CBSE 2022 · Region 2 · Set 2 · Q11 · 3 marks
The refractive index of an equilateral triangular prism kept in air is $\displaystyle \sqrt{2}$. Calculate :(a)the angle of minimum deviation.(b)the critical angle for the prism.
The refractive index of an equilateral triangular prism kept in air is $\displaystyle \sqrt{2}$. Calculate :
(a)
the angle of minimum deviation.
(b)
the critical angle for the prism.
Marking-scheme solution
(a)
$\displaystyle \mu=\frac{\sin \left(\dfrac{\delta_{m}+A}{2}\right.}{\sin (A / 2)}$
$$\sqrt{2}=\frac{\sin \left(\dfrac{\delta_{m}+$\displaystyle 60$^{\circ}}{$\displaystyle 2$}\right)}{\sin $\displaystyle 30$^{\circ}} \Rightarrow \sin \left(\frac{\delta_{m}+$\displaystyle 60$^{\circ}}{$\displaystyle 2$}\right)=\sqrt{2} \times \frac{1}{2}
$$\sin \left(\frac{\delta_{m}+$\displaystyle 60$^{\circ}}{$\displaystyle 2$}\right)=\frac{1}{\sqrt{2}}=>\left(\frac{\delta_{m}+$\displaystyle 60$^{\circ}}{$\displaystyle 2$}\right)=$\displaystyle 45$^{\circ}
$$\delta_{m}=$\displaystyle 90$^{\circ}-$\displaystyle 60$^{\circ} \Rightarrow \delta_{m}=$\displaystyle 30$^{\circ}
$$$$
\text { b) } \sin \theta_{c}=\frac{1}{\mu}=\frac{1}{\sqrt{2}}
$$\theta_{c}=$\displaystyle 45$^{\circ}
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CBSE Class 12 Physics past-paper question from the 2022board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.