Xét tia đỏ:
\(\begin{array}{l} + \sin {i_{d1}} = {n_d}{\mathop{\rm s}\nolimits} {\rm{in}}{{\rm{r}}_{d1}} \\\Rightarrow {\mathop{\rm s}\nolimits} {\rm{in}}{{\rm{r}}_{d1}} = \dfrac{{\sin {i_{d1}}}}{{{n_d}}} = \dfrac{{\sin {{50}^0}}}{{1,514}} = 0,506\\ \Rightarrow {{\rm{r}}_{d1}} = 30,{4^0}\\ + A = {{\rm{r}}_{d1}} + {{\rm{r}}_{d2}} \\\Rightarrow {{\rm{r}}_{d2}} = A - {{\rm{r}}_{d1}}\\ = 60 - 30,{4^0} = 29,{6^0}\\ + \sin {i_{d2}} = {n_d}{\mathop{\rm s}\nolimits} {\rm{in}}{{\rm{r}}_{d2}} \\= 1,514.\sin 29,{6^0} = 0,748 \\\Rightarrow {i_{d2}} = 48,{4^0}\\ + {D_d} = {i_{d1}} + {i_{d2}} - A \\= 50 + 48,{4^0} - 60 = 38,{4^0}\end{array}\)
Xét tia tím:
\(\begin{array}{l} + \sin {i_{t1}} = {n_t}{\mathop{\rm s}\nolimits} {\rm{in}}{{\rm{r}}_{t1}}\\ \Rightarrow {\mathop{\rm s}\nolimits} {\rm{in}}{{\rm{r}}_{t1}} = \dfrac{{\sin {i_{t1}}}}{{{n_t}}} = \dfrac{{\sin {{50}^0}}}{{1,5368}} = 0,498\\ \Rightarrow {{\rm{r}}_{t1}} = 29,{9^0}\\ + A = {{\rm{r}}_{t1}} + {{\rm{r}}_{t2}} \\\Rightarrow {{\rm{r}}_{t2}} = A - {{\rm{r}}_{t1}} = 60 - 29,{9^0} = 30,{1^0}\\ + \sin {i_{t2}} = {n_t}{\mathop{\rm s}\nolimits} {\rm{in}}{{\rm{r}}_{t2}} \\= 1,5368.\sin 30,{1^0} = 0,77 \\\Rightarrow {i_{d2}} = 50,{4^0}\\ + {D_t} = {i_{d1}} + {i_{d2}} - A \\= 50 + 50,{4^0} - 60 = 40,{4^0}\end{array}\)
\(TD = 2f(\tan \dfrac{{{D_t} - {D_d}}}{2})\\ = 2.1.(\tan \dfrac{{40,{4^0} - 38,{4^0}}}{2}) \\= {35.10^{ - 3}}m = 35mm\)
