CBSE 2026 · Region 2 · Set 1 · Q19 · 2 marks
A ray of light in air is incident at angle ∠i on a face of an equilateral glass prism and is refracted through the prism. As $\displaystyle \angle \mathrm{i}$ is varied, it is observed that the ray undergoes minimum deviation, when the $\displaystyle \angle \mathrm{i}$ is three-fourth of the angle of the prism. Calculate the speed of light in the prism.
Marking-scheme solution
Angle of incidence
$\displaystyle i=\dfrac{3}{4} A=45^{\circ}$
At minimum deviation
$\displaystyle r=\dfrac{A}{2}=30^{\circ}$
$\displaystyle n=\dfrac{\sin i}{\sin r}=\dfrac{\sin 45^{\circ}}{\sin 30^{\circ}}$
$\displaystyle n=\sqrt{2}$
$\displaystyle n=\dfrac{c}{v}$
$\displaystyle v=\dfrac{3}{\sqrt{2}} \times 10^{8} \mathrm{~m} / \mathrm{s}=2.12 \times 10^{8} \mathrm{~m} / \mathrm{s}$
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CBSE Class 12 Physics past-paper question from the 2026board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.