Find the value of $\theta$ in the given diagram.

• A

  $\sin ^{-1}\left(\frac{2}{\sqrt{3}}\right)$
• Correct answer
  B

  $\sin ^{-1}\left(\frac{1}{\sqrt{3}}\right)$
• C

  $\sin ^{-1}\left(\frac{1}{2}\right)$
• D

  $\sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)$

Solution

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 $\begin{aligned} & \Rightarrow \quad \text { angle }\left(90-\theta_1\right)=\text { critical angle } C \\ & \therefore \quad \sin C=\frac{1}{\mu} \\ & \sin \left(90-\theta_1\right)=\frac{\sqrt{3}}{2} \Rightarrow \cos \theta_1=\frac{\sqrt{3}}{2} \\ & \Rightarrow \quad \theta_1=30^{\circ} \\ & \end{aligned}$ Now, using Snells' law on lst interface, $\begin{array}{rlrl} \mathrm{n}_1 \sin \theta & =\mathrm{n}_2 \sin \theta_1 \\ \Rightarrow & & 1 \sin \theta=\frac{2}{\sqrt{3}} \sin 30^{\circ} & =\frac{2}{\sqrt{3}} \cdot \frac{1}{2} \\ \Rightarrow & & \sin \theta & =\frac{1}{\sqrt{3}} \\ \Rightarrow & & \theta & =\sin ^{-1}\left(\frac{1}{\sqrt{3}}\right) \end{array}$

Hence, the correct answer is option: 'B'.