Ta có ở I (Hình 28.3G):
nsinr1 = sin900 --> \(sinr_1 = \dfrac{1}{n}\)
Mặt khác: \(r_1+r_2 = A => r_2 = A -r_1\)
Ở J:
\(\begin{array}{l}n\sin {r_2} = \sin i'\\\Rightarrow n\sin (A - {r_1}) = \sin i'\\\Rightarrow \sin A\cos {r_1} - {\mathop{\rm s}\nolimits} {\rm{in}}{{\rm{r}}_1}{\rm{cosA = }}\dfrac{{\sin i'}}{n}\\\Rightarrow \sin A\sqrt {1 - {{\sin }^2}_{{r_1}}} - {\mathop{\rm s}\nolimits} {\rm{in}}{{\rm{r}}_1}{\rm{cosA = }}\dfrac{{\sin i'}}{n}\\\Rightarrow \sin A\dfrac{{\sqrt {{n^2} - 1} }}{n} - \dfrac{{{\rm{cosA}}}}{n} = \dfrac{{\sin i'}}{n}\end{array}\)
Do đó: \(\dfrac{{{\rm{cosA + sini'}}}}{{\sin A}} = \sqrt {{n^2} - 1} \)
