CBSE 2026 · Region 3 · Set 2 · Q23 · 3 marks
Define the term 'critical angle'. A small bulb is placed at the bottom of a tank containing a liquid of refractive index $\displaystyle \mu$ at a depth H. It is observed that the light emerges from a circular area of radius r of the surface. Obtain the expression for r in terms of H and $\displaystyle \mu$.
Marking-scheme solution
Critical angle: The angle of incidence corresponding to the angle of refraction $\displaystyle 90^{\circ}$ for a given pair of media when light goes from denser to rarer medium.
\[\begin{gathered}
\operatorname{Sin} i_{c}=\frac{1}{\mu}=\frac{r}{\sqrt{r^{2}+H^{2}}} \\
\mu^{2} r^{2}=r^{2}+H^{2} \\
\left(\mu^{2}-1\right) r^{2}=H^{2} \\
r=\frac{H}{\sqrt{\mu^{2}-1}}
\end{gathered}
\]Ray Optics and Optical InstrumentsTotal Internal ReflectionApplyshort_answermedium
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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.