A light beam is travelling from region I to region IV (refer figure). The refractive index in regions I, II, III and IV are $n_0,\frac{n_0}{2},\frac{n_0}{6}and\frac{n_0}{8}$ , respectively. The angle of incidence $\theta$  for which the beam just misses entering region IV is ${\sin }^{-1}(1/x)$ , then $x=$ 

(b) Since, light beam just misses entering region IV, it must strike the region IV interface at critical angle (C)
So, C =  ${\sin }^{-1}\left(\frac{{\mu }_r}{{\mu }_d}\right)={\sin }^{-1}\left(\frac{n_{0/8}}{n_{0/6}}\right)={\sin }^{-1}\left(\frac{3}{4}\right)$
   
Applying Snell’s law at interface 
               ${\mu }_1\sin i={\mu }_2\sin r={\mu }_3\sin r^{\text{'}}=$constant    

$$\begin{gathered} \Rightarrow {\mu }_1\sin i={\mu }_3\sin r^{\text{'}} \\ \Rightarrow n_0\sin \theta =\frac{n_0}{6}\sin C \\ \Rightarrow \sin \theta =\frac{1}{6}\times \frac{3}{4}=\frac{1}{8}\Rightarrow \theta ={\sin }^{-1}\left(\frac{1}{8}\right) \end{gathered}$$