Physics - Geometrical Optics Question with Solution | TestHub

Key Idea The angle of incidence in a denser medium for which the angle of refraction in rarer medium becomes $90^{\circ }$ is known as critical angle. $\mathrm{C}={\sin }^{-1}\frac{1}{\mu }$ where, $\mu =$ refractive index of denser to rarer medium.Since, BC grazes along sarface $\begin{array}{rl}& \Rightarrow \text{ angle }\left(90-{\theta }_1\right)=\text{ critical angle }C\\ & \therefore \sin C=\frac{1}{\mu }\\ & \sin \left(90-{\theta }_1\right)=\frac{\sqrt{3}}{2}\Rightarrow \cos {\theta }_1=\frac{\sqrt{3}}{2}\\ & \Rightarrow {\theta }_1=30^{\circ }\\ & \end{array}$ Now, using Snells' law on lst interface, $\begin{array}{rlrl}{\mathrm{n}}_1\sin \theta & ={\mathrm{n}}_2\sin {\theta }_1& & \\ \Rightarrow & & 1\sin \theta =\frac{2}{\sqrt{3}}\sin 30^{\circ }& =\frac{2}{\sqrt{3}}\cdot \frac{1}{2}\\ \Rightarrow & & \sin \theta & =\frac{1}{\sqrt{3}}\\ \Rightarrow & & \theta & ={\sin }^{-1}\left(\frac{1}{\sqrt{3}}\right)\end{array}$