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Physics · 2024 · 2 marks
Ray Optics and Optical InstrumentsRefraction at Spherical Surfaces and by Lenses2 marksApplyvery short answer
CBSE 2024 · Region 4 · Set 1 · Q21
(a) An air bubble is trapped at point B ( CB =20 cm ) in a glass sphere of radius 40 cm and refractive index 1.5 as shown in figure. Find the nature…
An air bubble is trapped at point $\displaystyle \mathrm{B}(\mathrm{CB}=20 \mathrm{~cm})$ in a glass sphere of radius $\displaystyle 40$ cm and refractive index $\displaystyle 1.5$ as shown in figure. Find the nature and position of the image of the bubble as seen by an observer at point $\displaystyle P$.
In normal adjustment, for a refracting telescope, the distance between objective and eye piece lens is $\displaystyle 1.00$ m . If the magnifying power of the telescope is $\displaystyle 19$ , find the focal length of the objective and the eyepiece lens.
Marking-scheme solution
Using the refraction formula at a spherical surface from denser to rarer medium:
$\displaystyle n_{1}=$ refractive index of rarer medium, $\displaystyle n_{2}=$ refractive index of denser medium
$\displaystyle \frac{n_{1}}{v}-\frac{n_{2}}{u}=\frac{n_{1}-n_{2}}{R}$
$\displaystyle u=-20 \mathrm{~cm}$, $\displaystyle R=-40 \mathrm{~cm}$, $\displaystyle n_{1}=1$, $\displaystyle n_{2}=1.5$
$\displaystyle \frac{1}{v}-\frac{1.5}{(-20)}=\frac{1-1.5}{(-40)}$
$\displaystyle v=-16 \mathrm{~cm}$
Nature of image is virtual.
Distance between objective and eyepiece:
$\displaystyle f_{0}+f_{e}=1.00 \mathrm{~m}=100 \mathrm{~cm}$
Magnifying power:
$\displaystyle m=\frac{f_{0}}{f_{e}}=19$
On solving:
$\displaystyle f_{0}=95 \mathrm{~cm}=0.95 \mathrm{~m}$
$\displaystyle f_{e}=5 \mathrm{~cm}=0.05 \mathrm{~m}$
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CBSE Class 12 Physics past-paper question from the 2024 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.