CBSE 2025 · Region 5 · Set 1 · Q27 · 3 marks
A rectangular glass slab ABCD (refractive index $\displaystyle 1.5$) is surrounded by a transparent liquid (refractive index $\displaystyle 1.25$) as shown in the figure. A ray of light is incident on face AB at an angle i such that it is refracted out grazing the face AD. Find the value of angle i.
Marking-scheme solution
For glass-liquid interface:
$\displaystyle \sin i_{c}=\frac{1}{n_{21}}$
$\displaystyle =\frac{1.25}{1.5}$
$\displaystyle =\frac{5}{6}$
$\displaystyle i_{c}+r=90^{\circ}$
$\displaystyle \sin r=\sqrt{1-\cos ^{2} r}=\frac{\sqrt{11}}{6}$
Since $\displaystyle \frac{\sin i}{\sin r}=n$
Therefore $\displaystyle \sin i=\frac{\sqrt{11}}{4}$ or $\displaystyle i=\sin ^{-1} \frac{\sqrt{11}}{4}$
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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.