CBSE 2026 · Region 3 · Set 1 · Q26 · 3 marks
A ray of light is travelling through a rectangular glass slab (refractive index $\displaystyle \frac{3}{2}$ ) and is incident on the horizontal glass-air surface at the critical angle for the two media. The slab is then brought in contact with water (refractive index $\displaystyle \frac{4}{3}$ ) such that a thin horizontal layer of water is formed on the surface of the slab. Find the angle at which the ray will emerge into air from the water-air surface.
Marking-scheme solution
$\displaystyle \sin i_{C}=\dfrac{1}{n}=\dfrac{2}{3}$
For the glass–water surface: $\displaystyle \dfrac{3}{2} \sin i_{C}=\dfrac{4}{3} \sin r$
$\displaystyle \dfrac{3}{2} \times \dfrac{2}{3}=\dfrac{4}{3} \sin r$
$\displaystyle \sin r=\dfrac{3}{4}$
For the water–air surface: $\displaystyle \dfrac{4}{3} \sin r=\sin e$
$\displaystyle \dfrac{4}{3} \times \dfrac{3}{4}=\sin e$
$\displaystyle \sin e=1$
$\displaystyle e=90^{\circ}$
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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.