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Science · 2024 · 5 marks
CBSE 2024 · Region 3 · Set 1 · Q36
(i)Draw a ray diagram to show the path of the refracted ray in each of the following cases : A ray of light incident on a concave lens(1)parallel to its principal axis, and(2)is directed towards its principal focus.(ii)A $\displaystyle 4$ cm tall object is placed perpendicular to the principal axis of convex lens of focal length $\displaystyle 24$ cm. The distance of object from the lens is $\displaystyle 16$ cm. Find the position and size of image formed.OR 36. (b) (i) Draw a ray diagram to show the path of the reflected ray in each of the following cases : A ray of light incident on a convex mirror(1)parallel to its principal axis, and(2)is directed towards its principal focus(ii)A $\displaystyle 1.5$ cm tall candle flame is placed perpendicular to the principal axis of a concave mirror of focal length $\displaystyle 12$ cm. If the distance of the flame from the pole of the mirror is $\displaystyle 18$ cm, use mirror formula to determine the position and size of the image formed.
(i)
Draw a ray diagram to show the path of the refracted ray in each of the following cases : A ray of light incident on a concave lens
(1)
parallel to its principal axis, and
(2)
is directed towards its principal focus.
(ii)
A $\displaystyle 4$ cm tall object is placed perpendicular to the principal axis of convex lens of focal length $\displaystyle 24$ cm. The distance of object from the lens is $\displaystyle 16$ cm. Find the position and size of image formed.
OR 36. (b) (i) Draw a ray diagram to show the path of the reflected ray in each of the following cases : A ray of light incident on a convex mirror
(1)
parallel to its principal axis, and
(2)
is directed towards its principal focus
(ii)
A $\displaystyle 1.5$ cm tall candle flame is placed perpendicular to the principal axis of a concave mirror of focal length $\displaystyle 12$ cm. If the distance of the flame from the pole of the mirror is $\displaystyle 18$ cm, use mirror formula to determine the position and size of the image formed.
Marking-scheme solution
(a) (i)
(1)
Fig.9.13(b)-Page-$\displaystyle 153$, NCERT.
Fig.9.14(b)-Page-$\displaystyle 154$, NCERT.
Fig.9.13(b)-Page-$\displaystyle 153$, NCERT.
Fig.9.14(b)-Page-$\displaystyle 154$, NCERT.(ii) Given \(\displaystyle u=-16 \mathrm{~cm}, \mathrm{f}=+24 \mathrm{~cm}, \mathrm{~h}=4 \mathrm{~cm}\)
\[\begin{aligned}
& \text { Formula used } \frac{1}{v}-\frac{1}{u}=\frac{1}{f} \\
& \qquad \frac{1}{v}-\frac{1}{(-16)}=\frac{1}{+24} \\
& \frac{1}{v}=\frac{-1}{48}
\end{aligned}
\]
(ii)
Here \(\displaystyle \mathrm{f}=-12 \mathrm{~cm}, u=-18 \mathrm{~cm}, v=?, \mathrm{~h}=1 \cdot 5 \mathrm{~cm}, \mathrm{~h}^{\prime}=\) ? Mirror formula \(\displaystyle \frac{1}{v}+\frac{1}{u}=\frac{1}{f}\)
\[\begin{aligned}
\therefore \frac{1}{v} & =\frac{1}{f}-\frac{1}{u} \\
& =\frac{1}{-12 \mathrm{~cm}}-\frac{1}{-18 \mathrm{~cm}} \\
& =\frac{-1}{36}
\end{aligned}
\]
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