Ta có \[\left\{ \begin{array}{l}{\rm{sini = nsinr}} \Leftrightarrow {\rm{n = }}\frac{{{\rm{sini}}}}{{{\rm{sin}}\,{\rm{r}}}}\\\,{\rm{i' + r}} = 90^\circ \Rightarrow {\rm{r}} = 90^\circ - {\rm{i'}}\\{\rm{i' = i}} \Rightarrow {\rm{r = 90^\circ - i}}\\{\rm{sin}}\,{\rm{r = sin}}\left( {{\rm{90^\circ - i}}} \right){\rm{ = cos}}\,{\rm{i}}\end{array} \right.\]\[ \Rightarrow {\rm{n = }}\frac{{{\rm{sini}}}}{{{\rm{cos}}\,{\rm{i}}}}{\rm{ = tani}} \Rightarrow {\rm{i}} = \arctan \,\frac{4}{3} = 53,1^\circ \]

b) Áp dụng định luật khúc xạ ánh sáng: \({{\rm{n}}_{\rm{1}}}{\rm{sini = }}{{\rm{n}}_{\rm{2}}}{\rm{sinr}} \Rightarrow {{\rm{n}}_1} = \frac{{{{\rm{n}}_2}{\rm{sinr}}}}{{{\rm{sini}}}} = \frac{{1.\sin 60^\circ }}{{\sin 40^\circ }} \approx 1,35\)