CBSE 2025 · Region 1 · Set 2 · Q19 · 2 marks
A spherical convex surface of radius of curvature $\displaystyle \mathrm{R}$ separates glass (refractive index $\displaystyle 1.5$) from air. Light from a point source placed in air at distance $\displaystyle \mathrm{R} / 2$ from the surface falls on it. Find the position and nature of the image formed.
Marking-scheme solution
Refraction from rarer to denser medium:
$\displaystyle \frac{n_{1}}{-u}+\frac{n_{2}}{v}=\frac{n_{2}-n_{1}}{R}$
$\displaystyle u=-\frac{R}{2}, n_{1}=1, n_{2}=1.5$
$\displaystyle \frac{2}{R}+\frac{1.5}{v}=\frac{1.5-1}{R}$
$\displaystyle \frac{1.5}{v}=\frac{0.5}{R}-\frac{2}{R}$
$\displaystyle \frac{1.5}{v}=-\frac{1.5}{R}$
$\displaystyle v=-R$
The image is virtual, in air, at distance R.
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CBSE Class 12 Physics past-paper question from the 2025board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.