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Science · 2023 · 4 marks
CBSE 2023 · Region 1 · Set 1 · Q39
Hold a concave mirror in your hand and direct its reflecting surface towards the sun. Direct the light reflected by the mirror on to a white card-board held close to the mirror. Move the card-board back and forth gradually until you find a bright, sharp spot of light on the board. This spot of light is the image of the sun on the sheet of paper; which is also termed as "Principal Focus" of the concave mirror.
(a)List two applications of concave mirror.(b)If the distance between the mirror and the principal focus is $\displaystyle 15$ cm, find the radius of curvature of the mirror.Draw a ray diagram to show the type of image formed when an object is placed between pole and focus of a concave mirror.An object $\displaystyle 10$ cm in size is placed at $\displaystyle 100$ cm in front of a concave mirror. If its image is formed at the same point where the object is located, find :(i)focal length of the mirror, and(ii)magnification of the image formed with sign as per Cartesian sign convention.
Hold a concave mirror in your hand and direct its reflecting surface towards the sun. Direct the light reflected by the mirror on to a white card-board held close to the mirror. Move the card-board back and forth gradually until you find a bright, sharp spot of light on the board. This spot of light is the image of the sun on the sheet of paper; which is also termed as "Principal Focus" of the concave mirror.
(a)
List two applications of concave mirror.
(b)
If the distance between the mirror and the principal focus is $\displaystyle 15$ cm, find the radius of curvature of the mirror.
Draw a ray diagram to show the type of image formed when an object is placed between pole and focus of a concave mirror.
An object $\displaystyle 10$ cm in size is placed at $\displaystyle 100$ cm in front of a concave mirror. If its image is formed at the same point where the object is located, find :
(i)
focal length of the mirror, and
(ii)
magnification of the image formed with sign as per Cartesian sign convention.
Marking-scheme solution
(a) Torches, search light, vehicles head lights, shaving mirrors, dentist's mirror, Solar furnaces.
(b) \(\displaystyle \mathrm{f}=15 \mathrm{~cm}\)
\[\begin{aligned}
& R=2 \mathrm{f} \\
& R=2 \times 15 \mathrm{~cm}=30 \mathrm{~cm}
\end{aligned}
\]
(c) 
OR
(i) \(\displaystyle \mathrm{h}=+10 \mathrm{~cm}\)
\[\mathrm{u}=-100 \mathrm{~cm}
\] \(\displaystyle \mathrm{v}=-100 \mathrm{~cm}\)
\[\frac{1}{v}+\frac{1}{u}=\frac{1}{f}
\] \(\displaystyle \frac{1}{100}-\frac{1}{100}==\frac{1}{f}\)
\[\frac{-2}{100}=\frac{1}{f}
\] \(\displaystyle \mathrm{f}=-50 \mathrm{~cm}\) Alternate answer for (i) Since u = v Therefore, object is placed at centre of curvature (C)
\[\mathrm{f}=\frac{R}{2}
\]
\[\begin{aligned}
& \mathrm{f}=\frac{-100}{2} \\
& \mathrm{f}=-50 \mathrm{~cm}
\end{aligned}
\](ii) \(\displaystyle \mathrm{m}=\frac{-v}{u}=\frac{-(-100)}{100}=-1\)
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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.