cm. Write the nature of the lens and find its power. If an object is placed at a distance of $\displaystyle 20$ cm from the optical centre of this lens, according to the New Cartesian Sign Convention, what will be the sign of magnification in this case ?
Marking-scheme solution
(a) \(\displaystyle \mathrm{h}_{1}=3 \mathrm{~cm}\)
\[\begin{aligned}
& f=-12 \mathrm{~cm} \\
& u=-18 \mathrm{~cm}
\end{aligned}
\]
(i) Image Distance
\[\begin{aligned}
& \frac{1}{f}=\frac{1}{v}+\frac{1}{u} \\
& \frac{1}{-12}=\frac{1}{v}+\frac{1}{-18} \\
& \frac{1}{-12}+\frac{1}{18}=\frac{1}{v} \\
& \frac{1}{v}=\frac{-3+2}{36}=\frac{-1}{36} \\
& v=-36 \mathrm{~cm}
\end{aligned}
\]
(ii)
Height of image
\[\begin{aligned}
& \mathrm{m}=\frac{\mathrm{h}_{2}}{\mathrm{~h}_{1}}=\frac{-\mathrm{v}}{\mathrm{u}} \\
& \frac{\mathrm{~h}_{2}}{3}=\frac{-[-36]}{-18} \\
& \mathrm{~h}_{2}=\frac{-36}{18} \times 3 \\
& \mathrm{~h}_{2}=-6 \mathrm{~cm}
\end{aligned}
\]
Degree of convergence or divergence of light/ Reciprocal of focal length of lens in metre.
It is diverging/concave lens
\(\displaystyle \mathrm{P}=\frac{1}{f(m)}=\frac{100}{f(c m)}\)
\[P=\frac{100}{-10 \mathrm{~cm}}=-10 \mathrm{D}
\]
Sign of magnification = + or positive
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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.