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Science · 2023 · 3 marks
CBSE 2023 · Region 2 · Set 1 · Q31
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 imageDefine 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 ?
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
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)
Here \(\displaystyle \mathrm{h}=3 \mathrm{~cm}\); f = - \(\displaystyle 12 \mathrm{~cm}, \mathrm{u}=-18 \mathrm{~cm}\),
(i)
v = ?, h' = ?
\[\frac{1}{f}=\frac{1}{v}+\frac{1}{u}
\]
\[\frac{1}{v}=\frac{1}{f}-\frac{1}{u}=\frac{1}{-12 \mathrm{~cm}}-\frac{1}{-18 \mathrm{~cm}}
\]
\[\therefore \mathrm{v}=-36 \mathrm{~cm}
\]
(ii) \[\begin{aligned}
& \mathrm{h}^{\prime}=-\frac{\mathrm{v}}{\mathrm{u}} \times \mathrm{h} \\
& \mathrm{~h}^{\prime}=(-) \frac{-36 \mathrm{~cm}}{-18 \mathrm{~cm}} \times 3 \mathrm{~cm}=-6 \mathrm{~cm}
\end{aligned}
\]
OR
(b) • Power of lens : Ability of a lens to converge or diverge light rays falling on it / Degree of convergence or divergence of light achieved by a lens / Reciprocal of focal length of lens in metre.
It is diverging/concave lens
\(\displaystyle \mathrm{P}=\frac{1}{f(m)}=\frac{100}{f(c m)} \mathrm{P}=\frac{100}{-10 \mathrm{~cm}}=-10 \mathrm{D}\)
Sign of magnification = + or positive
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CBSE Class 10 Science past-paper question from the 2023board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.